\documentclass[fullpage, 10pt, openany]{book} \usepackage{sectsty, fancyhdr, amsmath} %%%%%%%%%%%Package for writing the answers at the back of the book \usepackage{answers} %%%%%%%%%%Assorted packages \usepackage{makeidx} \usepackage{hangcaption} \usepackage{ntheorem} \usepackage[colorlinks=true,linkcolor=red, pdftitle={linear algebra}, pdfauthor={david santos}, bookmarksopen, dvips]{hyperref} \usepackage{pdflscape} \usepackage{thumbpdf} \usepackage[dvips]{graphicx} %%%%% Postscript drawing packages \usepackage{pstricks, pstricks-add} \usepackage{pst-plot} \usepackage{pst-3d} \usepackage{pst-eucl} %%%%%%%%%%%%%%CHAPTER HEADINGS \usepackage[Lenny]{fncychap} %%%%% FONTS \usepackage[latin1]{inputenc} \usepackage{t1enc} \usepackage{bookman} \usepackage{pifont, stmaryrd} %%for the dingautolists and the proofsymbol \usepackage{euler} %%Knuth's euler math font \mathversion{bold} \usepackage{mathrsfs} %%for the fancy cursive mathscr \usepackage{amsfonts} \usepackage{amssymb} \DeclareSymbolFont{EulerFraktur}{U}{euf}{m}{n} \SetSymbolFont{EulerFraktur}{bold}{U}{euf}{b}{n} \DeclareMathSymbol{!}\mathord {EulerFraktur}{"21} \DeclareMathSymbol{(}\mathopen {EulerFraktur}{"28} \DeclareMathSymbol{)}\mathclose{EulerFraktur}{"29} \DeclareMathSymbol{+}\mathbin {EulerFraktur}{"2B} \DeclareMathSymbol{-}\mathbin {EulerFraktur}{"2D} \DeclareMathSymbol{=}\mathrel {EulerFraktur}{"3D} \DeclareMathSymbol{[}\mathopen {EulerFraktur}{"5B} \DeclareMathSymbol{]}\mathclose{EulerFraktur}{"5D} \DeclareMathSymbol{"}\mathord {EulerFraktur}{"7D} \DeclareMathSymbol{&}\mathord {EulerFraktur}{"26} \DeclareMathSymbol{:}\mathrel {EulerFraktur}{"3A} \DeclareMathSymbol{;}\mathpunct{EulerFraktur}{"3B} \DeclareMathSymbol{?}\mathord {EulerFraktur}{"3F} \DeclareMathSymbol{^}\mathord {EulerFraktur}{"5E} \DeclareMathSymbol{`}\mathord {EulerFraktur}{"12} \DeclareMathDelimiter{(}{EulerFraktur}{"28}{largesymbols}{"00} \DeclareMathDelimiter{)}{EulerFraktur}{"29}{largesymbols}{"01} \DeclareMathDelimiter{[}{EulerFraktur}{"5B}{largesymbols}{"02} \DeclareMathDelimiter{]}{EulerFraktur}{"5D}{largesymbols}{"03} %%%%%%%%%%PAGE FORMATTING \topmargin -.7in \textheight 9.5in \oddsidemargin -.3in \evensidemargin -.3in \textwidth 7in %%%%%FLOAT PLACEMENT \renewcommand{\topfraction}{.85} \renewcommand{\bottomfraction}{.7} \renewcommand{\textfraction}{.15} \renewcommand{\floatpagefraction}{.66} \renewcommand{\dbltopfraction}{.66} \renewcommand{\dblfloatpagefraction}{.66} \setcounter{topnumber}{9} \setcounter{bottomnumber}{9} \setcounter{totalnumber}{20} \setcounter{dbltopnumber}{9} \newcount\n % recursion depth \newcount\x\newcount\y % pen location \newcount\xo\newcount\yo % old pen location \newcount\dx\newcount\dy % next step to take \newcount\t % temp for swap % \def\swp{\t=\dx\dx=\dy\dy=\t}% swap step deltas \def\L{\swp\multiply\dx by-1}% turn so step is pi/2 left of where it was \def\R{\swp\multiply\dy by-1}% turn so step is pi/2 right of where it was \def\S{%% take a step with pen down \xo=\x \yo=\y % \advance\x by\dx% \advance\y by\dy% \ln{\the\xo}{\the\yo}{\the\x}{\the\y}% } \def\ln#1#2#3#4{\qline(#1,#2)(#3,#4)}% draw a line from old to new pen position % represent angle, which is +-90 degrees, % as a [dx,dy] vector instead \def\h#1#2{\ifnum\n=0\relax\else% \advance\n by-1% #1\h#2#1\S#2\h#1#2\S\h#1#2#2\S\h#2#1#1% \advance\n by1\fi% } %%%%Title Page \makeatletter \def\thickhrulefill{\leavevmode \leaders \hrule height 1pt\hfill \kern \z@} \renewcommand{\maketitle}{\begin{titlepage}% \let\footnotesize\small \let\footnoterule\relax \parindent \z@ \reset@font \null\vfil \begin{flushleft} \@title \end{flushleft} \par \hrule height 4pt \par \begin{flushright} \@author \par \end{flushright} \vskip 60\p@ \vspace*{\stretch{2}} \psset{unit=2mm} \bigskip \makebox[\textwidth]{% \begin{pspicture}(0,0)(63,63) \x=0 \y=0 \dx=1 \dy=0 \n=6 \h\L\R \end{pspicture}% } \vskip 60\p@ \vspace*{\stretch{2}} \begin{center} \Large\textsf{\today\quad REVISION} \end{center} \end{titlepage}% \setcounter{footnote}{0}% } \makeatother \title{\textcolor{red}{\Large Linear Algebra Notes }} \author{\textcolor{blue}{David A. SANTOS} \\ \href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu}} %%%%%%% \setlength{\fboxrule}{1.5pt} %%%%%%%%%%%%%%%%%INTERVALS %%%%%%%% lo= left open, rc = right closed, etc. \def\lcrc#1#2{\left[#1 \ ; #2 \right]} \def\loro#1#2{ \left]#1 \ ; #2 \right[} \def\lcro#1#2{\left[#1 \ ; #2 \right. \left[ \right.} \def\lorc#1#2{\left. \right[#1 \ ; #2 \left.\right]} %%%%% Non-standard commands and symbols \newcommand{\BBZ}{\mathbb{Z}} \newcommand{\BBR}{\mathbb{R}} \newcommand{\BBN}{\mathbb{N}} \newcommand{\BBC}{\mathbb{C}} \newcommand{\BBQ}{\mathbb{Q}} \newcommand{\BBF}{\mathbb{F}} \newcommand{\dis}{\displaystyle} \def\fun#1#2#3#4#5{\everymath{\displaystyle}{{#1} : \vspace{1cm} \begin{array}{ccc}{#4} & \rightarrow & {#5}\\ {#2} & \mapsto & {#3} \\ \end{array}}} \def\binom#1#2{{#1\choose#2}} \def\T#1#2{\mathscr{T}_{#1}{#2}} \def\gl#1#2{{\bf GL}_{#1}(#2)} \def\algebra#1#2{\langle #1,#2\rangle} \def\field#1#2#3{\langle #1,#2, #3\rangle} \def\vecspace#1#2#3#4{\langle #1,#2, #3, #4\rangle} \def\span#1{{\rm span}\left(#1\right)} \def\rank#1{{\rm rank}\left(#1\right)} \def\sgn#1{{\rm signum}\left(#1\right)} \def\mat#1#2{{\bf M}_{#1}(#2)} \def\norm#1{{\left|\left|#1\right|\right|}} \newcommand{\bulletproduct}{{\scriptscriptstyle \stackrel{\bullet}{{}}}} \def\dotprod#1#2{\v{#1} \bulletproduct \v{#2}} \newcommand{\cross}{{\boldsymbol\times}} \def\crossprod#1#2{\v{#1}\cross \v{#2}} \def\sgn#1{{\rm sgn}(#1)} \def\tr#1{{\rm tr}\left(#1\right)} \def\adj#1{{\rm adj}\left(#1\right)} \def\proj#1#2{{\rm proj} _{\v{#2}} ^{\v{#1}} } \def\vol#1{{\rm volume}(#1)} \def\ball#1#2{B_{#2}({\bf #1})} \newcommand\zeropoint{O} \def\v#1{{\bf \overrightarrow{#1}}} \def\vect#1{\overrightarrow{#1}} \newcommand{\idefun}{{\bf Id\ }} \newcommand{\QUOTEME}[2]{\begin{quote}{\it\textbf{#1}}\nolinebreak[1] \blue\hspace*{\fill} \mbox{-\textsl{#2}} \hspace*{\fill}\end{quote}} \def\bipoint#1#2{[#1, #2]} \def\anglebetween#1#2{\widehat{(\v{#1}, \v{#2})}} \def\rank#1{{\rm rank}\left(#1\right)} \def\ker#1{{\rm ker}\left(#1\right)} \def\im#1{{\rm Im}\left(#1\right)} \def\lcm#1{{\rm lcm}\left(#1\right)} \def\rowrank#1{{\rm row\ rank}\left(#1\right)} \def\columnrank#1{{\rm column\ rank}\left(#1\right)} \setcounter{MaxMatrixCols}{15} %%%%Jim Hefferon's Linear Algebra macros. %--------grstep % For denoting a Gauss' reduction step. % Use as: \grstep{\rho_1+\rho_3} or \grstep[2\rho_5 \\ 3\rho_6]{\rho_1+\rho_3} \newcommand{\grstep}[2][\relax]{% \ensuremath{\mathrel{ \mathop{\rightsquigarrow}\limits^{#2\mathstrut}_{ \begin{subarray}{l} #1 \end{subarray}}}}} \newcommand{\swap}{\leftrightarrow} %-------------augmatrix % Augmented matrix. Usage (note the argument does not count the aug col): % \begin{augmatrix}{2} % 1 2 3 \\ 4 5 6 % \end{augmatrix} \newenvironment{augmatrix}[1]{% \left[\begin{array}{@{}*{#1}{c}|c@{}} }{% \end{array}\right] } %-------------bmat % For matrices with arguments. % Usage: \begin{bmat}{c|c|c} 1 &2 &3 \end{bmat} \newenvironment{bmat}[1]{ \left[\begin{array}{@{}#1@{}} }{\end{array}\right] } %------------colvec and rowvec % Column vector and row vector. Usage: % \colvec{1 \\ 2 \\ 3 \\ 4} and \rowvec{1 &2 &3} \newcommand{\colvec}[1]{\begin{bmatrix} #1 \end{bmatrix}} \newcommand{\colpoint}[1]{\begin{pmatrix} #1 \end{pmatrix}} \newcommand{\rowvec}[1]{\begin{bmatrix} #1 \end{bmatrix}} \renewcommand{\arraystretch}{1.2} %%%%% THEOREM-LIKE ENVIRONMENTS \newcommand{\proofsymbol}{\Pisymbol{pzd}{113}} \theorembodyfont{\small} \newtheorem{pro}{Problem}[section] \theorempreskipamount .5cm \theorempostskipamount .5cm \theoremstyle{change} \theoremheaderfont{\sffamily\bfseries} \theorembodyfont{\normalfont} \newtheorem{thm}{Theorem} \newtheorem{exa}[thm]{{\cal Example}} \newtheorem{cor}[thm]{Corollary} \newtheorem{df}[thm]{Definition} \newtheorem{lem}[thm]{Lemma} \Newassociation{answer}{Answer}{linearans} \newtheorem{rul}[thm]{Rule} \newenvironment{pf}[0]{\itshape\begin{quote}{\bf Proof: \ }}{\proofsymbol\end{quote}} \newenvironment{rem}[0]{\begin{quote}{\huge\textcolor{red}{\Pisymbol{pzd}{43}}}\itshape }{\end{quote}} \newenvironment{nota}[0]{\begin{quote}{\textcolor{red}{\bf Notation:\ }}\itshape }{\end{quote}} \newenvironment{solu}[0]{\begin{quote}{\bf Solution:\ }$\blacktriangleright$ \itshape }{$\blacktriangleleft$\end{quote}} \usepackage{multicol} %%%%%%%%%%%%%%%%%%%% \makeatletter \newcommand{\twocoltoc}{% \section*{\contentsname \@mkboth{% }{}}% \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \@starttoc{toc}% \end{multicols}} \makeatother %%%%%%%%%%%% \fontencoding{T1} \fontfamily{fmv} \selectfont \DeclareFixedFont{\bigghv}{T1}{phv}{b}{n}{1.2cm} \DeclareFixedFont{\bighv}{T1}{phv}{b}{n}{0.6cm} \DeclareFixedFont{\bigithv}{T1}{phv}{b}{sl}{0.8cm} \DeclareFixedFont{\smallit}{T1}{phv}{b}{it}{9pt} \makeindex %%%%%% And voilą the document !!! \begin{document} \Opensolutionfile{linearans}[linearansC1] \pagenumbering{roman} \renewcommand{\headrulewidth}{1pt} \renewcommand{\footrulewidth}{1pt} \lhead{\nouppercase{\textcolor{blue}{\rightmark}}} \rhead{\nouppercase{\textcolor{blue}{\leftmark}}} \lhead[\rm\thepage]{\textcolor{blue}{\it \rightmark}} \rhead[\it \textcolor{blue}{\leftmark}]{\rm\thepage} \cfoot{} \thispagestyle{empty} \pagestyle{fancy} % Front matter may follow here \begin{frontmatter} \maketitle {\small \twocoltoc{}} \end{frontmatter} \clearpage \begin{quote} Copyright \copyright{} 2007 David Anthony SANTOS. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled ``GNU Free Documentation License''. \end{quote} \clearpage \clearpage \chapter*{Preface} \markboth{}{} \addcontentsline{toc}{chapter}{Preface} \markright{Preface} These notes started during the Spring of 2002, when John MAJEWICZ and I each taught a section of Linear Algebra. I would like to thank him for numerous suggestions on the written notes. \bigskip The students of my class were: Craig BARIBAULT, Chun CAO, Jacky CHAN, Pho DO, Keith HARMON, Nicholas SELVAGGI, Sanda SHWE, and Huong VU. I must also thank my former student William CARROLL for some comments and for supplying the proofs of a few results. \bigskip John's students were David HERN\'{A}NDEZ, Adel JAILILI, Andrew KIM, Jong KIM, Abdelmounaim LAAYOUNI, Aju MATHEW, Nikita MORIN, Thomas NEGR\'{O}N, Latoya ROBINSON, and Saem SOEURN.\\ \bigskip Linear Algebra is often a student's first introduction to abstract mathematics. Linear Algebra is well suited for this, as it has a number of beautiful but elementary and easy to prove theorems. My purpose with these notes is to introduce students to the concept of proof in a gentle manner. \bigskip \hfill David A. Santos \href{mailto:dsantos@ccp.edu}{dsantos@ccp.edu} \vfill \section*{To the Student} \markboth{}{} \addcontentsline{toc}{chapter}{To the Student} \markright{To the Student} These notes are provided for your benefit as an attempt to organise the salient points of the course. They are a {\em very terse} account of the main ideas of the course, and are to be used mostly to refer to central definitions and theorems. The number of examples is minimal, and here you will find few exercises. The {\em motivation} or informal ideas of looking at a certain topic, the ideas linking a topic with another, the worked-out examples, etc., are given in class. Hence these notes are not a substitute to lectures: {\bf you must always attend to lectures}. The order of the notes may not necessarily be the order followed in the class. \bigskip There is a certain algebraic fluency that is necessary for a course at this level. These algebraic prerequisites would be difficult to codify here, as they vary depending on class response and the topic lectured. If at any stage you stumble in Algebra, seek help! I am here to help you! \bigskip Tutoring can sometimes help, but bear in mind that whoever tutors you may not be familiar with my conventions. Again, I am here to help! On the same vein, other books may help, but the approach presented here is at times unorthodox and finding alternative sources might be difficult. \bigskip Here are more recommendations: \begin{itemize} \item Read a section before class discussion, in particular, read the definitions. \item Class provides the informal discussion, and you will profit from the comments of your classmates, as well as gain confidence by providing your insights and interpretations of a topic. {\bf Don't be absent! } \item Once the lecture of a particular topic has been given, take a fresh look at the notes of the lecture topic. \item Try to understand a single example well, rather than ill-digest multiple examples. \item Start working on the distributed homework ahead of time. \item {\bf Ask questions during the lecture.} There are two main types of questions that you are likely to ask. \begin{enumerate} \item {\em Questions of Correction: Is that a minus sign there?} If you think that, for example, I have missed out a minus sign or wrote $P$ where it should have been $Q$,\footnote{My doctoral adviser used to say ``I said $A$, I wrote $B$, I meant $C$ and it should have been $D$!} then by all means, ask. No one likes to carry an error till line XLV because the audience failed to point out an error on line I. Don't wait till the end of the class to point out an error. Do it when there is still time to correct it! \item {\em Questions of Understanding: I don't get it!} Admitting that you do not understand something is an act requiring utmost courage. But if you don't, it is likely that many others in the audience also don't. On the same vein, if you feel you can explain a point to an inquiring classmate, I will allow you time in the lecture to do so. The best way to ask a question is something like: ``How did you get from the second step to the third step?'' or ``What does it mean to complete the square?'' Asseverations like ``I don't understand'' do not help me answer your queries. If I consider that you are asking the same questions too many times, it may be that you need extra help, in which case we will settle what to do outside the lecture. \end{enumerate} \item Don't fall behind! The sequence of topics is closely interrelated, with one topic leading to another. \item The use of calculators is allowed, especially in the occasional lengthy calculations. However, when graphing, you will need to provide algebraic/analytic/geometric support of your arguments. The questions on assignments and exams will be posed in such a way that it will be of no advantage to have a graphing calculator. \item Presentation is critical. Clearly outline your ideas. When writing solutions, outline major steps and write in complete sentences. As a guide, you may try to emulate the style presented in the scant examples furnished in these notes. \end{itemize} \clearpage \renewcommand{\arraystretch}{2.1} \renewcommand{\chaptermark}{\markboth{\chaptername\ \thechapter}} \renewcommand{\sectionmark}{\markright} \chapter{Preliminaries} \section{Sets and Notation} \pagenumbering{arabic} \setcounter{page}{1} \begin{df}We will mean by a {\em set} \index{sets} a collection of well defined members or {\em elements}.\end{df} \begin{df} The following sets have special symbols. \begin{center} \begin{tabular}{ll} $\BBN = \{0,1,2, 3, \ldots\}$ & denotes the set of natural numbers. \\ $\BBZ = \{\ldots, -3,-2,-1,0,1,2, 3, \ldots\}$ & denotes the set of integers. \\ $\BBQ$ & denotes the set of rational numbers. \\ $\BBR$ & denotes the set of real numbers. \\ $\BBC$ & denotes the set of complex numbers. \\ $\varnothing$ & denotes the empty set.\\ \end{tabular} \end{center} \end{df} \begin{df}[Implications] The symbol $\implies$ is read ``implies'', and the symbol $\Longleftrightarrow$ is read ``if and only if.''\end{df} \begin{exa} Prove that between any two rational numbers there is always a rational number. \end{exa} \begin{solu} Let $(a, c)\in\BBZ^2$, $(b, d)\in (\BBN\setminus\{0\})^2$, $\frac{a}{b}< \frac{c}{d}$. Then $da < bc$. Now $$ab + ad < ab + bc \implies a(b+d) < b(a+c) \implies \dfrac{a}{b} < \dfrac{a+c}{b+d}, $$ $$da + dc < cb + cd \implies d(a+c) < c(b+d) \implies \dfrac{a+c}{b+d} < \dfrac{c}{d}, $$ whence the rational number $\dfrac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$. \end{solu} \begin{rem} We can also argue that the average of two distinct numbers lies between the numbers and so if $r_1 2.$$ Determine, with proof, whether $\sim$ is reflexive, symmetric, and/or transitive. Is $\sim$ an equivalence relation? \end{exa} \begin{solu} Since $0^2 + 0^2 \ngtr 2$, we have $0\nsim 0$ and so $\sim$ is not reflexive. Now, $$\begin{array}{lll} a \sim b & \Leftrightarrow & a^2 + b^2 \\ & \Leftrightarrow & b^2 + a^2 \\ & \Leftrightarrow & b \sim a,\end{array}$$ so $\sim$ is symmetric. Also $0 \sim 3$ since $0^2 + 3^2 > 2$ and $3 \sim 1$ since $3^2 + 1^2 > 2$. But $0 \nsim 1$ since $0^2 + 1^2 \ngtr 2$. Thus the relation is not transitive. The relation, therefore, is not an equivalence relation. \end{solu} \begin{df} Let $\sim$ be an equivalence relation on a set $S$. Then the {\em equivalence class of $a$} is defined and denoted by $$[a] = \{x\in S: x\sim a\} .$$ \end{df} \begin{lem} Let $\sim$ be an equivalence relation on a set $S$. Then two equivalence classes are either identical or disjoint. \label{lem:equiv_classes}\end{lem} \begin{pf} We prove that if $(a, b)\in S^2$, and $[a]\cap [b] \neq \varnothing$ then $[a] = [b]$. Suppose that $x\in [a]\cap [b]$. Now $x\in [a]\implies x\sim a \implies a\sim x$, by symmetry. Similarly, $x\in [b]\implies x\sim b.$ By transitivity $$(a\sim x)\wedge (x\sim b) \implies a\sim b.$$Now, if $y\in [b]$ then $b\sim y$. Again by transitivity, $a\sim y$. This means that $y\in [a]$. We have shewn that $y\in [b]\implies y\in [a]$ and so $[b]\subseteq [a]$. In a similar fashion, we may prove that $[a]\subseteq [b]$. This establishes the result. \end{pf} \begin{thm} Let $S \neq \varnothing$ be a set. Any equivalence relation on $S$ induces a partition of $S$. Conversely, given a partition of $S$ into disjoint, non-empty subsets, we can define an equivalence relation on $S$ whose equivalence classes are precisely these subsets. \label{thm:equiv_relation_yields_partition}\end{thm} \begin{pf} By Lemma \ref{lem:equiv_classes}, if $\sim$ is an equivalence relation on $S$ then $$ S = \bigcup _{a\in S} [a],$$ and $[a]\cap [b] = \varnothing$ if $a\nsim b$. This proves the first half of the theorem. \bigskip Conversely, let $$ S = \bigcup _{\alpha} S_\alpha, \ \ S_\alpha \cap S_\beta = \varnothing\ \ {\rm if}\ \alpha \neq \beta,$$ be a partition of $S$. We define the relation $\approx$ on $S$ by letting $a\approx b$ if and only if they belong to the same $S_\alpha$. Since the $S_\alpha$ are mutually disjoint, it is clear that $\approx$ is an equivalence relation on $S$ and that for $a\in S_\alpha,$ we have $[a] = S_\alpha$. \end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} For $(a, b) \in(\BBQ\setminus \{0\} )^2$ define the relation $\sim$ as follows: $a\sim b \Leftrightarrow \frac{a}{b} \in \BBZ$. Determine whether this relation is reflexive, symmetric, and/or transitive. \begin{answer} $a \sim a$ since $\frac{a}{a} = 1\in \BBZ$, and so the relation is reflexive. The relation is not symmetric. For $2 \sim 1$ since $\frac{2}{1}\in \BBZ$ but $1 \nsim 2$ since $\frac{1}{2} \not\in \BBZ$. The relation is transitive. For assume $a\sim b$ and $b\sim c$. Then there exist $(m, n)\in \BBZ^2$ such that $\frac{a}{b} = m, \frac{b}{c} = n$. This gives $$\frac{a}{c} = \frac{a}{b}\cdot\frac{b}{c} = mn\in\BBZ,$$and so $a\sim c.$ \end{answer} \end{pro} \begin{pro} Give an example of a relation on $\BBZ\setminus \{0\} $ which is reflexive, but is neither symmetric nor transitive. \begin{answer} Here is one possible example: put $a\sim b \Leftrightarrow \frac{a^2 + a}{b}\in \BBZ$. Then clearly if $a\in\BBZ\setminus \{0\} $ we have $a\sim a$ since $\frac{a^2 + a}{a} = a + 1 \in \BBZ$. On the other hand, the relation is not symmetric, since $5\sim 2$ as $\frac{5^2 + 5}{2} = 15\in \BBZ$ but $2\not\sim 5$, as $\frac{2^2 + 2}{5} = \frac{6}{5}\not\in \BBZ$. It is not transitive either, since $\frac{5^2 + 5}{3}\in\BBZ\implies 5 \sim 3$ and $\frac{3^2 + 3}{12}\in\BBZ\implies 3 \sim 12$ but $\frac{5^2 + 5}{12}\not\in\BBZ$ and so $5\nsim 12$. \end{answer} \end{pro} \begin{pro} Define the relation $\sim$ in $\BBR$ by $x\sim y \iff xe^y = ye^x$. Prove that $\sim$ is an equivalence relation. \end{pro} \begin{pro} Define the relation $\sim$ in $\BBQ$ by $x\sim y \iff \exists h\in\BBZ$ such that $x = \dfrac{3y + h}{3}$. [A] Prove that $\sim$ is an equivalence relation. [B] Determine $[x]$, the equivalence of $x\in\BBQ$. [C] Is $\frac{2}{3}\sim\frac{4}{5}$? \begin{answer} [B] $[x] = x + \dfrac{1}{3}\BBZ$. [C] No. \end{answer} \end{pro} \end{multicols} \section{Binary Operations} \index{binary operation} \begin{df} Let $S, T$ be sets. A {\em binary operation} is a function $$\fun{\otimes}{(a, b)}{(a, b)}{S\times S}{T}.$$We usually use the ``infix'' notation $a \otimes b$ rather than the ``prefix'' notation $ \otimes (a, b)$. If $S = T$ then we say that the binary operation is {\em internal} or {\em closed} and if $S \neq T$ then we say that it is {\em external}.\index{binary operation!internal} If $$a\otimes b = b\otimes a$$ then we say that the operation $\otimes$ is {\em commutative}\index{binary operation!commutative} and if $$a\otimes (b \otimes c) = (a\otimes b)\otimes c,$$we say that it is {\em associative.} \index{binary operation!associative} If $\otimes$ is associative, then we can write $$a\otimes (b \otimes c) = (a \otimes b) \otimes c = a \otimes b \otimes c,$$without ambiguity. \end{df}\begin{rem}We usually omit the sign $\otimes$ and use juxtaposition to indicate the operation $\otimes$. Thus we write $ab$ instead of $a\otimes b$. \end{rem} \begin{exa} The operation $+$ (ordinary addition) on the set $\BBZ \times \BBZ$ is a commutative and associative closed binary operation. \end{exa} \begin{exa} The operation $-$ (ordinary subtraction) on the set $\BBN \times \BBN$ is a non-commutative, non-associative non-closed binary operation. \end{exa} \begin{exa} The operation $\otimes$ defined by $a \otimes b = 1 + ab$ on the set $\BBZ\times \BBZ$ is a commutative but non-associative internal binary operation. For $$a \otimes b = 1 + ab = 1 + ba = b a,$$proving commutativity. Also, $1 \otimes (2 \otimes 3) = 1 \otimes (7) = 8$ and $(1 \otimes 2) \otimes 3 = (3) \otimes 3 = 10$, evincing non-associativity. \end{exa} \begin{df} Let $S$ be a set and $ \otimes : S\times S \rightarrow S$ be a closed binary operation. The couple $\algebra{S}{ \otimes }$ is called an {\em algebra}.\index{algebra} \end{df} \begin{rem} When we desire to drop the sign $\otimes$ and indicate the binary operation by juxtaposition, we simply speak of the ``algebra $S$.'' \end{rem} \begin{exa} Both $\algebra{\BBZ}{+}$ and $\algebra{\BBQ}{\cdot}$ are algebras. Here $+$ is the standard addition of real numbers and $\cdot$ is the standard multiplication. \end{exa} \begin{exa} $\algebra{\BBZ}{-}$ is a non-commutative, non-associative algebra. Here $-$ is the standard subtraction operation on the real numbers \end{exa} \begin{exa}[Putnam Exam, 1972] Let $S$ be a set and let $*$ be a binary operation of $S$ satisfying the laws $\forall (x, y)\in S^2$ \begin{equation}x*(x*y) = y,\label{eq:law_1}\end{equation} \begin{equation}(y*x)*x = y.\label{eq:law_2}\end{equation} Shew that $*$ is commutative, but not necessarily associative. \end{exa} \begin{solu} By (\ref{eq:law_2}) $$x*y = ((x*y)*x)*x.$$ By (\ref{eq:law_2}) again $$((x*y)*x)*x = ((x*y)*((x*y)*y))*x.$$ By (\ref{eq:law_1}) $$((x*y)*((x*y)*y))*x = (y)*x = y*x,$$ which is what we wanted to prove. \bigskip To shew that the operation is not necessarily associative, specialise $S = \BBZ$ and $x*y = -x-y$ (the opposite of $x$ minus $y$). Then clearly in this case $*$ is commutative, and satisfies (\ref{eq:law_1}) and (\ref{eq:law_2}) but $$0*(0*1) = 0*(-0-1) = 0*(-1) = -0-(-1) = 1,$$ and $$(0*0)*1 = (-0-0)*1 = (0)*1 = -0 - 1 = -1,$$ evincing that the operation is not associative. \end{solu} \begin{df} Let $S$ be an algebra. Then $l\in S$ is called a {\em left identity} if $\forall s\in S$ we have $l s = s$. Similarly $r\in S$ is called a {\em right identity} if $\forall s\in S$ we have $s r = s$. \end{df} \begin{thm} If an algebra $S$ possesses a left identity $l$ and a right identity $r$ then $l = r.$ \end{thm} \begin{pf} Since $l$ is a left identity $$r = l r.$$Since $r$ is a right identity $$l = l r.$$Combining these two, we gather $$r = l r = l,$$whence the theorem follows. \end{pf} \begin{exa} In $\algebra{\BBZ}{+}$ the element $0\in \BBZ$ acts as an identity, and in $\algebra{\BBQ}{\cdot}$ the element $1\in \BBQ$ acts as an identity. \end{exa} \begin{df} Let $S$ be an algebra. An element $a\in S$ is said to be {\em left-cancellable} or {\em left-regular} if $\forall (x, y) \in S^2$ $$a x = a y \implies x = y.$$Similarly, element $b\in S$ is said to be {\em right-cancellable} or {\em right-regular} if $\forall (x, y) \in S^2$ $$x b = y b\implies x = y.$$ Finally, we say an element $c\in S$ is {\em cancellable} or {\em regular} if it is both left and right cancellable. \end{df} \begin{df} Let $\algebra{S}{ \otimes }$ and $\algebra{S}{\top}$ be algebras. We say that $\top$ is {\em left-distributive} with respect to $ \otimes $ if $$\forall (x, y, z)\in S^3, \ x\top (y \otimes z) = (x\top y)\otimes(x\top z).$$Similarly, we say that $\top$ is {\em right-distributive} with respect to $ \otimes $ if $$\forall (x, y, z)\in S^3, \ (y \otimes z)\top x = (y\top x)\otimes (z\top x).$$We say that $\top$ is {\em distributive} with respect to $ \otimes $ if it is both left and right distributive with respect to $ \otimes $. \end{df} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Let $$S = \{x\in\BBZ : \exists (a, b)\in \BBZ^2, x= a^3 + b^3 + c^3 -3abc \}.$$ Prove that $S$ is closed under multiplication, that is, if $x\in S$ and $y\in S$ then $xy \in S$. \begin{answer} Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then $\omega^2 + \omega + 1 = 0$ and $\omega^3=1$. Then $$x= a^3 + b^3 + c^3 -3abc = (a +b+c)(a + \omega b + \omega^2 c)(a + \omega^2b + c\omega), $$ $$y = u^3 + v^3 + w^3 - 3uvw = (u + v + w)(u + \omega v + \omega^2 w)(u + \omega^2v+ \omega w).$$ Then $$(a+b+c)(u+v+w) = au + av + aw + bu + bv + bw + cu + cv + cw, $$ $$\begin{array}{lll}(a + \omega b + \omega^2 c)(u + \omega v + \omega^2 w) & = & au + bw + cv \\ & & \qquad +\omega (av + bu + cw) \\ & & \qquad +\omega^2 (aw+bv+cu), \\ \end{array}$$ and $$\begin{array}{lll}(a + \omega^2 b + \omega c)(u + \omega^2 v + \omega w) & = & au + bw + cv \\ & & \qquad +\omega (aw + bv + cu) \\ & & \qquad +\omega^2 (av + bu + cw). \\ \end{array}$$ This proves that $$\begin{array}{lll}xy & = &(au + bw + cv)^3 + (aw + bv + cu)^3 + (av + bu + cw)^3\\ & & -3(au + bw + cv)(aw + bv + cu)(av + bu + cw), \\ \end{array} $$which proves that $S$ is closed under multiplication. \end{answer} \end{pro} \begin{pro} Let $\algebra{S}{ \otimes }$ be an associative algebra, let $a\in S$ be a fixed element and define the closed binary operation $\top$ by $$x\top y = x \otimes a \otimes y.$$Prove that $\top$ is also associative over $S\times S$. \begin{answer} We have $$x \top (y \top z) = x\top (y \otimes a \otimes z) = (x) \otimes (a) \otimes (y \otimes a \otimes z ) = x \otimes a \otimes y \otimes a \otimes z,$$where we may drop the parentheses since $ \otimes $ is associative. Similarly $$(x \top y) \top z = (x \otimes a \otimes y) \top z = (x \otimes a \otimes y) \otimes (a) \otimes (z) = x \otimes a \otimes y \otimes a \otimes z .$$ By virtue of having proved $$ x \top (y \top z) = (x \top y) \top z,$$associativity is established. \end{answer} \end{pro} \begin{pro} On $\BBQ \cap ]-1;1[$ define the a binary operation $ \otimes $ $$a \otimes b = \frac{a + b}{1 + ab},$$where juxtaposition means ordinary multiplication and $+$ is the ordinary addition of real numbers. Prove that \begin{dingautolist}{202} \item Prove that $ \otimes $ is a closed binary operation on $\BBQ \cap ]-1;1[$. \\ \item Prove that $ \otimes $ is both commutative and associative. \\ \item Find an element $e\in\BBR$ such that $(\forall a \in \BBQ \cap ]-1;1[)\ (e \otimes a = a)$. \\ \item Given $e$ as above and an arbitrary element $a\in \BBQ \cap ]-1;1[$, solve the equation $a \otimes b = e$ for $b$. \\ \end{dingautolist} \begin{answer} We proceed in order.\begin{dingautolist}{202} \item Clearly, if $a, b$ are rational numbers, $$|a| < 1, |b|<1 \implies |ab| < 1 \implies -1 < ab < 1 \implies 1 + ab > 0,$$whence the denominator never vanishes and since sums, multiplications and divisions of rational numbers are rational, $\dfrac{a + b}{1 + ab}$ is also rational. We must prove now that $-1 < \dfrac{a + b}{1 + ab} < 1$ for $(a, b)\in ]-1; 1[^2$. We have $$\begin{array}{lll} -1 < \dfrac{a + b}{1 + ab} < 1 & \Leftrightarrow & -1 - ab < a + b < 1 + ab \\ & \Leftrightarrow & -1 -ab - a - b < 0 < 1 + ab - a - b \\ & \Leftrightarrow & -(a + 1)(b + 1) < 0 < (a - 1)(b - 1). \end{array}$$ Since $(a, b)\in ]-1; 1[^2$, $(a + 1)(b + 1) > 0$ and so $-(a + 1)(b + 1) < 0$ giving the sinistral inequality. Similarly $a - 1 < 0$ and $b - 1 < 0$ give $(a - 1)(b - 1) > 0$, the dextral inequality. Since the steps are reversible, we have established that indeed $-1 < \dfrac{a + b}{1 + ab} < 1$. \item Since $a \otimes b = \dfrac{a + b}{1 + ab} = \dfrac{b + a}{1 + ba} = b \otimes a$, commutativity follows trivially. Now $$\begin{array}{lll}a \otimes (b \otimes c) & = & a \otimes \left(\dfrac{b + c}{1 + bc}\right) \vspace{2mm} \\ & = & \dfrac{a + \left(\dfrac{b + c}{1 + bc}\right)}{1 + a\left(\dfrac{b + c}{1 + bc}\right)} \vspace{2mm} \\ & = & \dfrac{a(1 + bc) + b + c}{1 + bc + a(b + c)} = \dfrac{a + b + c + abc}{1 + ab + bc + ca}. \end{array}$$One the other hand, $$\begin{array}{lll} (a \otimes b) \otimes c & = & \left(\dfrac{a + b}{1 + ab}\right) \otimes c \vspace{2mm} \\ & = & \dfrac{\left(\dfrac{a + b}{1 + ab}\right) + c}{1 + \left(\dfrac{a + b}{1 + ab}\right)c} \vspace{2mm} \\ & = & \dfrac{(a + b)+ c(1 + ab)}{1 + ab + (a + b)c} \vspace{2mm} \\ & = & \dfrac{a + b + c + abc}{1 + ab + bc + ca},\end{array}$$ whence $ \otimes $ is associative. \item If $a \otimes e = a$ then $\dfrac{a + e}{1 + ae} = a$, which gives $a + e = a + ea^2$ or $e(a^2 - 1) = 0$. Since $a \neq \pm 1$, we must have $e = 0$. \item If $a \otimes b = 0$, then $\dfrac{a + b}{1 + ab} = 0$, which means that $b = -a$. \end{dingautolist} \end{answer} \end{pro} \begin{pro} On $\BBR \setminus \{1\}$ define the a binary operation $ \otimes $ $$a \otimes b = a+b-ab,$$where juxtaposition means ordinary multiplication and $+$ is the ordinary addition of real numbers. Clearly $\otimes$ is a closed binary operation. Prove that \begin{dingautolist}{202} \item Prove that $ \otimes $ is both commutative and associative. \\ \item Find an element $e\in\BBR \setminus \{1\}$ such that $(\forall a \in \BBR \setminus \{1\})\ (e \otimes a = a)$. \\ \item Given $e$ as above and an arbitrary element $a\in \BBR \setminus \{1\}$, solve the equation $a \otimes b = e$ for $b$. \\ \end{dingautolist} \begin{answer} We proceed in order.\begin{dingautolist}{202} \item Since $a \otimes b = a+b-ab = b+a-ba = b \otimes a$, commutativity follows trivially. Now $$\begin{array}{lll}a \otimes (b \otimes c) & = & a\otimes (b + c -bc) \\ & = & a + b+c -bc -a(b+c-bc)\\ & = & a+b+c-ab-bc-ca + abc. \end{array}$$One the other hand, $$\begin{array}{lll} (a \otimes b) \otimes c & = & \left(a+b-ab\right) \otimes c \\ & = & a+b-ab + c - (a+b-ab)c \\ & = & a+b + c -ab-bc -ca + abc, \\ \end{array}$$ whence $ \otimes $ is associative. \item If $a \otimes e = a$ then $a+e-ae = a$, which gives $e(1-a) = 0$. Since $a \neq 1$, we must have $e = 0$. \item If $a \otimes b = 0$, then $a+b-ab = 0$, which means that $b(1-a) = -a$. Since $a\neq 1$ we find $b = -\dfrac{a}{1-a}$. \end{dingautolist} \end{answer} \end{pro} \begin{pro}[{\red\bf Putnam Exam, 1971}] Let $S$ be a set and let $\circ$ be a binary operation on $S$ satisfying the two laws $$(\forall x\in S) (x \circ x = x),$$and $$(\forall (x, y, z) \in S^3) ((x\circ y)\circ z = (y \circ z)\circ x). $$ Shew that $\circ$ is commutative. \begin{answer} We have $$\begin{array}{lll} x\circ y & = & (x\circ y)\circ (x\circ y) \\ & = & [y\circ (x \circ y)]\circ x \\ & = & [(x \circ y)\circ x] \circ y \\ & = & [(y \circ x)\circ x] \circ y \\ & = & [(x \circ x)\circ y] \circ y \\ & = & (y \circ y) \circ (x\circ x) \\ & = & y\circ x, \\ \end{array}$$proving commutativity. \end{answer} \end{pro} \begin{pro} Define the {\em symmetric difference} of the sets $A, B$ as $A \triangle B = (A\setminus B) \cup (B\setminus A)$. Prove that $\triangle$ is commutative and associative. \end{pro} \end{multicols} \section{$\BBZ_n$} \index{division algorithm} \begin{thm}[Division Algorithm] Let $n > 0$ be an integer. Then for any integer $a$ there exist unique integers $q$ (called the {\em quotient}) and $r$ (called the {\em remainder}) such that $a = qn + r$ and $0 \leq r < q$. \label{thm:division_algorithm} \end{thm} \begin{pf} In the proof of this theorem, we use the following property of the integers, called the {\em well-ordering principle}: any non-empty set of non-negative integers has a smallest element. \bigskip Consider the set $$S = \{ a -bn: b\in \BBZ \wedge a \geq bn\}.$$ Then $S$ is a collection of nonnegative integers and $S \neq \varnothing$ as $\pm a - 0\cdot n \in S$ and this is non-negative for one choice of sign. By the Well-Ordering Principle, $S $ has a least element, say $r$. Now, there must be some $q \in \BBZ$ such that $r = a - qn$ since $r \in S $. By construction, $r \geq 0.$ Let us prove that $r < n$. For assume that $r \geq n.$ Then $r > r - n = a - qn - n = a - (q + 1)n \geq 0$, since $r - n \geq 0.$ But then $a - (q + 1)n \in S $ and $a - (q + 1)n < r$ which contradicts the fact that $r$ is the smallest member of $S $. Thus we must have $0 \leq r < n.$ To prove that $r$ and $q$ are unique, assume that $q_1n + r_1 = a = q_2n + r_2$, $0 \leq r_1 < n$, $0 \leq r_2 < n. $ Then $r_2 - r_1 = n(q_1 - q_2)$, that is, $n$ divides $(r_2 - r_1)$. But $|r_2 - r_1| < n,$ whence $r_2 = r_1$. From this it also follows that $q_1 = q_2.$ This completes the proof. \end{pf} \begin{exa} If $n = 5$ the Division Algorithm says that we can arrange all the integers in five columns as follows: $$\begin{array}{rrrrr} \vdots & \vdots & \vdots & \vdots & \vdots \\ -10 & -9 & -8 & -7 & -6 \\ -5 & -4 & -3 & -2 & -1 \\ 0 & 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 & 9 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \end{array}$$ The arrangement above shews that any integer comes in one of $5$ flavours: those leaving remainder $0$ upon division by $5$, those leaving remainder $1$ upon division by $5$, etc. We let $$5\BBZ = \{\ldots, -15, -10, -5, 0 , 5, 10, 15, \ldots\} = \overline{0},$$ $$5\BBZ + 1 = \{\ldots, -14, -9, -4, 1 , 6, 11, 16, \ldots\} = \overline{1},$$ $$5\BBZ + 2= \{\ldots, -13, -8, -3, 2 , 7, 12, 17, \ldots\} = \overline{2},$$ $$5\BBZ + 3= \{\ldots, -12, -7, -2, 3 , 8, 13, 18, \ldots\}=\overline{3},$$ $$5\BBZ + 4 = \{\ldots, -11, -6, -1, 4 , 9, 14, 19, \ldots\}=\overline{4},$$ and $$\BBZ_5 = \{\overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4} \}.$$ \end{exa} Let $n $ be a fixed positive integer. Define the relation $\equiv$ by $x\equiv y$ if and only if they leave the same remainder upon division by $n$. Then clearly $\equiv$ is an equivalence relation. As such it partitions the set of integers $\BBZ$ into disjoint equivalence classes by Theorem \ref{thm:equiv_relation_yields_partition}. This motivates the following definition. \begin{df} Let $n $ be a positive integer. The $n$ {\em residue classes} upon division by $n$ are $$\overline{0} = n\BBZ, \ \ \overline{1} = n\BBZ + 1, \ \ \overline{2} = n\BBZ + 2, \ \ \ldots, \ \ \overline{n - 1} = n\BBZ + n - 1.$$ The {\em set of residue classes modulo} $n$ is $$\BBZ_n = \{\overline{0}, \overline{1}, \ldots , \overline{n - 1}\}.$$\index{residue classes} \end{df} Our interest is now to define some sort of ``addition'' and some sort of ``multiplication'' in $\BBZ_n$. \begin{thm}[Addition and Multiplication Modulo $n$] Let $n$ be a positive integer. For $(\overline{a\vphantom{b}}, \overline{b})\in (\BBZ_n)^2$ define $\overline{a\vphantom{b}} + \overline{b} = \overline{r}$, where $r$ is the remainder of $a + b$ upon division by $n$. and $\overline{a\vphantom{b}}\cdot \overline{b} = \overline{t}$, where $t$ is the remainder of $ab$ upon division by $n$. Then these operations are well defined. \index{addition modulo n}\index{multiplication modulo n} \end{thm} \begin{pf} We need to prove that given arbitrary representatives of the residue classes, we always obtain the same result from our operations. That is, if $\overline{a\vphantom{'}} = \overline{a'}$ and $\overline{b\vphantom{'}} = \overline{b'}$ then we have $\overline{a\vphantom{b}} + \overline{b} = \overline{a'} + \overline{b'}$ and $\overline{a\vphantom{b}}\cdot\overline{b} = \overline{a'\vphantom{b'}}\cdot\overline{b'}$. \bigskip Now $$\begin{array}{l}\overline{a} = \overline{a}'\implies \exists (q,q')\in\BBZ^2, r \in \BBN, a = qn + r, \ a' = q'n + r, \ 0 \leq r < n, \\ \overline{b} = \overline{b}'\implies \exists (q_1,q_1')\in\BBZ^2, r_1 \in \BBN, b = q_1n + r_1, \ b' = q_1'n + r_1, \ 0 \leq r_1 < n. \end{array}$$ Hence $$a+b = (q + q_1)n + r + r_1, \ \ \ a'+b' = (q' + q_1')n + r + r_1, $$meaning that both $a+b$ and $a'+b'$ leave the same remainder upon division by $n$, and therefore $$\overline{a\vphantom{b}} + \overline{b} = \overline{a+b} = \overline{a'+b'} = \overline{a'} + \overline{b'}.$$ \bigskip Similarly $$ab = (qq_1n + qr_1 + rq_1)n + rr_1, \ \ \ a'b' = (q'q_1'n + q'r_1 + rq_1')n + rr_1, $$and so both $ab$ and $a'b'$ leave the same remainder upon division by $n$, and therefore $$\overline{a\vphantom{b}}\cdot\overline{b} = \overline{ab} = \overline{a'b'} = \overline{a'\vphantom{b'}}\cdot\overline{b'}.$$ This proves the theorem. \end{pf} \begin{exa} Let $$\BBZ_6 = \{\overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5}\}$$be the residue classes modulo $6$. Construct the natural addition $+$ table for $\BBZ_6$. Also, construct the natural multiplication $\cdot$ table for $\BBZ_6$. \end{exa} \begin{solu}The required tables are given in tables \ref{tab:add_z6} and \ref{tab:mult_z6}. \end{solu} \begin{table}[h]\begin{minipage}{6cm}$$ \begin{array}{|c||c|c|c|c|c|c|} \hline + & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} \\ \hline \overline{1} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{0} \\ \hline \overline{2} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{0} & \overline{1} \\ \hline \overline{3} & \overline{3} & \overline{4} & \overline{5} & \overline{0} & \overline{1} & \overline{2} \\ \hline \overline{4} & \overline{4} & \overline{5} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \hline \overline{5} & \overline{5} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} \\ \hline \end{array} $$\footnotesize\hangcaption{Addition table for $\BBZ_6$.} \label{tab:add_z6}\end{minipage}\hfill \begin{minipage}{6cm}$$ \begin{array}{|c||c|c|c|c|c|c|} \hline \cdot & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} \\ \hline \overline{1} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} \\ \hline \overline{2} & \overline{0} & \overline{2} & \overline{4} & \overline{0} & \overline{2} & \overline{4} \\ \hline \overline{3} & \overline{0} & \overline{3} & \overline{0} & \overline{3} & \overline{0} & \overline{3} \\ \hline \overline{4} & \overline{0} & \overline{4} & \overline{2} & \overline{0} & \overline{4} & \overline{2} \\ \hline \overline{5} & \overline{0} & \overline{5} & \overline{4} & \overline{3} & \overline{2} & \overline{1} \\ \hline \end{array} $$\footnotesize\hangcaption{Multiplication table for $\BBZ_6$.} \label{tab:mult_z6}\end{minipage}\hfill \end{table} \bigskip We notice that even though $\overline{2} \neq \overline{0}$ and $\overline{3} \neq \overline{0}$ we have $\overline{2}\cdot \overline{3} = \overline{0}$ in $\BBZ_6$. This prompts the following definition. \begin{df}[Zero Divisor] An element $a\neq \overline{0}$ of $\BBZ_n$ is called a {\em zero divisor} if $ab = \overline{0}$ for some $b\in \BBZ_n$. \index{zero divisor!Zn} \end{df} We will extend the concept of zero divisor later on to various algebras. \begin{exa} Let $$\BBZ_7 = \{\overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5}, \overline{6}\}$$be the residue classes modulo $7$. Construct the natural addition $+$ table for $\BBZ_7$. Also, construct the natural multiplication $\cdot$ table for $\BBZ_7$\\ \end{exa} \begin{solu}The required tables are given in tables \ref{tab:add_z7} and \ref{tab:mult_z7}. \end{solu} \begin{table}[h] \begin{minipage}{6cm} $$ \begin{array}{|c||c|c|c|c|c|c|c|} \hline + & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} \\ \hline \overline{1} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{0} \\ \hline \overline{2} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{0} & \overline{1} \\ \hline \overline{3} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{0} & \overline{1} & \overline{2} \\ \hline \overline{4} & \overline{4} & \overline{5} & \overline{6} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \hline \overline{5} & \overline{5} & \overline{6} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} \\ \hline \overline{6} & \overline{6} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} \\ \hline \end{array} $$\footnotesize\hangcaption{Addition table for $\BBZ_7$.} \label{tab:add_z7}\end{minipage}\hfill \begin{minipage}{6cm}$$ \begin{array}{|c||c|c|c|c|c|c|c|} \hline \cdot & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} \\ \hline \overline{1} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} \\ \hline \overline{2} & \overline{0} & \overline{2} & \overline{4} & \overline{6} & \overline{1} & \overline{3} & \overline{5} \\ \hline \overline{3} & \overline{0} & \overline{3} & \overline{6} & \overline{2} & \overline{5} & \overline{1} & \overline{4} \\ \hline \overline{4} & \overline{0} & \overline{4} & \overline{1} & \overline{5} & \overline{2} & \overline{6} & \overline{3} \\ \hline \overline{5} & \overline{0} & \overline{5} & \overline{3} & \overline{1} & \overline{6} & \overline{4} & \overline{2} \\ \hline \overline{6} & \overline{0} & \overline{6} & \overline{5} & \overline{4} & \overline{3} & \overline{2} & \overline{1} \\ \hline \end{array} $$ \footnotesize\hangcaption{Multiplication table for $\BBZ_7$.} \label{tab:mult_z7} \end{minipage}\hfill \end{table} \begin{exa} Solve the equation $$\overline{5}x = \overline{3}$$in $\BBZ_{11}$. \label{exa:lineqmod11}\end{exa} \begin{solu}Multiplying by $\overline{9}$ on both sides $$\overline{45}x = \overline{27},$$that is, $$x = \overline{5}.$$ \end{solu} We will use the following result in the next section. \begin{df} Let $a$, $b$ be integers with one of them different from $0$. The greatest common divisor $d$ of $a, b$, denoted by $d = \gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. \end{df} \begin{thm}[Bachet-Bezout Theorem] The greatest common divisor of any two integers $a, b$ can be written as a linear combination of $a$ and $b$, i.e., there are integers $x, y$ with $$ \gcd(a, b) = ax + by.$$ \index{greatest common divisor} \label{thm:bachet_bezout} \index{theorem!Bachet-Bezout} \end{thm} \begin{pf} Let $A = \{ ax + by : ax + by > 0, x, y \in \BBZ\}$. Clearly one of $\pm a, \pm b$ is in $A$, as one of $a, b$ is not zero. By the Well Ordering Principle, $A$ has a smallest element, say $d$. Therefore, there are $x_0, y_0$ such that $d = ax_0 + by_0.$ We prove that $d = \gcd(a, b).$ To do this we prove that $d$ divides $a$ and $b$ and that if $t$ divides $a$ and $b$, then $t$ must also divide then $d$. \bigskip We first prove that $d$ divides $a.$ By the Division Algorithm, we can find integers $q, r, 0 \leq r < d$ such that $a = dq + r.$ Then $$ r = a - dq = a(1 - qx_0) - by_0.$$If $r > 0,$ then $r \in A$ is smaller than the smaller element of $A$, namely $d,$ a contradiction. Thus $r = 0.$ This entails $dq = a,$ i.e. $d$ divides $a.$ We can similarly prove that $d$ divides $b.$ \bigskip Assume that $t$ divides $a$ and $b$. Then $a = tm, b = tn$ for integers $m, n.$ Hence $d = ax_0 + bx_0 = t(mx_0 + ny_0),$ that is, $t$ divides $d.$ The theorem is thus proved. \end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Write the addition and multiplication tables of $\BBZ_{11}$ under natural addition and multiplication modulo $11$.\begin{answer} The tables appear in tables \ref{tab:add_z11} and \ref{tab:mult_z11}. \begin{table} \centering \begin{minipage}{7cm} $$ {\tiny \begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline + & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} \\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} \\ \hline \overline{1} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} & \overline{0} \\ \hline \overline{2} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8}& \overline{9} & \overline{10} & \overline{0} & \overline{1} \\ \hline \overline{3} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} & \overline{0} & \overline{1} & \overline{2} \\ \hline \overline{4} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} & \overline{0} & \overline{1} & \overline{2} & \overline{3} \\ \hline \overline{5} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} & \overline{0}& \overline{1} & \overline{2} & \overline{3} & \overline{4} \\ \hline \overline{6} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} \\ \hline \overline{7} & \overline{7} & \overline{0} & \overline{9} & \overline{10} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} \\ \hline \overline{8}& \overline{8} & \overline{9} & \overline{10} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} \\ \hline \overline{9} &\overline{9} & \overline{10} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5}& \overline{6} & \overline{7} & \overline{8} \\ \hline \overline{10} & \overline{10}& \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} \\ \hline \end{array}} $$ \footnotesize\hangcaption{Addition table for $\BBZ_{11}$.} \label{tab:add_z11} \end{minipage}\hspace{2cm} \begin{minipage}{7cm} $${\tiny \begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline \cdot & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6} & \overline{7} & \overline{8} & \overline{9} & \overline{10} \\ \hline \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} & \overline{0} \\ \hline \overline{1} & \overline{0} & \overline{1} & \overline{2} & \overline{3} & \overline{4} & \overline{5} & \overline{6}& \overline{7} & \overline{8} & \overline{9} & \overline{10} \\ \hline \overline{2} & \overline{0} & \overline{2} & \overline{4} & \overline{6} & \overline{8} & \overline{10} & \overline{1}& \overline{3} & \overline{5} & \overline{7} & \overline{9} \\ \hline \overline{3} & \overline{0} & \overline{3} & \overline{6} & \overline{9} & \overline{1} & \overline{4} & \overline{7} & \overline{10} & \overline{2} & \overline{5} & \overline{8} \\ \hline \overline{4} & \overline{0} & \overline{4} & \overline{8} & \overline{1} & \overline{5} & \overline{9} & \overline{2} & \overline{6} & \overline{10} & \overline{3} & \overline{7} \\ \hline \overline{5} & \overline{0} & \overline{5} & \overline{10} & \overline{4} & \overline{9} & \overline{3} & \overline{8} & \overline{2} & \overline{7} & \overline{1} & \overline{6} \\ \hline \overline{6} & \overline{0} & \overline{6} & \overline{1} & \overline{7} & \overline{2} & \overline{8} & \overline{3} & \overline{9} & \overline{4} & \overline{10} & \overline{5} \\ \hline \overline{7} & \overline{0} & \overline{7} & \overline{3} & \overline{10} & \overline{6} & \overline{2} & \overline{9} & \overline{5} & \overline{1} & \overline{8} & \overline{4} \\ \hline \overline{8} & \overline{0} & \overline{8} & \overline{5} & \overline{2} & \overline{10} & \overline{7} & \overline{4} & \overline{1} & \overline{9} & \overline{6} & \overline{3} \\ \hline \overline{9} & \overline{0} & \overline{9} & \overline{7} & \overline{5} & \overline{3} & \overline{1} & \overline{10}& \overline{8} & \overline{6} & \overline{4} & \overline{2} \\ \hline \overline{10} & \overline{0} & \overline{10} & \overline{9} & \overline{8} & \overline{7} & \overline{6} & \overline{5} & \overline{4} & \overline{3} & \overline{2} & \overline{1} \\ \hline \end{array}} $$ \footnotesize\hangcaption{Multiplication table $\BBZ_{11}$.}\label{tab:mult_z11} \end{minipage} \end{table} \end{answer} \end{pro} \begin{pro} Solve the equation $\overline{3}x^2 - \overline{5}x + \overline{1} = \overline{0}$ in $\BBZ_{11}$. \begin{answer} Observe that $$\overline{3}x^2 - \overline{5}x + \overline{1} = \overline{0} \implies \overline{4}(\overline{3}x^2 - \overline{5}x + \overline{1}) = \overline{4}\overline{0} \implies x^2 + \overline{2}x + \overline{1} + \overline{3} = \overline{0} \implies (x+\overline{1})^2 = \overline{8}.$$ We need to know whether $\overline{8}$ is a perfect square modulo $11$. Observe that $(\overline{11}-\overline{a})^2 = \overline{a}$, so we just need to check half the elements and see that $$\overline{1}^2 = \overline{1};\quad \overline{2}^2 = \overline{4};\quad \overline{3}^2 = \overline{9};\quad \overline{4}^2 = \overline{5};\quad \overline{5}^2 = \overline{3}, $$ whence $\overline{8}$ is not a perfect square modulo $11$ and so there are no solutions. \end{answer} \end{pro} \begin{pro}Solve the equation $$\overline{5}x^2 = \overline{3}$$in $\BBZ_{11}$. \begin{answer} From example \ref{exa:lineqmod11} $$x^2 = \overline{5}.$$ Now, the squares modulo $11$ are $\overline{0}^2 = \overline{0}$, $\overline{1}^2 = \overline{1}$, $\overline{2}^2 = \overline{4}$, $\overline{3}^2 = \overline{9}$, $\overline{4}^2 = \overline{5}$, $\overline{5}^2 = \overline{3}$. Also, $(\overline{11 - 4})^2 = \overline{7}^2 = \overline{5}$. Hence the solutions are $x = \overline{4}$ or $x = \overline{7}$. \end{answer} \end{pro} \begin{pro} Prove that if $n > 0$ is a composite integer, $\BBZ_n$ has zero divisors. \end{pro} \begin{pro} How many solutions does the equation $x^4+x^3+x^2+x+\overline{1}=\overline{0}$ have in $\BBZ_{11}$? \begin{answer} Put $f(x)= x^4+x^3+x^2+x+\overline{1}$. Then $$\begin{array} {l|l|l|} f(0) =1 \equiv 1 \mod 11 & f(1) =5 \equiv 5 \mod 11 & f(2) =31 \equiv 9 \mod 11\\ f(3) =121 \equiv 0 \mod 11 & f(4) =341 \equiv 0 \mod 11 & f(5) =781 \equiv 0 \mod 11\\ f(6) =1555 \equiv 4 \mod 11 & f(7) =2801 \equiv 7 \mod 11 & f(8) =4681 \equiv 6 \mod 11\\ f(9) =7381 \equiv 0 \mod 11 & f(10) =11111 \equiv 1 \mod 11 & \\ \end{array}$$ \end{answer} \end{pro} \end{multicols} \section{Fields} \index{field} \begin{df} Let $\BBF$ be a set having at least two elements $0_{\BBF }$ and $1_{\BBF }$ ($0_{\BBF } \neq 1_{\BBF }$) together with two operations $\cdot$ (multiplication, which we usually represent via juxtaposition) and $+$ (addition). A {\em field} $\field{ \BBF }{\cdot}{+}$ is a triplet satisfying the following axioms $\forall (a,b, c)\in \BBF ^3$: \begin{enumerate} \item[F1] Addition and multiplication are associative: \begin{equation}(a + b)+c = a+(b+c), \ \ \ (ab)c = a(bc)\label{fa:associative}\end{equation} \item[F2] Addition and multiplication are commutative: \begin{equation}\label{fa:commutative}a + b = b + a, \ \ \ ab = ba\end{equation} \item[F3] The multiplicative operation distributes over addition: \begin{equation}\label{fa:distributive}a(b+c) = ab + ac\end{equation} \item[F4] $0_{\BBF }$ is the additive identity: \begin{equation}\label{fa:existence_of_0}0_{\BBF } + a = a + 0_{\BBF } = a\end{equation} \item[F5] $1_{\BBF }$ is the multiplicative identity: \begin{equation}\label{fa:existence_of_1}1_{\BBF }a = a1_{\BBF } = a\end{equation} \item[F6] Every element has an additive inverse: \begin{equation}\label{fa:existence_of_additive_inverse}\exists -a\in \BBF, \ \ a + (-a) = (-a)+a = 0_{\BBF }\end{equation} \item[F7] Every non-zero element has a multiplicative inverse: if $a\neq 0_{\BBF }$\begin{equation}\label{fa:existence_of_multiplicative_inverse}\exists a^{-1}\in \BBF, \ \ \ aa^{-1} = a^{-1}a =1_{\BBF }\end{equation} \end{enumerate} The elements of a field are called {\em scalars}. \index{scalar} \end{df} An important property of fields is the following.\index{zero divisor!in a field} \begin{thm}\label{thm:field_no_zero_divisors} A field does not have zero divisors. \end{thm} \begin{pf} Assume that $ab = 0_{\BBF }$. If $a\neq 0_{\BBF }$ then it has a multiplicative inverse $a^{-1}$. We deduce $$a^{-1}ab = a^{-1}0_{\BBF } \implies b = 0_{\BBF }. $$This means that the only way of obtaining a zero product is if one of the factors is $0_{\BBF }$. \end{pf} \begin{exa} $\field{\BBQ}{\cdot}{+}$, $\field{\BBR}{\cdot}{+}$, and $\field{\BBC}{\cdot}{+}$ are all fields. The multiplicative identity in each case is $1$ and the additive identity is $0$. \end{exa} \begin{exa} Let $$\BBQ (\sqrt{2}) = \{a + \sqrt{2}b : (a, b)\in \BBQ^2\}$$ and define addition on this set as $$(a + \sqrt{2}b) + (c + \sqrt{2}d) = (a + c) + \sqrt{2}(b + d),$$and multiplication as $$(a + \sqrt{2}b) (c + \sqrt{2}d) = (ac + 2bd) + \sqrt{2}(ad + bc).$$Then $\field{\BBQ + \sqrt{2}\BBQ}{}{+}$ is a field. Observe $0_{\BBF } = 0$, $1_{\BBF } = 1$, that the additive inverse of $a + \sqrt{2}b$ is $-a - \sqrt{2}b$, and the multiplicative inverse of $a + \sqrt{2}b, (a, b) \neq (0, 0)$ is $$(a + \sqrt{2}b)^{-1} = \frac{1}{a + \sqrt{2}b} = \frac{a - \sqrt{2}b}{a^2 - 2b^2} = \frac{a}{a^2 - 2b^2} - \frac{\sqrt{2}b}{a^2 - 2b^2}. $$ Here $a^2 - 2b^2 \neq 0$ since $\sqrt{2}$ is irrational. \end{exa} \begin{thm}\label{thm:Zmodp_is_a_field} If $p$ is a prime, $\field{\BBZ_p}{\cdot}{+}$ is a field under $\cdot$ multiplication modulo $p$ and $+$ addition modulo $p$. \end{thm} \begin{pf} Clearly the additive identity is $\overline{0}$ and the multiplicative identity is $\overline{1}$. The additive inverse of $\overline{a}$ is $\overline{p - a}$. We must prove that every $\overline{a} \in \BBZ _p \setminus \{ \overline{0}\}$ has a multiplicative inverse. Such an $a$ satisfies $\gcd (a, p) =1$ and by the Bachet-Bezout Theorem \ref{thm:bachet_bezout}, there exist integers $x, y$ with $px + ay = 1.$ In such case we have $$\overline{1} = \overline{px + ay} = \overline{ay} = \overline{a}\cdot \overline{y},$$whence $(\overline{a})^{-1} = \overline{y}.$ \end{pf} \begin{df} A field is said to be of {\em characteristic} $p \neq 0$ if for some positive integer $p$ we have $\forall a\in \BBF, pa = 0_{\BBF }$, and no positive integer smaller than $p$ enjoys this property.\end{df} If the field does not have characteristic $p \neq 0$ then we say that it is of {\em characteristic } $0$. Clearly $\BBQ, \BBR$ and $\BBC$ are of characteristic $0$, while $\BBZ_p$ for prime $p$, is of characteristic $p$. \index{characteristic of a field} \begin{thm} The characteristic of a field is either $0$ or a prime. \end{thm} \begin{pf} If the characteristic of the field is $0$, there is nothing to prove. Let $p$ be the least positive integer for which $\forall a\in \BBF, pa = 0_{\BBF }$. Let us prove that $p$ must be a prime. Assume that instead we had $p = st$ with integers $s>1, t>1$. Take $a=1_{\BBF }$. Then we must have $(st)1_{\BBF } = 0_{\BBF }$, which entails $(s1_{\BBF })(t1_{\BBF }) = 0_{\BBF }$. But in a field there are no zero-divisors by Theorem \ref{thm:field_no_zero_divisors}, hence either $s1_{\BBF } = 0_{\BBF }$ or $t1_{\BBF } = 0_{\BBF }$. But either of these equalities contradicts the minimality of $p$. Hence $p$ is a prime. \end{pf} \section*{\psframebox{Homework}} \begin{pro} Consider the set of numbers $$\BBQ (\sqrt{2}, \sqrt{3}, \sqrt{6}) = \{a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}: (a, b, c, d)\in\BBQ^4\}. $$Assume that $\BBQ (\sqrt{2}, \sqrt{3}, \sqrt{6})$ is a field under ordinary addition and multiplication. What is the multiplicative inverse of the element $ \sqrt{2} + 2\sqrt{3} + 3\sqrt{6}$? \begin{answer} We have $$\begin{array}{lll} \dfrac{1}{\sqrt{2} + 2\sqrt{3} + 3\sqrt{6}} & = & \dfrac{ \sqrt{2} + 2\sqrt{3} - 3\sqrt{6}}{ (\sqrt{2} + 2\sqrt{3})^2 - (3\sqrt{6})^2} \\ & = & \dfrac{ \sqrt{2} + 2\sqrt{3} - 3\sqrt{6}}{2 + 12+ 4\sqrt{6} - 54} \\ & = & \dfrac{ \sqrt{2} + 2\sqrt{3} - 3\sqrt{6}}{-40 + 4\sqrt{6}} \\ & = & \dfrac{ (\sqrt{2} + 2\sqrt{3} - 3\sqrt{6})(-40 - 4\sqrt{6})}{40^2 - (4\sqrt{6})^2} \\ & = & \dfrac{ (\sqrt{2} + 2\sqrt{3} - 3\sqrt{6})(-40 - 4\sqrt{6})}{1504} \\ & = & -\dfrac{16\sqrt{2}+22\sqrt{3}-30\sqrt{6}-18}{376} \end{array} $$ \end{answer} \end{pro} \begin{pro} Let $\BBF$ be a field and $a, b$ two non-zero elements of $\BBF$. Prove that $$-(ab^{-1}) = (-a)b^{-1} = a(-b^{-1}).$$ \begin{answer} Since $$(-a)b^{-1} + ab^{-1} = (-a + a)b^{-1} = 0_{\BBF }b^{-1} = 0_{\BBF }, $$we obtain by adding $-(ab^{-1})$ to both sides that $$ (-a)b^{-1} = -(ab^{-1}). $$ Similarly, from $$a(-b^{-1}) + ab^{-1} = a(-b^{-1}+ b^{-1}) = a0_{\BBF } = 0_{\BBF }, $$we obtain by adding $-(ab^{-1})$ to both sides that $$ a(-b^{-1}) = -(ab^{-1}). $$ \end{answer} \end{pro} \begin{pro} Let $\BBF$ be a field and $a\neq 0_{\BBF }$. Prove that $$(-a)^{-1} = -(a^{-1}).$$ \end{pro} \begin{pro} Let $\BBF$ be a field and $a, b$ two non-zero elements of $\BBF$. Prove that $$ab^{-1} = (-a)(-b^{-1}).$$ \end{pro} \section{Functions} \begin{df} By a {\em function} or a {\em mapping} from one set to another, we mean a rule or mechanism that assigns to every input element of the first set a unique output element of the second set. We shall call the set of inputs the {\em domain} of the function, the set of {\em possible} outputs the {\em target set} of the function, and the set of {\em actual} outputs the {\em image} of the function. \end{df} We will generally refer to a function with the following notation: $$\fun{f}{x}{f(x)}{D}{T}.$$ Here $f$ is the {\em name of the function}, $D$ is its domain, $T$ is its target set, $x$ is the name of a typical input and $f(x)$ is the output or {\em image of $x$ under $f$}. We call the assignment $x \mapsto f(x)$ the {\em assignment rule} of the function. Sometimes $x$ is also called the {\em independent variable.} The set $f(D) = \{f(a)|a\in D\}$ is called the {\em image} of $f$. Observe that $f(D) \subseteq T.$ \vspace{1cm} \begin{figure}[htb] \centering \begin{minipage}{6cm} $$ \psset{unit=.75cm} \rput(-1.5,0){\rput(1.5, .8){\alpha} \rput(0, .5){1} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,.5)(2.9,.5) \rput(3,.5){2} \rput(0,0){2} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,0)(2.9,0) \rput(3,0){8} \rput(0, -.5){3} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(0.2,-.5)(2.9,-.5) \rput(3,-.5){4} \psellipse(1,2) \psellipse(3,0)(1,2)} $$\vspace{1cm} \footnotesize\hangcaption{An injection.} \label{injection} \end{minipage} \begin{minipage}{6cm}$$ \psset{unit=.75cm} \rput(-1.5,0){\rput(1.5, .8){\beta} \psellipse(1,2) \psellipse(3,0)(1,2) \rput(0,0){2} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,0)(2.9,0) \rput(3,0){2} \rput(0, .5){1} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,.5)(2.9,.5) \rput(3,.5){4} \rput(0, -.5){3} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(0.2,-.5)(2.9,.5)} $$\vspace{1cm} \footnotesize \hangcaption{Not an injection} \label{not_injection} \end{minipage} \end{figure} \begin{df} A function $\dis{\fun{f}{x}{f(x)}{X}{Y}}$ is said to be {\em injective} or {\em one-to-one} if $\forall (a, b) \in X^2,$ we have $$ a \neq b \implies f(a) \neq f(b).$$ This is equivalent to saying that $$f(a) = f(b) \implies a = b.$$ \end{df} \begin{exa} The function $\alpha$ in the diagram \ref{injection} is an injective function. The function $\beta$ represented by the diagram \ref{not_injection}, however, is not injective, $\beta (3) = \beta (1) = 4$, but $3 \neq 1$. \end{exa} \begin{exa} Prove that $$\fun{t}{x}{\frac{x + 1}{x - 1}}{\BBR \setminus \{1\}}{\BBR \setminus \{1\}} $$ is an injection.\end{exa} \begin{solu} Assume $t(a) = t(b).$ Then $${\everymath{\dis}\begin{array}{ccccccc} t(a) & = & t(b) & \implies & \frac{a + 1}{a - 1} & = & \frac{b + 1}{b - 1} \\ & & & \implies & (a + 1)(b - 1) & = & (b + 1)(a - 1) \\ & & & \implies & ab - a + b - 1 & = & ab - b + a - 1 \\ & & & \implies & 2a & = & 2b \\ & & & \implies & a & = & b \\ \end{array}} $$ We have proved that $t(a) = t(b) \implies a = b$, which shews that $t$ is injective. \end{solu}\vspace{1cm} \begin{figure}[h] \centering \begin{minipage}{6cm} $$ \psset{unit=.75cm} \rput(-1.5,0){\rput(0,0){2} \rput(0, .5){1} \rput(0, -.5){3} \rput(3,0){2} \rput(3,.5){4} \rput(1.5, .9){\beta} \psellipse(1,2) \psellipse(3,0)(1,2) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,0)(2.9,0) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,.5)(2.9,.5) \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(0.2,-.5)(2.9,.5)} $$\vspace{1cm} \footnotesize\hangcaption{A surjection} \label{surjection}\end{minipage} \begin{minipage}{6cm} $$ \psset{unit=.75cm} \rput(-1.5,0){\rput(3, -1){8} \rput(1.5, .8){\gamma} \psellipse(1,2) \psellipse(3,0)(1,2) \rput(0,0){2} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,0)(2.9,0) \rput(3,0){2} \rput(0, .5){1} \psline[linewidth=.4pt, labels=none, showpoints=true]{*->>}(.2,.5)(2.9,.5) \rput(3,.5){4}} $$\vspace{.3in} \footnotesize\hangcaption{Not a surjection} \label{not_surjection} \end{minipage}\end{figure} \begin{df} A function $f: A \rightarrow B$ is said to be {\em surjective} or {\em onto} if $(\forall b \in B) \ (\exists a \in A) : f(a) = b.$ That is, each element of $B$ has a pre-image in $A$. \end{df} \begin{rem}A function is surjective if its image coincides with its target set. It is easy to see that a graphical criterion for a function to be surjective is that every horizontal line passing through a point of the target set (a subset of the $y$-axis) of the function must also meet the curve. \end{rem} \begin{exa} The function $\beta$ represented by diagram \ref{surjection} is surjective. The function $\gamma$ represented by diagram \ref{not_surjection} is not surjective as $8$ does not have a preimage. \end{exa} \begin{exa} Prove that $\fun{t}{x}{x^3}{\BBR}{\BBR}$ is a surjection. \end{exa} \begin{solu} Since the graph of $t$ is that of a cubic polynomial with only one zero, every horizontal line passing through a point in $\BBR$ will eventually meet the graph of $g$, whence $t$ is surjective. To prove this analytically, proceed as follows. We must prove that $(\forall \ b \in\BBR) \ (\exists a)$ such that $t(a) = b.$ We choose $a$ so that $a = b^{1/3}$. Then $$t(a) = t(b^{1/3}) = (b^{1/3})^3 = b.$$Our choice of $a$ works and hence the function is surjective. \end{solu} \begin{df} A function is {\em bijective} if it is both injective and surjective. \end{df} \section*{\psframebox{Homework}} \begin{pro} Prove that $$\fun{h}{x}{x^3}{\BBR}{\BBR}$$ is an injection. \begin{answer} Assume $h(b) = h(a).$ Then $$\begin{array}{lllll} h(a) & = & h(b)& \implies & a^3 = b^3 \\ & & & \implies & a^3 - b^3 = 0 \\ & & & \implies & (a - b)(a^2 + ab + b^2) = 0 \\ \end{array} $$ Now, $$b^2 + ab + a^2 = \left(b + \frac{a}{2}\right)^2 + \frac{3a^2}{4}.$$ This shews that $b^2 + ab + a^2$ is positive unless both $a$ and $b$ are zero. Hence $a - b = 0$ in all cases. We have shewn that $h(b) = h(a) \implies a = b$, and the function is thus injective. \end{answer} \end{pro} \begin{pro} Shew that $$\fun{f}{x}{\frac{6x}{2x - 3}}{\BBR\setminus \left\{\frac{3}{2}\right\}}{\BBR\setminus \{3\}}$$is a bijection. \begin{answer} We have $$\begin{array}{lll} f(a) = f(b) & \iff & \dfrac{6a}{2a - 3} = \dfrac{6b}{2b - 3} \\ & \iff & 6a(2b - 3) = 6b(2a - 3) \\ & \iff & 12ab - 18a = 12ab - 18b \\ & \iff & -18a = -18b \\ & \iff & a = b, \end{array}$$ proving that $f$ is injective. Now, if $$f(x) = y, \ \ y\neq 3, $$then $$\frac{6x}{2x - 3} = y,$$that is $6x = y(2x - 3)$. Solving for $x$ we find $$x = \frac{3y}{2y - 6}.$$Since $2y - 6 \neq 0$, $x$ is a real number, and so $f$ is surjective. On combining the results we deduce that $f$ is bijective. \end{answer} \end{pro} \chapter{Matrices and Matrix Operations} \section{The Algebra of Matrices} \begin{df}\index{matrix} Let $ \field{ \BBF }{\cdot}{+}$ be a field. An $m\times n$ ($m$ by $n$) {\em matrix} $A$ with $m$ rows and $n$ columns with entries over $\BBF$ is a rectangular array of the form $$A = \begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \cr a_{21} & a_{22} & \cdots & a_{2n} \cr \vdots & \vdots & \cdots & \vdots \cr a_{m1} & a_{m2} & \cdots & a_{mn} \cr \end{bmatrix},$$ where $\forall (i, j) \in \{1, 2, \ldots , m \}\times \{1, 2, \ldots , n\}, \ \ a_{ij}\in \BBF$. \index{matrix} \end{df} \begin{rem} As a shortcut, we often use the notation $A = [a_{ij}]$ to denote the matrix $A$ with entries $a_{ij}$. Notice that when we refer to the matrix we put parentheses---as in ``$[a_{ij}]$,'' and when we refer to a specific entry we do not use the surrounding parentheses---as in ``$a_{ij}$.'' \end{rem} \begin{exa} $$A = \begin{bmatrix}0 & -1 & 1 \cr 1 & 2 & 3 \cr\end{bmatrix}$$is a $2\times 3$ matrix and $$B = \begin{bmatrix}-2 & 1 \cr 1 & 2 \cr 0 & 3 \cr\end{bmatrix}$$ is a $3\times 2$ matrix. \end{exa} \begin{exa} Write out explicitly the $4\times 4$ matrix $A = [a_{ij}]$ where $a_{ij} = i^2 - j^2$. \end{exa} \begin{solu} This is $$A = \begin{bmatrix} 1^2 - 1^1 & 1^2 - 2^2 & 1^2 - 3^2 & 1^2 - 4^2 \cr 2^2 - 1^2 & 2^2 - 2^2 & 2^2 - 3^2 & 2^2 - 4^2 \cr 3^2 - 1^2 & 3^2 - 2^2 & 3^2 - 3^2 & 3^2 - 4^2 \cr 4^2 - 1^2 & 4^2 - 2^2 & 4^2 - 3^2 & 4^2 - 4^2 \cr \end{bmatrix} = \begin{bmatrix} 0 & -3& -8 & -15 \cr 3 & 0 & -5 & -12 \cr 8 & 5 & 0 & -7 \cr 15 & 12 & 7 & 0 \cr\end{bmatrix}. $$ \end{solu} \begin{df} Let $ \field{ \BBF }{\cdot}{+}$ be a field. We denote by $\mat{m\times n}{ \BBF }$ the set of all $m\times n$ matrices with entries over $\BBF$. $\mat{n\times n}{ \BBF }$ is, in particular, the set of all square matrices of size $n$ with entries over $\BBF$. \end{df} \begin{df} The $m\times n$ {\em zero matrix} ${\bf 0}_{m\times n}\in\mat{m\times n}{ \BBF }$ is the matrix with $0_{\BBF }$'s everywhere, $${\bf 0}_{m\times n} = \begin{bmatrix}0_{\BBF } & 0_{\BBF } & 0_{\BBF } & \cdots & 0_{\BBF } \cr 0_{\BBF } & 0_{\BBF } & 0_{\BBF } & \cdots & 0_{\BBF } \cr 0_{\BBF } & 0_{\BBF } & 0_{\BBF } & \cdots & 0_{\BBF } \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0_{\BBF } & 0_{\BBF } & 0_{\BBF } & \cdots & 0_{\BBF } \cr\end{bmatrix}.$$ When $m = n$ we write ${\bf 0}_n$ as a shortcut for ${\bf 0}_{n\times n}$.\index{matrix!zero} \end{df} \begin{df} \index{matrix!identity} The $n\times n$ {\em identity matrix} ${\bf I}_n\in\mat{n\times n}{ \BBF }$ is the matrix with $1_{\BBF }$'s on the main diagonal and $0_{\BBF }$'s everywhere else, $${\bf I}_n = \begin{bmatrix}1_{\BBF } & 0_{\BBF } & 0_{\BBF } & \cdots & 0_{\BBF } \cr 0_{\BBF } & 1_{\BBF } & 0_{\BBF } & \cdots & 0_{\BBF } \cr 0_{\BBF } & 0_{\BBF } & 1_{\BBF } & \cdots & 0_{\BBF } \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0_{\BBF } & 0_{\BBF } & 0_{\BBF } & \cdots & 1_{\BBF } \cr\end{bmatrix}.$$ \end{df} \begin{df}[Matrix Addition and Multiplication of a Matrix by a Scalar]\index{matrix!addition}\index{matrix!scalar multiplication} Let $A = [a_{ij}] \in \mat{m\times n}{ \BBF }$, $B = [b_{ij}] \in \mat{m\times n}{ \BBF }$ and $\alpha\in \BBF$. The matrix $A + \alpha B$ is the matrix $C \in \mat{m\times n}{ \BBF }$ with entries $C = [c_{ij}]$ where $c_{ij} = a_{ij}+\alpha b_{ij}$. \end{df} \begin{exa} For $A = \begin{bmatrix} 1 & 1 \cr -1 & 1 \cr 0 & 2 \cr \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 1 \cr 2 & 1 \cr 0 & -1 \cr \end{bmatrix}$ we have $$A + 2B = \begin{bmatrix} -1 & 3 \cr 3 & 3 \cr 0 & 0. \cr \end{bmatrix}. $$ \end{exa} \begin{thm}Let $(A, B, C)\in (\mat{m\times n}{ \BBF })^3$ and $(\alpha, \beta) \in \BBF ^2$. Then \begin{enumerate} \item[M1] $\mat{m\times n}{ \BBF }$ is close under matrix addition and scalar multiplication \begin{equation}\label{m:closure} A + B \in \mat{m\times n}{ \BBF }, \ \ \ \ \alpha A \in \mat{m\times n}{ \BBF } \end{equation} \item[M2] Addition of matrices is commutative \begin{equation}\label{m:commutative} A + B = B + A \end{equation} \item[M3] Addition of matrices is associative \begin{equation}\label{m:associative} A + (B + C) = (A+B)+C \end{equation} \item[M4] There is a matrix ${\bf 0}_{m\times n}$ such that \begin{equation}\label{m:existence0}A+ {\bf 0}_{m\times n} \end{equation} \item[M5] There is a matrix $-A$ such that \begin{equation}\label{m:existence_additive_inverse} A + (-A) = (-A) + A = {\bf 0}_{m\times n} \end{equation} \item[M6] Distributive law \begin{equation}\label{m:distributive_law1} \alpha (A + B) = \alpha A + \alpha B \end{equation} \item[M7] Distributive law \begin{equation}\label{m:distributive_law2} (\alpha + \beta)A = \alpha A + \beta B \end{equation} \item[M8] \begin{equation}\label{m:1A} 1_{\BBF }A = A \end{equation} \item[M9] \begin{equation}\label{m:abA} \alpha (\beta A) = (\alpha\beta)A \end{equation} \end{enumerate} \end{thm} \begin{pf} The theorem follows at once by reducing each statement to an entry-wise and appealing to the field axioms. \end{pf} \section*{\psframebox{Homework}} \begin{pro} Write out explicitly the $3\times 3$ matrix $A = [a_{ij}]$ where $a_{ij} = i^j$. \begin{answer} $A = \begin{bmatrix} 1 & 1 & 1 \cr 2 & 4 & 8 \cr 3 & 9 & 27 \cr \end{bmatrix}.$ \end{answer} \end{pro} \begin{pro} Write out explicitly the $3\times 3$ matrix $A = [a_{ij}]$ where $a_{ij} = ij$. \begin{answer} $A = \begin{bmatrix} 1 & 2 & 3 \cr 2 & 4 & 6 \cr 3 & 6 & 9 \cr \end{bmatrix}.$ \end{answer} \end{pro} \begin{pro} Let $$M = \begin{bmatrix}a & -2a & c \cr 0 & -a & b \cr a + b & 0 & -1 \cr\end{bmatrix},\ \ \ \ N = \begin{bmatrix}1 & 2a & c \cr a & b - a & -b \cr a - b & 0 & -1 \cr \end{bmatrix}$$ be square matrices with entries over $\BBR$. Find $M+N$ and $2M$. \begin{answer} $M + N = \begin{bmatrix}a + 1 & 0 & 2c \cr a & b - 2a & 0 \cr 2a & 0 & -2 \cr\end{bmatrix}, \ \ \ 2M = \begin{bmatrix}2a & -4a & 2c \cr 0 & -2a & 2b \cr 2a + 2b & 0 & -2 \cr\end{bmatrix}.$ \end{answer} \end{pro} \begin{pro} Determine $x$ and $y$ such that $$\begin{bmatrix} 3 & x & 1 \cr 1 & 2 & 0 \cr \end{bmatrix} + 2\begin{bmatrix} 2 & 1 & 3 \cr 5 & x & 4 \cr\end{bmatrix} = \begin{bmatrix} 7 & 3 & 7 \cr 11 & y & 8 \cr\end{bmatrix}.$$ \begin{answer} $x = 1$ and $y = 4$. \end{answer} \end{pro} \begin{pro} Determine $2\times 2$ matrices $A$ and $B$ such that $$2A - 5B = \begin{bmatrix} 1 & -2 \cr 0 & 1 \cr \end{bmatrix}, \ \ \ -2A + 6B = \begin{bmatrix} 4 & 2 \cr 6 & 0 \cr\end{bmatrix}. $$ \begin{answer} $A = \begin{bmatrix}13 & -1 \cr 15 & 3 \cr \end{bmatrix}$, $B = \begin{bmatrix}5 & 0 \cr 6 & 1 \cr \end{bmatrix}$ \end{answer} \end{pro} \begin{pro} Let $A=[a_{ij}]\in\mat{n\times n}{\BBR}$. Prove that $$\min_j\max_ia_{ij}\geq\max_i\min_ja_{ij}.$$ \end{pro} \begin{pro} A person goes along the rows of a movie theater and asks the tallest person of each row to stand up. Then he selects the shortest of these people, who we will call the {\em shortest giant}. Another person goes along the rows and asks the shortest person to stand up and from these he selects the tallest, which we will call the {\em tallest midget}. Who is taller, the tallest midget or the shortest giant? \end{pro} \begin{pro}[{\red\bf Putnam Exam, 1959}] Choose five elements from the matrix $$\begin{bmatrix}11 & 17 & 25 & 19 & 16 \cr 24 & 10 & 13 & 15 & 3 \cr 12 & 5 & 14 & 2 & 18 \cr 23 & 4 & 1 & 8 & 22 \cr 6 & 20 & 7 & 21 & 9 \end{bmatrix},$$ no two coming from the same row or column, so that the minimum of these five elements is as large as possible. \begin{answer} The set of border elements is the union of two rows and two columns. Thus we may choose at most four elements from the border, and at least one from the central $3\times 3$ matrix. The largest element of this $3\times 3$ matrix is $15$, so any allowable choice of does not exceed $15$. The choice $25$, $15$, $18$,l $23$, $20$ shews that the largest minimum is indeed $15$.\end{answer} \end{pro} \section{Matrix Multiplication} \begin{df}\index{matrix!multiplication of} Let $A = [a_{ij}] \in \mat{m\times n}{ \BBF }$ and $B = [b_{ij}] \in \mat{n\times p}{ \BBF }$. Then the matrix product $AB$ is defined as the matrix $C = [c_{ij}]\in \mat{m\times p}{ \BBF }$ with entries $c_{ij} = \sum _{l=1} ^{n} a_{il} b_{lj}$: $$\begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \cr a_{21} & a_{22} & \cdots & a_{2n} \cr \vdots & \vdots & \cdots & \vdots \cr {\red a_{i1}} & {\red a_{i2}} & {\red \cdots} & {\red a_{in}}\cr \vdots & \vdots & \cdots & \vdots \cr a_{m1} & a_{m2} & \cdots & a_{mn} \cr \end{bmatrix}\begin{bmatrix}b_{11} &\cdots & {\blue b_{1j}} & \cdots & b_{1p} \cr b_{21} &\cdots & {\blue b_{2j}} & \cdots & b_{2p} \cr \vdots &\cdots & {\blue \vdots} & \cdots & \vdots \cr b_{n1} &\cdots & {\blue b_{nj}} & \cdots & b_{np} \cr \end{bmatrix} = \begin{bmatrix} c_{11} & \cdots & c_{1p} \cr c_{21} & \cdots & c_{2p} \cr \vdots & \cdots & \vdots \cr \cdots & {\green c_{ij}} & \cdots \cr \vdots & \cdots & \vdots \cr c_{m1} & \cdots & c_{mp} \cr\end{bmatrix}.$$ \end{df} \begin{rem} Observe that we use juxtaposition rather than a special symbol to denote matrix multiplication. This will simplify notation.In order to obtain the $ij$-th entry of the matrix $AB$ we multiply elementwise the $i$-th row of $A$ by the $j$-th column of $B$. Observe that $AB$ is a $m\times p$ matrix. \end{rem} \begin{exa} Let $M = \begin{bmatrix}1 & 2 \cr 3 & 4\cr\end{bmatrix}$ and $N = \begin{bmatrix}5 & 6 \cr 7& 8 \cr \end{bmatrix}$ be matrices over $\BBR$. Then $$MN =\begin{bmatrix}1 & 2 \cr 3 & 4\cr\end{bmatrix}\begin{bmatrix}5 & 6 \cr 7& 8 \cr \end{bmatrix} = \begin{bmatrix} 1\cdot 5 + 2\cdot 7 & 1\cdot 6 + 2\cdot 8 \cr 3\cdot 5 + 4\cdot 7 & 3\cdot 6 + 4 \cdot 8 \cr\end{bmatrix} = \begin{bmatrix}19 & 22 \cr 43 & 50 \cr \end{bmatrix},$$ and $$NM = \begin{bmatrix}5 & 6 \cr 7& 8 \cr \end{bmatrix}\begin{bmatrix}1 & 2 \cr 3 & 4\cr\end{bmatrix}= \begin{bmatrix}5\cdot 1 + 6\cdot 3 & 5\cdot 2 + 6\cdot 4 \cr 7\cdot 1 + 8\cdot 3 & 7\cdot 2 + 8\cdot 4\cr \end{bmatrix} = \begin{bmatrix}23 & 34 \cr 31& 46 \cr \end{bmatrix}.$$Hence, in particular, matrix multiplication is not necessarily commutative. \end{exa} \begin{exa} We have $$\begin{bmatrix} {1} & {1} & {1} \cr {1} & {1} & 1 \cr {1} & {1} & {1} \cr \end{bmatrix} \begin{bmatrix} {2} & -{1} & -{1} \\ \vspace{1mm} -{1} & {2} & -{1} \\ \vspace{1mm} -{1} & -{1} & {2} \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \cr \vspace{1mm} 0 & 0 & 0 \cr \vspace{1mm} 0 & 0 & 0 \cr \end{bmatrix},$$ over $\BBR$. Observe then that the product of two non-zero matrices may be the zero matrix. \end{exa} \begin{exa} Consider the matrix $$A = \begin{bmatrix} \overline{2} & \overline{1} & \overline{3} \cr \overline{0} & \overline{1} & \overline{1} \cr \overline{4} & \overline{4} & \overline{0} \cr \end{bmatrix}$$with entries over $\BBZ_5$. Then $$\begin{array}{lll}A^2 & = & \begin{bmatrix} \overline{2} & \overline{1} & \overline{3} \cr \overline{0} & \overline{1} & \overline{1} \cr \overline{4} & \overline{4} & \overline{0} \cr \end{bmatrix}\begin{bmatrix} \overline{2} & \overline{1} & \overline{3} \cr \overline{0} & \overline{1} & \overline{1} \cr \overline{4} & \overline{4} & \overline{0} \cr \end{bmatrix} \vspace{2mm}\\ & = & \begin{bmatrix} \overline{1} & \overline{0} & \overline{2} \cr \overline{4} & \overline{0} & \overline{1} \cr \overline{3} & \overline{3} & \overline{1} \cr \end{bmatrix}. \end{array} $$ \end{exa} \begin{rem} Even though matrix multiplication is not necessarily commutative, it is associative. \end{rem} \begin{thm} If $(A, B, C) \in \mat{m\times n }{ \BBF }\times\mat{n\times r}{ \BBF }\times\mat{r \times s}{ \BBF }$ we have $$(AB)C = A(BC),$$ i.e., matrix multiplication is associative.\end{thm}\begin{pf} To shew this we only need to consider the $ij$-th entry of each side, appeal to the associativity of the underlying field $\BBF$ and verify that both sides are indeed equal to $$\sum _{k = 1} ^{n}\sum _{k' = 1} ^{r} a_{ik}b_{kk'}c_{k'j}.$$ \end{pf} \begin{rem} By virtue of associativity, a square matrix commutes with its powers, that is, if $A\in\mat{n\times n}{ \BBF }$, and $(r, s)\in\BBN^2$, then $(A^r)(A^s) = (A^s)(A^r) = A^{r + s}$. \end{rem} \begin{exa} Let $A\in\mat{3\times 3}{\BBR}$ be given by$$A = \begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 1 \cr 1 & 1 & 1 \cr \end{bmatrix}.$$Demonstrate, using induction, that $A^n = 3^{n - 1}A$ for $n \in \BBN , n \geq 1$.\end{exa} \begin{solu} The assertion is trivial for $n = 1$. Assume its truth for $n -1$, that is, assume $A^{n - 1} = 3^{n - 2}A$. Observe that $$A^2 = \begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 1 \cr 1 & 1 & 1 \cr \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 1 \cr 1 & 1 & 1 \cr \end{bmatrix} = \begin{bmatrix} 3 & 3 & 3 \cr 3 & 3 & 3 \cr 3 & 3 & 3 \cr \end{bmatrix} = 3A.$$Now $$A^n = AA^{n - 1} = A(3^{n - 2}A) = 3^{n - 2}A^2 = 3^{n - 2}3A = 3^{n - 1}A, $$ and so the assertion is proved by induction. \end{solu} \begin{thm} Let $A\in\mat{n\times n}{ \BBF }$. Then there is a unique identity matrix. That is, if $E\in\mat{n\times n}{ \BBF }$ is such that $AE = EA = A$, then $E = {\bf I}_n$. \end{thm} \begin{pf} It is clear that for any $A\in\mat{n\times n}{ \BBF }$, $A{\bf I}_n = {\bf I}_nA = A$. Now because $E$ is an identity, $E{\bf I}_n = {\bf I}_n$. Because ${\bf I}_n$ is an identity, $E{\bf I}_n = E$. Whence $${\bf I}_n = E{\bf I}_n = E,$$demonstrating uniqueness. \end{pf} \begin{exa}Let $A = [a_{ij}]\in \mat{n\times n}{\BBR}$ be such that $a_{ij} = 0$ for $i > j$ and $a_{ij} = 1$ if $i \leq j$. Find $A^2$. \end{exa} \begin{solu} Let $A^2 = B = [b_{ij}]$. Then $$b_{ij} = \sum_{k = 1} ^n a_{ik}a_{kj}.$$Observe that the $i$-th row of $A$ has $i - 1$ $0$'s followed by $n - i + 1$ $1$'s, and the $j$-th column of $A$ has $j$ $1$'s followed by $n - j$ 0's. Therefore if $i - 1 > j$, then $b_{ij} = 0$. If $i \leq j + 1$, then $$b_{ij} = \sum_{k = i} ^{j} a_{ik}a_{kj} = j - i + 1.$$ This means that $$A^2 = \begin{bmatrix} 1 & 2 & 3 & 4 & \cdots & n - 1 & n \cr 0 & 1 & 2 & 3 & \cdots & n - 2 & n - 1 \cr 0 & 0 & 1 & 2 & \cdots & n - 3 & n - 2 \cr \vdots & \vdots & \vdots & \vdots & \cdots &\vdots & \vdots \cr 0 & 0 & 0 & 0 & \cdots & 1 & 2 \cr 0 & 0 & 0 & 0 & \cdots & 0 & 1\cr \end{bmatrix}.$$ \end{solu} \section*{\psframebox{Homework}} \begin{pro} Determine the product $$\begin{bmatrix} 1 & -1 \cr 1 & 1\cr\end{bmatrix}\begin{bmatrix}-2 & 1 \cr 0 & -1 \cr \end{bmatrix}\begin{bmatrix} 1 & 1 \cr 1 & 2 \cr \end{bmatrix}$$. \begin{answer}$\begin{bmatrix} 2 & 2 \cr 0 & -2 \cr \end{bmatrix}$ \end{answer} \end{pro} \begin{pro} Let $A = \begin{bmatrix}1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr \end{bmatrix}, \ \ B = \begin{bmatrix} a & b & c \cr c & a & b \cr b & c & a \cr \end{bmatrix}$. Find $AB$ and $BA$. \begin{answer} $$AB = \begin{bmatrix} a & b & c \cr c+a & a + b & b + c \cr a + b + c & a + b + c & a + b + c \cr \end{bmatrix}, \ BA = \begin{bmatrix} a + b + c & b + c & c \cr a + b + c & a + b & b \cr a + b + c & c + a & a \cr \end{bmatrix} $$ \end{answer} \end{pro} \begin{pro} Find $a+b+c$ if $\begin{bmatrix}1 & 2 & 3 \cr 2 & 3 & 1 \cr 3 & 1 & 2 \cr \end{bmatrix} \begin{bmatrix}1 & 1 & 1 \cr 2 & 2 & 2 \cr 3 & 3 & 3 \cr \end{bmatrix} = \begin{bmatrix}a & a & a \cr b & b & b \cr c & c & c \cr \end{bmatrix}$. \begin{answer} $\begin{bmatrix}1 & 2 & 3 \cr 2 & 3 & 1 \cr 3 & 1 & 2 \cr \end{bmatrix} \begin{bmatrix}1 & 1 & 1 \cr 2 & 2 & 2 \cr 3 & 3 & 3 \cr \end{bmatrix} = \begin{bmatrix}14 & 14 & 14 \cr 11 & 11 & 11 \cr 11 & 11 & 11 \cr \end{bmatrix}$, whence $a+b+c=36$. \end{answer} \end{pro} \begin{pro} Let $N = \begin{bmatrix}0 & -2 & -3 & -4 \cr 0 & 0 & -2 & -3 \cr 0 & 0 & 0 & -2 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}$. Find $N^{2008}$. \begin{answer} An easy computation leads to $N^2= \begin{bmatrix}0 & 0 & 4 & 12 \cr 0 & 0 & 0& 4 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}$, $N^3= \begin{bmatrix}0 & 0 & 0 & -8 \cr 0 & 0 & 0& 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}$ and $ N^4=\begin{bmatrix}0 & 0 & 0 & 0 \cr 0 & 0 & 0& 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}$. Hence any power---from the fourth on---is the zero matrix. \end{answer} \end{pro} \begin{pro} Let $$ A = \begin{bmatrix} \overline{2} & \overline{3} & \overline{4} & \overline{1} \cr \overline{1} & \overline{2} & \overline{3} & \overline{4} \cr \overline{4} & \overline{1} & \overline{2} & \overline{3} \cr \overline{3} & \overline{4} & \overline{1} & \overline{2} \cr \end{bmatrix}, \ \ \ B = \begin{bmatrix} \overline{1} & \overline{1} & \overline{1} & \overline{1} \cr \overline{1} & \overline{1} & \overline{1} & \overline{1} \cr \overline{1} & \overline{1} & \overline{1} & \overline{1} \cr \overline{1} & \overline{1} & \overline{1} & \overline{1} \cr \end{bmatrix}$$ be matrices in $\mat{4\times 4}{\BBZ_5}$ . Find the products $AB$ and $BA$. \begin{answer} $AB = {\bf 0}_4$ and $ BA = \begin{bmatrix} \overline{0} & \overline{2} & \overline{0} & \overline{3} \cr \overline{0} & \overline{2} & \overline{0} & \overline{3} \cr \overline{0} & \overline{2} & \overline{0} & \overline{3} \cr \overline{0} & \overline{2} & \overline{0} & \overline{3} \cr \end{bmatrix}.$ \end{answer} \end{pro} \begin{pro} Let $x$ be a real number, and put $$m(x) = \begin{bmatrix}1 & 0 & x\cr -x & 1 & -\dfrac{x^2}{2} \cr 0 &0 &1\cr \end{bmatrix}. $$If $a, b$ are real numbers, prove that \begin{enumerate} \item $m(a)m(b)=m(a+b)$. \item $m(a)m(-a)={\bf I}_3$, the $3\times 3$ identity matrix. \end{enumerate} \begin{answer} For the first part, observe that $$\begin{array}{lll}m(a)m(b) & = & \begin{bmatrix}1 & 0 & a\cr -a & 1 & -\dfrac{a^2}{2} \cr 0 &0 &1\cr \end{bmatrix}\begin{bmatrix}1 & 0 & b\cr -b & 1 & -\dfrac{b^2}{2} \cr 0 &0 &1\cr \end{bmatrix} \\ & = & \begin{bmatrix}1 & 0 & a+b\cr -a-b & 1 & -\dfrac{a^2}{2}-\dfrac{b^2}{2}+ab \cr 0 &0 &1\cr \end{bmatrix}\\ \\ & = & \begin{bmatrix}1 & 0 & a+b\cr -(a+b) & 1 & -\dfrac{(a+b)^2}{2} \cr 0 &0 &1\cr \end{bmatrix}\\ & = & m(a+b)\end{array}$$ For the second part, observe that using the preceding part of the problem, $$m(a)m(-a)= m(a-a) = m(0) =\begin{bmatrix}1 & 0 & 0\cr -0 & 1 & -\dfrac{0^2}{2} \cr 0 &0 &1\cr \end{bmatrix} ={\bf I}_3,$$giving the result. \end{answer} \end{pro} \begin{pro} A square matrix $X$ is called {\em idempotent} if $X^2=X$. Prove that if $AB=A$ and $BA=B$ then $A$ and $B$ are idempotent. \\ \begin{answer} Observe that $$A^2 = (AB)(AB) = A(BA)B = A(B)B = (AB)B = AB =A. $$ Similarly, $$B^2 = (BA)(BA) = B(AB)A = B(A)A = (BA)A = BA =B. $$ \end{answer} \end{pro} \begin{pro} Let $$ A = \begin{bmatrix} 0 & \dfrac{1}{2} & 0 \cr \dfrac{1}{2} & 0 & 0 \cr 0 & 0 & \dfrac{1}{2} \cr \end{bmatrix}. $$ Calculate the value of the infinite series $${\bf I}_3 + A + A^2 +A^3 + \cdots . $$ \begin{answer} For this problem you need to recall that if $|r|<1$, then $$a+ar + ar^2+ar^3+\cdots + \cdots =\dfrac{a}{1-r}. $$ This gives $$ 1 + \tfrac{1}{4} + \tfrac{1}{4^2}+ \tfrac{1}{4^3}+\cdots = \dfrac{1}{1-\tfrac{1}{4}} = \dfrac{4}{3}, $$ $$\tfrac{1}{2} + \tfrac{1}{2^3}+ \tfrac{1}{2^5}+\cdots = \dfrac{\tfrac{1}{2}}{1-\tfrac{1}{4}} = \dfrac{2}{3}, $$ and $$1 + \tfrac{1}{2} + \tfrac{1}{2^2}+ \tfrac{1}{2^3}+\cdots = \dfrac{1}{1-\tfrac{1}{2}} = 2. $$ By looking at a few small cases, it is easy to establish by induction that for $n\geq 1$ $$ A^{2n-1}= \begin{bmatrix} 0 & \dfrac{1}{2^{2n-1}} & 0 \cr \dfrac{1}{2^{2n-1}} & 0 & 0 \cr 0 & 0 & \dfrac{1}{2^{2n-1}} \cr \end{bmatrix}, \qquad A^{2n} = \begin{bmatrix}\dfrac{1}{2^{2n}} & 0 & 0 \cr 0 & \dfrac{1}{2^{2n}} & 0 \cr 0 & 0 & \dfrac{1}{2^{2n}} \cr \end{bmatrix}. $$ This gives $${\bf I}_3 + A + A^2 +A^3 + \cdots = \begin{bmatrix}1 + \tfrac{1}{4} + \tfrac{1}{4^2}+ \tfrac{1}{4^3}+\cdots & \tfrac{1}{2} + \tfrac{1}{2^3}+ \tfrac{1}{2^5}+\cdots & 0 \cr \tfrac{1}{2} + \tfrac{1}{2^3}+ \tfrac{1}{2^5}+\cdots & 1 + \tfrac{1}{4} + \tfrac{1}{4^2}+ \tfrac{1}{4^3}+\cdots & 0 \cr 0 & 0 & 1 + \tfrac{1}{2} + \tfrac{1}{2^2}+ \tfrac{1}{2^3}+\cdots \cr \end{bmatrix} = \begin{bmatrix} \tfrac{4}{3} & \tfrac{2}{3} & 0 \cr \tfrac{2}{3} & \tfrac{4}{3} & 0 \cr 0 & 0 & 2 \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Solve the equation $$\begin{bmatrix} -4 & x \cr -x & 4 \end{bmatrix}^2 = \begin{bmatrix} -1 & 0 \cr 0 & -1 \end{bmatrix}$$ over $\BBR$. \begin{answer} Observe that $$ \begin{bmatrix} -4 & x \cr -x & 4 \end{bmatrix}^2 =\begin{bmatrix} -4 & x \cr -x & 4 \end{bmatrix}\begin{bmatrix} -4 & x \cr -x & 4 \end{bmatrix} = \begin{bmatrix} 16 - x^2 & 0 \cr 0 & 16 - x^2 \end{bmatrix}, $$ and so we must have $16 - x^2 = -1$ or $x = \pm\sqrt{17}$. \end{answer} \end{pro} \begin{pro} Prove or disprove! If $(A,B)\in(\mat{n\times n}{ \BBF })^2$ are such that $AB = {\bf 0}_n$, then also $BA = {\bf 0}_n.$ \begin{answer} Disprove! Take $\dis{A = \left[\begin{array}{ll}1 & 0 \\ 0 & 0 \end{array}\right]}$ and $\dis{B = \left[\begin{array}{ll}0 & 1 \\ 0 & 0 \end{array}\right]}$. Then $AB = B$, but $BA = {\bf 0}_2$. \end{answer} \end{pro} \begin{pro} Prove or disprove! For all matrices $(A,B)\in(\mat{n\times n}{ \BBF })^2$, $$(A+B)(A-B) = A^2 - B^2.$$ \begin{answer} Disprove! Take for example $A = \begin{bmatrix} 0 & 0 \cr 1 & 1 \cr\end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \cr 1 & 0 \cr\end{bmatrix}$. Then $$A^2 - B^2 = \begin{bmatrix} -1 & 0 \cr 0 & 1 \cr\end{bmatrix} \neq \begin{bmatrix} -1 & 0 \cr -2 & 1 \cr\end{bmatrix}= (A+B)(A-B). $$ \end{answer} \end{pro} \begin{pro} Consider the matrix $A=\begin{bmatrix}1 & 2 \cr 3 & x \end{bmatrix}$, where $x$ is a real number. Find the value of $x$ such that there are non-zero $2\times 2 $ matrices $B$ such that $AB = \begin{bmatrix}0 & 0 \cr 0 & 0 \end{bmatrix}$. \begin{answer} $x=6$. \end{answer} \end{pro} \begin{pro} Prove, using mathematical induction, that $\begin{bmatrix}1 & 1 \cr 0 & 1 \cr \end{bmatrix}^n=\begin{bmatrix}1 & n \cr 0 & 1 \cr \end{bmatrix}$. \end{pro} \begin{pro} Let $M = \begin{bmatrix} 1 & -1 \cr -1 & 1 \cr\end{bmatrix}$. Find $M^6$. \begin{answer} $ \begin{bmatrix} 32 & -32 \cr -32 & 32 \cr\end{bmatrix}$.\end{answer} \end{pro} \begin{pro} Let $A = \begin{bmatrix} 0 & 3 \cr 2 & 0 \cr \end{bmatrix}$. Find, with proof, $A^{2003}$. \begin{answer} $A^{2003} = \begin{bmatrix} 0 & 2^{1001}3^{1002} \cr 2^{1002}3^{1001} & 0 \cr \end{bmatrix}$. \end{answer} \end{pro} \begin{pro} Let $(A, B, C)\in \mat{l\times m}{ \BBF }\times \mat{m\times n}{ \BBF }\times \mat{m\times n}{ \BBF }$ and $ \alpha\in \BBF$. Prove that $$A(B+C) = AB + AC,$$ $$(A+B)C = AC + BC,$$ $$\alpha (AB) = (\alpha A)B = A(\alpha B).$$ \end{pro} \begin{pro} Let $A \in \mat{2\times 2}{\BBR}$ be given by $$A = \begin{bmatrix} \cos \alpha & -\sin \alpha \cr \sin \alpha & \cos \alpha\end{bmatrix}.$$Demonstrate, using induction, that for $n \in \BBN , n \geq 1$. $$A^n = \begin{bmatrix} \cos n\alpha & -\sin n\alpha \cr \sin n\alpha & \cos n\alpha \cr \end{bmatrix}.$$\begin{answer}The assertion is clearly true for $n = 1.$ Assume that is it true for $n$, that is, assume $$A^{n} = \begin{bmatrix} \cos (n)\alpha & -\sin (n)\alpha \cr \sin (n)\alpha & \cos (n)\alpha \cr \end{bmatrix}.$$ Then $$\begin{array}{lll} A^{n+1} & = & AA^{n} \\ & = & \begin{bmatrix} \cos \alpha & -\sin \alpha \cr \sin \alpha & \cos \alpha \cr \end{bmatrix}\begin{bmatrix} \cos (n)\alpha & -\sin (n)\alpha \cr \sin (n)\alpha & \cos (n)\alpha \cr \end{bmatrix}\vspace{2mm} \\ & = & \begin{bmatrix} \cos\alpha\cos (n)\alpha - \sin\alpha\sin (n)\alpha& -\cos\alpha\sin (n)\alpha - \sin\alpha\cos (n)\alpha\cr \sin\alpha\cos (n)\alpha + \cos\alpha\sin (n)\alpha& -\sin\alpha\sin (n)\alpha + \cos\alpha\cos (n)\alpha \cr \end{bmatrix} \vspace{2mm} \\ & = & \begin{bmatrix} \cos (n+1)\alpha & -\sin (n+1)\alpha \cr \sin (n+1)\alpha & \cos (n+1)\alpha \cr \end{bmatrix}, \end{array}$$and the result follows by induction. \end{answer} \end{pro} \begin{pro} A matrix $A = [a_{ij}]\in \mat{n\times n}{\BBR}$ is said to be {\em checkered} if $a_{ij} = 0$ when $(j - i)$ is odd. Prove that the sum and the product of two checkered matrices is checkered. \begin{answer}Let $A = [a_{ij}], B = [b_{ij}]$ be checkered $n\times n$ matrices. Then $A + B = (a_{ij} + b_{ij})$. If $j - i$ is odd, then $a_{ij} + b_{ij} = 0 + 0 = 0$, which shows that $A + B$ is checkered. Furthermore, let $AB = [c_{ij}]$ with $c_{ij} = \sum _{k = 1} ^n a_{ik}b_{kj}$. If $i$ is even and $j$ odd, then $a_{ik} = 0$ for odd $k$ and $b_{kj} = 0$ for even $k$. Thus $c_{ij} = 0$ for $i$ even and $j$ odd. Similarly, $c_{ij} = 0$ for odd $i$ and even $j$. This proves that $AB$ is checkered. \end{answer} \end{pro} \begin{pro} Let $A\in\mat{3\times 3}{\BBR}$, $$A = \begin{bmatrix} 1 & 1 & 1 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix}. $$Prove that $$A^n = \begin{bmatrix} 1 & n & \frac{n(n + 1)}{2} \cr 0 & 1 & n \cr 0 & 0 & 1 \cr \end{bmatrix}.$$ \begin{answer} Put $$ J = \begin{bmatrix} 0 & 1 & 1 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr \end{bmatrix}. $$We first notice that $$J^2 = \begin{bmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr\end{bmatrix}, \ \ J^3 = {\bf 0}_3.$$ This means that the sum in the binomial expansion $$A^n = ({\bf I}_3 + J)^n = \sum _{k = 0} ^n \binom{n}{k}{\bf I}^{n - k}J^k$$ is a sum of zero matrices for $k \geq 3.$ We thus have $$\begin{array}{lll}A^n & = & {\bf I}^n _3 + n{\bf I}^{n - 1} _3J + \binom{n}{2}{\bf I}^{n - 2}_3J^2 \vspace{2mm}\\ & = & \begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr\end{bmatrix} + \begin{bmatrix} 0 & n & n \cr 0 & 0 & n \cr 0 & 0 & 0 \cr\end{bmatrix} + \begin{bmatrix} 0 & 0 & \binom{n}{2} \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr\end{bmatrix} \vspace{2mm}\\ & = & \begin{bmatrix} 1 & n & \frac{n(n + 1)}{2} \cr 0 & 1 & n \cr 0 & 0 & 1 \cr \end{bmatrix}, \end{array}$$giving the result, since $\binom{n}{2} = \frac{n(n - 1)}{2}$ and $n + \binom{n}{2} = \frac{n(n + 1)}{2}$. \end{answer} \end{pro} \begin{pro} Let $(A,B)\in(\mat{n\times n}{ \BBF })^2$ and $k$ be a positive integer such that $A^k = {\bf 0}_n$. If $AB = B$ prove that $B = {\bf 0}_n$. \begin{answer} Argue inductively, $$\begin{array}{l} A^2B = A(AB) = AB = B \\ A^3B = A(A^2B) = A(AB) = AB = B \\ \vdots \\ A^mB = AB = B. \end{array}$$Hence $B = A^mB = {\bf 0}_nB = {\bf 0}_n$. \end{answer} \end{pro} \begin{pro} \label{pro:cayley_hamilton_2x2} Let $A = \begin{bmatrix}a & b \cr c & d \cr \end{bmatrix}$. Demonstrate that $$A^2 - (a + d)A + (ad - bc){\bf I}_2= {\bf 0}_2 $$. \end{pro} \begin{pro} Let $A\in\mat{2}{ \BBF }$ and let $k\in\BBZ, k >2$. Prove that $A^k = {\bf 0}_2 $ if and only if $A^2 = {\bf 0}_2$. \begin{answer} Put $A = \begin{bmatrix}a & b \cr c & d \cr \end{bmatrix}.$ Using \ref{pro:cayley_hamilton_2x2}, deduce by iteration that $$A^k = (a +d)^{k-1}A. $$ \end{answer} \end{pro} \begin{pro}Find all matrices $A\in\mat{2\times 2}{\BBR}$ such that $A^2 = {\bf 0}_2$ \begin{answer} $\begin{bmatrix}a & b \cr c & -a \end{bmatrix}, \ bc = -a^2$\end{answer} \end{pro} \begin{pro}Find all matrices $A\in\mat{2\times 2}{\BBR}$ such that $A^2 = {\bf I}_2$ \begin{answer} $\pm {\bf I}_2, \begin{bmatrix}a & b \cr c & -a \end{bmatrix}, \ \ a^2 = 1 -bc$ \end{answer} \end{pro} \begin{pro} Find a solution $X\in\mat{2\times 2}{\BBR}$ for $$X^2 - 2X = \left[\begin{array}{rl} -1 & 0 \\ 6 & 3\end{array}\right].$$ \begin{answer} We complete squares by putting $\dis{Y = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = X - I}$. Then $$\begin{bmatrix} a^2 + bc & b(a + d) \\ c(a + d) & bc + d^2\end{bmatrix} = Y^2 = X^2 - 2X + I = (X - I)^2 = \begin{bmatrix} -1 & 0 \\ 6 & 3\end{bmatrix} + I = \begin{bmatrix} 0 & 0 \\ 6 & 4\end{bmatrix}.$$ This entails $a = 0, b = 0, cd = 6, d^2 = 4.$ Using $X = Y + I$, we find that there are two solutions, $$\begin{bmatrix} 1 & 0 \\ 3 & 3\end{bmatrix},\ \ \begin{bmatrix} 1 & 0 \\ -3 & -1\end{bmatrix}.$$ \end{answer} \end{pro} \begin{pro} Find, with proof, a $4\times 4$ {\bf non-zero} matrix $A$ such that $$A\begin{bmatrix}1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr \end{bmatrix} = \begin{bmatrix}1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 \cr \end{bmatrix}A = \begin{bmatrix}0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}. $$ \begin{answer} The matrix $$A=\begin{bmatrix}1 & -1 & 1 & -1 \cr -1 & 1 & -1 & 1 \cr -1 & 1 & -1 & 1 \cr 1 & -1 & 1 & -1 \cr \end{bmatrix} $$ clearly satisfies the conditions. \end{answer} \end{pro} \begin{pro} Let $X$ be a $2\times 2$ matrices with real number entries. Solve the equation $$X^2 + X = \begin{bmatrix} 1& 1 \cr 1 & 1 \cr \end{bmatrix}. $$ \begin{answer} Put $X = \begin{bmatrix} a & b \cr c & d \cr\end{bmatrix}$. Then $$X^2 + X = \begin{bmatrix} 1& 1 \cr 1 & 1 \cr \end{bmatrix} \iff \left\{\begin{array}{l} a^2 + bc + a = 1 \\ ab + bd + b =1\\ ca + dc + c =1\\ cb + d^2 + d = 1 \end{array}\right. \iff \left\{\begin{array}{l} a^2 + bc + a = 1 \\ b(a + d + 1) =1\\ c(a + d + 1) =1\\ (d-a)(a+d+1) = 0 \end{array}\right. \iff \left\{\begin{array}{l} d = a \neq \dfrac{1}{2} \\ c=b = \dfrac{1}{2a+1}\\ a^2 + \dfrac{1}{(2a+1)^2 +a =1}\\ \end{array}\right. $$The last equation holds $$\iff 4a^4 + 8a^3 +a^2-3a = 0 \iff a\in\{-\frac{3}{2},-1,0,\frac{1}{2}\}. $$Thus the set of solutions is $$ \{\begin{bmatrix} -1& -1 \cr -1 & -1 \cr \end{bmatrix}, \begin{bmatrix} 0& 1 \cr 1 & 0 \cr \end{bmatrix}, \begin{bmatrix} 1/2& 1/2 \cr 1/2 & 1/2 \cr \end{bmatrix}, \begin{bmatrix} -3/2& -1/2 \cr -1/2 & -3/2 \cr \end{bmatrix}\} $$ \end{answer} \end{pro} \begin{pro} Prove, by means of induction that for the following $n\times n$ we have $$\begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \cr 0 & 1 & 1 & \cdots & 1 \cr 0 & 0 & 1 & \cdots & 1 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & 1 \cr \end{bmatrix}^3 = \begin{bmatrix} 1 & 3 & 6 & \cdots & \frac{n(n+1)}{2} \cr 0 & 1 & 3 & \cdots & \frac{(n-1)n}{2} \cr 0 & 0 & 1 & \cdots & \frac{(n-2)(n-1)}{2} \cr \vdots & \vdots & \cdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & 1 \cr\end{bmatrix}. $$ \end{pro} \begin{pro} Let $$A = \begin{bmatrix} 1 & -1 & -1 \cr -1 & 1 & -1 \cr -1 & -1 & 1 \cr\end{bmatrix}.$$Conjecture a formula for $A^n$ and prove it using induction. \begin{answer} Observe that $A = 2{\bf I}_3 - J$, where ${\bf I}_3$ is the $3\times 3$ identity matrix and $$J = \begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 1 \cr 1 & 1 & 1 \cr\end{bmatrix}.$$ Observe that $J^k = 3^{k-1}J$ for integer $k \geq 1$. Using the binomial theorem we have $$\begin{array}{lll}A^n &= & (2{\bf I}_3 - J)^n \\ &= & \sum _{k=0} ^n \binom{n}{k} (2{\bf I}_3)^{n-k}(-1)^kJ^k \\ & = & 2^n{\bf I}_3+\dfrac{1}{3}J\sum _{k=1} ^n \binom{n}{k} 2^{n-k}(-1)^k3^{k} \\ & = & 2^n{\bf I}_3+ \dfrac{1}{3}J((-1)^n-2^n) \\ & = & \dfrac{1}{3}\begin{bmatrix} (-1)^n+2^{n+1} & (-1)^n-2^n & (-1)^n-2^n \cr (-1)^n-2^n& (-1)^n+2^{n+1} & (-1)^n-2^n \cr (-1)^n-2^n & (-1)^n-2^n & (-1)^n+2^{n+1} \cr\end{bmatrix}.\end{array}$$ \end{answer} \end{pro} \section{Trace and Transpose} \begin{df} Let $A = [a_{ij}] \in \mat{n\times n}{ \BBF }$. Then the {\em trace} of $A$, denoted by $\tr{A}$ is the sum of the diagonal elements of $A$, that is $$\tr{A} = \sum _{k = 1} ^n a_{kk}.$$ \index{trace} \end{df} \begin{thm} Let $A = [a_{ij}] \in \mat{n\times n}{ \BBF }, B = [b_{ij}]\in \mat{n\times n}{ \BBF }$. Then \begin{equation} \tr{A + B} = \tr{A} + \tr{B},\end{equation} \begin{equation} \tr{AB} = \tr{BA}.\end{equation} \label{tracethm}\end{thm} \begin{pf} The first assertion is trivial. To prove the second, observe that $AB = (\sum _{k = 1} ^n a_{ik}b_{kj})$ and $BA = (\sum _{k = 1} ^n b_{ik}a_{kj})$. Then $$\tr{AB} = \sum _{i = 1} ^n \sum _{k = 1} ^n a_{ik}b_{ki} = \sum_{k = 1} ^n \sum _{i = 1} ^n b_{ki}a_{ik} = \tr{BA}, $$whence the theorem follows. \end{pf} \begin{exa} Let $A\in \mat{n\times n}{\BBR}$. Shew that $A$ can be written as the sum of two matrices whose trace is different from $0$. \end{exa} \begin{solu} Write $$A = (A - \alpha {\bf I}_n) + \alpha {\bf I}_n.$$ Now, $\tr{A - \alpha {\bf I}_n} = \tr{A} - n\alpha$ and $\tr{\alpha {\bf I}_n} = n\alpha$. Thus it suffices to take $\alpha \neq \dfrac{\tr{A}}{n}, \alpha \neq 0$. Since $\BBR$ has infinitely many elements, we can find such an $\alpha$. \end{solu} \begin{exa} Let $A, B$ be square matrices of the same size and over the same field of characteristic $0$. Is it possible that $AB - BA = {\bf I}_n$? Prove or disprove! \end{exa} \begin{solu} This is impossible. For if, taking traces on both sides $$0 = \tr{AB}-\tr{BA} = \tr{AB - BA} = \tr{{\bf I}_n} = n$$ a contradiction, since $n >0$. \end{solu} \begin{df} \index{transpose} The {\em transpose} of a matrix of a matrix $A = [a_{ij}] \in \mat{m\times n}{ \BBF }$ is the matrix $A^T = B = [b_{ij}] \in \mat{n\times m}{ \BBF }$, where $b_{ij} = a_{ji}$. \index{matrix!transpose} \end{df} \begin{exa} If $$M = \begin{bmatrix}a & b & c \cr d & e & f \cr g & h & i \cr\end{bmatrix},$$ with entries in $\BBR$, then $$ M^T = \begin{bmatrix}a & d & g \cr b & e & h \cr c & f & i \cr\end{bmatrix}.$$ \end{exa} \begin{thm}\label{thm:properties_transpose} Let $$A = [a_{ij}] \in \mat{m\times n}{ \BBF },\ B = [b_{ij}] \in \mat{m\times n}{ \BBF },\ C = [c_{ij}] \in \mat{n\times r}{ \BBF },\ \alpha \in \BBF, u\in\BBN .$$Then \begin{equation} A^{TT} = A,\end{equation} \begin{equation} (A + \alpha B)^T =\ A^T + \alpha B^T,\end{equation} \begin{equation} (AC)^T =\ C^TA^T,\end{equation} \begin{equation} (A^{u})^T = (A^T)^u.\end{equation} \end{thm} \begin{pf} The first two assertions are obvious, and the fourth follows from the third by using induction. To prove the third put $A^T = (\alpha_{ij}), \ \alpha _{ij} = a_{ji}$, $C^T = (\gamma_{ij}), \ \gamma _{ij} = c_{ji}$, $AC = (u_{ij})$ and $C^TA^T = (v_{ij})$. Then $$u_{ij} = \sum _{k = 1} ^n a_{ik}c_{kj} = \sum _{k = 1} ^n \alpha _{ki}\gamma _{jk} = \sum _{k = 1} ^n \gamma _{jk}\alpha _{ki} = v_{ji},$$ whence the theorem follows. \end{pf} \begin{df}\index{matrix!skew-symmetric} \index{matrix!symmetric} A square matrix $A\in\mat{n\times n}{ \BBF }$ is {\em symmetric} if $A^T = A.$ A matrix $B\in\mat{n\times n}{ \BBF }$ is {\em skew-symmetric} if $B^T = -B.$ \end{df} \begin{exa} Let $A, B$ be square matrices of the same size, with $A$ symmetric and $B$ skew-symmetric. Prove that the matrix $A^2BA^2$ is skew-symmetric. \end{exa} \begin{solu} We have $$(A^2BA^2)^T = (A^2)^T (B)^T(A^2)^T = A^2(-B)A^2 = -A^2BA^2.$$ \end{solu} \begin{thm} Let $\BBF$ be a field of characteristic different from $2$. Then any square matrix $A$ can be written as the sum of a symmetric and a skew-symmetric matrix. \end{thm} \begin{pf} Observe that $$(A + A^T)^T = A^T + A^{TT} = A^T + A,$$and so $A + A^T$ is symmetric. Also, $$(A - A^T)^T = A^T - A^{TT} = -(A - A^T),$$and so $A - A^T$ is skew-symmetric. We only need to write $A$ as $$A = (2^{-1})(A + A^T) + (2^{-1})(A - A^T)$$to prove the assertion. \end{pf} \begin{exa} Find, with proof, a square matrix $A$ with entries in $\BBZ _2$ such $A$ is not the sum of a symmetric and and anti-symmetric matrix. \end{exa} \begin{solu}In $\BBZ_2$ every symmetric matrix is also anti-symmetric, since $-x = x$. Thus it is enough to take a non-symmetric matrix, for example, take $A = \begin{bmatrix} \overline{0} & \overline{1} \cr \overline{0} & \overline{0} \cr \end{bmatrix}$. \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Write $$ A = \begin{bmatrix} 1 & 2 & 3 \cr 2 & 3 & 1 \cr 3 & 1 & 2 \cr\end{bmatrix}\in\mat{3\times 3}{\BBR}$$ as the sum of two $3\times 3$ matrices ${\bf E}_1, {\bf E}_2$, with $\tr{{\bf E}_2} = 10$. \begin{answer} There are infinitely many solutions. Here is one: $$A = \begin{bmatrix} 1 & 2 & 3 \cr 2 & 3 & 1 \cr 3 & 1 & 2 \cr\end{bmatrix} = \begin{bmatrix} -9 & 2 & 3 \cr 2 & 3 & 1 \cr 3 & 1 & 2 \cr\end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr \end{bmatrix}.$$\end{answer} \end{pro} \begin{pro} Give an example of two matrices $A\in\mat{2\times 2}{\BBR}$ and $B\in \mat{2\times 2}{\BBR}$ that {\em simultaneously} satisfy the following properties: \begin{enumerate} \item $A\neq \begin{bmatrix}0 & 0 \cr 0 & 0 \cr \end{bmatrix}$ and $B\neq \begin{bmatrix}0 & 0 \cr 0 & 0 \cr \end{bmatrix}$. \item $AB=\begin{bmatrix}0 & 0 \cr 0 & 0 \cr \end{bmatrix}$ and $BA=\begin{bmatrix}0 & 0 \cr 0 & 0 \cr \end{bmatrix}$. \item $\tr{A}=\tr{B}=2$. \item $A=A^T$ and $B=B^T$. \end{enumerate} \begin{answer} There are infinitely many examples. One could take $A=\begin{bmatrix} 1 & 1 \cr 1 & 1 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & -1 \cr -1 & 1 \end{bmatrix}$. Another set is $A=\begin{bmatrix} 2 & 0 \cr 0 & 0 \end{bmatrix}$ and $B=\begin{bmatrix} 0 & 0 \cr 0 & 2 \end{bmatrix}$. \end{answer} \end{pro} \begin{pro} Shew that there are no matrices $(A, B, C, D)\in (\mat{n\times n}{\BBR})^4$ such that $$AC + DB = {\bf I}_n,$$ $$CA + BD = {\bf 0}_n.$$ \begin{answer}If such matrices existed, then by the first equation $$\tr{AC} + \tr{DB} = n.$$ By the second equation and by Theorem \ref{tracethm}, $$0 = \tr{CA} + \tr{BD} = \tr{AC} + \tr{DB} = n,$$ a contradiction, since $n \geq 1.$ \end{answer} \end{pro} \begin{pro} Let $(A, B) \in (\mat{2\times 2}{\BBR})^2$ be symmetric matrices. Must their product $AB$ be symmetric? Prove or disprove! \begin{answer}Disprove! This is not generally true. Take $A = \dis{\begin{bmatrix} 1 & 1 \cr 1 & 2 \end{bmatrix}}$ and $B = \dis{\begin{bmatrix} 3 & 0 \cr 0 & 1 \end{bmatrix}}$. Clearly $A^T = A$ and $B^T = B$. We have $$AB = \begin{bmatrix} 3 & 1 \cr 3 & 2\cr\end{bmatrix}$$but $$(AB)^T = \begin{bmatrix} 3 & 3 \cr 1 & 2\cr\end{bmatrix}.$$ \end{answer} \end{pro} \begin{pro} Given square matrices $(A, B) \in (\mat{7\times 7}{\BBR})^2$ such that $\tr{A^2} = \tr{B^2} = 1$, and $$ (A - B)^2 = 3{\bf I}_7, $$find $\tr{BA}$. \end{pro} \begin{pro} Consider the matrix $A = \begin{bmatrix} a & b \cr c & d \cr\end{bmatrix} \in \mat{2\times 2}{\BBR}$. Find necessary and sufficient conditions on $a, b, c, d$ so that $\tr{A^2} = (\tr{A})^2$. \begin{answer} We have $$\tr{A^2} = \tr{\begin{bmatrix} a & b \cr c & d \cr\end{bmatrix}\begin{bmatrix} a & b \cr c & d \cr\end{bmatrix}} = \tr{\begin{bmatrix} a^2 + bc & ab + bd \cr ca + cd & d^2 + cb \cr\end{bmatrix}} = a^2 + d^2 + 2bc $$and $$\left(\tr{\begin{bmatrix} a & b \cr c & d \cr\end{bmatrix}}\right)^2 = (a+d)^2. $$ Thus $$ \tr{A^2} = (\tr{A})^2 \iff a^2 + d^2 + 2bc = (a + d)^2 \iff bc = ad, $$is the condition sought. \end{answer} \end{pro} \begin{pro} Given a square matrix $A \in \mat{4\times 4}{\BBR}$ such that $\tr{A^2} = -4$, and $$ (A - {\bf I}_4)^2 = 3{\bf I}_4, $$find $\tr{A}$. \begin{answer} $$\begin{array}{lll}\tr{(A - {\bf I}_4)^2} & = & \tr{A^2 - 2A + {\bf I}_4} \\ & = & \tr{A^2} - 2\tr{A} + \tr{{\bf I}_4} \\ & = & -4 - 2\tr{A} + 4 \\ & = & -2\tr{A}, \end{array}$$ and $\tr{3{\bf I}_4} = 12. $ Hence $-2\tr{A} = 12$ or $\tr{A} = -6$. \end{answer} \end{pro} \begin{pro} Prove or disprove! If $A, B$ are square matrices of the same size, then it is always true that $\tr{AB} = \tr{A}\tr{B}$. \begin{answer} Disprove! Take $A = B = {\bf I}_n$ and $n>1$. Then $\tr{AB} = n< n^2 = \tr{A}\tr{B}$. \end{answer} \end{pro} \begin{pro} Prove or disprove! If $(A, B, C)\in(\mat{3\times 3}{ \BBF })^3$ then $\tr{ABC} = \tr{BAC}$. \begin{answer}Disprove! Take $A =\begin{bmatrix}1 & 0 \cr 0 & 0 \cr \end{bmatrix}$, $B =\begin{bmatrix}0 & 1 \cr 0 & 0 \cr \end{bmatrix}$, $C =\begin{bmatrix}0 & 0 \cr 1 & 0 \cr \end{bmatrix}$. Then $\tr{ABC} = 1$ but $\tr{BAC} = 0$. \end{answer} \end{pro} \begin{pro} Let $A$ be a square matrix. Prove that the matrix $AA^T$ is symmetric. \begin{answer} We have $$(AA^T)^T = (A^T)^TA^T = AA^T.$$ \end{answer} \end{pro} \begin{pro} Let $A, B$ be square matrices of the same size, with $A$ symmetric and $B$ skew-symmetric. Prove that the matrix $AB - BA$ is symmetric. \begin{answer} We have $$(AB - BA)^T = (AB)^T - (BA)^T = B^TA^T - A^TB^T = -BA - A(-B) = AB - BA.$$ \end{answer} \end{pro} \begin{pro} Let $A \in \mat{n\times n}{ \BBF }, A = [a_{ij}]$. Prove that $\tr{AA^T} = \sum _{i = 1} ^n \sum _{j = 1} ^n a_{ij} ^2$. \end{pro} \begin{pro} Let $X\in\mat{n\times n}{\BBR}$. Prove that if $XX^T = {\bf 0}_n$ then $X={\bf 0}_n$. \begin{answer} Let $X=[x_{ij}]$ and put $XX^T = [c_{ij}]$. Then $$0 = c_{ii} = \sum _{k = 1} ^n x_{ik}^2\implies x_{ik} = 0.$$ \end{answer} \end{pro} \begin{pro} Let $m,n,p$ be positive integers and $A\in\mat{m\times n}{\BBR}$, $B\in\mat{n\times p}{\BBR}$, $C\in\mat{p\times m}{\BBR}$. Prove that $(BA)^TA = (CA)^TA \implies BA = CA$. \end{pro} \end{multicols} \section{Special Matrices} \begin{df} The {\em main diagonal} of a square matrix $A = [a_{ij}]\in\mat{n\times n}{ \BBF }$ is the set $\{a_{ii}: i\leq n\}$. The {\em counter diagonal} of a square matrix $A = [a_{ij}]\in\mat{n\times n}{ \BBF }$ is the set $\{a_{(n -i + 1)i}: i\leq n\}$.\end{df} \begin{exa}The main diagonal of the matrix $$A = \begin{bmatrix} 0 & 1 & 5 \cr 3 & 2 & 4 \cr 9 & 8 & 7 \cr \end{bmatrix} $$is the set $\{0, 2, 7\}$. The counter diagonal of $A$ is the set $\{5, 2, 9\}$. \end{exa} \begin{df} A square matrix is a {\em diagonal} matrix if every entry off its main diagonal is $0_{\BBF }$. \index{matrix!diagonal} \end{df} \begin{exa}The matrix $$A = \begin{bmatrix} 1 & 0 & 0 \cr 0 & 2 & 0\cr 0 & 0 & 3 \cr \end{bmatrix} $$is a diagonal matrix. \end{exa} \begin{df} A square matrix is a {\em scalar} matrix if it is of the form $\alpha {\bf I}_n$ for some scalar $\alpha$. \index{matrix!scalar} \end{df} \begin{exa}The matrix $$A = \begin{bmatrix} 4 & 0 & 0 \cr 0 & 4 & 0\cr 0 & 0 & 4 \cr \end{bmatrix} = 4{\bf I}_3 $$is a scalar matrix. \end{exa} \begin{df} $A\in \mat{m\times n}{ \BBF }$ is said to be {\em upper triangular} if $$(\forall (i, j)\in \{1, 2, \cdots , n\}^2), (i > j, \ a_{ij} = 0_{\BBF }),$$ that is, every element below the main diagonal is $0_{\BBF }$. Similarly, $A$ is {\em lower triangular} if $$(\forall (i, j) \in \{1, 2, \cdots , n\}^2), (i < j, \ a_{ij} = 0_{\BBF }),$$ that is, every element above the main diagonal is $0_{\BBF }$. \end{df} \begin{exa} The matrix $A\in\mat{3\times 4}{\BBR}$ shewn is upper triangular and $B\in\mat{4\times 4}{\BBR}$ is lower triangular. $$A = \begin{bmatrix} 1 & a & b & c \cr 0 & 2 & 3 & 0 \cr 0 & 0 & 0 & 1 \cr \end{bmatrix} \ \ B = \begin{bmatrix} 1 & 0 & 0& 0 \cr 1 & a & 0& 0 \cr 0 & 2 & 3 & 0 \cr 1 & 1 & t & 1 \cr \end{bmatrix}$$ \end{exa} \begin{df}\index{Kronecker delta} The {\em Kronecker delta } $\delta _{ij}$ is defined by $$\delta _{ij} = \left\{\begin{array}{ll} 1_{\BBF } & {\rm if}\ i = j\\ 0_{\BBF } & {\rm if}\ i \neq j\end{array}\right. $$ \end{df} \begin{df}\index{matrix!elementary} The set of matrices ${\bf E}_{ij}\in\mat{m\times n}{ \BBF }, {\bf E}_{ij} = ( e_{rs})$ such that $e_{ij} = 1_{\BBF }$ and $e_{i'j'}=0_{\BBF }, (i',j')\neq (i,j)$ is called the set of {\em elementary matrices}. Observe that in fact $e_{rs} = \delta _{ir}\delta _{sj}$. \end{df} Elementary matrices have interesting effects when we pre-multiply and post-multiply a matrix by them. \begin{exa} Let $$A = \begin{bmatrix} 1 & 2 & 3 & 4\cr 5 & 6 & 7 & 8 \cr 9 & 10 & 11 & 12 \cr \end{bmatrix}, \ \ \ {\bf E}_{23} = \begin{bmatrix} 0 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr \end{bmatrix}. $$ Then $${\bf E}_{23}A =\begin{bmatrix} 0 & 0 & 0 & 0 \cr 9 & 10 & 11 & 12 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}, \ \ \ AE_{23} =\begin{bmatrix} 0 & 0 & 2 \cr 0 & 0 & 6 \cr 0 & 0 & 10\cr \end{bmatrix}. $$ \end{exa} \begin{thm}[Multiplication by Elementary Matrices]\label{thm:mult_elem_matrix} Let ${\bf E}_{ij}\in\mat{m\times n}{ \BBF }$ be an elementary matrix, and let $A\in\mat{n\times m}{ \BBF }$. Then ${\bf E}_{ij}A$ has as its $i$-th row the $j$-th row of $A$ and $0_{\BBF }$'s everywhere else. Similarly, $A{\bf E}_{ij}$ has as its $j$-th column the $i$-th column of $A$ and $0_{\BBF }$'s everywhere else. \end{thm} \begin{pf} Put $(\alpha _{uv}) = {\bf E}_{ij}A$. To obtain ${\bf E}_{ij}A$ we multiply the rows of ${\bf E}_{ij}$ by the columns of $A$. Now $$ \alpha _{uv} = \sum _{k = 1} ^n e_{uk}a_{kv} = \sum _{k = 1} ^n \delta _{ui}\delta _{kj}a_{kv} = \delta _{ui}a_{jv}.$$Therefore, for $u \neq i, \ \alpha _{uv} = 0_{\BBF }$, i.e., off of the $i$-th row the entries of ${\bf E}_{ij}A$ are $0_{\BBF }$, and $\alpha _{iv} = \alpha _{jv}$, that is, the $i$-th row of ${\bf E}_{ij}A$ is the $j$-th row of $A$. The case for $A{\bf E}_{ij}$ is similarly argued.\end{pf} The following corollary is immediate. \begin{cor} Let $({\bf E}_{ij}, {\bf E}_{kl})\in (\mat{n\times n}{ \BBF })^2$, be square elementary matrices. Then $${\bf E}_{ij}{\bf E}_{kl} = \delta _{jk}{\bf E}_{il}. $$ \end{cor} \begin{exa} Let $M\in\mat{n\times n}{ \BBF }$ be a matrix such that $AM = MA$ for all matrices $A\in\mat{n\times n}{ \BBF }$. Demonstrate that $M = a{\bf I}_n$ for some $a\in \BBF$, i.e. $M$ is a scalar matrix. \label{ex:scalarmatrix}\end{exa} \begin{solu} Assume $(s, t)\in\{1, 2, \ldots , n\}^2$. Let $M = (m_{ij})$ and ${\bf E}_{st}\in\mat{n\times n}{ \BBF }$. Since $M$ commutes with ${\bf E}_{st}$ we have $$ \begin{bmatrix} 0 & 0 & \ldots & 0 \cr \vdots & \vdots & \ldots & \vdots \cr m_{t1} & m_{t2} & \ldots & m_{tn} \cr \vdots & \vdots & \ldots & \vdots \cr 0 & 0 & \ldots & 0 \cr \end{bmatrix} = {\bf E}_{st}M = M{\bf E}_{st} = \begin{bmatrix} 0 & 0 & \ldots & m_{1s} & \ldots & 0\cr 0 & 0 & \vdots & m_{2s} & \vdots & 0\cr \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\cr 0 & 0 & \vdots & m_{(n - 1)s} & \vdots & 0\cr 0 & 0 & \vdots & m_{ns} & \vdots & 0\cr \end{bmatrix}$$ For arbitrary $s \neq t$ we have shown that $m_{st} = m_{ts} = 0$, and that $m_{ss} = m_{tt}$. Thus the entries off the main diagonal are zero and the diagonal entries are all equal to one another, whence $M$ is a scalar matrix. \end{solu} \begin{df} Let $\lambda\in \BBF$ and ${\bf E}_{ij}\in \mat{n\times n}{ \BBF }$. A square matrix in $\mat{n\times n}{ \BBF }$ of the form ${\bf I}_n + \lambda {\bf E}_{ij}$ is called a {\em transvection}. \end{df} \begin{exa} The matrix $$T = {\bf I}_3 + 4{\bf E}_{13} = \begin{bmatrix}1 & 0 & 4 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} $$is a transvection. Observe that if $$A = \begin{bmatrix}1 & 1 & 1 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix} $$then $$TA = \begin{bmatrix}1 & 0 & 4 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}\begin{bmatrix}1 & 1 & 1 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix} = \begin{bmatrix}5 & 9 & 13 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix},$$that is, pre-multiplication by $T$ adds $4$ times the third row of $A$ to the first row of $A$. Similarly, $$AT = \begin{bmatrix}1 & 1 & 1 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix}\begin{bmatrix}1 & 0 & 4 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} = \begin{bmatrix}1 & 1 & 5 \cr 5 & 6 & 27 \cr 1 & 2 & 7 \cr \end{bmatrix},$$that is, post-multiplication by $T$ adds $4$ times the first column of $A$ to the third row of $A$. \end{exa} In general, we have the following theorem. \begin{thm}[Multiplication by a Transvection Matrix]\label{thm:mult_by_transvection_matrix}\index{matrix!transvection} Let ${\bf I}_n + \lambda {\bf E}_{ij}\in \mat{n\times n}{ \BBF }$ be a transvection and let $A\in \mat{n\times m}{ \BBF }$. Then $({\bf I}_n + \lambda {\bf E}_{ij})A$ adds the $j$-th row of $A$ to its $i$-th row and leaves the other rows unchanged. Similarly, if $B\in \mat{p\times n}{ \BBF }$, $B({\bf I}_n + \lambda {\bf E}_{ij})$ adds the $i$-th column of $B$ to the $j$-th column and leaves the other columns unchanged. \end{thm} \begin{pf} Simply observe that $({\bf I}_n + \lambda {\bf E}_{ij})A = A + \lambda {\bf E}_{ij}A$ and $A({\bf I}_n + \lambda {\bf E}_{ij}) = A + \lambda A{\bf E}_{ij}$ and apply Theorem \ref{thm:mult_elem_matrix}. \end{pf} Observe that the particular transvection ${\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii}\in\mat{n\times n}{ \BBF }$ consists of a diagonal matrix with $1_{\BBF }$'s everywhere on the diagonal, except on the $ii$-th position, where it has a $\lambda$. \begin{df}\index{matrix!dilatation} If $\lambda \neq 0_{\BBF }$, we call the matrix ${\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii}$ a {\em dilatation matrix}. \end{df} \begin{exa} The matrix $$S = {\bf I}_3 + (4 - 1){\bf E}_{11} = \begin{bmatrix}4 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} $$is a dilatation matrix. Observe that if $$A = \begin{bmatrix}1 & 1 & 1 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix} $$then $$SA = \begin{bmatrix}4 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}\begin{bmatrix}1 & 1 & 1 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix} = \begin{bmatrix}4 & 4 & 4 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix},$$that is, pre-multiplication by $S$ multiplies by $4$ the first row of $A$. Similarly, $$AS = \begin{bmatrix}1 & 1 & 1 \cr 5 & 6 & 7 \cr 1 & 2 & 3 \cr \end{bmatrix}\begin{bmatrix}4 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} = \begin{bmatrix}4 & 1 & 1 \cr 20 & 6 & 7 \cr 4 & 2 & 3 \cr \end{bmatrix},$$that is, post-multiplication by $S$ multiplies by $4$ the first column of $A$. \end{exa} \begin{thm}[Multiplication by a Dilatation Matrix]\label{thm:mult_by_dilatation_matrix} Pre-multiplication of\\ the~matrix$A\in\mat{n\times m}{ \BBF }$ by the~dilatation~matrix ${\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii}\in\mat{n\times n}{ \BBF }$ multiplies the $i$-th row of $A$ by $\lambda$ and leaves the other rows of $A$ unchanged. Similarly, if $B\in \mat{p\times n}{ \BBF }$ post-multiplication of $B$ by ${\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii}$ multiplies the $i$-th column of $B$ by $\lambda$ and leaves the other columns of $B$ unchanged. \end{thm} \begin{pf} This follows by direct application of Theorem \ref{thm:mult_by_transvection_matrix}. \end{pf} \begin{df} We write ${\bf I}^{ij} _n$ for the matrix which permutes the $i$-th row with the $j$-th row of the identity matrix. We call ${\bf I}^{ij} _n$ a {\em transposition matrix}. \end{df} \begin{exa} We have $${\bf I}^{(23)} _4 = \begin{bmatrix}1 & 0 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 0 & 1 \cr \end{bmatrix}. $$ If $$A = \begin{bmatrix} 1 & 2 & 3 & 4\cr 5 & 6 & 7 & 8 \cr 9 & 10 & 11 & 12 \cr 13 & 14 & 15 & 16 \cr \end{bmatrix}, $$then $${\bf I}^{(23)} _4A = \begin{bmatrix} 1 & 2 & 3 & 4\cr 9 & 10 & 11 & 12 \cr 5 & 6 & 7 & 8 \cr 13 & 14 & 15 & 16 \cr \end{bmatrix}, $$and $$A{\bf I}^{(23)} _4 = \begin{bmatrix} 1 & 3 & 2 & 4\cr 5 & 7 & 6 & 8 \cr 9 & 11 & 10 & 12 \cr 13 & 15 & 14 & 16 \cr \end{bmatrix}.$$ \end{exa} \begin{thm}[Multiplication by a Transposition Matrix]\label{thm:mult_by_transposition_matrix}\index{matrix!transposition} If $A\in \mat{n\times m}{ \BBF }$, then ${\bf I}^{ij} _nA$ is the matrix obtained from $A$ permuting the the $i$-th row with the $j$-th row of $A$. Similarly, if $B\in \mat{p\times n}{ \BBF }$, then $B{\bf I}^{ij} _n$ is the matrix obtained from $B$ by permuting the $i$-th column with the $j$-th column of $B$.\end{thm} \begin{pf} We must prove that ${\bf I}^{ij} _n A$ exchanges the $i$-th and $j$-th rows but leaves the other rows unchanged. But this follows upon observing that $${\bf I}^{ij} _n = {\bf I}_n + {\bf E}_{ij} + {\bf E}_{ji} - {\bf E}_{ii} - {\bf E}_{jj}$$and appealing to Theorem \ref{thm:mult_elem_matrix}. \end{pf} \begin{df}\index{matrix!elimination} A square matrix which is either a transvection matrix, a dilatation matrix or a transposition matrix is called an {\em elimination matrix.} \end{df} \begin{rem} In a very loose way, we may associate pre-multiplication of a matrix $A$ by another matrix with an operation on the rows of $A$, and post-multiplication of a matrix $A$ by another with an operation on the columns of $A$. \end{rem} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Consider the matrices $$A = \begin{bmatrix} 1 & 0 & 1 & 0 \cr 0 & 1 & 0 & 1 \cr -1 & 1 & 1 & 1 \cr 1 & -1 & 1 & 1 \cr \end{bmatrix}, \ \ \ \ B = \begin{bmatrix} 4 & -2 & 4 & 2 \cr 0 & 1 & 0 & 1 \cr 1 & 1 & -1 & 1 \cr 1 & -1 & 1 & 1\cr \end{bmatrix}. $$Find a specific dilatation matrix $D$, a specific transposition matrix $P$, and a specific transvection matrix $T$ such that $B = TDAP$.\\ \begin{answer} Here is one possible approach. If we perform $C_1 \leftrightarrow C_3$ on $A$ we obtain $$A_1 = \begin{bmatrix} 1 & 0 & 1 & 0 \cr 0 & 1 & 0 & 1 \cr 1 & 1 & -1 & 1 \cr 1 & -1 & 1 & 1 \cr \end{bmatrix} \ \ \ \mathrm{so\ take\ } \ \ \ P = \begin{bmatrix} 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 \cr \end{bmatrix}. $$ Now perform $2R_1 \rightarrow R_1$ on $A_1$ to obtain $$A_2 = \begin{bmatrix} 2 & 0 & 2 & 0 \cr 0 & 1 & 0 & 1 \cr 1 & 1 & -1 & 1 \cr 1 & -1 & 1 & 1 \cr \end{bmatrix} \ \ \ \mathrm{so\ take\ } \ \ \ D = \begin{bmatrix} 2 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr \end{bmatrix}. $$ Finally, perform $R_1 + 2R_4 \rightarrow R_1$ on $A_2$ to obtain $$B = \begin{bmatrix} 4 & -2 & 4 & 2 \cr 0 & 1 & 0 & 1 \cr 1 & 1 & -1 & 1 \cr 1 & -1 & 1 & 1 \cr \end{bmatrix} \ \ \ \mathrm{so\ take\ } \ \ \ T = \begin{bmatrix} 1 & 0 & 0 & 2 \cr 0 & 1 & 0 & 0 \cr 0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 \cr \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} The matrix $$ A = \begin{bmatrix} a & b & c \cr d & e & f \cr g & h & i \cr \end{bmatrix} $$ is transformed into the matrix $$ B = \begin{bmatrix} h - g & g & i \cr e - d & d & f \cr 2b - 2a & 2a & 2c \cr \end{bmatrix} $$by a series of row and column operations. Find explicit permutation matrices $P, P'$, an explicit dilatation matrix $D$, and an explicit transvection matrix $T$ such that $$B = DPAP'T. $$ \begin{answer} Here is one possible approach. \begin{eqnarray*} \begin{bmatrix} a & b & c \cr d & e & f \cr g & h & i \cr \end{bmatrix} & \grstep{P: \rho_3 \swap \rho_1} & \begin{bmatrix} g & h & i \cr d & e & f \cr a & b & c \cr \end{bmatrix} \\ & \grstep{P': C_1 \swap C_2} & \begin{bmatrix} h & g & i \cr e & d & f \cr b & a & c \cr \end{bmatrix} \\ & \grstep{T: C_1 - C_2 \longrightarrow C_1} & \begin{bmatrix} h - g & g & i \cr e - d & d & f \cr b - a & a & c \cr \end{bmatrix} \\ & \grstep{D: 2\rho_3 \longrightarrow \rho_3} & \begin{bmatrix} h - g & g & i \cr e - d & d & f \cr 2b - 2a & 2a & 2c \cr \end{bmatrix} \end{eqnarray*} Thus we take $$P = \begin{bmatrix} 0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr \end{bmatrix}, \ \ P' = \begin{bmatrix} 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} , $$ $$T = \begin{bmatrix} 1 & 0 & 0 \cr -1 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}, \ \ \ D = \begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 2 \cr \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Let $A\in \mat{n\times n}{ \BBF }$. Prove that if $$(\forall X\in \mat{n\times n}{ \BBF }), \ (\tr{AX} = \tr{BX}),$$ then $A = B$. \begin{answer} Let ${\bf E}_{ij}\in\mat{n\times n}{ \BBF }$. Then $$ A{\bf E}_{ij} = \begin{bmatrix} 0 & 0 & \ldots & a_{1i} & \ldots & 0\cr 0 & 0 & \vdots & a_{2i} & \vdots & 0\cr \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\cr 0 & 0 & \vdots & a_{n - 1i} & \vdots & 0\cr 0 & 0 & \vdots & a_{ni} & \vdots & 0\cr\end{bmatrix}, $$where the entries appear on the $j$-column. Then we see that $\tr{A{\bf E}_{ij}} = a_{ji}$ and similarly, by considering $B{\bf E}_{ij}$, we see that $\tr{B{\bf E}_{ij}} = b_{ji}.$ Therefore $\forall i,j, \ \ a_{ji} = b_{ji}$, which implies that $A = B$. \end{answer} \end{pro} \begin{pro} Let $A\in \mat{n\times n}{\BBR}$ be such that $$(\forall X\in \mat{n\times n}{\BBR}),\ ((XA)^2 = {\bf 0}_{n}).$$ Prove that $A = {\bf 0}_{ n}$. \begin{answer} Let ${\bf E}_{st}\in\mat{n\times n}{\BBR}$. Then $$ {\bf E}_{ij}A = \begin{bmatrix} 0 & 0 & \ldots & 0 \cr \vdots & \vdots & \ldots & \vdots \cr a_{j1} & a_{j2} & \ldots & a_{jn} \cr \vdots & \vdots & \ldots & \vdots \cr 0 & 0 & \ldots & 0 \cr\end{bmatrix},$$where the entries appear on the $i$-th row. Thus $$ ({\bf E}_{ij}A)^2 = \begin{bmatrix} 0 & 0 & \ldots & 0 \cr \vdots & \vdots & \ldots & \vdots \cr a_{ji}a_{j1} & a_{ji}a_{j2} & \ldots & a_{ji}a_{jn} \cr \vdots & \vdots & \ldots & \vdots \cr 0 & 0 & \ldots & 0 \cr\end{bmatrix},$$which means that $\forall i, j, a_{ji}a_{jk} = 0$. In particular, $a_{ji} ^2 = 0,$ which means that $\forall i, j, a_{ji} = 0$, i.e., $A = {\bf 0}_{n}.$ \end{answer} \end{pro} \end{multicols} \section{Matrix Inversion} \begin{df} Let $A\in\mat{m\times n}{ \BBF }$. Then $A$ is said to be {\em left-invertible} if $\exists L\in\mat{n\times m}{ \BBF }$ such that $LA = {\bf I}_{n}$. $A$ is said to be {\em right-invertible} if $\exists R\in\mat{n\times m}{ \BBF }$ such that $AR = {\bf I}_{m}$. A matrix is said to be {\em invertible} if it possesses a right and a left inverse. A matrix which is not invertible is said to be {\em singular}. \end{df} \begin{exa} The matrix $A\in \mat{2\times 3}{\BBR}$ $$A = \begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}$$ has infinitely many right-inverses of the form $$ R_{(x, y)} = \begin{bmatrix} 1 & 0 \cr 0 & 1 \cr x & y \cr \end{bmatrix}. $$For $$\begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix} \begin{bmatrix} 1 & 0 \cr 0 & 1 \cr x & y \cr \end{bmatrix} = \begin{bmatrix} 1 & 0 \cr 0 & 1 \end{bmatrix},$$regardless of the values of $x$ and $y$. Observe, however, that $A$ does not have a left inverse, for $$\begin{bmatrix}a & b \cr c & d\cr f & g \cr \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix} = \begin{bmatrix}a & b & 0 \cr c & d & 0 \cr f & g & 0 \end{bmatrix},$$which will never give ${\bf I}_3$ regardless of the values of $a,b,c,d,f,g$. \end{exa} \begin{exa} If $\lambda \neq 0$, then the scalar matrix $\lambda {\bf I}_n$ is invertible, for $$ \left(\lambda {\bf I}_n\right)\left(\lambda^{-1} {\bf I}_n\right) = {\bf I}_n= \left(\lambda^{-1} {\bf I}_n\right)\left(\lambda {\bf I}_n\right).$$ \end{exa} \begin{exa} The zero matrix ${\bf 0}_n$ is singular. \end{exa} \begin{thm}\label{thm:inverse_square_matrices} Let $A\in\mat{n\times n}{ \BBF }$ a square matrix possessing a left inverse $L$ and a right inverse $R$. Then $L = R$. Thus an invertible square matrix possesses a unique inverse. \end{thm} \begin{pf} Observe that we have $LA = {\bf I}_n = AR$. Then $$L = L{\bf I}_n = L(AR) = (LA)R = {\bf I}_nR = R. $$ \end{pf} \begin{df} The subset of $\mat{n\times n}{ \BBF }$ of all invertible $n\times n$ matrices is denoted by $\gl{n}{ \BBF }$, read ``the linear group of rank $n$ over $\BBF$.'' \end{df} \begin{cor}\label{cor:inverse_of_product_of_matrices} Let $(A, B)\in (\gl{n}{ \BBF })^2$. Then $AB$ is also invertible and $$(AB)^{-1} = B^{-1}A^{-1}. $$ \end{cor} \begin{pf} Since $AB$ is a square matrix, it suffices to notice that $$B^{-1}A^{-1}(AB) = (AB)B^{-1}A^{-1} = {\bf I}_n$$ and that since the inverse of a square matrix is unique, we must have $B^{-1}A^{-1} = (AB)^{-1}.$ \end{pf} \begin{cor} If a square matrix $S\in\mat{n\times n}{ \BBF }$ is invertible, then $S^{-1}$ is also invertible and $(S^{-1})^{-1} = S$, in view of the uniqueness of the inverses of square matrices. \end{cor} \begin{cor} If a square matrix $A\in\mat{n\times n}{ \BBF }$ is invertible, then $A^T$ is also invertible and $(A^T)^{-1} = (A^{-1})^T$. \end{cor} \begin{pf} We claim that $(A^T)^{-1}=(A^{-1})^T$. For $$AA^{-1} = {\bf I}_n \implies (AA^{-1})^T = {\bf I}_n ^T \implies (A^{-1})^TA^T = {\bf I}_n, $$where we have used Theorem \ref{thm:properties_transpose}. \end{pf} The next few theorems will prove that elimination matrices are invertible matrices. \begin{thm}[Invertibility of Transvections] \label{thm:inverse_transvections} Let ${\bf I}_n + \lambda {\bf E}_{ij}\in \mat{n\times n}{ \BBF }$ be a transvection, and let $i \neq j$. Then $$({\bf I}_n + \lambda {\bf E}_{ij})^{-1} = {\bf I}_n -\lambda {\bf E}_{ij}.$$ \end{thm} \begin{pf} Expanding the product $$\begin{array}{lll} ({\bf I}_n + \lambda {\bf E}_{ij})({\bf I}_n -\lambda {\bf E}_{ij}) & = & {\bf I}_n + \lambda {\bf E}_{ij} -\lambda {\bf E}_{ij} -\lambda ^2 {\bf E}_{ij}{\bf E}_{ij} \\ & = & {\bf I}_n -\lambda^2\delta _{ij}{\bf E}_{ij}\\ & = & {\bf I}_n,\end{array}$$since $i \neq j$. \end{pf} \begin{exa} By Theorem \ref{thm:inverse_transvections}, we have $$\begin{bmatrix}1 & 0 & 3 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & -3 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{bmatrix}. $$ \end{exa} \begin{thm}[Invertibility of Dilatations]\label{thm:inver_dilatation} Let $\lambda \neq 0_{\BBF }$. Then $$({\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii})^{-1} = {\bf I}_n + (\lambda^{-1} - 1_{\BBF }) {\bf E}_{ii}.$$\end{thm} \begin{pf} Expanding the product $$\begin{array}{lll}({\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii})({\bf I}_n + (\lambda^{-1} - 1_{\BBF }) {\bf E}_{ii}) & = & {\bf I}_n + (\lambda - 1_{\BBF }) {\bf E}_{ii} \\ & & \quad + (\lambda^{-1} - 1_{\BBF }) {\bf E}_{ii} \\ & & \qquad + (\lambda - 1_{\BBF })(\lambda^{-1} - 1_{\BBF }){\bf E}_{ii} \\ & = & {\bf I}_n + (\lambda - 1_{\BBF }){\bf E}_{ii} \\ & & \quad + (\lambda^{-1} - 1_{\BBF }){\bf E}_{ii} \\ & & \qquad + (\lambda - 1_{\BBF })(\lambda^{-1} - 1_{\BBF })){\bf E}_{ii}\\ & = & {\bf I}_n + (\lambda - 1_{\BBF } + \lambda^{-1} - 1_{\BBF } + 1_{\BBF } \\ & & \qquad - \lambda - \lambda^{-1} - 1_{\BBF })){\bf E}_{ii}\\ & = & {\bf I}_n,\end{array}$$proving the assertion. \end{pf} \begin{exa} By Theorem \ref{thm:inver_dilatation}, we have $$\begin{bmatrix}1 & 0 & 0 \cr 0 & 2 & 0 \cr 0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \cr 0 & \frac{1}{2} & 0 \cr 0 & 0 & 1 \end{bmatrix} =\begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{bmatrix}. $$ \end{exa} Repeated applications of Theorem \ref{thm:inver_dilatation} gives the following corollary. \begin{cor} If $\lambda_1\lambda_2\lambda_3 \cdots \lambda_n \neq 0_{\BBF },$ then $$\begin{bmatrix}\lambda_1 & 0 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & 0 & \cdots & 0 \cr 0 & 0 & \lambda_3 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & 0 & \cdots & \lambda_n \cr \end{bmatrix} $$is invertible and $$\begin{bmatrix}\lambda_1 & 0 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & 0 & \cdots & 0 \cr 0 & 0 & \lambda_3 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & 0 & \cdots & \lambda_n \cr \end{bmatrix}^{-1} = \begin{bmatrix}\lambda_1 ^{-1} & 0 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 ^{-1} & 0 & 0 & \cdots & 0 \cr 0 & 0 & \lambda_3 ^{-1}& 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & 0 & \cdots & \lambda_n ^{-1} \cr \end{bmatrix} $$ \end{cor} \begin{thm}[Invertibility of Permutation Matrices]\label{thm:inverse_permutation_matrix} Let $\tau\in S_n$ be a permutation. Then$$ ({\bf I}^{ij} _n)^{-1} = ({\bf I}^{ij} _n)^T. $$ \end{thm} \begin{pf} By Theorem \ref{thm:mult_by_transposition_matrix} pre-multiplication of ${\bf I}^{ij} _n$ by ${\bf I}^{ij} _n$ exchanges the $i$-th row with the $j$-th row, meaning that they return to the original position in ${\bf I}_n$. Observe in particular that ${\bf I}^{ij} _n = ({\bf I}^{ij} _n)^T$, and so ${\bf I}^{ij} _n({\bf I}^{ij} _n)^T = {\bf I}_n$. \end{pf} \begin{exa} By Theorem \ref{thm:inverse_permutation_matrix}, we have $$\begin{bmatrix}1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \end{bmatrix} =\begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{bmatrix}. $$ \end{exa} \begin{cor} If a square matrix can be represented as the product of elimination matrices of the same size, then it is invertible. \end{cor} \begin{pf}This follows from Corollary \ref{cor:inverse_of_product_of_matrices}, and Theorems \ref{thm:inverse_transvections}, \ref{thm:inver_dilatation}, and \ref{thm:inverse_permutation_matrix}. \end{pf} \begin{exa} Observe that $$A = \begin{bmatrix} 1 & 0 & 0 \cr 0 & 3 & 4 \cr 0 & 0 & 1 \cr \end{bmatrix} $$is the transvection ${\bf I}_3 + 4{\bf E}_{23}$ followed by the dilatation of the second column of this transvection by $3$. Thus $$\begin{bmatrix} 1 & 0 & 0 \cr 0 & 3 & 4 \cr 0 & 0 & 1 \cr \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 4 \cr 0 & 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \cr 0 & 3 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}, $$and so $$\begin{array}{lll}\begin{bmatrix} 1 & 0 & 0 \cr 0 & 3 & 4 \cr 0 & 0 & 1 \cr \end{bmatrix}^{-1} & = & \begin{bmatrix} 1 & 0 & 0 \cr 0 & 3 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}^{-1}\begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 4 \cr 0 & 0 & 1 \cr \end{bmatrix}^{-1} \vspace{2mm} \\ & = & \begin{bmatrix} 1 & 0 & 0 \cr 0 & \frac{1}{3} & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & -4 \cr 0 & 0 & 1 \cr \end{bmatrix}\vspace{2mm}\\ & = & \begin{bmatrix} 1 & 0 & 0 \cr 0 & \frac{1}{3} & -\frac{4}{3} \cr 0 & 0 & 1 \cr \end{bmatrix}. \end{array}$$ \end{exa} \begin{exa} We have $$\begin{bmatrix}1 & 1 & 1 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix} = \begin{bmatrix}1 & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}\begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix}, $$hence $$\begin{array}{lll}\begin{bmatrix}1 & 1 & 1 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix}^{-1} & = & \begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix}^{-1}\begin{bmatrix}1 & 1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}^{-1}\vspace{2mm}\\ & = & \begin{bmatrix}1 & 0 & 0 \cr 0 & 1 & -1 \cr 0 & 0 & 1 \cr \end{bmatrix}\begin{bmatrix}1 & -1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}\vspace{2mm}\\ & = &\begin{bmatrix}1 & -1 & 0 \cr 0 & 1 & -1 \cr 0 & 0 & 1 \cr \end{bmatrix}. \end{array} $$ \end{exa}\bigskip In the next section we will give a general method that will permit us to find the inverse of a square matrix when it exists. \begin{exa}Let $\dis{T = \begin{bmatrix}a & b \cr c & d \cr\end{bmatrix}\in\mat{2\times 2}{\BBR}}$. Then $$ \begin{bmatrix}a & b \cr c & d \cr\end{bmatrix}\begin{bmatrix} d & -b \cr -c & a \cr\end{bmatrix} = (ad - bc)\begin{bmatrix}1 & 0 \cr 0 & 1 \cr\end{bmatrix}$$Thus if $ad - bc \neq 0$ we see that $$T^{-1} = \begin{bmatrix}\frac{d}{ad - bc} & -\frac{b}{ad - bc} \cr -\frac{c}{ad - bc} & \frac{a}{ad - bc} \cr \end{bmatrix}.$$ \end{exa} \begin{exa} If $$A = \begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix}, $$ then $A$ is invertible, for an easy computation shews that $$ A^2 = \begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix}^2 = 4{\bf I}_4, $$whence the inverse sought is $$ A^{-1} = \frac{1}{4}A = \frac{1}{4}\begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix} = \begin{bmatrix} 1/4 & 1/4 & 1/4 & 1/4 \cr 1/4 & 1/4 & -1/4 & -1/4 \cr 1/4 & -1/4 & 1/4 & -1/4 \cr 1/4 & -1/4 & -1/4 & 1/4 \cr\end{bmatrix}. $$ \end{exa} \begin{exa}\index{matrix!nilpotent} A matrix $A\in\mat{n\times n}{\BBR}$ is said to be {\em nilpotent} of index $k$ if satisfies $A \neq {\bf 0}_n, A^2 \neq {\bf 0}_n, \ldots , A^{k - 1} \neq {\bf 0}_n$ and $A^k = {\bf 0}_n$ for integer $k \geq 1.$ Prove that if $A$ is nilpotent, then ${\bf I}_n - A$ is invertible and find its inverse. \end{exa} \begin{solu} To motivate the solution, think that instead of a matrix, we had a real number $x$ with $|x| < 1$. Then the inverse of $1 - x$ is $$(1 - x)^{-1} = \frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots .$$Notice now that since $A^{k} = {\bf 0}_n,$ then $A^p = {\bf 0}_n$ for $p \geq k$. We conjecture thus that $$({\bf I}_n - A)^{-1} = {\bf I}_n + A + A^2 + \cdots + A^{k - 1}.$$The conjecture is easily verified, as $$\begin{array}{lll}({\bf I}_n - A)({\bf I}_n + A + A^2 + \cdots + A^{k - 1}) & = & {\bf I}_n + A + A^2 + \cdots + A^{k - 1} \\ & & \qquad - (A + A^2 + A^3 + \cdots + A^{k}) \\ & = & {\bf I}_n\end{array}$$ and $$\begin{array}{lll}({\bf I}_n + A + A^2 + \cdots + A^{k - 1})({\bf I}_n - A) & =& {\bf I}_n - A + A - A^2 + A^3 - A^4 + \cdots \\ & & \qquad \cdots + A^{k - 2} - A^{k - 1} + A^{k - 1} - A^{k} \\ & = &{\bf I}_n. \end{array}$$ \end{solu} \begin{exa} The inverse of $A\in\mat{3\times 3}{\BBZ_5}$, $$A = \begin{bmatrix}\overline{2} & \overline{0} & \overline{0} \cr \overline{0} & \overline{3} & \overline{0} \cr \overline{0} & \overline{0} & \overline{4} \cr \end{bmatrix}$$ is $$A^{-1} = \begin{bmatrix}\overline{3} & \overline{0} & \overline{0} \cr \overline{0} & \overline{2} & \overline{0} \cr \overline{0} & \overline{0} & \overline{4} \cr \end{bmatrix},$$as $$AA^{-1} = \begin{bmatrix}\overline{2} & \overline{0} & \overline{0} \cr \overline{0} & \overline{3} & \overline{0} \cr \overline{0} & \overline{0} & \overline{4} \cr \end{bmatrix}\begin{bmatrix}\overline{3} & \overline{0} & \overline{0} \cr \overline{0} & \overline{2} & \overline{0} \cr \overline{0} & \overline{0} & \overline{4} \cr \end{bmatrix} = \begin{bmatrix}\overline{1} & \overline{0} & \overline{0} \cr \overline{0} & \overline{1} & \overline{0} \cr \overline{0} & \overline{0} & \overline{1} \cr \end{bmatrix}$$ \end{exa} \begin{exa}[Putnam Exam, 1991] Let $A$ and $B$ be different $n \times n$ matrices with real entries. If $A^3 = B^3$ and $A^2B = B^2A$, prove that $A^2 + B^2$ is not invertible. \end{exa} \begin{solu}Observe that $$ (A^2 + B^2)(A - B) = A^3 - A^2B + B^2A - B^3 = {\bf 0}_n. $$If $A^2 + B^2$ were invertible, then we would have $$ A - B = (A^2 + B^2)^{-1}(A^2 + B^2)(A - B) = {\bf 0}_n,$$contradicting the fact that $A$ and $B$ are different matrices. \end{solu} \begin{lem}\label{lem:inverse_and_zero_row} If $A\in\mat{n\times n}{ \BBF }$ has a row or a column consisting all of $0_{\BBF }$'s, then $A$ is singular. \end{lem} \begin{pf} If $A$ were invertible, the $(i,i)$-th entry of the product of its inverse with $A$ would be $1_{\BBF }$. But if the $i$-th row of $A$ is all $0_{\BBF }$'s, then $\sum _{k = 1} ^n a_{ik}b_{ki} = 0_{\BBF }$, so the $(i,i)$ entry of any matrix product with $A$ is $0_{\BBF }$, and never $1_{\BBF }$. \end{pf} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} The inverse of the matrix $A = \begin{bmatrix}1 & 1& 1 \cr 1 & 1 & 2 \cr 1 & 2 & 3 \cr \end{bmatrix}$ is the matrix $A^{-1} = \begin{bmatrix} a & 1& -1 \cr 1 & b & 1 \cr -1 & 1 & 0 \cr \end{bmatrix}$. Determine $a$ and $b$. \begin{answer} $a = 1$, $b=-2$. \end{answer} \end{pro} \begin{pro} A square matrix $A$ satisfies $A^3 \neq {\bf 0}_n$ but $A^4 = {\bf 0}_n$. Demonstrate that ${\bf I}_n + A$ is invertible and find, with proof, its inverse. \\ \begin{answer} Claim: $A^{-1} = {\bf I}_n - A + A^2 - A^3.$ For observe that $$({\bf I}_n + A)({\bf I}_n - A + A^2 - A^3) = {\bf I}_n - A + A^2 - A^3 - A + A^2 - A^3 +A^4 = {\bf I}_n, $$ proving the claim. \end{answer} \end{pro} \begin{pro} Prove or disprove! If $(A,B,A+B)\in(\gl{n}{\BBR})^3$ then $(A+B)^{-1} = A^{-1} + B^{-1}$. \begin{answer} Disprove! It is enough to take $A=B = 2{\bf I}_n$. Then $(A+ B)^{-1} = (4{\bf I}_n)^{-1} = \frac{1}{4}{\bf I}_n$ but $A^{-1} + B^{-1} = \frac{1}{2}{\bf I}_n + \frac{1}{2}{\bf I}_n = {\bf I}_n$. \end{answer} \end{pro} \begin{pro} Let $S\in\gl{n}{ \BBF }$, $(A,B)\in(\mat{n\times n}{ \BBF })^2$, and $k$ a positive integer. Prove that if $B = SAS^{-1}$ then $B^k =SA^kS^{-1}$. \end{pro} \begin{pro} Let $A\in\mat{n\times n}{ \BBF }$ and let $k$ be a positive integer. Prove that $A$ is invertible if and only if $A^k$ is invertible. \end{pro} \begin{pro} Let $S\in\gl{n}{\BBC}$, $A\in\mat{n\times n}{\BBC}$ with $A^k = {\bf 0}_n$ for some positive integer $k$. Prove that both ${\bf I}_n - SAS^{-1}$ and ${\bf I}_n - S^{-1}AS$ are invertible and find their inverses. \end{pro} \begin{pro} Let $A$ and $B$ be square matrices of the same size such that both $A - B$ and $A+B$ are invertible. Put $C = (A-B)^{-1} + (A+B)^{-1}$. Prove that $$ACA - ACB +BCA - BCB = 2A. $$ \end{pro} \begin{pro} Let $A, B, C$ be non-zero square matrices of the same size over the same field and such that $ABC = {\bf 0}_n$. Prove that at least two of these three matrices are not invertible. \begin{answer} We argue by contradiction. If exactly one of the matrices is not invertible, the identities $$ A =A{\bf I}_n = (ABC)(BC)^{-1} = {\bf 0}_n, $$ $$ B ={\bf I}_nB{\bf I}_n = (A)^{-1}(ABC)C^{-1} = {\bf 0}_n, $$ $$ C ={\bf I}_nC = (AB)^{-1}(ABC) = {\bf 0}_n, $$ shew a contradiction depending on which of the matrices are invertible. If all the matrices are invertible then $${\bf 0}_n = {\bf 0}_nC^{-1}B^{-1}A^{-1}= (ABC)C^{-1}B^{-1}A^{-1} = {\bf I}_n,$$ also gives a contradiction. \end{answer} \end{pro} \begin{pro} Let $(A,B)\in (\mat{n\times n}{ \BBF })^2$ be such that $A^2 = B^2 = (AB)^2 = {\bf I}_n$. Prove that $AB = BA$. \begin{answer} Observe that $A, B, AB$ are invertible. Hence $$\begin{array}{lll} A^2B^2 = {\bf I}_n = (AB)^2& \implies & AABB = ABAB \\ & \implies & AB = BA, \end{array}$$by cancelling $A$ on the left and $B$ on the right. One can also argue that $A = A^{-1}$, $B= B^{-1}$, and so $AB = (AB)^{-1} = B^{-1}A^{-1} = BA$. \end{answer} \end{pro} \begin{pro} Let $A = \begin{bmatrix}a & b & b & \cdots & b \cr b & a & b & \cdots & b \cr b & b & a & \cdots & b \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr b & b & b & \cdots & a \cr \end{bmatrix}\in\mat{n\times n}{ \BBF }$, $n > 1$, $(a, b)\in \BBF ^2$. Determine when $A$ is invertible and find this inverse when it exists. \begin{answer} Observe that $A = (a - b){\bf I}_n + bU$, where $U$ is the $n\times n$ matrix with $1_{\BBF }$'s everywhere. Prove that $$A^2 = (2(a-b) + nb)A - ((a-b)^2 + nb(a-b)){\bf I}_n.$$ \end{answer} \end{pro} \begin{pro} \index{magic square matrices} Let $(A,B)\in (\mat{n\times n}{ \BBF })^2$ be matrices such that $A+B =AB$. Demonstrate that $A - {\bf I}_n$ is invertible and find this inverse. \begin{answer} Compute $(A - {\bf I}_n)(B - {\bf I}_n)$. \end{answer} \end{pro} \begin{pro} Let $S\in\gl{n}{ \BBF }$ and $A\in\mat{n\times n}{ \BBF }$. Prove that $\tr{A} = \tr{SAS^{-1}}$. \begin{answer}By Theorem \ref{tracethm} we have $\tr{SAS^{-1}} = \tr{S^{-1}SA} = \tr{A}$. \end{answer} \end{pro} \begin{pro} Let $A\in\mat{n\times n}{\BBR}$ be a skew-symmetric matrix. Prove that ${\bf I}_n + A$ is invertible. Furthermore, if $B = ({\bf I}_n - A)({\bf I}_n + A)^{-1}$, prove that $B^{-1} = B^T$. \end{pro} \begin{pro} A matrix $A\in\mat{n\times n}{ \BBF }$ is said to be a {\em magic square} if the sum of each individual row equals the sum of each individual column. Assume that $A$ is a magic square and invertible. Prove that $A^{-1}$ is also a magic square. \end{pro} \end{multicols} \section{Block Matrices} \begin{df} Let $A\in\mat{m\times n}{ \BBF }$, $B\in\mat{m\times s}{ \BBF }$, $C\in\mat{r\times n}{ \BBF }$, $D\in\mat{r\times s}{ \BBF }$. We use the notation $$L = \begin{bmat}{c|c}A & B \cr\hline C & D \end{bmat} $$for the {\em block matrix} $L\in\mat{(m + r)\times (n + s)}{ \BBF }$. \end{df} \begin{rem} If $(A,A')\in (\mat{m}{ \BBF })^2$, $(B,B')\in (\mat{m\times n}{ \BBF })^2$, $(C,C')\in (\mat{n\times m}{ \BBF })^2$, $(D,D')\in (\mat{m}{ \BBF })^2$, and $$S = \begin{bmat}{c|c}A & B \cr\hline C & D \end{bmat}, \ \ \ T = \begin{bmat}{c|c}A' & B' \cr\hline C' & D' \end{bmat}, $$then it is easy to verify that $$ ST = \begin{bmat}{c|c}AA' + BC' & AB' + BD' \cr\hline CA' + DC' & CB' + DD' \end{bmat}. $$ \end{rem} \begin{lem}\label{lem:inverse_block_matrices} Let $L\in\mat{(m + r)\times (m + r)}{ \BBF }$ be the square block matrix $$L = \begin{bmat}{c|c}A & C \cr\hline {\bf 0}_{r\times m} & B \end{bmat}, $$with square matrices $A\in\mat{m}{ \BBF }$ and $B\in\mat{r\times r}{ \BBF }$, and a matrix $C\in\mat{m\times r}{ \BBF }$. Then $L$ is invertible if and only if $A$ and $B$ are, in which case $$L^{-1} = \begin{bmat}{c|c}A^{-1} & -A^{-1}CB^{-1} \cr\hline {\bf 0}_{r\times m} & B^{-1} \end{bmat} $$ \end{lem} \begin{pf} Assume first that $A$, and $B$ are invertible. Direct calculation yields $$\begin{array}{lll}\begin{bmat}{c|c}A & C \cr\hline {\bf 0}_{r\times m} & B \end{bmat}\begin{bmat}{c|c}A^{-1} & -A^{-1}CB^{-1} \cr\hline {\bf 0}_{r\times m} & B^{-1} \end{bmat} & = & \begin{bmat}{c|c}AA^{-1} & -AA^{-1}CB^{-1} + CB^{-1} \cr\hline {\bf 0}_{r\times m} & BB^{-1} \cr \end{bmat} \vspace{2mm} \\ & = & \begin{bmat}{c|c}{\bf I}_m & {\bf 0}_{m\times r} \cr\hline {\bf 0}_{r\times m} & {\bf I}_{r} \cr \end{bmat}\\ & = & {\bf I}_{m + r}. \end{array} $$ Assume now that $L$ is invertible, $L^{-1} =\dis{\begin{bmat}{c|c}E & H \cr\hline J& K \end{bmat}} $, with $E\in\mat{m}{ \BBF }$ and $K\in\mat{r\times r}{ \BBF }$, but that, say, $B$ is singular. Then $$\begin{array}{lll} \begin{bmat}{c|c}{\bf I}_m & {\bf 0}_{m\times r} \cr\hline {\bf 0}_{r\times m} & {\bf I}_{r} \end{bmat} & = & LL^{-1} \\ & = & \begin{bmat}{c|c}A & C \cr\hline {\bf 0}_{r\times m} & B \end{bmat}\begin{bmat}{c|c}E & H \cr\hline J& K \end{bmat} \vspace{2mm} \\ & = & \begin{bmat}{c|c}AE + CJ& AH + BK \cr\hline BJ& BK \end{bmat}, \end{array} $$which gives $BK = {\bf I}_r$, i.e., $B$ is invertible, a contradiction. \end{pf} \section{Rank of a Matrix} \begin{df} Let $(A, B)\in (\mat{m\times n}{ \BBF })^2$. We say that $A$ is {\em row-equivalent} to $B$ if there exists a matrix $R\in\gl{m}{ \BBF }$ such that $B = RA$. Similarly, we say that $A$ is {\em column-equivalent} to $B$ if there exists a matrix $C\in\gl{m}{ \BBF }$ such that $B = AC$. We say that $A$ and $B$ are {\em equivalent} if $\exists (P,Q)\in\gl{m}{ \BBF }\times \gl{n}{ \BBF }$ such that $B = PAQ$. \end{df} \begin{thm} Row equivalence, column equivalence, and equivalence are equivalence relations. \end{thm} \begin{pf} We prove the result for row equivalence. The result for column equivalence, and equivalence are analogously proved. \bigskip Since ${\bf I}_m\in\gl{m}{ \BBF }$ and $A = {\bf I}_mA$, row equivalence is a reflexive relation. Assume $(A, B)\in (\mat{m\times n}{ \BBF })^2$ and that $\exists P\in\gl{m}{ \BBF }$ such that $B = PA$. Then $A = P^{-1}B$ and since $P^{-1}\in\gl{m}{ \BBF }$, we see that row equivalence is a symmetric relation. Finally assume $(A, B, C)\in (\mat{m\times n}{ \BBF })^3$ and that $\exists P\in\gl{m}{ \BBF }, \ \exists P'\in\gl{m}{ \BBF }$ such that $A = PB, B = P'C$. Then $A = PP'C$. But $PP'\in\gl{m}{ \BBF }$ in view of Corollary \ref{cor:inverse_of_product_of_matrices}. This completes the proof. \end{pf} \begin{thm}\label{thm:normal_form}\index{Hermite normal form} Let $A\in\mat{m\times n}{ \BBF }$. Then $A$ can be reduced, by means of pre-multiplication and post-multiplication by elimination matrices, to a unique matrix of the form \begin{equation}D_{m,n,r} = \begin{bmat}{c|c}{\bf I}_r & {\bf 0}_{r\times (n - r)} \cr\hline {\bf 0}_{(m - r) \times r} & {\bf 0}_{(m - r)\times (n - r)} \cr \end{bmat},\end{equation}called the {\em Hermite normal form} of $A$. Thus there exist $P\in \gl{m}{ \BBF }, Q\in \gl{n}{ \BBF }$ such that $D_{m,n,r} = PAQ$. The integer $r\geq 0$ is called the {\em rank} of the matrix $A$ which we denote by $\rank{A}$. \index{rank} \end{thm} \begin{pf} If $A$ is the $m\times n$ zero matrix, then the theorem is obvious, taking $r = 0$. Assume hence that $A$ is not the zero matrix. We proceed as follows using the {\em Gau\ss-Jordan Algorithm}. \begin{enumerate} \item[{\bf GJ-1}] Since $A$ is a non-zero matrix, it has a non-zero column. By means of permutation matrices we move this column to the first column. \item[{\bf GJ-2}] Since this column is a non-zero column, it must have an entry $a \neq 0_{\BBF }$. Again, by means of permutation matrices, we move the row on which this entry is to the first row. \item[{\bf GJ-3}] By means of a dilatation matrix with scale factor $a^{-1}$, we make this new $(1,1)$ entry into a $1_{\BBF }$. \item[{\bf GJ-4}] By means of transvections (adding various multiples of row $1$ to the other rows) we now annihilate every entry below the entry $(1,1)$. \end{enumerate} This process ends up in a matrix of the form \begin{equation}P_1AQ_1 = \begin{bmat}{c|cccc} 1_{\BBF } & * & * & \cdots & * \cr\hline 0_{\BBF } & b_{22} & b_{23} & \cdots & b_{2n} \cr 0_{\BBF } & b_{32} & b_{33} & \cdots & b_{3n} \cr 0_{\BBF } & \vdots & \vdots & \cdots & \vdots \cr 0_{\BBF } & b_{m2} & b_{m3} & \cdots & b_{mn} \cr \end{bmat}. \end{equation} Here the asterisks represent unknown entries. Observe that the $b$'s form a $(m - 1)\times (n - 1)$ matrix. \begin{enumerate} \item[{\bf GJ-5}] Apply {\bf GJ-1} through {\bf GJ-4} to the matrix of the $b$'s. \end{enumerate} Observe that this results in a matrix of the form \begin{equation}P_2AQ_2 =\begin{bmat}{cc|ccc} 1_{\BBF } & * & * & \cdots & * \cr 0_{\BBF } & 1_{\BBF } & * & \cdots & * \cr\hline 0_{\BBF } & 0_{\BBF } & c_{33} & \cdots & c_{3n} \cr 0_{\BBF } & \vdots & \vdots & \cdots & \vdots \cr 0_{\BBF } & 0_{\BBF } & c_{m3} & \cdots & c_{mn} \cr \end{bmat}. \end{equation} \begin{enumerate} \item[{\bf GJ-6}] Add the appropriate multiple of column 1 to column 2, that is, apply a transvection, in order to make the entry in the $(1, 2)$ position $0_{\BBF }$. \end{enumerate} This now gives a matrix of the form \begin{equation}P_3AQ_3 =\begin{bmat}{cc|ccc} 1_{\BBF } & 0_{\BBF } & * & \cdots & * \cr 0_{\BBF } & 1_{\BBF } & * & \cdots & * \cr\hline 0_{\BBF } & 0_{\BBF } & c_{33} & \cdots & c_{3n} \cr 0_{\BBF } & \vdots & \vdots & \cdots & \vdots \cr 0_{\BBF } & 0_{\BBF } & c_{m3} & \cdots & c_{mn} \cr \end{bmat}. \end{equation} The matrix of the $c$'s has size $(m - 2)\times (n - 2)$. \begin{enumerate} \item[{\bf GJ-7}] Apply {\bf GJ-1} through {\bf GJ-6} to the matrix of the $c$'s, etc. \end{enumerate} Observe that this process eventually stops, and in fact, it is clear that $\rank{A} \leq \min (m, n)$. \bigskip Suppose now that $A$ were equivalent to a matrix $D_{m,n,s}$ with $s>r$. Since matrix equivalence is an equivalence relation, $D_{m,n,s}$ and $D_{m,n,r}$ would be equivalent, and so there would be $R\in\gl{m}{ \BBF }$, $S\in\gl{n}{ \BBF }$, such that $RD_{m,n,r}S = D_{m,n,s}$, that is, $RD_{m,n,r} = D_{m,n,s}S^{-1}$. Partition $R$ and $S^{-1}$ as follows $$ R = \begin{bmat}{c|c} R_{11} & R_{12} \cr\hline R_{21} & R_{22} \end{bmat}, \ \ \ S^{-1} = \begin{bmat}{c|cc} S_{11} & S_{12} & S_{13} \cr \hline S_{21} & S_{22} & S_{23} \cr S_{31} & S_{32} & S_{33} \cr \end{bmat}, $$with $(R_{11},S_{11})^2\in(\mat{r\times r}{ \BBF })^2, S_{22}\in\mat{(s - r)\times (s - r)}{ \BBF }$. We have $$RD_{m,n,r} = \begin{bmat}{c|c} R_{11} & R_{12} \cr\hline R_{21} & R_{22} \end{bmat}\begin{bmat}{c|c} {\bf I}_r & {\bf 0}_{(m - r)\times r} \cr\hline {\bf 0}_{(m - r)\times r} & {\bf 0}_{r\times (m - r)} \cr\end{bmat}= \begin{bmat}{c|c} R_{11} & {\bf 0}_{(m - r)\times r} \cr\hline R_{21} & {\bf 0}_{r\times (m - r)} \end{bmat}, $$ and $$\begin{array}{lll}D_{m,n,s}S^{-1} & = & \begin{bmat}{c|cc} {\bf I}_r & {\bf 0}_{ r\times (s - r) } & {\bf 0}_{ r \times (n - s)} \cr\hline {\bf 0}_{(s - r) \times r} & {\bf I}_{s - r} & {\bf 0}_{(s - r) \times (n - s)} \cr {\bf 0}_{(m - s) \times r} & {\bf 0}_{(m - s) \times (s - r)} & {\bf 0}_{(m - s) \times (n - s)} \cr\end{bmat}\begin{bmat}{c|cc} S_{11} & S_{12} & S_{13} \cr \hline S_{21} & S_{22} & S_{23} \cr S_{31} & S_{32} & S_{33} \cr \end{bmat}\vspace{2mm} \\ & = & \begin{bmat}{c|cc} S_{11} & S_{12} & S_{13} \cr\hline S_{21} & S_{22} & S_{23} \cr {\bf 0}_{(m - s) \times r} & {\bf 0}_{(m - s) \times (s - r)} & {\bf 0}_{(m - s) \times (n - s)} \cr\end{bmat} .\end{array}$$ Since we are assuming $$ \begin{bmat}{c|c} R_{11} & {\bf 0}_{(m - r)\times r} \cr\hline R_{21} & {\bf 0}_{r\times (m - r)} \end{bmat} = \begin{bmat}{c|cc} S_{11} & S_{12} & S_{13} \cr\hline S_{21} & S_{22} & S_{23} \cr {\bf 0}_{(m - s) \times r} & {\bf 0}_{(m - s) \times (s - r)} & {\bf 0}_{(m - s) \times (n - s)} \cr\end{bmat}, $$we must have $S_{12} = {\bf 0}_{r \times (s - r)}$, $S_{13} = {\bf 0}_{r \times (n - s)}$, $S_{22} = {\bf 0}_{(s - r) \times (s - r)}$, $S_{23} = {\bf 0}_{(s - r)\times (n - s)}$. Hence $$S^{-1} = \begin{bmat}{c|cc} S_{11} & {\bf 0}_{r \times (s - r)}& {\bf 0}_{r \times (n - s)} \cr\hline S_{21} & {\bf 0}_{(s - r) \times (s - r)} & {\bf 0}_{(s - r)\times (n - s)} \cr S_{31} & S_{32} & S_{33} \cr\end{bmat}. $$The matrix $$ \begin{bmat}{c|cc} {\bf 0}_{(s - r) \times (s - r)} & {\bf 0}_{(s - r)\times (n - s)} \cr\hline S_{32} & S_{33} \cr\end{bmat} $$ is non-invertible, by virtue of Lemma \ref{lem:inverse_and_zero_row}. This entails that $S^{-1}$ is non-invertible by virtue of Lemma \ref{lem:inverse_block_matrices}. This is a contradiction, since $S$ is assumed invertible, and hence $S^{-1}$ must also be invertible. \end{pf} \begin{rem} Albeit the rank of a matrix is unique, the matrices $P$ and $Q$ appearing in Theorem \ref{thm:normal_form} are not necessarily unique. For example, the matrix $$\begin{bmatrix} 1 & 0 \cr 0 & 1 \cr 0 & 0 \end{bmatrix} $$ has rank $2$, the matrix $$\begin{bmatrix} 1 & 0 & x \cr 0 & 1 & y \cr 0 & 0 & 1 \cr\end{bmatrix}$$is invertible, and an easy computation shews that $$\begin{bmatrix} 1 & 0 & x \cr 0 & 1 & y \cr 0 & 0 & 1 \cr\end{bmatrix}\begin{bmatrix} 1 & 0 \cr 0 & 1 \cr 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \cr 0 & 1 \cr \end{bmatrix} = \begin{bmatrix} 1 & 0 \cr 0 & 1 \cr 0 & 0 \cr\end{bmatrix}, $$regardless of the values of $x$ and $y$. \end{rem} \begin{cor}\label{cor:row_rank_is_column_rank} Let $A\in\mat{m\times n}{ \BBF }$. Then $\rank{A} = \rank{A^T}$. \end{cor} \begin{pf} Let $P,Q,D_{m,n,r}$ as in Theorem \ref{thm:normal_form}. Observe that $P^T, Q^T$ are invertible. Then $$PAQ = D_{m,n,r} \implies Q^TA^TP^T = D_{m,n,r}^T = D_{n,m,r}, $$and since this last matrix has the same number of $1_{\BBF }$'s as $D_{m,n,r}$, the corollary is proven. \end{pf} \begin{exa} Shew that $$A = \begin{bmatrix} 0 & 2 & 3 \cr 0 & 1 & 0 \cr \end{bmatrix} $$ has $\rank{A} = 2$ and find invertible matrices $P\in\gl{2}{\BBR}$ and $Q\in\gl{3}{\BBR}$ such that $$PAQ = \begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}. $$ \end{exa} \begin{solu}We first transpose the first and third columns by effecting $$\begin{bmatrix} 0 & 2 & 3 \cr 0 & 1 & 0 \cr \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr \end{bmatrix} = \begin{bmatrix} 3 & 2 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}. $$We now subtract twice the second row from the first, by effecting $$ \begin{bmatrix} 1 & -2 \cr 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 3 & 2 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix} = \begin{bmatrix} 3 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}. $$ Finally, we divide the first row by $3$, $$ \begin{bmatrix} 1/3 & 0 \cr 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 3 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}.$$ We conclude that $$ \begin{bmatrix} 1/3 & 0 \cr 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 1 & -2 \cr 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 0 & 2 & 3 \cr 0 & 1 & 0 \cr \end{bmatrix}\begin{bmatrix}0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}, $$ from where we may take $$P = \begin{bmatrix} 1/3 & 0 \cr 0 & 1 \cr \end{bmatrix}\begin{bmatrix} 1 & -2 \cr 0 & 1 \cr \end{bmatrix} = \begin{bmatrix} 1/3 & -2/3 \cr 0 & 1 \cr \end{bmatrix} $$ and $$Q = \begin{bmatrix}0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr \end{bmatrix}. $$ \end{solu} In practice it is easier to do away with the multiplication by elimination matrices and perform row and column operations on the {\em augmented $(m + n)\times (m + n)$ matrix} $$\begin{bmat}{c|c} {\bf I}_n & {\bf 0}_{n\times m} \cr\hline A & {\bf I}_m \cr \end{bmat}. $$ \begin{df} Denote the rows of a matrix $A\in \mat{m\times n}{ \BBF }$ by $R_1, R_2, \ldots , R_m$, and its columns by $C_1, C_2, \ldots , C_n$. The elimination operations will be denoted as follows. \begin{itemize} \item Exchanging the $i$-th row with the $j$-th row, which we denote by $R_i \leftrightarrow R_j$, and the $s$-th column by the $t$-th column by $C_s \leftrightarrow C_t$. \item A dilatation of the $i$-th row by a non-zero scalar $\alpha\in \BBF \setminus \{0_{\BBF }\}$, we will denote by $\alpha R_i \rightarrow R_i $. Similarly, $\beta C_j \rightarrow C_j $ denotes the dilatation of the $j$-th column by the non-zero scalar $\beta$. \item A transvection on the rows will be denoted by $R_i + \alpha R_j \rightarrow R_i$, and one on the columns by $C_s + \beta C_t \rightarrow C_s$. \end{itemize} \end{df} \begin{exa} Find the Hermite normal form of $$A = \begin{bmatrix}-1 & 0 \cr 0 & 0 \cr 1 & 1 \cr 1 & 2 \cr \end{bmatrix}. $$ \end{exa} \begin{solu}First observe that $\rank{A} \leq \min (4,2) = 2$, so the rank can be either $1$ or $2$ (why not $0$?). Form the augmented matrix $$\begin{bmat}{cc|cccc}1 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr \hline -1 & 0 & 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 1 & 1 & 0 & 0 & 1 & 0 \cr 1 & 2 & 0 & 0 & 0 & 1\cr \end{bmat}.$$ Perform $R_5 + R_3\rightarrow R_5$ and $R_6 + R_3\rightarrow R_6$ successively, obtaining $$\begin{bmat}{cc|cccc}1 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr \hline -1 & 0 & 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 & 1 & 0 \cr 0 & 2 & 1 & 0 & 0 & 1\cr \end{bmat}.$$ Perform $R_6 - 2R_5 \rightarrow R_6$ $$\begin{bmat}{cc|cccc}1 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr \hline -1 & 0 & 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 & 1 & 0 \cr 0 & 0 & -1 & 0 & -2 & 1\cr \end{bmat}.$$Perform $R_4 \leftrightarrow R_5$ $$\begin{bmat}{cc|cccc}1 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr \hline -1 & 0 & 1 & 0 & 0 & 0 \cr 0 & 1 & 1 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 0 & 0 & -1 & 0 & -2 & 1\cr \end{bmat}.$$Finally, perform $-R_3\rightarrow R_3$ $$\begin{bmat}{cc|cccc}1 & 0 & 0 & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 & 0 & 0 \cr \hline 1 & 0 & -1 & 0 & 0 & 0 \cr 0 & 1 & 1 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1 & 0 & 0 \cr 0 & 0 & -1 & 0 & -2 & 1\cr \end{bmat}.$$ We conclude that $$ \begin{bmatrix}-1 & 0 & 0 & 0 \cr 1 & 0& 1& 0 \cr 0 & 1 & 0 & 0 \cr -1 & 0 & -2 & 1 \cr \end{bmatrix} \begin{bmatrix}-1 & 0 \cr 0 & 0 \cr 1 & 1 \cr 1 & 2 \cr \end{bmatrix}\begin{bmatrix}1 & 0 \cr 0 & 1\cr\end{bmatrix} = \begin{bmatrix}1 & 0 \cr 0 & 1 \cr 0 & 0 \cr 0 & 0 \cr \end{bmatrix}. $$ \end{solu} \begin{thm}\label{thm:rank_of_product} Let $A\in\mat{m\times n}{ \BBF }$, $B\in\mat{n\times p}{ \BBF }$. Then $$\rank{AB} \leq \min (\rank{A}, \rank{B}). $$ \end{thm} \begin{pf} We prove that $\rank{A} \geq \rank{AB}$. The proof that $\rank{B} \geq \rank{AB}$ is similar and left to the reader. Put $r = \rank{A}, s = \rank{AB}$. There exist matrices $P\in\gl{m}{ \BBF }$, $Q\in\gl{ n}{ \BBF }$, $S\in\gl{m}{ \BBF }$, $T\in\gl{p}{ \BBF }$ such that $$PAQ = D_{m,n,r}, \ \ \ SABT = D_{m,p,s}. $$ \bigskip Now $$D_{m,p,s} = SABT = SP^{-1}D_{m,n,r}Q^{-1}BT, $$from where it follows that $$ PS^{-1}D_{m,p,s} = D_{m,n,r}Q^{-1}BT.$$Now the proof is analogous to the uniqueness proof of Theorem \ref{thm:normal_form}. Put $U = PS^{-1}\in\gl{m}{\BBR}$ and $V = Q^{-1}BT \in \mat{n\times p}{ \BBF }$, and partition $U$ and $V$ as follows: $$ U = \begin{bmat}{c|c} U_{11} & U_{12} \cr\hline U_{21} & U_{22} \end{bmat}, \ \ \ V = \begin{bmat}{c|c} V_{11} & V_{12} \cr\hline V_{21} & V_{22} \end{bmat}, $$with $U_{11}\in\mat{s}{ \BBF }, V_{11}\in\mat{r\times r}{ \BBF }$. Then $$UD_{m,p,s} = \begin{bmat}{c|c} U_{11} & U_{12} \cr\hline U_{21} & U_{22} \end{bmat}\begin{bmat}{c|c} {\bf I}_s & {\bf 0}_{s\times (p - s)}\cr\hline {\bf 0}_{(m - s)\times s} & {\bf 0}_{(m - s)\times (p - s)} \end{bmat}\in\mat{m\times p}{ \BBF }, $$ and $$D_{m,p,s}V = \begin{bmat}{c|c} {\bf I}_r & {\bf 0}_{r\times (n - r)}\cr\hline {\bf 0}_{(m - r)\times r} & {\bf 0}_{(m - r)\times (n - r)} \end{bmat}\begin{bmat}{c|c} V_{11} & V_{12} \cr\hline V_{21} & V_{22} \end{bmat} \in\mat{m\times p}{ \BBF }. $$ From the equality of these two $m\times p$ matrices, it follows that $$\begin{bmat}{c|c} U_{11} & {\bf 0}_{s\times (p - s)}\cr\hline U_{21}& {\bf 0}_{(m - s)\times (p - s)} \end{bmat} = \begin{bmat}{c|c} V_{11} & V_{12}\cr\hline {\bf 0}_{(m - r)\times r} & {\bf 0}_{(m - r)\times (n - r)} \end{bmat}.$$If $s > r$ then (i) $U_{11}$ would have at least one row of $0_{\BBF }$'s meaning that $U_{11}$ is non-invertible by Lemma \ref{lem:inverse_and_zero_row}. (ii) $U_{21} = {\bf 0}_{(m - s)\times s}$. Thus from (i) and (ii) and from Lemma \ref{lem:inverse_block_matrices}, $U$ is not invertible, which is a contradiction. \end{pf} \begin{cor} Let $A\in\mat{m\times n}{ \BBF }$, $B\in\mat{n\times p}{ \BBF }$. If $A$ is invertible then $\rank{AB} = \rank{B}$. If $B$ is invertible then $\rank{AB} = \rank{A}$. \end{cor} \begin{pf} Using Theorem \ref{thm:rank_of_product}, if $A$ is invertible $$\rank{AB } \leq \rank{B} = \rank{A^{-1}AB} \leq \rank{AB},$$ and so $\rank{B} = \rank{AB}$. A similar argument works when $B$ is invertible. \end{pf} \begin{exa} Study the various possibilities for the rank of the matrix $$A = \begin{bmatrix} 1 & 1 & 1 \cr b + c & c + a & a + b \cr bc & ca & ab\end{bmatrix} .$$ \end{exa} \begin{solu}Performing $R_2 - (b + c)R_1 \rightarrow R_2$ and $R_3 - bcR_1 \rightarrow R_3$, we find $$\begin{bmatrix} 1 & 1 & 1 \cr 0 & a - b & a - c \cr 0 & 0 & (b - c)(a - c)\end{bmatrix} .$$ Performing $C_2 - C_1 \rightarrow C_2$ and $C_3 - C_1 \rightarrow C_3$, we find $$\begin{bmatrix} 1 & 0 & 0 \cr 0 & a - b & a - c \cr 0 & 0 & (b - c)(a - c)\end{bmatrix} .$$ We now examine the various ways of getting rows consisting only of $0$'s. If $a = b = c,$ the last two rows are $0$-rows and so $\rank{A} = 1$. If exactly two of $a, b, c$ are equal, the last row is a $0$-row, but the middle one is not, and so $\rank{A} = 2$ in this case. If none of $a, b, c$ are equal, then the rank is clearly $3$. \end{solu} \section*{\psframebox{Homework}} \begin{pro} On a symmetric matrix $A\in\mat{n\times n}{\BBR}$ with $n \geq 3$, $$R_3 - 3R_1 \rightarrow R_3$$ successively followed by $$C_3 - 3C_1 \rightarrow C_3$$ are performed. Is the resulting matrix still symmetric? \end{pro} \begin{pro} Find the rank of $$\begin{bmatrix}a+1 & a+2 & a+3 & a+4 & a+5 \cr a+2 & a+3 & a+4 & a+5 & a+6 \cr a+3 & a+4 & a+5 & a+6 & a+7 \cr a+4 & a+5 & a+6 & a+7 & a+8 \cr \end{bmatrix} \in\mat{5\times 5}{\BBR}.$$ \begin{answer} The rank is $2$. \end{answer} \end{pro} \begin{pro} Let $A, B$ be arbitrary $n \times n$ matrices over $\BBR$. Prove or disprove! $\rank{AB} = \rank{BA}.$\begin{answer} If $B$ is invertible, then $\rank{AB} = \rank{A} = \rank{BA}$. Similarly, if $A$ is invertible $\rank{AB} = \rank{B} = \rank{BA}$. Now, take $\dis{A = \left[\begin{array}{ll}1 & 0 \\ 0 & 0 \end{array}\right]}$ and $\dis{B = \left[\begin{array}{ll}0 & 1 \\ 0 & 0 \end{array}\right]}$. Then $AB = B,$ and so $\rank{AB} = 1.$ But $BA = \left[\begin{array}{ll}0 & 0 \\ 0 & 0 \end{array}\right]$, and so $\rank{BA} = 0.$ \end{answer} \end{pro} \begin{pro} Determine the rank of the matrix $ \begin{bmatrix}1 & 1 & 0 & 0 \cr 0 & 0 & 1& 1 \cr 2 & 2 & 2 & 2 \cr 2 & 0 & 0 & 2 \cr \end{bmatrix}$ . \begin{answer} Observe that $$ \begin{bmatrix}1 & 1 & 0 & 0 \cr 0 & 0 & 1& 1 \cr 2 & 2 & 2 & 2 \cr 2 & 0 & 0 & 2 \cr \end{bmatrix} \grstep[R_4-2R_1\rightarrow R_4]{R_3-2(R_1+R_2) \rightarrow R_3} \begin{bmatrix}1 & 1 & 0 & 0 \cr 0 & 0 & 1& 1 \cr 0 & 0 & 0 & 0 \cr 0 & -2 & 0 & 2 \cr \end{bmatrix},$$whence the matrix has three pivots and so rank $3$. \end{answer} \end{pro} \begin{pro} Suppose that the matrix $\begin{bmatrix} 4 & 2 \cr x^2 & x \end{bmatrix}\in \mat{2\times 2}{\BBR}$ has rank $1$. How many possible values can $x$ assume? \begin{answer} The maximum rank of this matrix could be $2$. Hence, for the rank to be $1$, the rows must be proportional, which entails $$ \dfrac{x^2}{4}=\dfrac{x}{2} \implies x^2-2x=0 \implies x\in \{0,2\}. $$ \end{answer} \end{pro} \begin{pro} Demonstrate that a non-zero $n\times n$ matrix $A$ over a field $\BBF$ has rank $1$ if and only if $A$ can be factored as $A=XY$, where $X\in \mat{n\times 1}{\BBF}$ and $Y\in \mat{1\times n}{\BBF}$. \begin{answer} Assume first that the non-zero $n\times n$ matrix $A$ over a field $\BBF$ has rank $1$. By permuting the rows of the matrix we may assume that every other row is a scalar multiple of the first row, which is non-zero since the rank is $1$. Hence $A$ must be of the form $$A = \begin{bmatrix} a_1 & a_2 & \cdots & a_n \cr \lambda _1 a_1 & \lambda _1 a_2 & \cdots & \lambda _1 a_n \cr \vdots & \vdots & \cdots & \vdots \cr \lambda _{n-1} a_1 & \lambda _{n-1} a_2 & \cdots & \lambda _{n-1} a_n \cr \end{bmatrix} = \colvec{a_1 \\ a_2 \\ \vdots \\ a_n}\colvec{1 & \lambda _1 & \cdots & \lambda _{n-1}}:=XY, $$ which means that the claimed factorisation indeed exists. \bigskip Conversely, assume that $A$ can be factored as $A=XY$, where $X\in \mat{n\times 1}{\BBF}$ and $Y\in \mat{1\times n}{\BBF}$. Since $A$ is non-zero, we must have $\rank{A}\geq 1$. Similarly, neither $X$ nor $Y$ could be all zeroes, because otherwise $A$ would be zero. This means that $\rank{X}=1=\rank{Y}$. Now, since $$ \rank{A}\leq \min (\rank{X}, \rank{Y}) = 1, $$we deduce that $\rank{A}\leq 1$, proving that $\rank{A}=1$. \end{answer} \end{pro} \begin{pro} Study the various possibilities for the rank of the matrix $$ \begin{bmatrix}1 & a & 1 & b \cr a & 1 & b & 1 \cr 1 & b & 1 & a \cr b & 1 & a & 1 \cr \end{bmatrix} $$ when $(a,b)\in\BBR^2$. \begin{answer} Effecting $R_3 - R_1 \rightarrow R_3$; $aR_4 - bR_2 \rightarrow R_4$ successively, we obtain $$ \begin{bmatrix}1 & a & 1 & b \cr a & 1 & b & 1 \cr 1 & b & 1 & a \cr b & 1 & a & 1 \cr \end{bmatrix} \rightsquigarrow \begin{bmatrix}1 & a & 1 & b \cr a & 1 & b & 1 \cr 0 & b - a & 0 & a - b \cr 0 & a - b & a^2 - b^2 & a - b \cr \end{bmatrix} . $$ Performing $R_2 - aR_1 \rightarrow R_2$; $R_4 + R_3 \rightarrow R_4$ we have $$\rightsquigarrow \begin{bmatrix}1 & a & 1 & b \cr 0 & 1 - a^2 & b - a & 1 - ab \cr 0 & b - a & 0 & a - b \cr 0 & 0 & a^2 - b^2 & 2(a - b) \cr \end{bmatrix}. $$ Performing $(1 - a^2)R_3 - (b - a)R_2\rightarrow R_3$ we have $$\rightsquigarrow \begin{bmatrix}1 & a & 1 & b \cr 0 & 1 - a^2 & b - a & 1 - ab \cr 0 & 0 & -a^2 + 2ab - b^2 & 2a-2b-a^3+ab^2 \cr 0 & 0 & a^2 - b^2 & 2(a - b) \cr \end{bmatrix}. $$ Performing $R_3 - R_4\rightarrow R_3$ we have $$\rightsquigarrow \begin{bmatrix}1 & a & 1 & b \cr 0 & 1 - a^2 & b - a & 1 - ab \cr 0 & 0 & -2a(a - b) & -a(a^2 - b^2) \cr 0 & 0 & a^2 - b^2 & 2(a - b) \cr \end{bmatrix}. $$ Performing $2aR_4 + (a + b)R_3\rightarrow R_4$ we have $$\begin{bmatrix}1 & a & 1 & b \cr 0 & 1 - a^2 & b - a & 1 - ab \cr 0 & 0 & -2a(a - b) & -a(a^2 - b^2) \cr 0 & 0 & 0 & 4a^2-4ab-a^4+a^2b^2-ba^3+ab^3 \cr \end{bmatrix}. $$ Factorising, this is $$= \begin{bmatrix}1 & a & 1 & b \cr 0 & 1 - a^2 & b - a & 1 - ab \cr 0 & 0 & -2a(a - b) & -a(a - b)(a + b) \cr 0 & 0 & 0 & -a(a+2+b)(a-b)(a-2+b) \cr \end{bmatrix}.$$ Thus the rank is $4$ if $(a+2+b)(a-b)(a-2+b) \neq 0$. The rank is $3$ if $a + b = 2$ and $(a,b) \neq (1,1)$ or if $a + b = -2$ and $(a,b) \neq (-1,-1)$. The rank is $2$ if $a = b \neq 1$ and $a \neq -1$. The rank is $1$ if $a = b = \pm 1$. \end{answer} \end{pro} \begin{pro} Find the rank of $\begin{bmatrix}1 &-1 & 0 & 1 \cr m & 1 &-1 &-1 \cr 1 &-m & 1 & 0 \cr 1 &-1 & m& 2 \cr\end{bmatrix}$ as a function of $m\in\BBC$. \begin{answer}$\rank{A}=4$ if $m^3 + m^2 + 2 \neq 0$, and $\rank{A}=3$ otherwise. \end{answer} \end{pro} \begin{pro} Determine the rank of the matrix $\begin{bmatrix}a^2 & ab & ab & b^2 \cr ab & a^2 & b^2 & ab \cr ab & b^2 & a^2 & ab \cr b^2 & ab & ab & a^2 \cr \end{bmatrix}.$ \begin{answer} The rank is $4$ if $a\neq \pm b$. The rank is $1$ is $a = \pm b \neq 0$. The rank is $0$ if $a=b=0$. \end{answer} \end{pro} \begin{pro} Determine the rank of the matrix $\begin{bmatrix}1 & 1 & 1 & 1 \cr a & b & a & b \cr c & c & d & d \cr ac & bc & ad & bd \cr \end{bmatrix}.$ \begin{answer} The rank is $4$ if $(a-b)(c-d)\neq 0$. The rank is $2$ is $a = b, c \neq d$ or if $a \neq b, c = d$. The rank is $1$ if $a=b$ and $c=d$. \end{answer} \end{pro} \begin{pro} Let $A\in\mat{3\times 2}{\BBR}$, $B\in \mat{2\times 2}{\BBR}$, and $C\in\mat{2\times 3}{\BBR}$ be such that $ABC = \begin{bmatrix} 1&1&2 \cr -2&x&1 \cr 1&-2&1 \cr\end{bmatrix}$. Find $x$. \begin{answer}Observe that $\rank{ABC}\leq \rank{B}\leq 2$. Now, $$\begin{bmatrix} 1&1&2 \cr -2&x&1 \cr 1&-2&1 \cr\end{bmatrix} \grstep[R_3-R_1\rightarrow R_3]{R_2+2R_1\to R_2} \begin{bmatrix} 1&1&2 \cr 0&x+2&5 \cr 0&-3&-1 \cr\end{bmatrix}, $$has rank at least $2$, since the first and third rows are not proportional. This means that it must have rank exactly two, and the last two rows must be proportional. Hence $$ \dfrac{x+2}{-3}= \dfrac{5}{-1} \implies x = 13. $$ \end{answer} \end{pro} \begin{pro} Let $B$ be the matrix obtained by adjoining a row (or column) to a matrix $A$. Prove that either $\rank{B} = \rank{A}$ or $\rank{B} = \rank{A}+1$. \end{pro} \begin{pro} Let $A\in\mat{n\times n}{\BBR}$. Prove that $\rank{A} = \rank{AA^T}$. Find a counterexample in the case $A\in\mat{n\times n}{\BBC}$. \begin{answer} For the counterexample consider $A = \begin{bmatrix}1& i \cr i & -1 \cr \end{bmatrix}$. \end{answer} \end{pro} \begin{pro} Prove that the rank of a skew-symmetric matrix with real number entries is an even number. \end{pro} \section{Rank and Invertibility} \begin{thm}\label{thm:rank_and_invertibility} A matrix $A\in\mat{m\times n}{ \BBF }$ is left-invertible if and only if $\rank{A} = n$. A matrix $A\in\mat{m\times n}{ \BBF }$ is right-invertible if and only if $\rank{A} = m$. \end{thm} \begin{pf} Observe that we always have $\rank{A} \leq n$. If $A$ is left invertible, then $\exists L\in\mat{n\times m}{ \BBF }$ such that $LA = {\bf I}_n$. By Theorem \ref{thm:rank_of_product}, $$ n = \rank{{\bf I}_n} = \rank{LA} \leq \rank{A}, $$whence the two inequalities give $\rank{A} = n$. \bigskip Conversely, assume that $\rank{A} = n$. Then $\rank{A^T} = n$ by Corollary \ref{cor:row_rank_is_column_rank}, and so by Theorem \ref{thm:normal_form} there exist $P\in\gl{m}{ \BBF }$, $Q\in\gl{n}{ \BBF }$, such that $$PAQ = \begin{bmatrix}{\bf I}_n \cr {\bf 0}_{(m - n)\times n} \cr \end{bmatrix}, \ \ \ Q^TA^TP^T = \begin{bmatrix}{\bf I}_n & {\bf 0}_{n\times (m - n)} \cr \end{bmatrix}. $$ This gives $$\begin{array}{lll}Q^TA^TP^TPAQ = {\bf I}_n & \implies & A^TP^TPA = (Q^T)^{-1}Q^{-1} \\ & \implies & ((Q^T)^{-1}Q^{-1})^{-1}A^TP^TPA = {\bf I}_n,\end{array}$$and so $((Q^T)^{-1}Q^{-1})^{-1}A^TP^TP$ is a left inverse for $A$. \bigskip The right-invertibility case is argued similarly. \end{pf} By combining Theorem \ref{thm:rank_and_invertibility} and Theorem \ref{thm:inverse_square_matrices}, the following corollary is thus immediate. \begin{cor}\label{cor:left_inverse_is_right} If $A\in\mat{m\times n}{ \BBF }$ possesses a left inverse $L$ and a right inverse $R$ then $m = n$ and $L=R$.\end{cor} We use Gau\ss-Jordan Reduction to find the inverse of $A\in\gl{n}{ \BBF }$. We form the {\em augmented matrix} $ T = [A|{\bf I}_n]$ which is obtained by putting $A$ side by side with the identity matrix ${\bf I}_n$. We perform permissible row operations on $T$ until instead of $A$ we obtain ${\bf I}_n$, which will appear if the matrix is invertible. The matrix on the right will be $A^{-1}$. We finish with $[{\bf I}_n|A^{-1}]$. \begin{rem} If $A \in\mat{n\times n}{\BBR}$ is non-invertible, then the left hand side in the procedure above will not reduce to ${\bf I}_n$. \end{rem} \begin{exa} Find the inverse of the matrix $B\in\mat{3\times 3}{\BBZ_7}$, $$B = \begin{bmatrix} \overline{6} & \overline{0}& \overline{1} \cr \overline{3} & \overline{2} & \overline{0} \cr \overline{1} & \overline{0} & \overline{1}\cr \end{bmatrix}.$$ \end{exa} \begin{solu}We have \begin{eqnarray*} \begin{bmat}{ccc|ccc} \overline{6} & \overline{0}& \overline{1} & \overline{1} & \overline{0} & \overline{0} \cr \overline{3} & \overline{2} & \overline{0} & \overline{0} & \overline{1} & \overline{0} \cr \overline{1} & \overline{0} & \overline{1} & \overline{0} & \overline{0} & \overline{1}\cr \end{bmat} & \grstep{R_1 \leftrightarrow R_3} & \begin{bmat}{ccc|ccc} \overline{1} & \overline{0} & \overline{1} & \overline{0} & \overline{0} & \overline{1}\cr \overline{3} & \overline{2} & \overline{0} & \overline{0} & \overline{1} & \overline{0} \cr \overline{6} & \overline{0}& \overline{1} & \overline{1} & \overline{0} & \overline{0} \cr \end{bmat} \\ & \grstep[R_2 - \overline{3}R_1 \rightarrow R_2]{R_3 - \overline{6}R_1 \rightarrow R_3} & \begin{bmat}{ccc|ccc} \overline{1} & \overline{0} & \overline{1} & \overline{0} & \overline{0} & \overline{1}\cr \overline{0} & \overline{2} & \overline{4} & \overline{0} & \overline{1} & \overline{4} \cr \overline{0} & \overline{0}& \overline{2} & \overline{1} & \overline{0} & \overline{1} \cr \end{bmat} \\ & \grstep[5R_1 + R_3 \rightarrow R_1]{R_2 - \overline{2}R_3 \rightarrow R_2} & \begin{bmat}{ccc|ccc} \overline{5} & \overline{0} & \overline{0} & \overline{1} & \overline{0} & \overline{6}\cr \overline{0} & \overline{2} & \overline{0} & \overline{5} & \overline{1} & \overline{2} \cr \overline{0} & \overline{0}& \overline{2} & \overline{1} & \overline{0} & \overline{1} \cr \end{bmat} \\ & \grstep[\overline{4}R_2 \rightarrow R_2]{\overline{3}R_1 \rightarrow R_1;\ \overline{4}R_3 \rightarrow R_3} & \begin{bmat}{ccc|ccc} \overline{1} & \overline{0} & \overline{0} & \overline{3} & \overline{0} & \overline{4}\cr \overline{0} & \overline{1} & \overline{0} & \overline{6} & \overline{4} & \overline{1} \cr \overline{0} & \overline{0}& \overline{1} & \overline{4} & \overline{0} & \overline{4} \cr \end{bmat}. \\ \end{eqnarray*} We conclude that $$\begin{bmatrix} \overline{6} & \overline{0}& \overline{1} \cr \overline{3} & \overline{2} & \overline{0} \cr \overline{1} & \overline{0} & \overline{1}\cr \end{bmatrix}^{-1} = \begin{bmatrix} \overline{3} & \overline{0}& \overline{4} \cr \overline{6} & \overline{4} & \overline{1} \cr \overline{4} & \overline{0} & \overline{4}\cr \end{bmatrix}.$$ \end{solu} %%%%%I use Jim Hefferon's macros for Gauss-Jordan Reduction \begin{exa} Use Gau\ss-Jordan reduction to find the inverse of the matrix $A = \left[\begin{array}{lll} 0 & 1 & -1 \\ \vspace{1mm} 4 & -3 & 4 \\ \vspace{1mm} 3 & -3 & 4 \\ \end{array}\right] .$ Also, find $A^{2001}$. \end{exa} \begin{solu}Operating on the augmented matrix \begin{eqnarray*} \begin{bmat}{ccc|ccc} 0 & 1 & -1 & 1 & 0 & 0 \\ 4 & -3 & 4 & 0 & 1 & 0 \\ 3 & -3 & 4 & 0 & 0 & 1\\ \end{bmat} &\grstep{R_2 - R_3 \rightarrow R_2} &\begin{bmat}{ccc|ccc} 0 & 1 & -1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 & -1\\ 3 & -3 & 4 & 0 & 0 & 1\\ \end{bmat}\\ &\grstep{R_3 - 3R_2\rightarrow R_3} &\begin{bmat}{ccc|ccc} 0 & 1 & -1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 & -1\\ 0 & -3 & 4 & 0 & -3 & 4\\ \end{bmat} \\ &\grstep{R_3 + 3R_1\rightarrow R_3} &\begin{bmat}{ccc|ccc} 0 & 1 & -1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 3 & -3 & 4\\ \end{bmat} \\ &\grstep{R_1 + R_3 \rightarrow R_1} &\begin{bmat}{ccc|ccc} 0 & 1 & 0 & 4 & -3 & 4\\ 1 & 0 & 0 & 0 & 1 & -1\\ 0 & 0 & 1 & 3 & -3 & 4\\ \end{bmat} \\ &\grstep{R_1 \leftrightarrow R_2} &\begin{bmat}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1\\ 0 & 1 & 0 & 4 & -3 & 4\\ 0 & 0 & 1 & 3 & -3 & 4\\ \end{bmat}. \end{eqnarray*} Thus we deduce that $$A^{-1} = \begin{bmatrix} 0 & 1 & -1\cr 4 & -3 & 4\cr 3 & -3 & 4\cr \end{bmatrix} = A. $$From $A^{-1} = A$ we deduce $A^2 = {\bf I}_n$. Hence $A^{2000} = (A^2)^{1000} = {\bf I}_n ^{1000} = {\bf I}_n$ and $A^{2001} = A(A^{2000}) = A{\bf I}_n = A$. \end{solu} \begin{exa} Find the inverse of the triangular matrix $A\in\mat{n\times n}{\BBR}$, $$ A = \begin{bmatrix}1 & 1 & 1 & \cdots & 1 \cr 0 & 1 & 1 & \cdots & 1 \cr 0 & 0 & 1 & \cdots & 1 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & 1 \cr \end{bmatrix}. $$ \end{exa} \begin{solu}Form the augmented matrix $$\begin{bmat}{ccccc|ccccc}1 & 1 & 1 & \cdots & 1 & 1 & 0 & 0 & \cdots & 0 \cr 0 & 1 & 1 & \cdots & 1 & 0 & 1 & 0 & \cdots & 0\cr 0 & 0 & 1 & \cdots & 1 & 0 & 0 & 1 & \cdots & 0\cr \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & 1 & 0 & 0 & 0 & \cdots & 1\cr \end{bmat},$$ and perform $R_k - R_{k + 1} \rightarrow R_k$ successively for $k = 1, 2, \ldots, n - 1$, obtaining $$\begin{bmat}{ccccc|ccccc}1 & 0 & 0 & \cdots & 0 & 1 & -1 & 0 & \cdots & 0 \cr 0 & 1 & 0 & \cdots & 0 & 0 & 1 & -1 & \cdots & 0\cr 0 & 0 & 1 & \cdots & 0 & 0 & 0 & 1 & \cdots & 0\cr \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & 1 & 0 & 0 & 0 & \cdots & 1\cr \end{bmat},$$ whence $$A^{-1} = \begin{bmatrix} 1 & -1 & 0 & \cdots & 0 \cr 0 & 1 & -1 & \cdots & 0\cr 0 & 0 & 1 & \cdots & 0\cr \vdots & \vdots & \vdots & \cdots & \vdots \cr0 & 0 & 0 & \cdots & 1\cr \end{bmatrix}, $$ that is, the inverse of $A$ has $1$'s on the diagonal and $-1$'s on the superdiagonal. \end{solu} \begin{thm}\label{thm:inverse_triangular_matrices} Let $A\in \mat{n\times n}{ \BBF }$ be a triangular matrix such that $a_{11}a_{22} \cdots a_{nn} \neq 0_{\BBF }$. Then $A$ is invertible. \end{thm} \begin{pf} Since the entry $a_{kk} \neq 0_{\BBF }$ we multiply the $k$-th row by $a_{kk} ^{-1}$ and then proceed to subtract the appropriate multiples of the preceding $k - 1$ rows at each stage. \end{pf} \begin{exa}[Putnam Exam, 1969] Let $A$ and $B$ be matrices of size $3\times 2$ and $2\times 3$ respectively. Suppose that their product $AB$ is given by $$AB = \begin{bmatrix}8 & 2 & -2 \cr 2 & 5 & 4 \cr -2 & 4 & 5 \cr\end{bmatrix}.$$Demonstrate that the product $BA$ is given by $$BA = \begin{bmatrix}9 & 0 \cr 0 & 9 \cr\end{bmatrix}.$$ \end{exa} \begin{solu}Observe that $$ (AB)^2 = \begin{bmatrix}8 & 2 & -2 \cr 2 & 5 & 4 \cr -2 & 4 & 5 \cr\end{bmatrix}\begin{bmatrix}8 & 2 & -2 \cr 2 & 5 & 4 \cr -2 & 4 & 5 \cr\end{bmatrix} = \begin{bmatrix}72 & 18& -18\cr 18 & 45 & 36 \cr -18 & 36 & 45 \cr \end{bmatrix}= 9AB. $$ Performing $R_3 + R_2 \rightarrow R_3$, $R_1 - 4R_2 \rightarrow R_2$, and $2R_3 + R_1 \rightarrow R_3$ in succession we see that $$ \begin{bmatrix}8 & 2 & -2 \cr 2 & 5 & 4 \cr -2 & 4 & 5 \cr\end{bmatrix}\rightsquigarrow \begin{bmatrix}0 & -18 & -18 \cr 2 & 5 & 4 \cr 0 & 0 & 0 \cr\end{bmatrix} \rightsquigarrow \begin{bmatrix}0 & -18 & 0 \cr 2 & 5 & -1 \cr 0 & 0 & 0 \cr\end{bmatrix} \rightsquigarrow \begin{bmatrix}0 & -18 & 0 \cr 0 & 5 & -1 \cr 0 & 0 & 0 \cr\end{bmatrix}, $$ and so $\rank{AB} = 2$. This entails that $\rank{(AB)^2} = 2$. Now, since $BA$ is a $2\times 2$ matrix, $\rank{BA} \leq 2$. Also $$ 2 = \rank{(AB)^2} = \rank{ABAB} \leq \rank{ABA} \leq \rank{BA}, $$and we must conclude that $\rank{BA} =2$. This means that $BA$ is invertible and so $$\begin{array}{lll} (AB)^2 = 9AB & \implies & A(BA - 9{\bf I}_{2})B = {\bf 0}_3 \\ & \implies & BA(BA - 9{\bf I}_{2})BA = B{\bf 0}_3A \\ & \implies & BA(BA - 9{\bf I}_{2})BA = {\bf 0}_2 \\ & \implies & (BA)^{-1}BA(BA - 9{\bf I}_{2})BA(BA)^{-1} = (BA)^{-1}{\bf 0}_2(BA)^{-1} \\ & \implies & BA - 9{\bf I}_{2} = {\bf 0}_2 \\ \end{array} $$ \end{solu} \section*{\psframebox{Homework}} \begin{pro}\label{pro:inverse_matrix_mod7} Find the inverse of the matrix $$\begin{bmatrix} \overline{1} & \overline{2} & \overline{3} \cr \overline{2} & \overline{3} & \overline{1} \cr \overline{3} & \overline{1} & \overline{2} \cr \end{bmatrix} \in \mat{3\times 3}{\BBZ_7}.$$ \begin{answer} We form the augmented matrix $$ \begin{bmat}{ccc|ccc} \overline{1} & \overline{2} & \overline{3} & \overline{1} & \overline{0} & \overline{0} \cr \overline{2} & \overline{3} & \overline{1} & \overline{0} & \overline{1} & \overline{0}\cr \overline{3} & \overline{1} & \overline{2} & \overline{0} & \overline{0} & \overline{1}\cr \end{bmat} $$ From $R_2 -\overline{2}R_1 \rightarrow R_2 $ and $R_3 -\overline{3}R_1 \rightarrow R_3 $ we obtain $$\rightsquigarrow \begin{bmat}{ccc|ccc} \overline{1} & \overline{2} & \overline{3}& \overline{1} & \overline{0} & \overline{0} \cr \overline{0} & \overline{6} & \overline{2} & \overline{5} & \overline{1} & \overline{0} \cr \overline{0} & \overline{2} & \overline{0} & \overline{4} & \overline{0} & \overline{1} \cr \end{bmat} . $$ From $R_2 \leftrightarrow R_3$ we obtain $$\rightsquigarrow \begin{bmat}{ccc|ccc} \overline{1} & \overline{2} & \overline{3} & \overline{1} & \overline{0} & \overline{0} \cr \overline{0} & \overline{2} & \overline{0} & \overline{4} & \overline{0} & \overline{1} \cr \overline{0} & \overline{6} & \overline{2} & \overline{5} & \overline{1} & \overline{0} \cr \end{bmat}. $$ Now, from $R_1 - R_2 \rightarrow R_1$ and $R_3 - \overline{3}R_2 \rightarrow R_3$, we obtain $$\rightsquigarrow \begin{bmat}{ccc|ccc} \overline{1} & \overline{0} & \overline{3} & \overline{4} & \overline{0} & \overline{6} \cr \overline{0} & \overline{2} & \overline{0} & \overline{4} & \overline{0} & \overline{1} \cr \overline{0} & \overline{0} & \overline{2} & \overline{0} & \overline{1} & \overline{4} \cr \end{bmat}. $$ From $4R_2 \rightarrow R_2$ and $4R_3 \rightarrow R_3$, we obtain $$\rightsquigarrow \begin{bmat}{ccc|ccc} \overline{1} & \overline{0} & \overline{3} & \overline{4} & \overline{0} & \overline{6} \cr \overline{0} & \overline{1} & \overline{0} & \overline{2} & \overline{0} & \overline{4} \cr \overline{0} & \overline{0} & \overline{1} & \overline{0} & \overline{4} & \overline{2} \cr \end{bmat}. $$ Finally, from $R_1 - \overline{3}R_3 \rightarrow R_3$ we obtain $$\rightsquigarrow \begin{bmat}{ccc|ccc} \overline{1} & \overline{0} & \overline{0} & \overline{4} & \overline{2} & \overline{0} \cr \overline{0} & \overline{1} & \overline{0} & \overline{2} & \overline{0} & \overline{4} \cr \overline{0} & \overline{0} & \overline{1} & \overline{0} & \overline{4} & \overline{2} \cr \end{bmat}. $$ We deduce that $$ \begin{bmatrix} \overline{1} & \overline{2} & \overline{3} \cr \overline{2} & \overline{3} & \overline{1} \cr \overline{3} & \overline{1} & \overline{2} \cr \end{bmatrix}^{-1} = \begin{bmatrix} \overline{4} & \overline{2} & \overline{0} \cr \overline{2} & \overline{0} & \overline{4} \cr \overline{0} & \overline{4} & \overline{2} \cr \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Let $(A, B)\in\mat{3\times 3}{\BBR}$ be given by $$A = \begin{bmatrix} a & b & c \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}, \ \ \ B = \begin{bmatrix} 0 & 0 & -1 \cr 0 & -1 & a \cr -1 & a & b \cr\end{bmatrix}.$$Find $B^{-1}$ and prove that $A^T = BAB^{-1}$. \begin{answer} To find the inverse of $B$ we consider the augmented matrix $$\begin{bmat}{ccc|ccc}0 & 0 & -1 & 1 & 0 & 0 \cr 0 & -1 & a & 0 & 1 & 0\cr -1 & a & b & 0 & 0 & 1\cr\end{bmat}. $$ Performing $R_1 \leftrightarrow R_3$, $-R_3\rightarrow R_3$, in succession, $$\begin{bmat}{ccc|ccc} -1 & a & b & 0 & 0 & 1\cr0 & -1 & a & 0 & 1 & 0\cr 0 & 0 & 1 & -1 & 0 & 0 \cr\end{bmat}. $$Performing $R_1 + aR_2 \rightarrow R_1$ and $R_2 - aR_3 \rightarrow R_2$ in succession, $$\begin{bmat}{ccc|ccc} -1 & 0 & b + a^2 & 0 & a & 1\cr0 & -1 & 0 & a & 1 & 0\cr 0 & 0 & 1 & -1 & 0 & 0 \cr\end{bmat}. $$ Performing $R_1 - (b + a^2)R_3 \rightarrow R_3$, $-R_1 \rightarrow R_1$ and $-R_2 \rightarrow R_2$ in succession, we find $$\begin{bmat}{ccc|ccc} 1 & 0 & 0 & -b - a^2 & -a & -1\cr0 & 1 & 0 & -a & -1 & 0\cr 0 & 0 & 1 & -1 & 0 & 0 \cr\end{bmat}, $$whence $$B^{-1} = \begin{bmatrix} -b - a^2 & -a & -1\cr -a & -1 & 0\cr -1 & 0 & 0 \cr\end{bmatrix}. $$ Now, $$\begin{array}{lll} BAB^{-1} & = & \begin{bmatrix} 0 & 0 & -1 \cr 0 & -1 & a \cr -1 & a & b \cr\end{bmatrix}\begin{bmatrix} a & b & c \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr \end{bmatrix}\begin{bmatrix} -b - a^2 & -a & -1\cr -a & -1 & 0\cr -1 & 0 & 0 \cr\end{bmatrix} \\ & = & \begin{bmatrix} 0 & -1 & 0 \cr -1 & a& 0\cr 0 & 0& -c \cr\end{bmatrix}\begin{bmatrix} -b - a^2 & -a & -1\cr -a & -1 & 0\cr -1 & 0 & 0 \cr\end{bmatrix} \\ & = & \begin{bmatrix} a & 1& 0\cr b & 0& 1\cr c & 0 & 0\cr \end{bmatrix} \\ & = & A^T, \end{array} $$which is what we wanted to prove. \end{answer} \end{pro} \begin{pro} Let $A=\begin{bmatrix} 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & x \cr \end{bmatrix}$ where $x\neq 0$ is a real number. Find $A^{-1}$. \begin{answer} First, form the augmented matrix: $$\begin{bmat}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \cr 1 & 1 & 0 & 0 & 1 & 0 \cr 1 & 1 & x & 0 & 0 & 1 \cr\end{bmat}.$$ Perform $R_2-R_1 \rightarrow R_2$ and $R_3-R_1 \rightarrow R_3$: $$\begin{bmat}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \cr 0 & 1 & 0 & -1 & 1 & 0 \cr 0 & 1 & x & -1 & 0 & 1 \cr\end{bmat}.$$ Performing $R_3-R_2 \rightarrow R_3$: $$\begin{bmat}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \cr 0 & 1 & 0 & -1 & 1 & 0 \cr 0 & 0 & x & 0 & -1 & 1 \cr\end{bmat}.$$ Finally, performing $\dfrac{1}{x}R_3\rightarrow R_3$: $$\begin{bmat}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0 \cr 0 & 1 & 0 & -1 & 1 & 0 \cr 0 & 0 & 1 & 0 & -\dfrac{1}{x} & \dfrac{1}{x} \cr\end{bmat}.$$ \end{answer} \end{pro} \begin{pro} If the inverse of the matrix $M = \begin{bmatrix} 1 & 0 & 1 \cr -1 & 0 & 0 \cr 0 & 1 & 1 \cr\end{bmatrix}$ is the matrix $M^{-1} = \begin{bmatrix} 0 & -1 & 0 \cr -1 & -1 & a \cr 1 & 1 & b \cr\end{bmatrix}$, find $(a, b)$. \begin{answer} Since $MM^{-1}={\bf I}_3$, multiplying the first row of $M$ times the third column of $M^{-1}$, and again, the third row of $M$ times the third column of $M^{-1}$, we gather that $$ 1\cdot 0 + 0\cdot a + 1\cdot b = 0, \qquad 0 \cdot 0 + 1\cdot a + 1\cdot b =1 \implies b=0, a=1. $$ \end{answer} \end{pro} \begin{pro} Let $A=\begin{bmatrix} 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr \end{bmatrix}$ and let $n>0$ be an integer. Find $(A^n)^{-1}$. \begin{answer} It is easy to prove by induction that $A^n = \begin{bmatrix} 1 & 0 & 0 \cr n & 1 & 0 \cr \dfrac{n(n+1)}{2} & n & 1 \cr\end{bmatrix}$. Row-reducing, $(A^n)^{-1} = \begin{bmatrix} 1 & 0 & 0 \cr -n & 1 & 0 \cr \dfrac{(n-1)n}{2} & -n & 1 \cr\end{bmatrix}$. \end{answer} \end{pro} \begin{pro} Give an example of a $2\times 2$ invertible matrix $A$ over $\BBR$ such that $A+A^{-1}$ is the zero matrix. \begin{answer} Take, for example, $A = \begin{bmatrix} 0 & -1 \cr 1 & 0 \end{bmatrix}=-A^{-1}$. \end{answer} \end{pro} \begin{pro} Find all the values of the parameter $a$ for which the matrix $B$ given below is not invertible. $$B = \left[ \begin{array}{lll} -1 & a + 2 & 2 \\ 0 & a & 1 \\ 2 & 1 & a \end{array} \right]$$ \begin{answer} Operating formally, and using elementary row operations, we find $$ B^{-1} = \left[\begin{array}{lll} -\frac{a^2 - 1}{a^2 - 5+ 2a} & \frac{a^2 + 2a - 2}{a^2 - 5 + 2a} & \frac{a-2}{a^2- 5 + 2a} \\ -\frac{2}{a^2 - 5 + 2a} & \frac{a + 4}{a^2- 5 + 2a} & -\frac{1}{a^2 - 5 + 2a} \\ \frac{2a}{a^2 - 5 + 2a} & -\frac{2a + 5}{a^2 - 5 + 2a} & \frac{a}{a^2- 5 + 2a} \end{array} \right] . $$ Thus $B$ is invertible whenever $a \neq -1 \pm \sqrt{6}.$ \end{answer} \end{pro} \begin{pro} Find the inverse of the triangular matrix $$\begin{bmatrix} a & 2a & 3a \cr 0 & b & 2b \cr 0 & 0 & c \cr \end{bmatrix}\in\mat{3\times 3}{\BBR} $$assuming that $abc \neq 0$. \begin{answer} Form the augmented matrix $$\begin{bmat}{ccc|ccc} a & 2a & 3a & 1 & 0 & 0 \cr 0 & b & 2b & 0 & 1 & 0 \cr 0 & 0 & c & 0 & 0 & 1 \cr \end{bmat}. $$Perform $\dfrac{1}{a}R_1 \rightarrow R_1$, $\dfrac{1}{b}R_2 \rightarrow R_2$, $\dfrac{1}{a}R_3 \rightarrow R_3$, in succession, obtaining $$\begin{bmat}{ccc|ccc} 1 & 2 & 3 & 1/a & 0 & 0 \cr 0 & 1 & 2 & 0 & 1/b & 0 \cr 0 & 0 & 1 & 0 & 0 & 1/c \cr \end{bmat}. $$ Now perform $R_1 - 2R_2\rightarrow R_1$ and $R_2 - 2R_3\rightarrow R_2$ in succession, to obtain $$\begin{bmat}{ccc|ccc} 1 & 0 & -1 & 1/a & -2/a & 0 \cr 0 & 1 & 0 & 0 & 1/b & -2/c \cr 0 & 0 & 1 & 0 & 0 & 1/c \cr \end{bmat}. $$ Finally, perform $R_1 + R_3\rightarrow R_1$ to obtain $$\begin{bmat}{ccc|ccc} 1 & 0 & 0 & 1/a & -2/b & 1/c \cr 0 & 1 & 0 & 0 & 1/b & -2/c \cr 0 & 0 & 1 & 0 & 0 & 1/c \cr \end{bmat}. $$Whence $$\begin{bmatrix} a & 2a & 3a \cr 0 & b & 2b \cr 0 & 0 & c \cr \end{bmatrix}^{-1} = \begin{bmatrix} 1/a & -2/b & 1/c \cr 0 & 1/b & -2/c \cr 0 & 0 & 1/c \cr \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro}\label{pro:inverse_with_abc} Under what conditions is the matrix $$\begin{bmatrix} b & a & 0 \cr c & 0 & a \cr 0 & c & b \cr \end{bmatrix} $$ invertible? Find the inverse under these conditions. \begin{answer}To compute the inverse matrix we proceed formally as follows. The augmented matrix is $$ \begin{bmat}{ccc|ccc} b & a & 0 & 1 & 0 & 0 \cr c & 0 & a & 0 & 1 & 0 \cr 0 & c & b & 0 & 0 & 1\cr\end{bmat}. $$ Performing $bR_2 - cR_1 \rightarrow R_2$ we find $$ \begin{bmat}{ccc|ccc} b & a & 0 & 1 & 0 & 0 \cr 0 & -ca & ab & -c & b & 0 \cr 0 & c & b & 0 & 0 & 1\cr\end{bmat}. $$ Performing $aR_3 + R_2 \rightarrow R_3$ we obtain $$ \begin{bmat}{ccc|ccc} b & a & 0 & 1 & 0 & 0 \cr 0 & -ca & ab & -c & b & 0 \cr 0 & 0 & 2ab & -c & b & a\cr\end{bmat}. $$ Performing $2R_2 - R_3 \rightarrow R_2$ we obtain $$ \begin{bmat}{ccc|ccc} b & a & 0 & 1 & 0 & 0 \cr 0 & -2ca & 0 & -c & b & -a\cr 0 & 0 & 2ab & -c & b & a\cr\end{bmat}. $$ Performing $2cR_1 + R_2 \rightarrow R_1$ we obtain $$ \begin{bmat}{ccc|ccc} 2bc & 0 & 0 & c & b & -a \cr 0 & -2ca & 0 & -c & b & -a\cr 0 & 0 & 2ab & -c & b & a\cr\end{bmat}. $$ From here we easily conclude that $$ \begin{bmatrix} b & a & 0 \cr c & 0 & a \cr 0 & c & b \cr \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{2b} & \frac{1}{2c} & -\frac{a}{2bc} \cr \frac{1}{2a} & -\frac{b}{2ac} & \frac{1}{2c} \cr -\frac{c}{2ba} & \frac{1}{2a} & \frac{1}{2b}\cr \end{bmatrix} $$ as long as $abc \neq 0$. \end{answer} \end{pro} \begin{pro} Let $A$ and $B$ be $n\times n$ matrices over a field $\BBF$ such that $AB$ is invertible. Prove that both $A$ and $B$ must be invertible. \begin{answer} Since $AB$ is invertible, $\rank{AB}=n$. Thus $$ n = \rank{AB}\leq \rank{A}\leq n\implies \rank{A}=n, $$ $$ n = \rank{AB}\leq \rank{B}\leq n\implies \rank{B}=n, $$ whence $A$ and $B$ are invertible. \end{answer} \end{pro} \begin{pro} Find the inverse of the matrix $$ \begin{bmatrix}1+a & 1 & 1 \cr 1 & 1 +b & 1 \cr 1 & 1 & 1 + c \cr \end{bmatrix} $$ \begin{answer}Form the expanded matrix $$ \begin{bmat}{ccc|ccc} 1 + a & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 + b & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 + c & 0 & 0 & 1\\ \end{bmat} .$$ Perform $bcR_1\rightarrow R_1$, $abR_3 \rightarrow R_3$, $caR_2 \rightarrow R_2$. The matrix turns into $$ \begin{bmat}{ccc|ccc} bc +abc& bc &bc & bc & 0 & 0 \\ ca & ca+abc & ca & 0 & ca & 0 \\ ab & ab & ab+abc & 0 & 0 & ab\\ \end{bmat}. $$ Perform $R_1 + R_2 + R_3 \rightarrow R_1$ the matrix turns into $$ \begin{bmat}{ccc|ccc} ab + bc+ca +abc& ab + bc+ca +abc & ab + bc+ca +abc & bc & ca & ab \\ ca & ca+abc & ca & 0 & ca & 0 \\ ab & ab & ab+abc & 0 & 0 & ab\\ \end{bmat}. $$ Perform $\frac{1}{ab + bc+ca +abc}R_1 \rightarrow R_1$. The matrix turns into $$ \begin{bmat}{ccc|ccc} 1& 1 & 1 & \frac{bc}{ab + bc+ca +abc} & \frac{ca}{ab + bc+ca +abc} & \frac{ab}{ab + bc+ca +abc} \\ ca & ca+abc & ca & 0 & ca & 0 \\ ab & ab & ab+abc & 0 & 0 & ab\\ \end{bmat}. $$ Perform $R_2-caR_1 \rightarrow R_2$ and $R_3 - abR_3 \rightarrow R_3$. We get $$ \begin{bmat}{ccc|ccc} 1& 1 & 1 & \frac{bc}{ab + bc+ca +abc} & \frac{ca}{ab + bc+ca +abc} & \frac{ab}{ab + bc+ca +abc} \\ 0 & abc & 0 & -\frac{abc^2}{ab + bc+ca +abc} & ca - \frac{c^2a^2}{ab + bc+ca +abc} & -\frac{a^2bc}{ab + bc+ca +abc} \\ 0 & 0 & abc & -\frac{ab^2c}{ab + bc+ca +abc} & -\frac{a^2bc}{ab + bc+ca +abc} & ab - \frac{a^2b^2}{ab + bc+ca +abc}\\ \end{bmat}. $$Perform $\frac{1}{ABC}R_2\rightarrow R_2$ and $\frac{1}{ABC}R_3\rightarrow R_3$. We obtain $$ \begin{bmat}{ccc|ccc} 1& 1 & 1 & \frac{bc}{ab + bc+ca +abc} & \frac{ca}{ab + bc+ca +abc} & \frac{ab}{ab + bc+ca +abc} \\ 0 & 1 & 0 & -\frac{c}{ab + bc+ca +abc} & \frac{1}{b} - \frac{ca}{b(ab + bc+ca +abc)} & -\frac{a}{ab + bc+ca +abc} \\ 0 & 0 & 1 & -\frac{b}{ab + bc+ca +abc} & -\frac{a}{ab + bc+ca +abc} & \frac{1}{c} - \frac{ab}{c(ab + bc+ca +abc)}\\ \end{bmat}. $$Finally we perform $R_1 - R_2 - R_3 \rightarrow R_1$, getting $$ \begin{bmat}{ccc|ccc} 1& 0 & 0 & \frac{a + b + bc}{ab + bc+ca +abc} & -\frac{c}{ab + bc+ca +abc} & -\frac{b}{ab + bc+ca +abc} \\ 0 & 1 & 0 & -\frac{c}{ab + bc+ca +abc} & \frac{1}{b} - \frac{ca}{b(ab + bc+ca +abc)} & -\frac{a}{ab + bc+ca +abc} \\ 0 & 0 & 1 & -\frac{b}{ab + bc+ca +abc} & -\frac{a}{ab + bc+ca +abc} & \frac{1}{c} - \frac{ab}{c(ab + bc+ca +abc)}\\ \end{bmat}. $$ We conclude that the inverse is $$\begin{bmatrix} \frac{b + c + bc}{ab+bc+ca + abc} & -\frac{c}{ab+bc+ca + abc}& -\frac{b}{ab+bc+ca + abc} \cr -\frac{c}{ab+bc+ca + abc} & \frac{c + a + ca}{ab+bc+ca + abc}& -\frac{a}{ab+bc+ca + abc} \cr -\frac{b}{ab+bc+ca + abc} & -\frac{a}{ab+bc+ca + abc}& \frac{a + b + ab}{ab+bc+ca + abc} \cr \end{bmatrix}$$ \end{answer} \end{pro} \begin{pro} Prove that for the $n\times n$ ($n > 1$) matrix $$ \begin{bmatrix}0 & 1 & 1 & \ldots & 1 \cr 1 & 0 & 1 & \ldots & 1 \cr 1 & 1 & 0 & \ldots & 1 \cr \vdots & \vdots & \vdots & \ldots & \vdots \cr 1 & 1& 1& \ldots & 0 \end{bmatrix}^{-1} = \dfrac{1}{n-1}\begin{bmatrix}2-n & 1 & 1 & \ldots & 1 \cr 1 & 2-n & 1 & \ldots & 1 \cr 1 & 1 & 2-n & \ldots & 1 \cr \vdots & \vdots & \vdots & \ldots & \vdots \cr 1 & 1& 1& \ldots & 2-n \cr \end{bmatrix} $$ \end{pro} \begin{pro} Prove that the $n\times n$ ($n > 1$) matrix $$ \begin{bmatrix}1+a & 1 & 1 & \ldots & 1 \cr 1 & 1+a & 1 & \ldots & 1 \cr 1 & 1 & 1+a & \ldots & 1 \cr \vdots & \vdots & \vdots & \ldots & \vdots \cr 1 & 1& 1& \ldots & 1+a \end{bmatrix}$$ has inverse $$-\dfrac{1}{a(n+a)}\begin{bmatrix}1-n-a & 1 & 1 & \ldots & 1 \cr 1 & 1-n-a & 1 & \ldots & 1 \cr 1 & 1 & 1-n-a & \ldots & 1 \cr \vdots & \vdots & \vdots & \ldots & \vdots \cr 1 & 1& 1& \ldots & 1-n-a \cr \end{bmatrix} $$ \end{pro} \begin{pro} Prove that $$\begin{bmatrix}1 & 3 & 5 & 7 & \cdots & (2n - 1) \cr (2n -1) & 1 & 3 & 5 & \cdots & (2n - 3) \cr (2n - 3) & (2n - 1) & 1 & 3 & \cdots & (2n - 5) \cr \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \cr 3 & 5 & 7 & 9 & \cdots & 1 \cr \end{bmatrix}$$has inverse $$ \frac{1}{2n^3} \begin{bmatrix} 2 - n^2 & 2 + n^2 & 2 & 2 & \cdots & 2 \cr 2 & 2 - n^2 & 2 + n^2 & 2 & \cdots & 2 \cr 2 & 2 & 2 - n^2 & 2 + n^2 & \cdots & 2 \cr \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \cr 2 + n^2 & 2 & 2 & 2 & \cdots & 2 - n^2\cr \end{bmatrix}.$$ \end{pro} \begin{pro} Prove that the $n\times n$ ($n > 1$) matrix $$ \begin{bmatrix}1+a_1 & 1 & 1 & \ldots & 1 \cr 1 & 1+a_2 & 1 & \ldots & 1 \cr 1 & 1 & 1+a_3 & \ldots & 1 \cr \vdots & \vdots & \vdots & \ldots & \vdots \cr 1 & 1& 1& \ldots & 1+a_n \end{bmatrix}$$ has inverse $$ -\dfrac{1}{s}\begin{bmatrix}\dfrac{1-a_1s}{a_1 ^2} & \dfrac{1}{a_1a_2} & \dfrac{1}{a_1a_3} & \ldots & \dfrac{1}{a_1a_n} \cr \dfrac{1}{a_2a_1}&\dfrac{1-a_2s}{a_2 ^2} & \dfrac{1}{a_2a_3} & \ldots & \dfrac{1}{a_2a_n} \cr \dfrac{1}{a_3a_1} & \dfrac{1}{a_3a_2} & \dfrac{1-a_3s}{a_3 ^2} & \ldots & \dfrac{1}{a_3a_1n} \cr \vdots & \vdots & \vdots & \ldots & \vdots \cr \dfrac{1}{a_na_1} & \dfrac{1}{a_na_2}& \dfrac{1}{a_na_3}& \ldots & \dfrac{1-a_ns}{a_n ^2} \cr \end{bmatrix}, $$where $s = 1 + \frac{1}{a_1} + \frac{1}{a_2}+\cdots + \frac{1}{a_n}$. \end{pro} \begin{pro}Let $A\in\mat{5\times 5}{\BBR}$. Shew that if $\rank{A^2} <5$, then $\rank{A} <5$. \begin{answer} Since $\rank{A^2} < 5$, $A^2$ is not invertible. But then $A$ is not invertible and hence $\rank{A}<5$. \end{answer} \end{pro} \begin{pro} Let $p$ be an odd prime. How many invertible $2\times 2$ matrices are there with entries all in $\BBZ_p$? \begin{answer} Each entry can be chosen in $p$ ways, which means that there are $p^2$ ways of choosing the two entries of an arbitrary row. The first row cannot be the zero row, hence there are $p^2-1$ ways of choosing it. The second row cannot be one of the $p$ multiples of the first row, hence there are $p^2-p$ ways of choosing it. In total, this gives $(p^2-1)(p^2-p)$ invertible matrices in $\BBZ _p$. \end{answer} \end{pro} \begin{pro} Let $A, B$ be matrices of the same size. Prove that $\rank{A+B} \leq \rank{A} + \rank{B}$. \begin{answer} Assume that both $A$ and $B$ are $m\times n$ matrices. Let $C=[A\quad B]$ be the $m\times (2n)$ obtained by juxtaposing $A$ to $B$. $\rank{C}$ is the number of linearly independent columns of $C$, which is composed of the columns of $A$ and $B$. By column-reducing the first $n$ columns, we find $\rank{A}$ linearly independent columns. By column-reducing columns $n+1$ to $2n$, we find $\rank{B}$ linearly independent columns. These $\rank{A}+\rank{B}$ columns are distinct, and are a subset of the columns of $C$. Since $C$ has at most $\rank{C}$ linearly independent columns, it follows that $\rank{C}\leq \rank{A}+\rank{B}$. Furthermore, by adding the $n+k$-column ($1\leq k \leq n$) of $C$ to the $k$-th column, we see that $C$ is column-equivalent to $[A+B\ B]$. But clearly $$\rank{A+B}\leq \rank{[A+B\ B]}= \rank{C}, $$since $[A+B\ B]$ is obtained by adding columns to $A+B$. We deduce $$\rank{A+B}\leq \rank{[A+B\ B]}= \rank{C} \leq \rank{A}+\rank{B}, $$ as was to be shewn. \end{answer} \end{pro} \begin{pro} Let $A\in\mat{3,2}{\BBR}$ and $B\in\mat{2,3}{\BBR}$ be matrices such that $AB = \begin{bmatrix} 0 & -1 & -1 \cr -1 & 0 & -1 \cr 1 & 1 & 2\end{bmatrix}$. Prove that $BA = {\bf I}_2$. \begin{answer} Since the first two columns of $AB$ are not proportional, and since the last column is the sum of the first two, $\rank{AB}=2$. Now, $$(AB)^2 = \begin{bmatrix} 0 & -1 & -1 \cr -1 & 0 & -1 \cr 1 & 1 & 2\end{bmatrix}^2 = \begin{bmatrix} 0 & -1 & -1 \cr -1 & 0 & -1 \cr 1 & 1 & 2\end{bmatrix}=AB.$$ Since $BA$ is a $2\times 2$ matrix, $\rank{BA}\leq 2$. Also, $$2 = \rank{AB}=\rank{(AB)^2} = \rank{A(BA)B}\leq \rank{BA}, $$ whence $\rank{BA}=2$, which means $BA$ is invertible. Finally, $$(AB)^2-AB= {\bf 0}_3 \implies A(BA-{\bf I}_2)B= {\bf 0}_3\implies BA(BA-{\bf I}_2)BA= B{\bf 0}_3A \implies BA-{\bf I}_2= {\bf 0}_2, $$ since $BA$ is invertible and we may cancel it. \end{answer} \end{pro} \chapter{Linear Equations} \section{Definitions} We can write a system of $m$ linear equations in $n$ variables over a field $\BBF$ $$a_{11}x_1 + a_{12}x_2 + a_{13}x_3 + \cdots + a_{1n}x_n = y_1,$$ $$a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + \cdots + a_{2n}x_n = y_2,$$ $$\vdots$$ $$a_{m1}x_1 + a_{m2}x_2 + a_{m3}x_3 + \cdots + a_{mn}x_n = y_m,$$in matrix form as \begin{equation}\label{eq:matrix_equation}\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \cr a_{21} & a_{22} & \cdots & a_{2n} \cr \vdots & \vdots & \vdots & \vdots \cr a_{m1} & a_{m2} & \cdots & a_{mn} \cr \end{bmatrix}\colvec{x_1 \\ x_2 \\ \vdots \\ x_n} = \colvec{y_1 \\ y_2 \\ \vdots \\ y_m}.\end{equation} We write the above matrix relation in the abbreviated form \begin{equation} AX = Y, \end{equation} where $A$ is the matrix of coefficients, $X$ is the matrix of variables and $Y$ is the matrix of constants. Most often we will dispense with the matrix of variables $X$ and will simply write the {\em augmented matrix} of the system as \begin{equation}[A |Y] = \begin{bmat}{cccc|c} a_{11} & a_{12} & \cdots & a_{1n} & y_1\cr a_{21} & a_{22} & \cdots & a_{2n} & y_2 \cr \vdots & \vdots & \vdots & \vdots & \vdots \cr a_{m1} & a_{m2} & \cdots & a_{mn} & y_m \cr \end{bmat}.\end{equation} \begin{df} Let $AX = Y$ be as in \ref{eq:matrix_equation}. If $Y = {\bf 0}_{m\times 1}$, then the system is called {\em homogeneous}, otherwise it is called {\em inhomogeneous}. The set $$\{X\in\mat{n\times 1}{ \BBF }: AX = {\bf 0}_{m\times 1}\} $$is called the {\em kernel} or {\em nullspace} of $A$ and it is denoted by $\ker{A}$. \end{df} \begin{rem} Observe that we always have ${\bf 0}_{n\times 1}\in\ker{A}\in\mat{m\times n}{ \BBF }$. \end{rem} \begin{df} A system of linear equations is {\em consistent} if it has a solution. If the system does not have a solution then we say that it is {\em inconsistent}. \end{df} \begin{df} If a row of a matrix is non-zero, we call the first non-zero entry of this row a {\em pivot} for this row. \index{pivot} \end{df} \begin{df} A matrix $M\in\mat{m\times n}{ \BBF }$ is a {\em row-echelon} matrix if \begin{itemize} \item All the zero rows of $M$, if any, are at the bottom of $M$. \item For any two consecutive rows $R_i$ and $R_{i + 1}$, either $R_{i + 1}$ is all $0_{\BBF }$'s or the pivot of $R_{i + 1}$ is immediately to the right of the pivot of $R_i$. \end{itemize} The variables accompanying these pivots are called the {\em leading variables}. Those variables which are not leading variables are the {\em free parameters.} \end{df} \begin{exa} The matrices $$\begin{bmatrix}\pscirclebox[doubleline=true, linecolor=red]{1} & 0 & 1 & 1 \cr 0 & 0 & \pscirclebox[doubleline=true, linecolor=red]{2} & 2 \cr 0 & 0 & 0 & \pscirclebox[doubleline=true, linecolor=red]{3} \cr 0 & 0 & 0 & 0\end{bmatrix}, \ \ \ \begin{bmatrix}\pscirclebox[doubleline=true, linecolor=red]{1} & 0 & 1 & 1 \cr 0 & 0 & 0 & \pscirclebox[doubleline=true, linecolor=red]{1} \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0\end{bmatrix}, \ \ \ $$ are in row-echelon form, with the pivots circled, but the matrices $$\begin{bmatrix}1 & 0 & 1 & 1 \cr 0 & 0 & 1 & 2 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 0 & 0\end{bmatrix}, \ \ \ \begin{bmatrix}1 & 0 & 1 & 1 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 1 \cr 0 & 0 & 0 & 0 \cr \end{bmatrix}, \ \ \ $$are not in row-echelon form. \end{exa} \begin{rem} Observe that given a matrix $A\in\mat{m\times n}{ \BBF }$, by following Gau\ss-Jordan reduction \`{a} la Theorem \ref{thm:normal_form}, we can find a matrix $P\in\gl{m}{ \BBF }$ such that $PA = B$ is in row-echelon form. \end{rem} \begin{exa} Solve the system of linear equations $$\begin{bmatrix}1 & 1 & 1 & 1 \cr 0 & 2 & 1 & 0 \cr 0 & 0 & 1 & -1 \cr 0 & 0 & 0 & 2 \cr \end{bmatrix} \colvec{x\\ y \\ z \\ w} = \colvec{-3 \\ -1 \\ 4 \\ - 6}. $$ \end{exa} \begin{solu}Observe that the matrix of coefficients is already in row-echelon form. Clearly every variable is a leading variable, and by back substitution $$2w = -6 \implies w = -\frac{6}{2} = -3,$$ $$z - w = 4 \implies z = 4 + w = 4 - 3 = 1,$$ $$2y + z = -1 \implies y = -\frac{1}{2} - \frac{1}{2}z = -1,$$ $$x + y + z + w = -3 \implies x = -3 - y - z - w = 0.$$ The (unique) solution is thus $$\colvec{x \\ y \\ z \\ w} = \colvec{0 \\ -1 \\ 1 \\ -3}.$$ \end{solu} \begin{exa} Solve the system of linear equations $$\begin{bmatrix}1 & 1 & 1 & 1 \cr 0 & 2 & 1 & 0 \cr 0 & 0 & 1 & -1 \cr \end{bmatrix} \colvec{x\\ y \\ z \\ w} = \colvec{-3 \\ -1 \\ 4 }. $$ \end{exa} \begin{solu}The system is already in row-echelon form, and we see that $x, y, z$ are leading variables while $w$ is a free parameter. We put $w = t$. Using back substitution, and operating from the bottom up, we find $$z - w = 4 \implies z = 4 + w = 4 + t,$$ $$2y + z = -1 \implies y = -\frac{1}{2} - \frac{1}{2}z = -\frac{1}{2} - 2 - \frac{1}{2}t = -\frac{5}{2} - \frac{1}{2}t,$$ $$x + y + z + w = -3 \implies x = -3 - y - z - w = -3 + \frac{5}{2} + \frac{1}{2}t - 4 - t - t = -\frac{9}{2} - \frac{3}{2}t.$$ The solution is thus $$\colvec{x \\ y \\ z \\ w} = \colvec{-\frac{9}{2} - \frac{3}{2}t \\-\frac{5}{2} - \frac{1}{2}t \\ 4 + t \\ t }, \ t \in \BBR. $$ \end{solu} \begin{exa} Solve the system of linear equations $$\begin{bmatrix}1 & 1 & 1 & 1 \cr 0 & 2 & 1 & 0 \cr \end{bmatrix} \colvec{x\\ y \\ z \\ w} = \colvec{-3 \\ -1 }. $$ \end{exa} \begin{solu}We see that $x, y$ are leading variables, while $z, w$ are free parameters. We put $z = s, w = t$. Operating from the bottom up, we find $$2y + z = -1 \implies y = -\frac{1}{2} - \frac{1}{2}z = -\frac{1}{2} - \frac{1}{2}s,$$ $$x + y + z + w = -3 \implies x = -3 - y - z - w = -\frac{5}{2} - \frac{3}{2}s - t.$$ The solution is thus $$\colvec{x \\ y \\ z \\ w} = \colvec{-\frac{5}{2} - \frac{3}{2}s - t \\-\frac{1}{2} - \frac{1}{2}s \\ s \\ t }, \ (s, t) \in \BBR^2. $$ \end{solu} \begin{exa} Find all the solutions of the system $$ x + \overline{2}y + \overline{2}z = \overline{0}, $$ $$ y + \overline{2}z = \overline{1}, $$working in $\BBZ_{3}$. \end{exa} \begin{solu}The augmented matrix of the system is $$\begin{bmat}{ccc|c} \overline{1} & \overline{2} & \overline{2} & \overline{0} \cr \overline{0} & \overline{1} & \overline{2} & \overline{1} \cr \end{bmat}. $$ The system is already in row-echelon form and $x, y$ are leading variables while $z$ is a free parameter. We find $$ y = \overline{1} - \overline{2}z = \overline{1} + \overline{1}z, $$ and $$ x = -\overline{2}y - \overline{2}z = \overline{1} + \overline{2}z. $$Thus $$\colvec{x\\ y \\ z} = \colvec{\overline{1} + \overline{2}z\\ \overline{1} + \overline{1}z \\ z}, \ \ z\in\BBZ_3.$$ Letting $z = \overline{0}, \overline{1}, \overline{2}$ successively, we find the three solutions $$\colvec{x\\ y \\ z} = \colvec{\overline{1} \\ \overline{1} \\ \overline{0}},$$ $$\colvec{x\\ y \\ z} = \colvec{\overline{0} \\ \overline{2} \\ \overline{1}},$$and $$\colvec{x\\ y \\ z} = \colvec{\overline{2} \\ \overline{0} \\ \overline{2}}.$$ \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Find all the solutions in $\BBZ_3$ of the system $$\begin{array}{l}x + y + z + w = \overline{0}, \\ \overline{2}y + w = \overline{2}. \end{array} $$ \begin{answer} The free variables are $z$ and $w$. We have $$ \overline{2}y + w = \overline{2} \implies \overline{2}y = \overline{2} - w \implies y = \overline{1} + w, $$and $$x + y + z + w = \overline{0} \implies x = -y-z-w = \overline{2}y + \overline{2}z + \overline{2}w. $$Hence $$\colvec{x\\ y \\ z \\ w} = \colvec{\overline{0} \\ \overline{1} \\ \overline{0}\\ \overline{0}} + z\colvec{\overline{0} \\ \overline{0} \\ \overline{1}\\ \overline{0}} + w\colvec{\overline{0} \\ \overline{0} \\ \overline{0}\\ \overline{1}}. $$ This gives the $9$ solutions. \end{answer} \end{pro} \begin{pro} In $\BBZ_7$, given that $$ \begin{bmatrix} \overline{1} & \overline{2} & \overline{3} \cr \overline{2} & \overline{3} & \overline{1} \cr \overline{3} & \overline{1} & \overline{2} \cr \end{bmatrix}^{-1} = \begin{bmatrix} \overline{4} & \overline{2} & \overline{0} \cr \overline{2} & \overline{0} & \overline{4} \cr \overline{0} & \overline{4} & \overline{2} \cr \end{bmatrix},$$find all solutions of the system $$\overline{1}x + \overline{2}y + \overline{3}z = \overline{5};$$ $$\overline{2}x + \overline{3}y + \overline{1}z = \overline{6};$$ $$\overline{3}x + \overline{1}y + \overline{2}z = \overline{0}.$$ \begin{answer} We have $$ \begin{bmatrix} \overline{1} & \overline{2} & \overline{3} \cr \overline{2} & \overline{3} & \overline{1} \cr \overline{3} & \overline{1} & \overline{2} \cr \end{bmatrix}\begin{bmatrix}x\cr y \cr z \end{bmatrix} = \begin{bmatrix}\overline{5}\cr \overline{6}\cr \overline{0}\end{bmatrix},$$ Hence $$ \begin{bmatrix}x\cr y \cr z \end{bmatrix} = \begin{bmatrix} \overline{1} & \overline{2} & \overline{3} \cr \overline{2} & \overline{3} & \overline{1} \cr \overline{3} & \overline{1} & \overline{2} \cr \end{bmatrix}^{-1}\begin{bmatrix}\overline{5}\cr \overline{6}\cr \overline{0}\end{bmatrix} = \begin{bmatrix} \overline{4} & \overline{2} & \overline{0} \cr \overline{2} & \overline{0} & \overline{4} \cr \overline{0} & \overline{4} & \overline{2} \cr \end{bmatrix}\begin{bmatrix}\overline{5}\cr \overline{6}\cr \overline{0}\end{bmatrix}= \begin{bmatrix}\overline{4}\cr \overline{3} \cr\overline{3} \end{bmatrix} . $$ \end{answer} \end{pro} \begin{pro} Solve in $\BBZ_{13}$: $$ x-\overline{2}y+z=\overline{5}, \qquad \overline{2}x+\overline{2}y=\overline{7}, \qquad \overline{5}x-\overline{3}y +\overline{4}z=\overline{1}.$$ \begin{answer} The augmented matrix of the system is \begin{eqnarray*} \begin{bmat}{lll|l} \overline{1} & -\overline{2} & \overline{1} & \overline{5}\\ \overline{2} & \overline{2} & \overline{0} & \overline{7}\\ \overline{5} & -\overline{3} & \overline{4} & \overline{1}\\ \end{bmat} & \grstep[R_2-\overline{2}R_1\to R_2 ]{R_3-\overline{5}R_1\to R_3} & \begin{bmat}{lll|l} \overline{1} & -\overline{2} & \overline{1} & \overline{5}\\ \overline{0} & \overline{6} & -\overline{2} & -\overline{3}\\ \overline{0} & \overline{7} & -\overline{1} & \overline{2}\\ \end{bmat} \cr & \grstep{\overline{6}R_3-\overline{7}R_2\to R_3} & \begin{bmat}{lll|l} \overline{1} & -\overline{2} & \overline{1} & \overline{5}\\ \overline{0} & \overline{6} & -\overline{2} & -\overline{3}\\ \overline{0} & \overline{0} & \overline{8} & \overline{7}\\ \end{bmat} \cr \end{eqnarray*} Backward substitution yields $$\overline{8}z = \overline{7} \implies \overline{5}\cdot \overline{8}z = \overline{5}\cdot \overline{7} \implies z = \overline{35}=\overline{9}, $$ $$ \overline{6}y=\overline{2}z-\overline{3} =\overline{2}\cdot \overline{9}-\overline{3}= \overline{15}=\overline{2} \implies \overline{11}\cdot \overline{6}y = \overline{11}\cdot \overline{2}\implies y = \overline{22}=\overline{9}, $$ $$ x=\overline{2}y-\overline{1}z+\overline{5} =\overline{2}\cdot \overline{9}-\overline{1}\cdot \overline{9}+\overline{5}=\overline{14}=\overline{1}. $$ Conclusion : $$\psframebox{ x= \overline{1}, \qquad y = \overline{9}, \qquad z = \overline{9}.} $$ Check: $$\overline{1} - \overline{2}\cdot \overline{9} + \overline{9} = -\overline{8}\stackrel{\checkmark}{=} \overline{5}, $$ $$\overline{2}\cdot \overline{1} + \overline{2}\cdot \overline{9} = \overline{20}\stackrel{\checkmark}{=} \overline{7}, $$ $$\overline{5}\cdot \overline{1} - \overline{3}\cdot \overline{9} + \overline{4}\cdot \overline{9} = \overline{14}\stackrel{\checkmark}{=} \overline{1}. $$ \end{answer} \end{pro} \begin{pro} Find, with proof, a polynomial $p(x)$ with real number coefficients and degree $3$ such that $$p(-1) = -10, \quad p(0) = -1, \quad p(1) = 2, \quad p(2)= 23.$$ \begin{answer} We need to solve the system $$ a-b+c-d=p(-1)=-10, $$ $$ a=p(0)=-1, $$ $$ a+b+c+d=p(1)=2, $$ $$a+2b+4c+8d=p(2)=23. $$Using row reduction or otherwise, we find $a=-1$, $b=2$, $c=-3$, $d=4$, and so the polynomial is $$p(x) =4x^3-3x^2+2x-1.$$ \end{answer} \end{pro} \begin{pro} This problem introduces Hill block ciphers, which are a way of encoding information with an {\em encoding matrix} $A\in\mat{n\times n}{\BBZ _{26}}$, where $n$ is a strictly positive integer. Split a plaintext into blocks of $n$ letters, creating a series of $n\times 1$ matrices $P_k$, and consider the numerical equivalent ($A=0$, $B=1$, $C=2$, \ldots , $Z=25$) of each letter. The encoded message is the translation to letters of the $n\times 1$ matrices $C_k =AP_k \mod 26$. \bigskip For example, suppose you want to encode the message {\bf COMMUNISTS EAT OFFAL} with the encoding matrix $$A = \begin{bmatrix} 0 & 1 & 0 \cr 3 & 0 & 0 \cr 0 & 0 & 2 \cr \end{bmatrix}, $$a $3\times 3$ matrix. First, split the plaintext into groups of three letters: $$ {\bf COM} \ {\bf MUN} \ {\bf IST} \ {\bf SEA} \ {\bf TOF} \ {\bf FAL}. $$ Form $3\times 1$ matrices with each set of letters and find their numerical equivalent, for example, $$P_1=\colvec{{\bf C}\\ {\bf O}\\ {\bf M}} = \colvec{2\\ 14 \\ 12}. $$ Find the product $AP_1$ modulo $26$, and translate into letters: $$ AP_1= \begin{bmatrix} 0 & 1 & 0 \cr 3 & 0 & 0 \cr 0 & 0 & 2 \cr \end{bmatrix} \colvec{2\\ 14 \\ 12}= \colvec{14\\ 6 \\ 24} = \colvec{{\bf O}\\ {\bf G}\\ {\bf Y}},$$ hence {\bf COM} is encoded into {\bf OGY}. Your task is to complete the encoding of the message. \begin{answer} Using the encoding chart $$\begin{array}{|c||c||c||c||c||c||c||c||c||c||c||c|c|} \hline 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline A & B & C & D & E &F & G & H & I & J & K & L & M \\ \hline 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \\ \hline N & O & P & Q & R & S & T & U & V & W & X & Y & Z \\ \hline \end{array}$$ we find $$P_2=\colvec{{\bf M}\\ {\bf U}\\ {\bf N}} = \colvec{12\\ 20 \\ 13}, \quad P_3=\colvec{{\bf I}\\ {\bf S}\\ {\bf T}} = \colvec{8\\ 18 \\ 19}, \quad P_4=\colvec{{\bf S}\\ {\bf E}\\ {\bf A}} = \colvec{18\\ 4 \\ 0}, \quad P_5=\colvec{{\bf T}\\ {\bf O}\\ {\bf F}} = \colvec{19\\ 14 \\ 5}, \quad P_6=\colvec{{\bf F}\\ {\bf A}\\ {\bf L}} = \colvec{5\\ 0 \\ 11}. $$ Thus $$AP_2 = \colvec{20\\ 10 \\ 0} = \colvec{{\bf U}\\ {\bf K}\\ {\bf A}},\ AP_3 = \colvec{18\\ 24 \\ 12} = \colvec{{\bf S}\\ {\bf Y}\\ {\bf M}},\ AP_4 = \colvec{4\\ 2 \\ 0} = \colvec{{\bf E}\\ {\bf C}\\ {\bf A}},\ AP_5 = \colvec{14\\ 5 \\ 10} = \colvec{{\bf O}\\ {\bf F}\\ {\bf K}},\ AP_6 = \colvec{0\\ 15 \\ 22} = \colvec{{\bf A}\\ {\bf P}\\ {\bf W}}. $$ Finally, the message is encoded into $$ {\bf OGY} \ {\bf UKA} \ {\bf SYM} \ {\bf ECA} \ {\bf OFK} \ {\bf APW}. $$ \end{answer} \end{pro} \begin{pro} Find all solutions in $\BBZ _{103}$, if any, to the system $$ \begin{array}{rll} x_0+x_1 & = & 0, \\ x_0+x_2 & = & 1, \\ x_0+x_3 & = & 2, \\ \vdots & \vdots & \vdots \\ x_{0}+x_{100} & = & 99, \\ x_0+x_1 +x_2+\cdots + x_{100}& = & 4949. \\ \end{array} $$ Hints: $0+1+2+\cdots + 99= 4950$, \quad $99\cdot 77 -103\cdot 74= 1$. \begin{answer}Observe that since $103$ is prime, $\BBZ_{103}$ is a field. Adding the first hundred equations, $$100x_0 + x_1+x_2+\cdots + x_{100}=4950 \implies 99x_0 = 4950- 4949=1 \implies x_0 = 77 \mod 103. $$ Now, for $1\leq k \leq 100$, $$x_k = k-1 - x_0 = k-78=k+25. $$ This gives $$x_1=26, x_2=27, \ldots , x_{77}=102, x_{78}=0, x_{79}=1, x_{80}=2,\ldots , x_{100}=22. $$ \end{answer} \end{pro} \end{multicols} \section{Existence of Solutions} We now answer the question of deciding when a system of linear equations is solvable. \begin{lem}\label{lem:row_echelon_homo_system} Let $A\in\mat{m\times n}{ \BBF }$ be in row-echelon form, and let $X\in\mat{n\times 1}{ \BBF }$ be a matrix of variables. The homogeneous system $AX = {\bf 0}_{m\times 1}$ of $m$ linear equations in $n$ variables has (i) a unique solution if $m = n$, (ii) multiple solutions if $m < n$. \end{lem} \begin{pf} If $m = n$ then $A$ is a square triangular matrix whose diagonal elements are different from $0_{ \BBF }$. As such, it is invertible by virtue of Theorem \ref{thm:inverse_triangular_matrices}. Thus $$AX = {\bf 0}_{n\times 1} \implies X = A^{-1}{\bf 0}_{n\times 1} = {\bf 0}_{n\times 1}$$so there is only the unique solution $X = {\bf 0}_{n\times 1}$, called the {\em trivial solution}. \bigskip If $m < n$ then there are $n - m$ free variables. Letting these variables run through the elements of the field, we obtain multiple solutions. Thus if the field has infinitely many elements, we obtain infinitely many solutions, and if the field has $k$ elements, we obtain $k^{n - m}$ solutions. Observe that in this case there is always a non-trivial solution. \end{pf} \begin{thm} Let $A\in\mat{m\times n}{ \BBF }$, and let $X\in\mat{n\times 1}{ \BBF }$ be a matrix of variables. The homogeneous system $AX = {\bf 0}_{m\times 1}$ of $m$ linear equations in $n$ variables always has a non-trivial solution if $m < n$. \end{thm} \begin{pf} We can find a matrix $P\in\gl{m}{ \BBF }$ such that $B = PA$ is in row-echelon form. Now $$AX = {\bf 0}_{m\times 1} \iff PAX = {\bf 0}_{m\times 1} \iff BX = {\bf 0}_{m\times 1}. $$ That is, the systems $AX = {\bf 0}_{m\times 1}$ and $BX = {\bf 0}_{m\times 1}$ have the same set of solutions. But by Lemma \ref{lem:row_echelon_homo_system} there is a non-trivial solution. \end{pf} \begin{thm}[Kronecker-Capelli] \index{theorem!Kronecker-Capelli} Let $A\in\mat{m\times n}{ \BBF }, Y\in \mat{m\times 1}{ \BBF }$ be constant matrices and $X\in\mat{n\times 1}{ \BBF }$ be a matrix of variables. The matrix equation $AX = Y$ is solvable if and only if $$\rank{A} = \rank{[A|Y]}.$$ \end{thm} \begin{pf} Assume first that $AX = Y$, $$ X = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}. $$Let the columns of $[A|X]$ be denoted by $C_i, 1 \leq i \leq n$. Observe that that $[A|X]\in \mat{m\times (n+1)}{ \BBF }$ and that the $(n+1)$-th column of $[A|X]$ is $$C_{n + 1} = AX = \begin{bmatrix} x_1a_{11} + x_2a_{12} + \cdots + x_na_{1n}\\ x_1a_{21} + x_2a_{22} + \cdots + x_na_{2n}\\ \vdots \\ x_1a_{n1} + x_2a_{n2} + \cdots + x_na_{nn}\end{bmatrix} = \sum _{i=1} ^n x_iC_i. $$ By performing $C_{n+1} - \sum_{j = 1} ^nx_jC_j \rightarrow C_{n + 1} $ on $[A|Y] = [A|AX]$ we obtain $[A|0_{n\times 1}]$. Thus $\rank{[A|Y]} = \rank{[A|0_{n\times 1}]} = \rank{A}$. \bigskip Now assume that $r = \rank{A} = \rank{[A|Y]}$. This means that adding an extra column to $A$ does not change the rank, and hence, by a sequence column operations $[A|Y]$ is equivalent to $[A|0_{n\times 1}]$. Observe that none of these operations is a permutation of the columns, since the first $n$ columns of $[A|Y]$ and $[A|0_{n\times 1}]$ are the same. This means that $Y$ can be obtained from the columns $C_i, 1 \leq i \leq n$ of $A$ by means of transvections and dilatations. But then $$Y= \sum _{i = 1} ^nx_iC_i. $$The solutions sought is thus $$ X = \begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}. $$ \end{pf} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Let $A\in\mat{n\times p}{ \BBF }$, $B\in\mat{n\times q}{ \BBF }$ and put $C = [A\ \ B]\in\mat{n\times (p+q)}{ \BBF }$ Prove that $\rank{A} = \rank{C} \iff \exists P\in\mat{p}{q}$ such that $B = AP$. \end{pro} \end{multicols} \section{Examples of Linear Systems} \begin{exa} Use row reduction to solve the system $$\begin{array}{lllllllll} x & + & 2y & + & 3z & + & 4w & = & 8 \\ x & + & 2y & + & 4z & + & 7w & = & 12 \\ 2x & + & 4y & + & 6z & + & 8w & = & 16 \\ \end{array}$$ \end{exa} \begin{solu} Form the expanded matrix of coefficients and apply row operations to obtain \begin{eqnarray*} \begin{bmat}{llll|l} 1 & 2 & 3 & 4 & 8 \\ 1 & 2 & 4 & 7 & 12 \\ 2 & 4 & 6 & 8 & 16 \\ \end{bmat} & \grstep[R_2 - R_1 \rightarrow R_2]{R_3 - 2R_1 \rightarrow R_3} & \begin{bmat}{llll|l} 1 & 2 & 3 & 4 & 8 \\ 0 & 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmat} . \end{eqnarray*} The matrix is now in row-echelon form. The variables $x$ and $z$ are the pivots, so $w$ and $y$ are free. Setting $w = s, y = t$ we have $$z = 4 - 3s,$$$$ x = 8 - 4w - 3z - 2y = 8 - 4s - 3(4 - 3s) - 2t = -4 + 5s - 2t.$$ Hence the solution is given by $$\colvec{x\\ y\\ z\\ w} = \colvec{-4 + 5s - 2t\\ t\\ 4 - 3s\\ s}.$$ \end{solu} \begin{exa} Find $\alpha \in \BBR$ such that the system $$x + y - z = 1,$$ $$2x + 3y + \alpha z = 3,$$ $$x + \alpha y + 3z = 2,$$ posses (i) no solution, (ii) infinitely many solutions, (iii) a unique solution. \end{exa} \begin{solu}The augmented matrix of the system is $$\begin{augmatrix}{3} 1 & 1 & -1 & 1 \\ 2 & 3 & \alpha & 3 \\ 1 & \alpha & 3 & 2 \\ \end{augmatrix}.$$ By performing $R_2 - 2R_1 \rightarrow R_2$ and $R_3 - R_1 \rightarrow R_3$ we obtain $$\rightsquigarrow\begin{augmatrix}{3} 1 & 1 & -1 & 1 \\ 0 & 1 & \alpha + 2 & 1 \\ 0 & \alpha - 1 & 4 & 1 \end{augmatrix}.$$ By performing $R_3 - (\alpha - 1)R_2 \rightarrow R_3$ on this last matrix we obtain $$\rightsquigarrow\begin{augmatrix}{3} 1 & 1 & - 1 & 1 \\ 0 & 1 & \alpha + 2 & 1 \\ 0 & 0 & (\alpha - 2)(\alpha + 3) & \alpha - 2\end{augmatrix}.$$ If $\alpha = -3,$ we obtain no solution. If $\alpha = 2$, there is an infinity of solutions $$\colvec{x \\ y \\ z} = \colvec{5t \\ 1 - 4t \\ t}, \ \ \ t\in\BBR.$$ If $\alpha \neq 2$ and $\alpha \neq 3$, there is a unique solution $$\colvec{x \\ y \\ z} = \colvec{1 \\ \dfrac{1}{\alpha + 3} \\ \dfrac{1}{\alpha + 3}}.$$ \end{solu} \begin{exa} Solve the system $$\begin{bmatrix} \overline{6} & \overline{0}& \overline{1} \cr \overline{3} & \overline{2} & \overline{0} \cr \overline{1} & \overline{0} & \overline{1}\cr \end{bmatrix}\colvec{x\\ y\\ z} = \colvec{\overline{1} \\ \overline{0} \\ \overline{2}},$$ for $(x, y, z)\in (\BBZ_7)^3$. \end{exa}\begin{solu}Performing operations on the augmented matrix we have \begin{eqnarray*} \begin{bmat}{ccc|c} \overline{6} & \overline{0}& \overline{1} & \overline{1} \cr \overline{3} & \overline{2} & \overline{0} & \overline{0} \cr \overline{1} & \overline{0} & \overline{1}& \overline{2}\cr \end{bmat} & \grstep{R_1 \leftrightarrow R_3}& \begin{bmat}{ccc|c} \overline{1} & \overline{0} & \overline{1}& \overline{2}\cr \overline{3} & \overline{2} & \overline{0} & \overline{0} \cr \overline{6} & \overline{0}& \overline{1} & \overline{1} \cr \end{bmat} \\ & \grstep[R_2 - \overline{3}R_1 \rightarrow R_2]{R_3 - \overline{6}R_1 \rightarrow R_3} & \begin{bmat}{ccc|c} \overline{1} & \overline{0} & \overline{1}& \overline{2}\cr \overline{0} & \overline{2} & \overline{4} & \overline{1} \cr \overline{0} & \overline{0}& \overline{2} & \overline{3} \cr \end{bmat} \end{eqnarray*} This gives $$\overline{2}z = \overline{3} \implies z = \overline{5},$$ $$\overline{2}y = \overline{1} - \overline{4}z = \overline{2} \implies y = \overline{1},$$ $$x = \overline{2} - z = \overline{4}.$$ The solution is thus $$(x, y, z) = (\overline{4}, \overline{1}, \overline{5}).$$ \end{solu} \section*{\psframebox{Homework}} \begin{pro} Find the general solution to the system $$ \begin{bmatrix}1 & 1 & 1 & 1 & 1 \cr 1 & 0 & 1 & 0 & 1 \cr 2 & 1 & 2 & 1 & 2 \cr 4 & 2 & 4 & 2 & 4 \cr 1 & 0 & 0 & 0 & 1 \cr \end{bmatrix}\colvec{a \\ b \\ c \\ d \\ f} = \colvec{1 \\ -1 \\ 0 \\ 0 \\ 0} $$ or shew that there is no solution. \begin{answer} Observe that the third row is the sum of the first two rows and the fourth row is twice the third. So we have \begin{eqnarray*} \begin{bmat}{ccccc|c}1 & 1 & 1 & 1 & 1 & 1\cr 1 & 0 & 1 & 0 & 1 & -1\cr 2 & 1 & 2 & 1 & 2 & 0\cr 4 & 2 & 4 & 2 & 4 & 0\cr 1 & 0 & 0 & 0 & 1 & 0\cr \end{bmat} & \grstep[R_4 - 2R_1 - 2R_2 \rightarrow R_4]{R_3-R_1-R_2\rightarrow R_3} & \begin{bmat}{ccccc|c}1 & 1 & 1 & 1 & 1 & 1\cr 1 & 0 & 1 & 0 & 1& -1 \cr 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0\cr 1 & 0 & 0 & 0 & 1 & 0\cr \end{bmat}\\ & \grstep[R_1 - R_5 \rightarrow R_1]{R_2-R_5 \rightarrow R_2} & \begin{bmat}{ccccc|c}0 & 1 & 1 & 1 & 0 & 1\cr 0 & 0 & 1 & 0 & 0 & -1\cr 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0\cr 1 & 0 & 0 & 0 & 1 & 0\cr \end{bmat}\\ \end{eqnarray*} Rearranging the rows we obtain $$ \begin{bmat}{ccccc|c} 1 & 0 & 0 & 0 & 1 & 0\cr 0 & 1 & 1 & 1 & 0 & 1\cr 0 & 0 & 1 & 0 & 0 & -1\cr 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0\cr \end{bmat} . $$Hence $d$ and $f$ are free variables. We obtain $$c = -1,$$ $$ b = 1 -c -d = 2-d, $$ $$a = -f. $$ The solution is $$\colvec{a\\ b\\ c\\ d\\ f} = \colvec{0\\ 2\\ -1 \\ 0 \\ 0} + d\colvec{0 \\ -1 \\ 0 \\ 1 \\ 0} + f\colvec{-1\\ 0 \\ 0 \\ 0 \\ 1}. $$ \end{answer} \end{pro} \begin{pro} Find all solutions of the system $$ \begin{bmatrix}1 & 1 & 1 & 1 & 1 \cr 1 & 1 & 1 & 1 & 2 \cr 1 & 1 & 1 & 3 & 3 \cr 1 & 1 & 4 & 4 & 4 \cr 1 & 2 & 3 & 4 & 5 \cr \end{bmatrix}\colvec{a \\ b \\ c \\ d \\ f} = \colvec{3 \\ 4 \\ 7 \\ 6 \\ 9}, $$ if any. \begin{answer} The unique solution is $\colvec{1\\ 1\\ -1\\ 1\\ 1}$. \end{answer} \end{pro} \begin{pro} Study the system $$x + 2my + z = 4m;$$ $$2mx + y + z = 2;$$$$x + y + 2mz = 2m^2,$$with real parameter $m$. You must determine, with proof, for which $m$ this system has (i) no solution, (ii) exactly one solution, and (iii) infinitely many solutions. \begin{answer} The augmented matrix of the system is $$\begin{bmat}{ccc|c}2m & 1 & 1 & 2 \cr 1 & 2m & 1 & 4m \cr 1 & 1 & 2m & 2m^2 \cr \end{bmat}. $$ Performing $R_1 \leftrightarrow R_2$. $$\begin{bmat}{ccc|c} 1 & 2m & 1 & 4m \cr 2m & 1 & 1 & 2 \cr 1 & 1 & 2m & 2m^2 \cr \end{bmat}. $$Performing $R_2 \leftrightarrow R_3$. $$\begin{bmat}{ccc|c} 1 & 2m & 1 & 4m \cr 1 & 1 & 2m & 2m^2 \cr 2m & 1 & 1 & 2 \cr \end{bmat}. $$ Performing $R_2 - R_1 \rightarrow R_1$ and $R_3 - 2mR_1 \rightarrow R_3$ we obtain $$\begin{bmat}{ccc|c} 1 & 2m & 1 & 4m \cr 0 & 1 - 2m & 2m - 1 & 2m^2 - 4m \cr 0 & 1 - 4m^2 & 1 - 2m & 2 - 8m^2 \cr \end{bmat}. $$ If $m = \frac{1}{2}$ the matrix becomes $$ \begin{bmat}{ccc|c} 1 & 1 & 1 & 2 \cr 0 & 0 & 0 & -\frac{3}{2} \cr 0 & 0 & 0 & 0 \cr \end{bmat} $$ and hence it does not have a solution. If $m \neq \frac{1}{2}$, by performing $\frac{1}{1 - 2m}R_2 \rightarrow R_2$ and $\frac{1}{1 - 2m}R_3 \rightarrow R_3$, the matrix becomes $$\begin{bmat}{ccc|c} 1 & 2m & 1 & 4m \cr 0 & 1 & - 1 & \frac{2m(m - 2)}{1 - 2m} \cr 0 & 1 + 2m & 1 & 2(1 + 2m) \cr \end{bmat}. $$ Performing $R_3 - (1 + 2m)R_2 \rightarrow R_3$ we obtain $$\begin{bmat}{ccc|c} 1 & 2m & 1 & 4m \cr 0 & 1 & - 1 & \frac{2m(m - 2)}{1 - 2m} \cr 0 & 0 & 2 + 2m & \frac{2(1 + 2m)(1 - m^2)}{1 -2m} \cr \end{bmat}. $$ If $m = -1$ then the matrix reduces to $$\begin{bmat}{ccc|c} 1 & -2 & 1 & -4 \cr 0 & 1 & - 1 & 2 \cr 0 & 0 & 0 & 0 \cr \end{bmat}. $$The solution in this case is $$\begin{bmatrix} x \cr y \cr z \cr \end{bmatrix} = \begin{bmatrix} z \cr 2 + z \cr z \end{bmatrix}. $$ If $m \neq -1, m \neq -\frac{1}{2}$ we have the solutions $$\begin{bmatrix} x \cr y \cr z \cr \end{bmatrix} = \begin{bmatrix} \frac{m - 1}{1 - 2m} \cr \frac{1 - 3m}{1 - 2m} \cr \frac{(1 + 2m)(1 - m)}{1 - 2m} \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Study the following system of linear equations with parameter $a$. $$(2a - 1)x + ay - (a + 1)z = 1,$$ $$ ax + y - 2z = 1, $$ $$2x + (3 - a)y + (2a - 6)z = 1.$$ You must determine for which $a$ there is: (i) no solution, (ii) a unique solution, (iii) infinitely many solutions. \begin{answer} By performing the elementary row operations, we obtain the following triangular form: $$ ax + y - 2z = 1,$$ $$ (a - 1)^2y + (1 - a)(a - 2)z = 1 - a,$$ $$(a - 2)z = 0. $$ If $a = 2,$ there is an infinity of solutions: $$\colvec{x\\ y\\ z} = \colvec{1 + t\\ -1\\ t} \ t\in\BBR.$$ Assume $a \neq 2$. Then $z = 0$ and the system becomes $$ax + y = 1,$$ $$ (a - 1)^2y = 1 - a,$$ $$2x + (3 - a)y = 1.$$ We see that if $a = 1$, the system becomes $$x + y = 1,$$ $$2x + 2y = 1,$$and so there is no solution. If $(a - 1)(a - 2) \neq 0$, we obtain the unique solution $$\colvec{x\\ y \\ z} = \colvec{\frac{1}{a - 1}\\ -\frac{1}{a - 1}\\ 0}.$$ \end{answer} \end{pro} \begin{pro} Determine the values of the parameter $m$ for which the system $$\begin{array}{lllllll} x &+& y &+& (1-m)z &=& m+2 \cr (1+m)x &-& y &+& 2z &=& 0 \cr 2x &-& my &+& 3z &=& m+2 \cr \end{array} $$ is solvable. \begin{answer} The system is solvable if $m \neq 0, m \neq \pm 2$. If $m \neq 2$ there is the solution $\colvec{x\\ y \\ z} = \colvec{\frac{1}{m-2} \\ \frac{m+3}{m-2} \\ \frac{m+2}{m-2} }.$ \end{answer} \end{pro} \begin{pro} Determine the values of the parameter $m$ for which the system $$\begin{array}{lllllllll} x &+& y &+& z &+& t &=& 4a \cr x &-& y &-& z &+& t &=& 4b \cr -x &-& y &+& z &+& t &=& 4c \cr x &-& y &+&z &-&t &=& 4d \cr \end{array} $$ is solvable. \begin{answer} There is the unique solution $\colvec{x \\ y \\ z\\ t} = \colvec{a + d + b - c\\ -c - d - b + a\\ d + c - b + a\\ c - d + b+ a}. $ \end{answer} \end{pro} \begin{pro} It is known that the system $$ay + bx = c;$$ $$cx + az = b;$$ $$bz + cy = a$$possesses a unique solution. What conditions must $(a, b, c)\in\BBR^3$ fulfill in this case? Find this unique solution. \begin{answer} The system can be written as $$ \begin{bmatrix} b & a & 0 \cr c & 0 & a \cr 0 & c & b \cr \end{bmatrix}\begin{bmatrix} x \cr y \cr z \end{bmatrix} = \begin{bmatrix} c \cr b \cr a \end{bmatrix}. $$ The system will have the unique solution $$\begin{array}{lll} \begin{bmatrix} x \cr y \cr z \end{bmatrix} & = & \begin{bmatrix} b & a & 0 \cr c & 0 & a \cr 0 & c & b \cr \end{bmatrix}^{-1} \begin{bmatrix} c \cr b \cr a \cr \end{bmatrix}\vspace{3mm}\\ & = &\begin{bmatrix} \frac{1}{2b} & \frac{1}{2c} & -\frac{a}{2bc} \cr \frac{1}{2a} & -\frac{b}{2ac} & \frac{1}{2c} \cr -\frac{c}{2ba} & \frac{1}{2a} & \frac{1}{2b}\cr \end{bmatrix} \begin{bmatrix} c \cr b \cr a \cr \end{bmatrix}\vspace{3mm}\\ & = & \begin{bmatrix} \dfrac{b^2 + c^2 - a^2}{2bc} \cr \dfrac{a^2 + c^2 - b^2}{2ac}\cr \dfrac{a^2 + b^2 - c^2}{2ab}\cr\end{bmatrix}\end{array}, $$as long as the inverse matrix exists, which is as long as $abc\neq 0$ \end{answer} \end{pro} \begin{pro} For which values of the real parameter $a$ does the following system have (i) no solutions, (ii) exactly one solution, (iii) infinitely many solutions? $$ \begin{array}{rrrrrrr} (1-a)x & + & (2a+1)y & + & (2a+2)z & = & a, \\ ax & + & ay & & & = & 2a+2, \\ 2x & + & (a+1)y & + & (a-1)z & = & a^2-2a+9. \\ \end{array} $$ \begin{answer} We first form the augmented matrix, \begin{eqnarray*} \begin{bmat}{lll|l} 1-a & 2a+1 & 2a+2 & a \\ a & a & 0 & 2a+2 \\ 2 & a+1 & a-1 & a^2-2a+9 \\ \end{bmat} & \grstep{R_1 + R_2 \rightarrow R_1} & \begin{bmat}{lll|l} 1 & 3a+1 & 2a+2 & 3a+2 \\ a & a & 0 & 2a+2 \\ 2 & a+1 & a-1 & a^2-2a+9 \\ \end{bmat} \\ & \grstep[R_3-2R_1\rightarrow R_3]{R_2 - aR_1 \rightarrow R_2} & \begin{bmat}{lll|l} 1 & 3a+1 & 2a+2 & 3a+2 \\ 0 & -3a^2 & -2a^2-2a & -3a^2+2 \\ 0 & -5a-1 & -3a-5 & a^2-8a+5 \\ \end{bmat}\\ . \end{eqnarray*} After $(-5a-1)R_2 +3a^2R_3 \rightarrow R_2$, this las matrix becomes $$ \begin{bmat}{lll|l} 1 & 3a+1 & 2a+2 & 3a+2 \\ 0 & 0 & a^3-3a^2+2a & 3a^4-9a^3+18a^2-10a-2 \\ 0 & -5a-1 & -3a-5 & a^2-8a+5 \\ \end{bmat}. $$Exchanging the last two rows and factoring, $$ \begin{bmat}{lll|l} 1 & 3a+1 & 2a+2 & 3a+2 \\ 0 & -5a-1 & -3a-5 & a^2-8a+5 \\ 0 & 0 & a(a-1)(a-2) & (a-1)(3a^3-6a^2+12a+2) \\ \end{bmat}. $$ Thus we must examine $a\in \{1,2,3\}$ and $a\not\in \{0,1,2\}$. \bigskip Clearly, if $a(a-1)(a-2)\neq 0$, then there is the unique solution $$ \left\{ z=\frac {2+12a-6a^{2}+3a^{3}}{a \left( a-2 \right) },\qquad y=-\frac {2a^{3}-3\,a^{2}+6a+10}{a \left( a-2 \right) },\qquad x={\frac {2\,a^{3}-a^{2}+4a+6}{a \left( a-2 \right) }} \right\}. $$ If $a=0$, the system becomes $$x+y+2z=0, \quad 0 = 2, \quad 2x+y-z=0, $$ which is inconsistent (no solutions). \bigskip If $a=1$, the system becomes $$ 3y+4z=1, \qquad x+y=1, \qquad 2x+2y=8, $$ which has infinitely many solutions, $$\left\{ y=\dfrac{1}{3}-\dfrac{4}{3}z,\qquad x=\dfrac{2}{3}+\dfrac{4}{3}z,\qquad z=z \right\}. $$ \bigskip If $a=2$, the system becomes $$ -x+5y+6z=2, \quad 2x+2y=6, \qquad 2x+3y + z = 9, $$ which is also inconsistent, as can be seen by observing that $$(-x+5y+6z)-6(2x+3y + z)=2-18 \implies -13x-13y=-18, $$which contradicts the equation $2x+2y=6$. \end{answer} \end{pro} \begin{pro} Find strictly positive real numbers $x, y, z$ such that $$\begin{array}{lcl} x^3y^2z^6 &=& 1 \\ x^4y^5z^{12} &=& 2 \\ x^2y^2z^5 &=& 3.\\ \end{array} $$ \begin{answer} $$\begin{array}{lcl}x &=& 2^{-2} 3^6 \\ y &=& 2^{-3}3^{12} \\ z &=& 2^2 3^{-7}. \\ \end{array}$$ \end{answer} \end{pro} \begin{pro}[Leningrad Mathematical Olympiad, 1987, Grade 5] The numbers $1$, $2$, $\ldots$ , $16$ are arranged in a $4\times 4$ matrix $A$ as shewn below. We may add $1$ to all the numbers of any row or subtract $1$ from all numbers of any column. Using only the allowed operations, how can we obtain $A^T$? $$ A = \begin{bmatrix} 1 & 2 & 3 & 4 \cr 5 & 6 & 7 & 8 \cr 9 & 10 & 11 & 12 \cr 13 & 14 & 15 & 16 \cr \end{bmatrix} $$ \begin{answer} Denote the addition operations applied to the rows by $a_1$, $a_2$, $a_3$, $a_4$ and the subtraction operations to the columns by $b_1$, $b_2$, $b_3$, $b_4.$ Comparing $A$ and $A^T$ we obtain $7$ equations in $8$ unknowns. By inspecting the diagonal entries, and the entries of the first row of $A$ and $A^T$, we deduce the following equations $$a_1 = b_1,$$ $$a_2 = b_2,$$$$a_3 = b_3,$$$$a_4 = b_4,$$$$a_1 - b_2 = 3,$$ $$a_1 - b_3 = 6,$$ $$a_1 - b_4 = 9.$$This is a system of $7$ equations in $8$ unknowns. We may let $a_4 = 0$ and thus obtain $a_1 = b_1 = 9,$ $a_2 = b_2 = 6,$ $a_3 = b_3 = 3,$ $a_4 = b_4 = 0.$ \end{answer} \end{pro} \begin{pro}[International Mathematics Olympiad, 1963] Find all solutions $x_1, x_2, x_3, x_4, x_5$ of the system $$x_5 + x_2 = yx_1;$$$$x_1 + x_3 = yx_2;$$$$x_2 + x_4 = yx_3;$$$$x_3 + x_5 = yx_4;$$$$x_4 + x_1 = yx_5,$$where $y$ is a parameter.\begin{answer} The augmented matrix of this system is $$ \begin{bmat}{ccccc|c} -y & 1 & 0 & 0 & 1 & 0\cr 1 & -y & 1 & 0 & 0 & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 1 & 0 & 0 & 1 & -y & 0 \cr \end{bmat}. $$ Permute the rows to obtain $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 1 & -y & 1 & 0 & 0 & 0 \cr -y & 1 & 0 & 0 & 1 & 0\cr \end{bmat}.$$ Performing $R_5 + yR_1 \rightarrow R_5$ and $R_4 - R_1 \rightarrow R_4$ we get $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 0 & -y & 1 & -1 & y & 0 \cr 0 & 1 & 0 & y & 1 - y^2& 0\cr \end{bmat}. $$ Performing $R_5 - R_2 \rightarrow R_5$ and $R_4 + yR_2 \rightarrow R_4$ we get $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 0 & 0 & 1 - y^2 & y-1 & y & 0 \cr 0 & 0 & y & y - 1 & 1 - y^2& 0\cr \end{bmat}. $$ Performing $R_5 - yR_3 \rightarrow R_5$ and $R_4 + (y^2 -1)R_3 \rightarrow R_4$ we get $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 0 & 0 & 0 & -y^3 + 2y-1 & y^2 + y - 1 & 0 \cr 0 & 0 & 0 & y^2 + y - 1 & 1 - y - y^2& 0\cr \end{bmat}. $$ Performing $R_5 + R_4\rightarrow R_5$ we get $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 0 & 0 & 0 & -y^3 + 2y-1 & y^2 +y - 1 & 0 \cr 0 & 0 & 0 & -y^3 +y^2 + 3y - 2 & 0& 0\cr \end{bmat}. $$Upon factoring, the matrix is equivalent to $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 0 & 0 & 0 & -(y-1)(y^2+y-1) & y^2 +y - 1 & 0 \cr 0 & 0 & 0 & -(y-2)(y^2+y-1) & 0& 0\cr \end{bmat}. $$ Thus $(y-2)(y^2+y-1)x_4 = 0$. If $y = 2$ then the system reduces to $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -2 & 0 \cr 0 & 1 & -2 & 1 & 0 & 0 \cr 0 & 0 & 1 & -2 & 1 & 0 \cr 0 & 0 & 0 & -5 & 5 & 0 \cr 0 & 0 & 0 & 0 & 0& 0\cr \end{bmat}. $$ In this case $x_5$ is free and by backwards substitution we obtain $$\begin{bmatrix} x_1 \cr x_2 \cr x_3 \cr x_4 \cr x_5\end{bmatrix} = \begin{bmatrix} t \cr t \cr t \cr t \cr t\end{bmatrix}, \ \ \ t\in\BBR.$$ If $y^2 + y - 1 = 0$ then the system reduces to $$ \begin{bmat}{ccccc|c}1 & 0 & 0 & 1 & -y & 0 \cr 0 & 1 & -y & 1 & 0 & 0 \cr 0 & 0 & 1 & -y & 1 & 0 \cr 0 & 0 & 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 & 0& 0\cr \end{bmat}. $$ In this case $x_4, x_5$ are free, and $$\begin{bmatrix} x_1 \cr x_2 \cr x_3 \cr x_4 \cr x_5\end{bmatrix} = \begin{bmatrix} yt - s \cr y^2s - yt - s \cr ys - t \cr s \cr t\end{bmatrix}, \ \ \ (s,t)\in\BBR^2.$$ Since $y^2s - s = (y^2 + y - 1)s - ys$, this last solution can be also written as $$\begin{bmatrix} x_1 \cr x_2 \cr x_3 \cr x_4 \cr x_5\end{bmatrix} = \begin{bmatrix} yt - s \cr -ys - yt \cr ys - t \cr s \cr t\end{bmatrix}, \ \ \ (s,t)\in\BBR^2.$$ Finally, if $(y - 2)(y^2 + y - 1) \neq 0$, then $x_4 = 0$, and we obtain $$\begin{bmatrix} x_1 \cr x_2 \cr x_3 \cr x_4 \cr x_5\end{bmatrix} = \begin{bmatrix} 0 \cr 0 \cr 0 \cr 0 \cr 0\end{bmatrix}.$$ \end{answer} \end{pro} \chapter{Vector Spaces} \section{Vector Spaces} \begin{df} A {\em vector space} $\vecspace{V}{+}{\cdot}{ \BBF }$ over a field $\field{ \BBF }{+}{}$ is a non-empty set $V$ whose elements are called {\em vectors}, possessing two operations $+$ (vector addition), and $\cdot$ (scalar multiplication) which satisfy the following axioms. \index{vector space}\index{vectors!in a vector space} $$\forall (\v{a}, \v{b}, \v{c}) \in V^3,\ \forall (\alpha, \beta)\in \BBF ^2,$$ \begin{enumerate} \item[VS1] {\bf Closure\ under \ vector\ addition} : \begin{equation}\v{a} + \v{b}\in V ,\label{vs:closure_under_addtion}\end{equation} \item[VS2] {\bf Closure\ under \ scalar\ multiplication} \begin{equation}\alpha\v{a} \in V , \label{vs:closure_under_scalar_multiplication} \end{equation} \item[VS3] {\bf Commutativity} \begin{equation} \v{a} + \v{b} = \v{b} + \v{a} \label{vs:commutativity}\end{equation} \item[VS4] {\bf Associativity}\begin{equation} {(\v{a} + \v{b}) + \v{c} = \v{a} + (\v{b} + \v{c})} \label{vs:associativity}\end{equation} \item[VS5] {\bf Existence\ of\ an\ additive\ identity} \begin{equation}\exists \ \v{0}\in V: \v{a} + \v{0} = \v{a} + \v{0}= \v{a} \label{vs:additive_identity}\end{equation} \item[VS6] {\bf Existence\ of\ additive\ inverses} \begin{equation} \exists \ -\v{a}\in V: \v{a} + (-\v{a})=(-\v{a}) + \v{a} = \v{0} \label{vs:additive_inverses}\end{equation} \item[VS7] {\bf Distributive\ Law} \begin{equation} \alpha(\v{a} + \v{b}) = \alpha\v{a} + \alpha\v{b} \label{vs:distributive_law_1}\end{equation} \item[VS8]{\bf Distributive\ Law} \begin{equation}(\alpha + \beta)\v{a} = \alpha \v{a} + \beta\v{a} \label{vs:distributive_law_2} \end{equation} \item[VS9] \begin{equation}1_{\BBF }\v{a} = \v{a} \label{vs:1v_is_v}\end{equation} \item[VS10] \begin{equation} (\alpha \beta) \v{a} = \alpha (\beta \v{a}) \label{vs:associative_scalar_product} \end{equation} \end{enumerate} \end{df} \begin{exa} If $n$ is a positive integer, then $\vecspace{\BBF ^n}{+}{\cdot}{ \BBF }$ is a vector space by defining $$(a_1, a_2, \ldots ,a_n) + (b_1, b_2, \ldots ,b_n) = (a_1 + b_1, a_2 + b_2, \ldots ,a_n + b_n),$$ $$\lambda(a_1, a_2, \ldots ,a_n) = (\lambda a_1, \lambda a_2, \ldots , \lambda a_n).$$ In particular, $\vecspace{\BBZ^2 _2}{+}{\cdot}{\BBZ_2}$ is a vector space with only four elements and we have seen the two-dimensional and tridimensional spaces $\vecspace{\BBR^2}{+}{\cdot}{\BBR}$ and $\vecspace{\BBR^3}{+}{\cdot}{\BBR}$. \end{exa} \begin{exa} $\vecspace{\mat{m\times n}{ \BBF }}{+}{\cdot}{ \BBF }$ is a vector space under matrix addition and scalar multiplication of matrices. \end{exa} \begin{exa} If $$\BBF[x] = \{a_0 + a_1x + a_2x + \cdots + a_nx^n: a_i\in \BBF, \ \ n\in \BBN\}$$ denotes the set of polynomials with coefficients in a field $\field{ \BBF }{+}{}$ then $\vecspace{\BBF[x]}{+}{\cdot}{ \BBF }$ is a vector space, under polynomial addition and scalar multiplication of a polynomial. \end{exa} \begin{exa} If $$\BBF_n[x] = \{a_0 + a_1x + a_2x + \cdots + a_kx^k: a_i\in \BBF, \ \ n\in \BBN, k \leq n\}$$ denotes the set of polynomials with coefficients in a field $\field{ \BBF }{+}{}$ and degree at most $n$, then $\vecspace{\BBF_n[x]}{+}{\cdot}{ \BBF }$ is a vector space, under polynomial addition and scalar multiplication of a polynomial. \end{exa} \begin{exa} Let $k\in\BBN$ and let $C^k(\BBR^{[a; b]})$ denote the set of $k$-fold continuously differentiable real-valued functions defined on the interval $[a; b]$. Then $C^k(\BBR^{[a; b]})$ is a vector space under addition of functions and multiplication of a function by a scalar. \end{exa} \begin{exa} Let $p \in ]1;+\infty[$. Consider the set of sequences $\{a_n\}_{n = 0} ^\infty, \ \ a_n\in\BBC$, $$ l^p =\left\{\{a_n\}_{n = 0} ^\infty: \sum _{n = 0} ^\infty |a_n|^p < + \infty\right\}.$$ Then $l^p$ is a vector space by defining addition as termwise addition of sequences and scalar multiplication as termwise multiplication: $$ \{a_n\}_{n = 0} ^\infty + \{b_n \}_{n = 0} ^\infty= \{(a_n + b_n)\}_{n = 0} ^\infty , $$ $$ \lambda \{a_n \}_{n = 0} ^\infty = \{\lambda a_n\}_{n = 0} ^\infty , \ \ \lambda\in\BBC. $$ All the axioms of a vector space follow trivially from the fact that we are adding complex numbers, except that we must prove that in $l^p$ there is closure under addition and scalar multiplication. Since $ \sum _{n = 0} ^\infty |a_n|^p < + \infty \implies \sum _{n = 0} ^\infty |\lambda a_n|^p < + \infty$ closure under scalar multiplication follows easily. To prove closure under addition, observe that if $z\in \BBC$ then $|z| \in \BBR_+$ and so by the Minkowski Inequality Theorem \ref{thm:minkowski_inequality} we have \begin{equation}\begin{array}{lll}\left( \sum _{n = 0} ^N |a_n + b_n|^p \right)^{1/p} & \leq & \left(\sum _{n = 0} ^N |a_n|^p \right)^{1/p}+ \left(\sum _{n = 0} ^N |b_n|^p \right)^{1/p} \\ & \leq & \left(\sum _{n = 0} ^\infty |a_n|^p\right)^{1/p} + \left(\sum _{n = 0} ^\infty |b_n|^p\right)^{1/p}.\end{array} \label{eq:lp_1} \end{equation} This in turn implies that the series on the left in (\ref{eq:lp_1}) converges, and so we may take the limit as $N\rightarrow +\infty$ obtaining \begin{equation} \left(\sum _{n = 0} ^\infty |a_n + b_n|^p \right)^{1/p} \leq \left(\sum _{n = 0} ^\infty |a_n|^p\right)^{1/p} + \left(\sum _{n = 0} ^\infty |b_n|^p\right)^{1/p}.\label{eq:lp_2}\end{equation} Now (\ref{eq:lp_2}) implies that the sum of two sequences in $l^p$ is also in $l^p$, which demonstrates closure under addition. \end{exa} \begin{exa} The set $$V = \{a + b\sqrt{2} + c\sqrt{3}: (a, b, c) \in \BBQ^3\}$$ with addition defined as $$(a + b\sqrt{2} + c\sqrt{3}) + (a' + b'\sqrt{2} + c'\sqrt{3}) = (a + a') + (b + b')\sqrt{2} + (c + c')\sqrt{3},$$and scalar multiplication defined as $$\lambda (a + b\sqrt{2} + c\sqrt{3}) = (\lambda a) + (\lambda b)\sqrt{2} + (\lambda c)\sqrt{3},$$constitutes a vector space over $\BBQ$. \end{exa} \begin{thm} In any vector space $\vecspace{V}{+}{\cdot}{ \BBF }$, $$\forall \ \ \alpha \in \BBF, \ \ \ \alpha\v{0} = \v{0}. $$ \end{thm} \begin{pf} We have $$\alpha\v{0} = \alpha(\v{0} + \v{0}) = \alpha\v{0} + \alpha\v{0}.$$Hence $$\alpha\v{0} - \alpha\v{0} = \alpha\v{0},$$or $$\v{0} = \alpha\v{0},$$proving the theorem. \end{pf} \begin{thm} In any vector space $\vecspace{V}{+}{\cdot}{ \BBF }$, $$\forall \ \ \v{v} \in V, \ \ \ 0_{\BBF }\v{v} = \v{0}. $$ \label{thm:zeroscalar_times_vector}\end{thm} \begin{pf} We have $$0_{\BBF }\v{v} = (0_{\BBF } + 0_{\BBF })\v{v} = 0_{\BBF }\v{v} + 0_{\BBF }\v{v}.$$Therefore $$0_{\BBF }\v{v} - 0_{\BBF }\v{v} = 0_{\BBF }\v{v},$$or $$\v{0} = 0_{\BBF }\v{v},$$proving the theorem. \end{pf} \begin{thm} In any vector space $\vecspace{V}{+}{\cdot}{ \BBF }$, $\alpha\in \BBF , \ \ \v{v} \in V,$ $$\alpha\v{v}= \v{0} \ \ \ \implies\ \ \ \alpha = 0_{\BBF } \ \ \ \vee\ \ \ \v{v} = \v{0}.$$ \end{thm} \begin{pf} Assume that $\alpha \neq 0_{\BBF }$. Then $\alpha$ possesses a multiplicative inverse $\alpha^{-1}$ such that $\alpha^{-1}\alpha = 1_{\BBF }$. Thus $$\alpha\v{v} = \v{0} \implies \alpha^{-1}\alpha\v{v} = \alpha^{-1}\v{0}.$$By Theorem \ref{thm:zeroscalar_times_vector}, $\alpha^{-1}\v{0} = \v{0}$. Hence $$\alpha^{-1}\alpha\v{v} = \v{0}.$$Since by Axiom \ref{vs:1v_is_v}, we have $\alpha^{-1}\alpha\v{v} = 1_{\BBF }\v{v} = \v{v}$, and so we conclude that $\v{v} = \v{0}$. \end{pf} \begin{thm} In any vector space $\vecspace{V}{+}{\cdot}{ \BBF }$, $$\forall \alpha \in \BBF, \ \ \ \forall \ \ \v{v} \in V, \ \ \ (-\alpha)\v{v} = \alpha(-\v{v}) = -(\alpha\v{v}). $$ \end{thm} \begin{pf} We have $$0_{\BBF }\v{v} = (\alpha + (-\alpha))\v{v} = \alpha \v{v} + (-\alpha) \v{v},$$whence $$-(\alpha \v{v}) + 0_{\BBF } \v{v} = (-\alpha) \v{v},$$that is$$-(\alpha \v{v}) = (-\alpha) \v{v}.$$Similarly, $$\v{0} = \alpha (\v{v} - \v{v}) = \alpha\v{v} + \alpha(-\v{v}),$$whence $$-(\alpha \v{v}) + \v{0} = \alpha (-\v{v}),$$that is$$-(\alpha \v{v})= \alpha (-\v{v}),$$proving the theorem. \end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Is $\BBR^2$ with vector addition and scalar multiplication defined as $$\colvec{x_1 \\ x_2 } + \colvec{ y_1 \\ y_2} = \colvec{x_1 + y_1 \\ x_2 + y_2}, \ \ \ \lambda\colvec{x_1\\ x_2} = \colvec{\lambda x_1 \\ 0} $$a vector space? \begin{answer} No, since $1_{\BBF }\v{v} = \v{v}$ is not fulfilled. For example $$1\cdot \colvec{1\\ 1} = \colvec{1\cdot 1 \\ 0 } \neq \colvec{1\\ 1}. $$ \end{answer} \end{pro} \begin{pro} Demonstrate that the commutativity axiom \ref{vs:commutativity} is redundant. \begin{answer} We expand $(1_{\BBF } + 1_{\BBF })(\v{a} + \v{b})$ in two ways, first using \ref{vs:distributive_law_1} first and then \ref{vs:distributive_law_2}, obtaining $$(1_{\BBF } + 1_{\BBF })(\v{a} + \v{b}) = (1_{\BBF } + 1_{\BBF })\v{a} + (1_{\BBF } + 1_{\BBF }) \v{b} = \v{a} + \v{a} + \v{b} + \v{b},$$and then using \ref{vs:distributive_law_2} first and then \ref{vs:distributive_law_1}, obtaining $$(1_{\BBF } + 1_{\BBF })(\v{a} + \v{b}) = 1_{\BBF }(\v{a} + \v{b}) + 1_{\BBF }(\v{a} + \v{b})= \v{a} + \v{b} + \v{a} + \v{b}.$$ We thus have the equality $$\v{a} + \v{a} + \v{b} + \v{b} = \v{a} + \v{b} + \v{a} + \v{b}.$$Cancelling $\v{a}$ from the left and $\v{b}$ from the right, we obtain $$\v{a} + \v{b} = \v{b} + \v{a},$$which is what we wanted to shew. \end{answer} \end{pro} \begin{pro} Let $V = \BBR^+ = ]0; +\infty[$, the positive real numbers and $F = \BBR$, the real numbers. Demonstrate that $V$ is a vector space over $\BBF$ if vector addition is defined as $a \oplus b = ab$, $(a, b)\in (\BBR^+)^2$ and scalar multiplication is defined as $\alpha \otimes a = a^{\alpha}$, $(\alpha, a)\in (\BBR, \BBR^+)$. \begin{answer} We must prove that each of the axioms of a vector space are satisfied. Clearly if $(x,y, \alpha)\in\BBR^+\times\BBR^+\times \BBR$ then $x\oplus y = xy > 0$ and $\alpha\otimes x = x^\alpha > 0$, so $V$ is closed under vector addition and scalar multiplication. Commutativity and associativity of vector addition are obvious. \bigskip Let $A$ be additive identity. Then we need $$x\oplus A = x \implies xA = x \implies A = 1.$$Thus the additive identity is $1$. Suppose $I$ is the additive inverse of $x$. Then $$x\oplus I = 1 \implies xI = 1 \implies I = \dfrac{1}{x}. $$Hence the additive inverse of $x$ is $\dfrac{1}{x}$. \bigskip Now $$\alpha \otimes (x\oplus y) = (xy)^{\alpha} = x^{\alpha}y^{\alpha} = x^{\alpha} \oplus y^{\alpha} = (\alpha \otimes x)\oplus (\alpha \otimes y), $$ and $$(\alpha + \beta)\otimes x = x^{\alpha + \beta} = x^\alpha x^{\beta} = (x^\alpha)\oplus (x^{\beta}) = (\alpha \otimes x) \oplus (\beta \otimes x), $$whence the distributive laws hold. \bigskip Finally, $$1\otimes x = x^1 = x, $$and $$\alpha \otimes (\beta \otimes x) = (\beta \otimes x)^\alpha = (x^\beta)^\alpha = x^{\alpha\beta} = (\alpha\beta)\otimes x, $$and the last two axioms also hold. \end{answer} \end{pro} \begin{pro} Let $\BBC$ denote the complex numbers and $\BBR$ denote the real numbers. Is $\BBC$ a vector space over $\BBR$ under ordinary addition and multiplication? Is $\BBR$ a vector space over $\BBC$? \begin{answer} $\BBC$ is a vector space over $\BBR$, the proof is trivial. But $\BBR$ is not a vector space over $\BBC$, since, for example taking $i$ as a scalar (from $\BBC$) and $1$ as a vector (from $\BBR$) the scalar multiple $i\cdot 1 = i \not\in\BBR$ and so there is no closure under scalar multiplication. \end{answer} \end{pro} \begin{pro} Construct a vector space with exactly $8$ elements. \begin{answer} One example is $$(\BBZ_2)^3 = \left\{\colvec{\overline{0}\\ \overline{0} \\ \overline{0}}, \colvec{\overline{0}\\ \overline{0} \\ \overline{1}}, \colvec{\overline{0}\\ \overline{1} \\ \overline{0}}, \colvec{\overline{0}\\ \overline{1} \\ \overline{1}}, \colvec{\overline{1}\\ \overline{0} \\ \overline{0}}, \colvec{\overline{1}\\ \overline{0} \\ \overline{1}}, \colvec{\overline{1}\\ \overline{1} \\ \overline{0}}, \colvec{\overline{1}\\ \overline{1} \\ \overline{1}}\right\}. $$Addition is the natural element-wise addition and scalar multiplication is ordinary element-wise scalar multiplication. \end{answer} \end{pro} \begin{pro} Construct a vector space with exactly $9$ elements. \begin{answer} One example is $$(\BBZ_3)^2 = \left\{\colvec{\overline{0}\\ \overline{0}}, \colvec{ \overline{0} \\ \overline{1}}, \colvec{\overline{0}\\ \overline{2}}, \colvec{\overline{1}\\ \overline{0}}, \colvec{\overline{1}\\ \overline{1} }, \colvec{\overline{1}\\ \overline{2} }, \colvec{\overline{2}\\ \overline{0} }, \colvec{\overline{2}\\ \overline{1} }, \colvec{\overline{2}\\ \overline{2} }\right\}. $$Addition is the natural element-wise addition and scalar multiplication is ordinary element-wise scalar multiplication. \end{answer} \end{pro} \end{multicols} \section{Vector Subspaces} \begin{df}Let $\vecspace{V}{+}{\cdot}{ \BBF }$ be a vector space. A non-empty subset $U \subseteq V$ which is also a vector space under the inherited operations of $V$ is called a {\em vector subspace of $V$.} \end{df} \begin{exa} Trivially, $X_1 = \{\v{0}\}$ and $X_2 = V$ are vector subspaces of $V$. \end{exa} \begin{thm} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ be a vector space. Then $U \subseteq V, \ \ U \neq \varnothing$ is a subspace of $V$ if and only if $\forall \alpha\in \BBF$ and $\forall (\v{a}, \v{b})\in U^2$ it is verified that $$\v{a} + \alpha\v{b} \in U.$$ \label{thm:condition_for_subspace}\end{thm} \begin{pf} Observe that $U$ inherits commutativity, associativity and the distributive laws from $V$. Thus a non-empty $U \subseteq V$ is a vector subspace of $V$ if (i) $U$ is closed under scalar multiplication, that is, if $\alpha\in \BBF$ and $\v{v}\in U$, then $\alpha \v{v}\in U$; (ii) $U$ is closed under vector addition, that is, if $(\v{u}, \v{v})\in U^2$, then $\v{u} + \v{v}\in U$. Observe that (i) gives the existence of inverses in $U$, for take $\alpha = -1_{\BBF }$ and so $\v{v}\in U \implies -\v{v}\in U$. This coupled with (ii) gives the existence of the zero-vector, for $\v{0} = \v{v} - \v{v}\in U$. Thus we need to prove that if a non-empty subset of $V$ satisfies the property stated in the Theorem then it is closed under scalar multiplication and vector addition, and vice-versa, if a non-empty subset of $V$ is closed under scalar multiplication and vector addition, then it satisfies the property stated in the Theorem. But this is trivial. \end{pf} \begin{exa} Shew that $\dis{X =\left\{A \in\mat{n\times n}{ \BBF }: \tr{A} = 0_{\BBF }\right\}}$ is a subspace of $\mat{n\times n}{ \BBF }$. \end{exa} \begin{solu}Take $A, B \in X, \alpha \in {\BBR}$. Then $$\tr{A + \alpha B} = \tr{A} + \alpha\tr{B} = 0_{\BBF } + \alpha (0_{\BBF }) = 0_{\BBF }.$$ Hence $A + \alpha B \in X$, meaning that $X$ is a subspace of $\mat{n\times n}{ \BBF }$. \end{solu} \begin{exa} Let $U \in \mat{n\times n}{ \BBF }$ be an arbitrary but fixed. Shew that $$ \mathscr{C}_U = \left\{A \in \mat{n\times n}{ \BBF }: AU = UA\right\}$$ is a subspace of $\mat{n\times n}{ \BBF }$. \end{exa} \begin{solu}Take $(A, B) \in (\mathscr{C}_U)^2$. Then $AU = UA$ and $BU = UB.$ Now $$(A + \alpha B)U = AU + \alpha BU = UA + \alpha UB = U(A + \alpha B),$$ meaning that $A + \alpha B \in \mathscr{C}_U$. Hence $\mathscr{C}_U$ is a subspace of $\mat{n\times n}{ \BBF }$. $\mathscr{C}_U$ is called the {\em commutator} of $U$. \end{solu} \begin{thm} Let $X \subseteq V, \ \ Y \subseteq V$ be vector subspaces of a vector space $\vecspace{V}{+}{\cdot}{ \BBF }$. Then their intersection $X \cap Y$ is also a vector subspace of $V$. \end{thm} \begin{pf} Let $\alpha\in \BBF$ and $(\v{a}, \v{b})\in (X\cap Y)^2$. Then clearly $(\v{a}, \v{b})\in X$ and $(\v{a}, \v{b})\in Y$. Since $X$ is a vector subspace, $\v{a} + \alpha \v{b}\in X$ and since $Y$ is a vector subspace, $\v{a} + \alpha \v{b}\in Y$. Thus $$\v{a} + \alpha \v{b}\in X \cap Y$$ and so $X \cap Y$ is a vector subspace of $V$ by virtue of Theorem \ref{thm:condition_for_subspace}. \end{pf} \begin{rem} We we will soon see that the only vector subspaces of $\vecspace{\BBR^2}{+}{\cdot}{\BBR}$ are the set containing the zero-vector, any line through the origin, and $\BBR^2$ itself. The only vector subspaces of $\vecspace{\BBR^3}{+}{\cdot}{\BBR}$ are the set containing the zero-vector, any line through the origin, any plane containing the origin and $\BBR^3$ itself. \end{rem} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Prove that $$ X = \left\{\colvec{a \\ b \\ c \\ d}\in\BBR^4: a - b - 3d = 0\right\} $$ is a vector subspace of $\BBR^4$. \begin{answer} Take $\alpha\in\BBR$ and $$\v{x} = \colvec{a \\ b \\ c \\ d}\in X, \ \ a - b - 3d = 0, \ \ \ \v{y} = \colvec{a' \\ b' \\ c' \\ d'}\in X, \ \ a' - b' - 3d' = 0. $$ Then $$\v{x} + \alpha \v{y} = \colvec{a \\ b \\ c \\ d} + \alpha\colvec{a' \\ b' \\ c' \\ d'} = \colvec{a + \alpha a' \\ b + \alpha b' \\ c + \alpha c' \\ d + \alpha d'}. $$ Observe that $$ (a + \alpha a') - (b + \alpha b') - 3(d + \alpha d') = (a - b - 3d) + \alpha (a' - b' - 3d') = 0 + \alpha 0 = 0,$$ meaning that $\v{x} + \alpha\v{y}\in X$, and so $X$ is a vector subspace of $\BBR^4$. \end{answer} \end{pro} \begin{pro} Prove that $$X = \left\{\begin{bmatrix}a\cr 2a - 3b\cr 5b\cr a + 2b\cr a\end{bmatrix}:a, b \in \BBR\right\}$$is a vector subspace of $\BBR^5$. \label{exa:subspace_in_r5}\begin{answer} Take $$\v{u} = \begin{bmatrix}a_1\cr 2a_1 - 3b_1\cr 5b_1\cr a_1 + 2b_1\cr a_1\end{bmatrix}, \v{v} = \begin{bmatrix}a_2\cr 2a_2 - 3b_2\cr 5b_2\cr a_2 + 2b_2\cr a_2\end{bmatrix}, \ \ \alpha \in \BBR .$$Put $s = a_1 + \alpha a_2, t = b_1 + \alpha b_2$. Then $$ \v{u} + \alpha\v{v} = \begin{bmatrix}a_1 + \alpha a_2\cr 2(a_1 + \alpha a_2) - 3(b_1 + \alpha b_2)\cr 5(b_1 + \alpha b_2)\cr (a_1 + \alpha a_2) + 2(b_1 + \alpha b_2)\cr a_1 + \alpha a_2\end{bmatrix} = \begin{bmatrix} s\cr 2s - 3t\cr 5t\cr s + 2t\cr s\end{bmatrix} \in X,$$since this last matrix has the basic shape of matrices in $X$. This shews that $X$ is a vector subspace of $\BBR^5$. \end{answer} \end{pro} \begin{pro} Let $A\in \mat{m\times n}{\BBF}$ be a fixed matrix. Demonstrate that $$ S = \{X\in\mat{n\times 1}{\BBF}: AX= {\bf 0}_{m\times 1}\}$$is a subspace of $\mat{n\times 1}{\BBF} $. \end{pro} \begin{pro} Prove that the set $X\subseteq \mat{n\times n}{ \BBF }$ of upper triangular matrices is a subspace of $\mat{n\times n}{ \BBF }$. \end{pro} \begin{pro} Prove that the set $X\subseteq \mat{n\times n}{ \BBF }$ of symmetric matrices is a subspace of $\mat{n\times n}{ \BBF }$. \end{pro} \begin{pro} Prove that the set $X\subseteq \mat{n\times n}{ \BBF }$ of skew-symmetric matrices is a subspace of $\mat{n\times n}{ \BBF }$. \end{pro} \begin{pro} Prove that the following subsets are {\bf not} subspaces of the given vector space. Here you must say which of the axioms for a vector space fail.\begin{dingautolist}{202} \item $\dis{\left\{\colvec{a\\ b\\ 0}: a, b \in {\BBR}, a^2 + b^2 = 1\right\} \subseteq {\BBR}^3}$ \item $\dis{\left\{\colvec{a\\ b\\ 0}: a, b \in {\BBR}^2, ab = 0\right\} \subseteq {\BBR}^3}$ \item $\dis{\left\{ \begin{bmatrix} a & b \cr 0 & 0\end{bmatrix}: (a, b)\in {\BBR}^2, a + b^2 = 0\right\} \ \subseteq \mat{2\times 2}{\BBR}}$ \end{dingautolist} \begin{answer} We shew that some of the properties in the definition of vector subspace fail to hold in these sets. \begin{dingautolist}{202}\item Take $\v{x} = \colvec{0\\ 1\\ 0}, \ \alpha = 2.$ Then $\v{x} \in V$ but $2\v{x} = \colvec{0\\ 2\\ 0} \not\in V$ as $0^2 + 2^2 = 4 \neq 1.$ So $V$ is not closed under scalar multiplication. \item Take $\v{x} = \colvec{0\\ 1\\ 0}, \v{y} = \colvec{1\\ 0\\ 0}$. Then $\v{x} \in W, \v{y} \in W$ but $\v{x} + \v{y} = \colvec{1\\ 1\\ 0} \not\in W$ as $1\cdot 1 = 1 \neq 0.$ Hence $W$ is not closed under vector addition. \item Take $\v{x} = \dis{ \begin{bmatrix} -1 & 1 \cr 0 & 0 \end{bmatrix}.}$ Then $\v{x} \in Z$ but $-\v{x} = -\dis{ \begin{bmatrix} -1 & 1 \cr 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \cr 0 & 0 \end{bmatrix} \not\in Z}$ as $1 + (-1)^2 = 2 \neq 0.$ So $Z$ is not closed under scalar multiplication. \end{dingautolist} \end{answer} \end{pro} \begin{pro} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ be a vector space, and let $U_1 \subseteq V$ and $U_2 \subseteq V$ be vector subspaces. Prove that if $U_1 \cup U_2$ is a vector subspace of $V$, then either $U_1 \subseteq U_2$ or $U_2 \subseteq U_1$. \begin{answer} Assume $U_1\nsubseteq U_2$ and $U_2\nsubseteq U_1$. Take $\v{v}\in U_2 \setminus U_1$ (which is possible because $U_2\nsubseteq U_1$) and $\v{u}\in U_1 \setminus U_2$ (which is possible because $U_1\nsubseteq U_2$). If $\v{u} + \v{v} \in U_1$, then---as $-\v{u}$ is also in $U_1$---the sum of two vectors in $U_1$ must also be in $U_1$ giving $$ \v{u} + \v{v} - \v{u} = \v{v} \in U_1, $$a contradiction. Similarly if $\v{u} + \v{v} \in U_2$, then---as $-\v{v}$ also in $U_2$---the sum of two vectors in $U_2$ must also be in $U_1$ giving $$ \v{u} + \v{v} - \v{u} = \v{u} \in U_2, $$another contradiction. Hence either $U_1\subseteq U_2$ or $U_2\subseteq U_1$ (or possibly both). \end{answer} \end{pro} \begin{pro} Let $V$ a vector space over a field $\BBF$. If $\BBF$ is infinite, show that $V$ is {\bf not} the set-theoretic union of a finite number of {\bf proper} subspaces. \begin{answer} Assume contrariwise that $V = U_1 \bigcup U_2 \bigcup \cdots \bigcup U_k$ is the shortest such list. Since the $U_j$ are proper subspaces, $k > 1.$ Choose $\v{x} \in U_1, \v{x} \not\in U_2 \bigcup \cdots \bigcup U_k$ and choose $\v{y} \not\in U_1.$ Put $L = \{\v{y} + \alpha\v{x}| \alpha \in \BBF\}$. Claim: $L \bigcap U_1 = \varnothing$. For if $\v{u} \in L \bigcap U_1$ then $\exists a_0 \in \BBF$ with $\v{u} = \v{y} + a_0\v{x}$ and so $\v{y} = \v{u} - a_0\v{x} \in U_1$, a contradiction. So $L$ and $U_1$ are disjoint. \bigskip We now shew that $L$ has at most one vector in common with $U_j, 2 \leq j \leq k.$ For, if there were two elements of $\BBF$, $a \neq b$ with $\v{y} + a\v{x}, \v{y} + b\v{x} \in U_j, j \geq 2$ then $$(a - b)\v{x} = (\v{y} + a\v{x}) - (\v{y} + b\v{x}) \in U_j,$$ contrary to the choice of $\v{x}.$ \bigskip Conclusion: since $\BBF$ is infinite, $L$ is infinite. But we have shewn that $L$ can have at most one element in common with the $U_j$. This means that there are not enough $U_j$ to go around to cover the whole of $L$. So $V$ cannot be a finite union of proper subspaces. \end{answer} \end{pro} \begin{pro} Give an example of a finite vector space $V$ over a finite field $\BBF$ such that $$V = V_1 \cup V_2 \cup V_3,$$where the $V_k$ are proper subspaces. \begin{answer} Take $F = \BBZ _2, V = F\times F.$ Then $V$ has the four elements $$\begin{bmatrix}\overline{0}\cr \overline{0}\end{bmatrix}, \begin{bmatrix}\overline{0}\cr \overline{1}\end{bmatrix}, \begin{bmatrix}\overline{1}\cr \overline{0}\end{bmatrix}, \begin{bmatrix}\overline{1}\cr \overline{1}\end{bmatrix},$$with the following subspaces $$V_1 = \left\{\begin{bmatrix}\overline{0}\cr \overline{0}\end{bmatrix}, \begin{bmatrix}\overline{0}\cr \overline{1}\end{bmatrix}\right\}, \ \ V_2 = \left\{\begin{bmatrix}\overline{0}\cr \overline{0}\end{bmatrix}, \begin{bmatrix}\overline{1}\cr \overline{0}\end{bmatrix}\right\},\ \ V_3 = \left\{\begin{bmatrix}\overline{0}\cr \overline{0}\end{bmatrix}, \begin{bmatrix}\overline{1}\cr \overline{1}\end{bmatrix}\right\}.$$ It is easy to verify that these subspaces satisfy the conditions of the problem. \end{answer} \end{pro} \end{multicols} \section{Linear Independence} \begin{df} Let $(\lambda_1, \lambda_2, \cdots , \lambda_n)\in \BBF ^n$. Then the vectorial sum $$\sum_{j = 1} ^n \lambda_j\v{ a}_j$$ is said to be a {\em linear combination} of the vectors $\v{ a}_i \in V, 1 \leq i \leq n.$ \end{df} \begin{exa} Any matrix $\dis{\begin{bmatrix}a & b \cr c & d \cr\end{bmatrix}\in\mat{2\times 2}{\BBR}}$ can be written as a linear combination of the matrices $$\dis{\begin{bmatrix}1 & 0 \cr 0 & 0 \cr\end{bmatrix}},\ \dis{\begin{bmatrix}0 & 1 \cr 0 & 0 \cr\end{bmatrix}},\ \dis{\begin{bmatrix}0 & 0 \cr 1 & 0 \cr\end{bmatrix}}, \ \dis{\begin{bmatrix}0 & 0 \cr 0 & 1 \cr\end{bmatrix}},$$ for $$\begin{bmatrix}a & b \cr c & d \cr\end{bmatrix} = a\begin{bmatrix}1 & 0 \cr 0 & 0 \cr\end{bmatrix} + b\begin{bmatrix}0 & 1 \cr 0 & 0 \cr\end{bmatrix} + c\begin{bmatrix}0 & 0 \cr 1 & 0 \cr\end{bmatrix} + d\begin{bmatrix}0 & 0 \cr 0 & 1 \cr\end{bmatrix}.$$ \end{exa} \begin{exa} Any polynomial of degree at most $2$, say $a + bx + cx^2 \in\BBR_2[x]$ can be written as a linear combination of $1$, $x - 1$, and $x^2 - x + 2$, for $$ a + bx + cx^2 = (a - c)(1) + (b + c)(x - 1) + c(x^2 - x + 2). $$ \end{exa} Generalising the notion of two parallel vectors, we have \begin{df} The vectors $\v{ a}_i \in V, 1 \leq i \leq n,$ are {\em linearly dependent} or {\em tied} if $$\exists (\lambda_1, \lambda_2, \cdots , \lambda_n) \in \BBF ^n \setminus \{{\bf 0}\} \ \ {\rm such \ that\ } \sum_{j = 1} ^n \lambda_j\v{ a}_j = \v{0},$$that is, if there is a non-trivial linear combination of them adding to the zero vector. \end{df}\begin{df} The vectors $\v{ a}_i \in V, 1 \leq i \leq n,$ are {\em linearly independent} or {\em free} if they are not linearly dependent. That is, if $(\lambda_1, \lambda_2, \cdots , \lambda_n) \in \BBF ^n $ then $$\sum_{j = 1} ^n \lambda_j\v{ a}_j = \v{0} \implies \lambda_1 = \lambda_2= \cdots = \lambda_n = 0_{\BBF }.$$ \end{df} \begin{rem} A family of vectors is linearly independent if and only if the only linear combination of them giving the zero-vector is the trivial linear combination. \end{rem} \begin{exa} $$\left\{\colvec{1 \\ 2 \\ 3}, \colvec{4 \\ 5 \\ 6}, \colvec{7 \\ 8 \\ 9}\right\}$$is a tied family of vectors in $\BBR^3$, since $$(1)\colvec{1 \\ 2 \\ 3} + (-2) \colvec{4 \\ 5 \\ 6} + (1)\colvec{7 \\ 8 \\ 9} = \colvec{0 \\ 0 \\ 0}.$$ \end{exa} \begin{exa} Let $\v{u}, \v{v}$ be linearly independent vectors in some vector space over a field $\BBF$ with characteristic different from $2$. Shew that the two new vectors $\v{x} = \v{u} - \v{v}$ and $\v{y} = \v{u} + \v{v}$ are also linearly independent. \end{exa}\begin{solu}Assume that $a(\v{u} - \v{v}) + b(\v{u} + \v{v}) = \v{0}.$ Then $$(a + b)\v{u} + (a - b)\v{v} = \v{0}.$$Since $\v{u}, \v{v}$ are linearly independent, the above coefficients must be 0, that is, $a + b = 0_{\BBF }$ and $a - b = 0_{\BBF }$. But this gives $2a = 2b = 0_{\BBF },$ which implies $a = b = 0_{\BBF },$ if the characteristic of the field is not $2$. This proves the linear independence of $\v{u} - \v{v}$ and $\v{u} + \v{v}$. \end{solu} \begin{thm}\label{thm:lin_ind_columns} Let $A\in\mat{m\times n}{ \BBF }$. Then the columns of $A$ are linearly independent if and only the only solution to the system $AX = {\bf 0}_m$ is the trivial solution. \end{thm} \begin{pf} Let $A_1, \ldots , A_n$ be the columns of $A$. Since $$x_1A_1 + x_2A_2 + \cdots + x_nA_n = AX, $$the result follows. \end{pf} \begin{thm} Any family $$\{\v{0}, \v{u}_1, \v{u}_2, \ldots , \v{u}_k\}$$ containing the zero-vector is linearly dependent. \end{thm} \begin{pf} This follows at once by observing that $$1_{\BBF }\v{0} + 0_{\BBF }\v{u}_1 + 0_{\BBF }\v{u}_2 + \ldots , + 0_{\BBF }\v{u}_k = \v{0}$$is a non-trivial linear combination of these vectors equalling the zero-vector. \end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Shew that $$\left\{\colvec{1 \\ 0 \\ 0}, \colvec{1 \\ 1 \\ 0}, \colvec{1 \\ 1 \\ 1}\right\}$$forms a free family of vectors in $\BBR^3$. \begin{answer} If $$a\colvec{1 \\ 0 \\ 0} + b\colvec{1 \\ 1 \\ 0} + c\colvec{1 \\ 1 \\ 1} = \v{0},$$then $$\colvec{a + b + c \\ b + c \\ c} = \colvec{0 \\ 0 \\ 0}.$$This clearly entails that $c = b = a = 0,$ and so the family is free. \end{answer} \end{pro} \begin{pro} Prove that the set $$\left\{\begin{bmatrix}1\cr 1\cr 1\cr 1\end{bmatrix}, \begin{bmatrix}1\cr 1\cr -1\cr -1\end{bmatrix}, \begin{bmatrix}1\cr -1\cr 1\cr -1\end{bmatrix}, \begin{bmatrix}1\cr 1\cr 0\cr 1\end{bmatrix}\right\}$$ is a linearly independent set of vectors in $\BBR^4$ and shew that $X = \begin{bmatrix}1\cr 2\cr 1\cr 1\end{bmatrix}$ can be written as a linear combination of these vectors. \begin{answer} Assume$$a\begin{bmatrix}1\cr 1\cr 1\cr 1\end{bmatrix} + b\begin{bmatrix}1\cr 1\cr -1\cr -1\end{bmatrix} + c\begin{bmatrix}1\cr -1\cr 1\cr -1\end{bmatrix} + d\begin{bmatrix}1\cr 1\cr 0\cr 1\end{bmatrix} = \begin{bmatrix}0\cr 0\cr 0\cr 0\end{bmatrix}. $$ Then $$a + b + c + d = 0,$$$$a + b - c + d = 0,$$$$a - b + c = 0,$$ $$a - b - c + d = 0.$$Subtracting the second equation from the first, we deduce $2c = 0$, that is, $c = 0.$ Subtracting the third equation from the fourth, we deduce $-2c + d = 0$ or $d = 0.$ From the first and third equations, we then deduce $a + b = 0$ and $a - b = 0,$ which entails $a = b = 0.$ In conclusion, $a = b = c = d = 0$. \bigskip Now, put $$x\begin{bmatrix}1\cr 1\cr 1\cr 1\end{bmatrix} + y\begin{bmatrix}1\cr 1\cr -1\cr -1\end{bmatrix} + z\begin{bmatrix}1\cr -1\cr 1\cr -1\end{bmatrix} + w\begin{bmatrix}1\cr 1\cr 0\cr 1\end{bmatrix} = \begin{bmatrix}1\cr 2\cr 1\cr 1\end{bmatrix}. $$ Then $$x + y + z + w = 1,$$$$x + y - z + w = 2,$$$$x - y + z = 1,$$ $$x - y - z + w = 1.$$Solving as before, we find $$2\begin{bmatrix}1\cr 1\cr 1\cr 1\end{bmatrix} + \frac{1}{2}\begin{bmatrix}1\cr 1\cr -1\cr -1 \end{bmatrix} - \frac{1}{2}\begin{bmatrix}1\cr -1\cr 1\cr -1\end{bmatrix} - \begin{bmatrix}1\cr 1\cr 0\cr 1\end{bmatrix} = \begin{bmatrix}1\cr 2\cr 1\cr 1\end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Let $(\v{u}, \v{v})\in (\BBR^n)^2$. Prove that $|\dotprod{u}{v}| = \norm{\v{u}}\norm{\v{v}}$ if and only if $\v{u}$ and $\v{v}$ are linearly dependent. \end{pro} \begin{pro} Prove that $$\left\{\begin{bmatrix}1 & 0 \cr 0 & 1 \cr \end{bmatrix}, \begin{bmatrix}1 & 0 \cr 0 & -1 \cr \end{bmatrix}, \begin{bmatrix}0 & 1 \cr 1 & 0 \cr \end{bmatrix}, \begin{bmatrix}0 & 1 \cr -1 & 0 \cr \end{bmatrix}\right\}$$is a linearly independent family over $\BBR$. Write $\begin{bmatrix}1 & 1 \cr 1 & 1 \cr \end{bmatrix}$ as a linear combination of these matrices. \end{pro} \begin{pro} Let $\{\v{v}_1, \v{v}_2, \v{v}_3, \v{v}_4 \}$ be a linearly independent family of vectors. Prove that the family $$\{\v{v}_1 + \v{v}_2, \v{v}_2 + \v{v}_3, \v{v}_3 + \v{v}_4, \v{v}_4 + \v{v}_1\}$$is not linearly independent. \begin{answer} We have $$(\v{v}_1 + \v{v}_2) - (\v{v}_2 + \v{v}_3) + (\v{v}_3 + \v{v}_4) - (\v{v}_4 + \v{v}_1) = \v{0},$$a non-trivial linear combination of these vectors equalling the zero-vector. \end{answer} \end{pro} \begin{pro} Let $\{\v{v}_1, \v{v}_2, \v{v}_3\}$ be linearly independent vectors in $\BBR^5$. Are the vectors $$ \v{b}_1 = 3\v{v}_1+ 2\v{v}_2 + 4\v{v}_3, $$ $$ \v{b}_2 = \v{v}_1+ 4\v{v}_2 + 2\v{v}_3, $$ $$ \v{b}_3 = 9\v{v}_1+ 4\v{v}_2 + 3\v{v}_3, $$ $$ \v{b}_4 = \v{v}_1+ 2\v{v}_2 + 5\v{v}_3, $$linearly independent? Prove or disprove! \end{pro} \begin{pro} Is the family $\{1, \sqrt{2}\}$ linearly independent over $\BBQ$? \begin{answer}Yes. Suppose that $a + b\sqrt{2} = 0$ is a non-trivial linear combination of $1$ and $\sqrt{2}$ with rational numbers $a$ and $b$. If one of $a, b$ is different from $0$ then so is the other. Hence $$a + b \sqrt{2} = 0 \implies \sqrt{2} = -\dfrac{b}{a}.$$The sinistral side of the equality $\sqrt{2} = -\dfrac{b}{a}$ is irrational whereas the dextral side is rational, a contradiction.\end{answer} \end{pro} \begin{pro} Is the family $\{1, \sqrt{2}\}$ linearly independent over $\BBR$? \begin{answer}No. The representation $2\cdot 1 + (-\sqrt{2}) \sqrt{2} = 0$ is a non-trivial linear combination of $1$ and $\sqrt{2}$. \end{answer} \end{pro} \begin{pro} Consider the vector space $$V = \{a + b\sqrt{2} + c\sqrt{3}: (a, b, c) \in \BBQ^3\}.$$ \begin{enumerate} \item Shew that $\{1, \sqrt{2}, \sqrt{3}\}$ are linearly independent over $\BBQ$. \\ \item Express $$\frac{1}{1 - \sqrt{2}} + \frac{2}{\sqrt{12} - 2}$$ as a linear combination of $\{1, \sqrt{2}, \sqrt{3}\}$.\\ \end{enumerate} \begin{answer} \begin{enumerate}\item Assume that $$a + b\sqrt{2} + c\sqrt{3} = 0, \ \ a, b, c, \in \BBQ, a^2 + b^2 + c^2 \neq 0. $$If $ac \neq 0,$ then $$b\sqrt{2} = -a - c\sqrt{3} \Leftrightarrow 2b^2 = a^2 + 2ac\sqrt{3} + 3c^2 \Leftrightarrow \frac{2b^2 - a^2 - 3c^2}{2ac} = \sqrt{3}.$$ The dextral side of the last implication is irrational, whereas the sinistral side is rational. Thus it must be the case that $ac = 0.$ If $a = 0, c \neq 0$ then $$b\sqrt{2} + c\sqrt{3} = 0 \Leftrightarrow -\frac{b}{c} = \sqrt{\frac{3}{2}},$$and again the dextral side is irrational and the sinistral side is rational. Thus if $a = 0$ then also $c = 0.$ We can similarly prove that $c = 0$ entails $a = 0.$ Thus we have $$b\sqrt{2} = 0,$$ which means that $b = 0.$ Therefore $$a + b\sqrt{2} + c\sqrt{3} = 0, a, b, c, \in \BBQ , \ \Leftrightarrow a = b = c = 0.$$This proves that $\{1, \sqrt{2}, \sqrt{3}\}$ are linearly independent over $\BBQ$. \item Rationalising denominators, $$\begin{array}{lll}\dfrac{1}{1 - \sqrt{2}} + \dfrac{2}{\sqrt{12} - 2} & = & \dfrac{1 + \sqrt{2}}{1 - 2} + \dfrac{2\sqrt{12} + 4}{12 - 4} \\ & = & -1 - \sqrt{2} + \dfrac{1}{2}\sqrt{3} + \dfrac{1}{2} \\ & = & -\dfrac{1}{2} - \sqrt{2} + \dfrac{1}{2}\sqrt{3}. \end{array}$$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $f, g, h$ belong to $C^\infty (\BBR^\BBR)$ (the space of infinitely continuously differentiable real-valued functions defined on the real line) and be given by $$f(x) = e^x, g(x) = e^{2x}, h(x) = e^{3x}.$$ Shew that $f, g, h$ are linearly independent over $\BBR$. \begin{answer} Assume that $$ae^x + be^{2x} + ce^{3x} = 0.$$Then $$c = -ae^{-2x} - be^{-x}. $$Letting $x\rightarrow +\infty$, we obtain $c = 0$. Thus$$ae^x + be^{2x} = 0,$$and so $$b = -ae^{-x}. $$Again, letting $x\rightarrow +\infty$, we obtain $b = 0$. This yields $$ae^x = 0.$$Since the exponential function never vanishes, we deduce that $a = 0$. Thus $a = b = c = 0$ and the family is linearly independent over $\BBR$. \end{answer} \end{pro} \begin{pro} Let $f, g, h$ belong to $C^\infty(\BBR^\BBR)$ be given by $$f(x) = \cos^2x, g(x) = \sin^2x, h(x) = \cos 2x.$$ Shew that $f, g, h$ are linearly dependent over $\BBR$. \begin{answer} This follows at once from the identity $$\cos 2x = \cos^2x - \sin^2x,$$which implies $$\cos 2x -\cos^2x + \sin^2x = 0.$$ \end{answer} \end{pro} \end{multicols} \section{Spanning Sets} \begin{df} A family $\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \} \subseteq V$ is said to {\em span} or {\em generate} $V$ if every $\v{v}\in V$ can be written as a linear combination of the $\v{u}_j$'s. \index{spanning set} \end{df} \begin{thm} If $\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \} \subseteq V$ spans $V$, then any superset $$\{\v{w}, \v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \} \subseteq V$$ also spans $V$. \end{thm} \begin{pf} This follows at once from $$ \sum _{i = 1} ^l \lambda _i \v{u}_i = 0_{\BBF }\v{w} + \sum _{i = 1} ^l \lambda _i \v{u}_i.$$ \end{pf} \begin{exa} The family of vectors $$\left\{\v{i} = \colvec{1\\ 0 \\ 0}, \v{j} = \colvec{0\\ 1\\ 0}, \v{k} = \colvec{0\\ 0\\ 1}\right\}$$ spans $\BBR^3$ since given $\colvec{a\\ b\\ c}\in\BBR^3$ we may write $$\colvec{a\\ b\\ c} = a\v{i} + b\v{j} + c\v{k} .$$ \label{exa:ijk}\end{exa} \begin{exa} Prove that the family of vectors $$\left\{\v{t}_1 = \colvec{1\\ 0 \\ 0}, \v{t}_2 = \colvec{1\\ 1\\ 0}, \v{t}_3 = \colvec{1\\ 1\\ 1}\right\}$$ spans $\BBR^3$.\label{triagbasisr3}\end{exa} \begin{solu}This follows from the identity $$\colvec{a\\ b\\ c} = (a - b)\colvec{1\\ 0\\ 0} + (b - c)\colvec{1\\ 1\\ 0} + c\colvec{1\\ 1\\ 1} = (a - b)\v{t}_1 + (b - c)\v{t}_2 + c\v{t}_3.$$ \end{solu} \begin{exa} Since $$\begin{bmatrix}a & b \cr c & d \cr\end{bmatrix} = a\begin{bmatrix}1 & 0 \cr 0 & 0 \cr\end{bmatrix} + b\begin{bmatrix}0 & 1 \cr 0 & 0 \cr\end{bmatrix} + c\begin{bmatrix}0 & 0 \cr 1 & 0 \cr\end{bmatrix} + d\begin{bmatrix}0 & 0 \cr 0 & 1 \cr\end{bmatrix}$$the set of matrices $\dis{\begin{bmatrix}1 & 0 \cr 0 & 0 \cr\end{bmatrix}}$, $\dis{\begin{bmatrix}0 & 1 \cr 0 & 0 \cr\end{bmatrix}}$, $\dis{\begin{bmatrix}0 & 0 \cr 1 & 0 \cr\end{bmatrix}}$, $\dis{\begin{bmatrix}0 & 0 \cr 0 & 1 \cr\end{bmatrix}}$ is a spanning set for $\mat{2\times 2}{\BBR}$. \end{exa} \begin{exa} The set $$\{1, x, x^2, x^3, \ldots , x^n, \ldots\}$$spans $\BBR[x]$, the set of polynomials with real coefficients and indeterminate $x$. \end{exa} \begin{df}\index{vector!span} The {\em span} of a family of vectors $\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \}$ is the set of all finite linear combinations obtained from the $u_i$'s. We denote the span of $\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \}$ by $$\span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, }.$$ \end{df} \begin{thm} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ be a vector space. Then $$\span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, } \subseteq V$$ is a vector subspace of $V$. \end{thm} \begin{pf} Let $\alpha\in \BBF$ and let $$\v{x} = \sum _{k = 1} ^l a_k\v{u}_k , \ \ \v{y} = \sum _{k = 1} ^l b_k\v{u}_k ,$$be in $\span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, }$ (some of the coefficients might be $0_{\BBF }$). Then $$\v{x} + \alpha \v{y} = \sum _{k = 1} ^l (a_k + \alpha b_k)\v{u}_k \in \span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, },$$and so $\span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, }$ is a subspace of $V$. \end{pf} \begin{cor} $\span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, } \subseteq V$ is the smallest vector subspace of $V$ (in the sense of set inclusion) containing the set $$\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \}.$$ \end{cor} \begin{pf} If $W \subseteq V$ is a vector subspace of $V$ containing the set $$\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, \}$$ then it contains every finite linear combination of them, and hence, it contains $\span{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots, }.$ \end{pf} \begin{exa} If $A\in\mat{2\times 2}{\BBR}$, $\dis{A\in \span{\begin{bmatrix}1 & 0 \cr 0 & 0 \cr \end{bmatrix}, \begin{bmatrix}0 & 0 \cr 0 & 1 \cr \end{bmatrix}, \begin{bmatrix}0 & 1 \cr 1 & 0 \cr \end{bmatrix}}}$ then $A$ has the form $$a\begin{bmatrix}1 & 0 \cr 0 & 0 \cr \end{bmatrix} + b\begin{bmatrix}0 & 0 \cr 0 & 1 \cr \end{bmatrix} + c \begin{bmatrix}0 & 1 \cr 1 & 0 \cr \end{bmatrix} = \begin{bmatrix}a & c \cr c & b \cr \end{bmatrix},$$i.e., this family spans the set of all symmetric $2\times 2$ matrices over $\BBR$. \end{exa} \begin{thm} Let $V$ be a vector space over a field $\BBF$ and let $(\v{v}, \v{w})\in V^2$, $\gamma\in \BBF\setminus\{0_{\BBF }\}$. Then $$\span{\v{v}, \v{w}} = \span{\v{v},\gamma\v{w}}.$$ \label{thm:linear_combinations_and_span_1}\end{thm} \begin{pf} The equality $$a\v{v} + b\v{w} = a\v{v} + (b\gamma^{-1})(\gamma\v{w}), $$proves the statement. \end{pf} \begin{thm} Let $V$ be a vector space over a field $\BBF$ and let $(\v{v}, \v{w})\in V^2$, $\gamma\in \BBF$. Then $$\span{\v{v}, \v{w}} = \span{\v{w}, \v{v} + \gamma\v{w}}.$$ \label{thm:linear_combinations_and_span_2}\end{thm} \begin{pf} This follows from the equality $$a\v{v} + b\v{w} = a(\v{v} + \gamma\v{w}) + (b - a\gamma)\v{w}. $$ \end{pf} \section*{\psframebox{Homework}} \begin{pro} Let $\BBR_3[x]$ denote the set of polynomials with degree at most $3$ and real coefficients. Prove that the set $$\{1, 1 + x, (1 + x)^2, (1 + x)^3\}$$spans $\BBR_3[x]$. \begin{answer} Given an arbitrary polynomial $$p(x) = a + bx + cx^2 + dx^3,$$we must shew that there are real numbers $s, t, u, v$ such that $$p(x) = s + t(1 + x) + u(1 + x)^2 + v(1 + x)^3.$$In order to do this we find the Taylor expansion of $p$ around $x = -1$. Letting $x = -1$ in this last equality, $$s = p(-1) = a - b + c - d\in\BBR.$$ Now, $$p'(x) = b + 2cx + 3dx^2 = t + 2u(1 + x) + 3v(1 + x)^2. $$Letting $x = -1$ we find $$t = p'(-1) = b -2c +3d\in\BBR.$$Again, $$p''(x) = 2c + 6dx = 2u + 6v(1 + x).$$Letting $x = -1$ we find $$u = p''(-1) = c - 3d \in \BBR.$$Finally, $$p'''(x) = 6d = 6v,$$so we let $v = d\in\BBR$. In other words, we have $$p(x) = a + bx + cx^2 + dx^3 = (a - b + c - d) + (b - 2c + 3d)(1 + x) + (c - 3d)(1 + x)^2 + d(1 + x)^3.$$ \end{answer} \end{pro} \begin{pro} Shew that $\dis{\colvec{1\\ 1 \\ -1}}\not\in\span{\colvec{1\\ 0 \\ -1}, \colvec{0\\ 1 \\ -1}}$.\begin{answer} Assume contrariwise that $$\colvec{1\\ 1 \\ -1} = a\colvec{1\\ 0 \\ -1} + b\colvec{0\\ 1 \\ -1}. $$ Then we must have $$a = 1,$$ $$ b = 1, $$ $$ -a - b = -1,$$ which is impossible. Thus $\colvec{1\\ 1 \\ -1}$ is not a linear combination of $\colvec{1\\ 0 \\ -1}, \colvec{0\\ 1 \\ -1}$ and hence is not in $\span{\colvec{1\\ 0 \\ -1}, \colvec{0\\ 1 \\ -1}}$. \end{answer} \end{pro} \begin{pro} What is $\dis{ \span{\begin{bmatrix}1 & 0 \cr 0 & 0 \cr \end{bmatrix}, \begin{bmatrix}0 & 0 \cr 0 & 1 \cr \end{bmatrix}, \begin{bmatrix}0 & 1 \cr -1 & 0 \cr \end{bmatrix}}}?$ \begin{answer}It is $$a\begin{bmatrix}1 & 0 \cr 0 & 0 \cr \end{bmatrix} + b\begin{bmatrix}0 & 0 \cr 0 & 1 \cr \end{bmatrix} + c \begin{bmatrix}0 & 1 \cr -1 & 0 \cr \end{bmatrix} = \begin{bmatrix}a & c \cr -c & b \cr \end{bmatrix},$$i.e., this family spans the set of all skew-symmetric $2\times 2$ matrices over $\BBR$.\end{answer} \end{pro} \begin{pro} Prove that$$\span{\begin{bmatrix}1 & 0 \cr 0 & 1 \cr \end{bmatrix}, \begin{bmatrix}1 & 0 \cr 0 & -1 \cr \end{bmatrix}, \begin{bmatrix}0 & 1 \cr 1 & 0 \cr \end{bmatrix}, \begin{bmatrix}0 & 1 \cr -1 & 0 \cr \end{bmatrix}} = \mat{2\times 2}{\BBR}.$$ \end{pro} \begin{pro} For the vectors in $\BBR^3$, $$\v{a} = \colvec{1\\ 2\\ 1},\ \v{b} = \colvec{1\\ 3\\ 2},\ \v{c} = \colvec{1\\ 1\\ 0},\ \v{d} = \colvec{3\\ 8\\ 5},$$prove that $$\span{\v{a},\v{b}} = \span{\v{c},\v{d}}.$$ \end{pro} \section{Bases} \begin{df} A family $\{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots\} \subseteq V$ is said to be a {\em basis} of $V$ if (i) they are linearly independent, (ii) they span $V$. \index{basis} \end{df} \begin{exa} The family $$ \v{e}_i = \begin{bmatrix}0_{\BBF }\cr \vdots \cr 0_{\BBF } \cr 1_{\BBF } \cr 0_{\BBF } \cr \vdots \cr 0_{\BBF } \end{bmatrix},$$where there is a $1_{\BBF }$ on the $i$-th slot and $0_{\BBF }$'s on the other $n - 1$ positions, is a basis for $\BBF ^n$. \end{exa} \begin{thm} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ be a vector space and let $$U = \{\v{u}_1, \v{u}_2, \ldots , \v{u}_k, \ldots\} \subseteq V$$ be a family of linearly independent vectors in $V$ which is maximal in the sense that if $U'$ is any other family of vectors of $V$ properly containing $U$ then $U'$ is a dependent family. Then $U$ forms a basis for $V$. \label{thm:maximal_spanning_set}\end{thm} \begin{pf} Since $U$ is a linearly independent family, we need only to prove that it spans $V$. Take $\v{v}\in V$. If $\v{v}\in U$ then there is nothing to prove, so assume that $\v{v}\in V\setminus U$. Consider the set $U' = U \cup \{\v{v}\}$. This set properly contains $U$, and so, by assumption, it forms a dependent family. There exists scalars $\alpha_0, \alpha_1, \ldots, \alpha_n$ such that $$\alpha_0\v{v} + \alpha_1\v{u}_1 + \cdots + \alpha_n\v{u}_n = \v{0}. $$ Now, $\alpha_0 \neq 0_{\BBF }$, otherwise the $\v{u}_i$ would be linearly dependent. Hence $\alpha_0 ^{-1}$ exists and we have $$\v{v} = -\alpha_0 ^{-1}( \alpha_1\v{u}_1 + \cdots + \alpha_n\v{u}_n), $$and so the $\v{u}_i$ span $V$. \end{pf} \begin{rem} From Theorem \ref{thm:maximal_spanning_set} it follows that to shew that a vector space has a basis it is enough to shew that it has a maximal linearly independent set of vectors. Such a proof requires something called Z\"{o}rn's Lemma, and it is beyond our scope. We dodge the whole business by taking as an axiom that every vector space possesses a basis. \end{rem} \begin{thm}[Steinitz Replacement Theorem] \index{theorem!Steinitz}Let $\vecspace{V}{+}{\cdot}{ \BBF }$ be a vector space and let $U = \{\v{u}_1, \v{u}_2, \ldots \} \subseteq V$. Let $W = \{\v{w}_1, \v{w}_2, \ldots, \v{w}_k \}$ be an independent family of vectors in $\span{U}$. Then there exist $k$ of the $\v{u}_i$'s, say $\{\v{u}_1, \v{u}_2, \ldots, \v{u}_k \}$ which may be replaced by the $\v{w}_i$'s in such a way that $$\span{\v{w}_1, \v{w}_2, \ldots, \v{w}_k, \v{u}_{k + 1}, \ldots } = \span{U}. $$ \label{thm:steinitz_replacement} \end{thm} \begin{pf} We prove this by induction on $k$. If $k = 1$, then $$\v{w}_1 = \alpha_1\v{u}_1 + \alpha_2\v{u}_2 + \cdots + \alpha_n\v{u}_n $$for some $n$ and scalars $\alpha_i$. There is an $\alpha_i \neq 0_{\BBF }$, since otherwise $\v{w}_1 = \v{0}$ contrary to the assumption that the $\v{w}_i$ are linearly independent. After reordering, we may assume that $\alpha_1 \neq 0_{\BBF }$. Hence $$\v{u}_1 = \alpha_1 ^{-1}(\v{w}_1 -( \alpha_2\v{u}_2 + \cdots + \alpha_n\v{u}_n)), $$and so $\v{u}_1 \in \span{\v{w}_1, \v{u}_2, \ldots, }$ and $$\span{\v{w}_1, \v{u}_2, \ldots, } = \span{\v{u}_1, \v{u}_2, \ldots, }.$$ \bigskip Assume now that the theorem is true for any set of fewer than $k$ independent vectors. We may thus assume that that $\{\v{u}_1, \ldots\}$ has more than $k - 1$ vectors and that $$\span{\v{w}_1, \v{w}_2, \ldots, \v{w}_{k - 1}, \v{u}_k, , \ldots } = \span{U}.$$ Since $\v{w}_k\in U$ we have $$\v{w}_k = \beta_1\v{w}_1 + \beta_2\v{w}_2 + \cdots + \beta_{k - 1}\v{w}_{k - 1} + \gamma_k\v{u}_k + \gamma_{k+1}\v{u}_{k+1} + \gamma_m\v{u}_m. $$If all the $\gamma_i = 0_{\BBF }$, then the $\{\v{w}_1, \v{w}_2, \ldots, \v{w}_k \}$ would be linearly dependent, contrary to assumption. Thus there is a $\gamma_i \neq 0_{\BBF }$, and after reordering, we may assume that $\gamma_k \neq 0_{\BBF }$. We have therefore $$\v{u}_k = \gamma_k ^{-1}(\v{w}_k - (\beta_1\v{w}_1 + \beta_2\v{w}_2 + \cdots + \beta_{k - 1}\v{w}_{k - 1} + \gamma_{k+1}\v{u}_{k+1} + \gamma_m\v{u}_m)). $$ But this means that $$ \span{\v{w}_1, \v{w}_2, \ldots, \v{w}_k, \v{u}_{k + 1}, \ldots } = \span{U}. $$This finishes the proof. \end{pf} \begin{cor} Let $\{\v{w}_1, \v{w}_2, \ldots , \v{w}_n\}$ be an independent family of vectors with $ V = \span{\v{w}_1, \v{w}_2, \ldots , \v{w}_n}$. If we also have $ V = \span{\v{u}_1, \v{u}_2, \ldots , \v{u}_\nu}$, then \begin{enumerate} \item $n \leq \nu$, \item $n = \nu$ if and only if the $\{\v{y}_1, \v{y}_2, \ldots , \v{y}_\nu\}$ are a linearly independent family. \item Any basis for $V$ has exactly $n$ elements. \end{enumerate} \end{cor} \begin{pf} \begin{enumerate} \item In the Steinitz Replacement Theorem \ref{thm:steinitz_replacement} replace the first $n$ $\v{u}_i$'s by the $\v{w}_i$'s and $n \leq \nu$ follows. \item If $\{\v{u}_1, \v{u}_2, \ldots , \v{u}_\nu\}$ are a linearly independent family, then we may interchange the r\^{o}le of the $\v{w}_i$ and $\v{u}_i$ obtaining $\nu \leq n$. Conversely, if $\nu = n$ and if the $\v{u}_i$ are dependent, we could express some $\v{u}_i$ as a linear combination of the remaining $\nu - 1$ vectors, and thus we would have shewn that some $\nu - 1$ vectors span $V$. From (1) in this corollary we would conclude that $n \leq \nu - 1$, contradicting $n = \nu$. \item This follows from the definition of what a basis is and from (2) of this corollary. \end{enumerate} \end{pf} \begin{df} The {\em dimension} of a vector space $\vecspace{V}{+}{\cdot}{ \BBF }$ is the number of elements of any of its bases, and we denote it by $\dim V$. \end{df} \begin{thm} Any linearly independent family of vectors $$\{\v{x}_1, \v{x}_2, \ldots , \v{x}_k\}$$ in a vector space $V$ can be completed into a family $$\{\v{x}_1, \v{x}_2, \ldots , \v{x}_k, \v{y}_{k + 1}, \v{y}_{k + 2}, \ldots \}$$ so that this latter family become a basis for $V$. \label{thm:enlargement_of_basis}\end{thm} \begin{pf} Take any basis $\{\v{u}_1, \ldots, \v{u}_k, \v{u}_{k+1}, \ldots, \}$ and use Steinitz Replacement Theorem \ref{thm:steinitz_replacement}. \end{pf} \begin{cor} If $U \subseteq V$ is a vector subspace of a finite dimensional vector space $V$ then $\dim U \leq \dim V$. \end{cor} \begin{pf} Since any basis of $U$ can be extended to a basis of $V$, it follows that the number of elements of the basis of $U$ is at most as large as that for $V$. \end{pf} \begin{exa}Find a basis and the dimension of the space generated by the set of symmetric matrices in $\mat{n\times n}{\BBR}$. \end{exa} \begin{solu} Let ${\bf E}_{ij}\in\mat{n\times n}{\BBR}$ be the $n\times n$ matrix with a $1$ on the $ij$-th position and $0$'s everywhere else. For $1 \leq i < j \leq n$, consider the $\binom{n}{2} = \dfrac{n(n - 1)}{2}$ matrices $A_{ij} = {\bf E}_{ij} + {\bf E}_{ji}$. The $A_{ij}$ have a $1$ on the $ij$-th and $ji$-th position and $0$'s everywhere else. They, together with the $n$ matrices ${\bf E}_{ii}, 1 \leq i \leq n$ constitute a basis for the space of symmetric matrices. The dimension of this space is thus $$ \frac{n(n - 1)}{2} + n = \frac{n(n + 1)}{2}.$$ \end{solu} \begin{thm}\label{thm:base_matrix_is_invertible} Let $\{\v{u}_1, \ldots , \v{u}_n\}$ be vectors in $\BBR^n$. Then the $\v{u}$'s form a basis if and only if the $n\times n$ matrix $A$ formed by taking the $\v{u}$'s as the columns of $A$ is invertible. \end{thm} \begin{pf} Since we have the right number of vectors, it is enough to prove that the $\v{u}$'s are linearly independent. But if $X = \colvec{x_1 \\ x_2 \\ \vdots \\ x_n}$, then $$x_1\v{u}_1 + \cdots + x_n\v{u}_n = AX. $$If $A$ is invertible, then $AX = {\bf 0}_n \implies X = A^{-1}{\bf 0}_n = {\bf 0}_n$, meaning that $x_1 = x_2 = \cdots x_n = 0$, so the $\v{u}$'s are linearly independent. \bigskip Conversely, assume that the $\v{u}$'s are linearly independent. Then the equation $AX = {\bf 0}_n$ has a unique solution. Let $r = \rank{A}$ and let $(P, Q)\in (\gl{n}{\BBR})^2$ be matrices such that $A = P^{-1}D_{n,n,r}Q^{-1}$, where $D_{n,n,r}$ is the Hermite normal form of $A$. Thus $$ AX = {\bf 0}_n \implies P^{-1}D_{n,n,r}Q^{-1}X = {\bf 0}_n \implies D_{n,n,r}Q^{-1}X = {\bf 0}_n. $$ Put $Q^{-1}X = \colvec{y_1 \\ y_2 \\ \vdots \\ y_n}$. Then $$D_{n,n,r}Q^{-1}X = {\bf 0}_n \implies y_1\v{e}_1 + \cdots + y_r\v{e}_r = {\bf 0}_n, $$ where $\v{e}_j$ is the $n$-dimensional column vector with a $1$ on the $j$-th slot and $0$'s everywhere else. If $r< n$ then $y_{r+1}, \ldots , y_n$ can be taken arbitrarily and so there would not be a unique solution, a contradiction. Hence $r = n$ and $A$ is invertible. \end{pf} \section*{\psframebox{Homework}} \begin{pro} In problem \ref{exa:subspace_in_r5} we saw that $$X = \left\{\begin{bmatrix}a\cr 2a - 3b\cr 5b\cr a + 2b\cr a\end{bmatrix}:a, b \in \BBR\right\}$$is a vector subspace of $\BBR^5$. Find a basis for this subspace. \begin{answer} We have $$\begin{bmatrix}a\cr 2a - 3b\cr 5b\cr a + 2b\cr a\end{bmatrix} = a\begin{bmatrix}1\cr 2\cr 0\cr 1\cr 1\end{bmatrix} + b\begin{bmatrix}0\cr -3\cr 5\cr 2\cr 0\end{bmatrix},$$so clearly the family $$\left\{\begin{bmatrix}1\cr 2\cr 0\cr 1\cr 1\end{bmatrix},\ \begin{bmatrix}0\cr -3\cr 5\cr 2\cr 0\end{bmatrix} \right\}$$ spans the subspace. To shew that this is a linearly independent family, assume that $$ a\begin{bmatrix}1\cr 2\cr 0\cr 1\cr 1\end{bmatrix} + b\begin{bmatrix}0\cr -3\cr 5\cr 2\cr 0\end{bmatrix} = \begin{bmatrix}0\cr 0\cr 0\cr 0\cr 0\end{bmatrix}.$$ Then it follows clearly that $a = b = 0$, and so this is a linearly independent family. Conclusion: $$\left\{\begin{bmatrix}1\cr 2\cr 0\cr 1\cr 1\end{bmatrix},\ \begin{bmatrix}0\cr -3\cr 5\cr 2\cr 0\end{bmatrix} \right\}$$ is a basis for the subspace. \end{answer} \end{pro} \begin{pro} Let $\{\v{v}_1, \v{v}_2, \v{v}_3, \v{v}_4 , \v{v}_5 \}$ be a basis for a vector space $V$ over a field $\BBF$. Prove that $$\{\v{v}_1 + \v{v}_2, \v{v}_2 + \v{v}_3, \v{v}_3 + \v{v}_4, \v{v}_4 + \v{v}_5, \v{v}_5 + \v{v}_1\}$$is also a basis for $V$. \begin{answer} Suppose $$\begin{array}{lll}\v{0} & = & a(\v{v}_1 + \v{v}_2) + b(\v{v}_2 + \v{v}_3) + c(\v{v}_3 + \v{v}_4) + d(\v{v}_4 + \v{v}_5) + f(\v{v}_5 + \v{v}_1) \\ & = & (a + f)\v{v}_1 + (a + b )\v{v}_2 + (b + c)\v{v}_3 + (c + d)\v{v}_4 + (d + f )\v{v}_5. \end{array}$$ Since $\{\v{v}_1, \v{v}_2, \ldots , \v{v}_5 \}$ are linearly independent, we have $$a + f = 0,$$$$a + b = 0$$$$b + c = 0$$$$c + d = 0$$$$d + f = 0.$$Solving we find $a = b = c = d = f = 0,$ which means that the $$\{\v{v}_1 + \v{v}_2, \v{v}_2 + \v{v}_3, \v{v}_3 + \v{v}_4, \v{v}_4 + \v{v}_5, \v{v}_5 + \v{v}_1\}$$ are linearly independent. Since the dimension of $V$ is 5, and we have 5 linearly independent vectors, they must also be a basis for $V$. \end{answer} \end{pro} \begin{pro} Find a basis for the solution space of the system of $n + 1$ linear equations of $2n$ unknowns $$x_1 + x_2 + \cdots + x_n = 0,$$ $$x_2 + x_3 + \cdots + x_{n + 1} = 0,$$ $$\vdots \vdots \vdots$$ $$x_{n + 1} + x_{n + 2} + \cdots + x_{2n} = 0.$$ \begin{answer} The matrix of coefficients is already in echelon form. The dimension of the solution space is $n - 1$ and the following vectors in $\BBR^{2n}$ form a basis for the solution space $$a_1 = \begin{bmatrix}-1\cr 1\cr 0\cr \vdots \cr 0\cr -1\cr 1\cr 0\cr \vdots \cr 0 \cr\end{bmatrix},\ \ \ a_2 = \begin{bmatrix}-1\cr 0\cr 1\cr \vdots\cr 0\cr -1\cr 0\cr 1\cr \vdots \cr 0\cr\end{bmatrix},\ldots , \ \ \ a_{n - 1} = \begin{bmatrix}-1\cr 0\cr \ldots \cr 1\cr -1\cr 0\cr \vdots \cr 0\cr 1 \cr\end{bmatrix}.$$ (The ``second'' $-1$ occurs on the $n$-th position. The $1$'s migrate from the $2$nd and $n + 1$-th position on $a_1$ to the $n - 1$-th and $2n$-th position on $a_{n - 1}$.) \end{answer} \end{pro} \begin{pro} Prove that the set $V$ of skew-symmetric $n\times n$ matrices is a subspace of $\mat{n\times n}{\BBR}$ and find its dimension. Exhibit a basis for $V$. \begin{answer}Let $A^T=-A$ and $B^T=-B$ be skew symmetric $n\times n$ matrices. Then if $\lambda\in\BBR$ is a scalar, then $$ (A+\lambda B)^T = -(A+\lambda B), $$so $A+\lambda B$ is also skew-symmetric, proving that $V$ is a subspace. Now consider the set of $$ 1+2+\cdots + (n-1)= \dfrac{(n-1)n}{2} $$matrices $A_k$, which are $0$ everywhere except in the $ij$-th and $ji$-spot, where $1 \leq i < j \leq n,$ $a_{ij}=1=-a_{ji}$ and $i+j=k$, $3\leq k \leq 2n-1$. (In the case $n=3$, they are $$\begin{bmatrix} 0 & 1 & 0 & \cr -1 & 0 & 0 \cr 0 & 0 & 0 \end{bmatrix}, \quad \begin{bmatrix} 0 & 0 & 1 & \cr 0 & 0 & 0 \cr -1 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & \cr 0 & 0 & 1 \cr 0 & -1 & 0 \end{bmatrix}, $$ for example.) It is clear that these matrices form a basis for $V$ and hence $V$ has dimension $\dfrac{(n-1)n}{2}$. \end{answer} \end{pro} \begin{pro}Prove that the set $$X = \left\{(a, b, c, d)| b + 2c = 0\right\} \subseteq \BBR^4$$ is a vector subspace of $\BBR^4$. Find its dimension and a basis for $X$. \begin{answer} Take $(\v{u}, \v{v})\in X^2$ and $\alpha \in \BBR$. Then $$ \v{u} = \colvec{a \\ b \\ c \\ d}, \ \ b + 2c = 0, \ \ \ \v{v} = \colvec{a' \\ b' \\ c' \\ d'}, \ \ b' + 2c' = 0. $$We have $$\v{u} + \alpha\v{v} =\colvec{a + \alpha a' \\ b+ \alpha b' \\ c+ \alpha c' \\ d+ \alpha d'}, $$and to demonstrate that $\v{u} + \alpha\v{v}\in X$ we need to shew that $(b+ \alpha b') + 2(c+ \alpha c') = 0$. But this is easy, as $$ (b+ \alpha b') + 2(c+ \alpha c') = (b + 2c) + \alpha (b' + 2c') = 0 + \alpha 0 = 0. $$Now $$\begin{bmatrix}a\cr b\cr c\cr d\end{bmatrix} = \begin{bmatrix}a\cr -2c\cr c\cr d\end{bmatrix} = a\begin{bmatrix}1\cr 0\cr 0\cr 0\end{bmatrix} + c\begin{bmatrix}0\cr -2\cr 1\cr 0\end{bmatrix} + d\begin{bmatrix}0\cr 0\cr 0\cr 1\end{bmatrix}$$ It is clear that $$\begin{bmatrix}1\cr 0\cr 0\cr 0\end{bmatrix}, \begin{bmatrix}0\cr -2\cr 1\cr 0\end{bmatrix}, \begin{bmatrix}0\cr 0\cr 0\cr 1\end{bmatrix}$$ are linearly independent and span $X$. They thus constitute a basis for $X$. \end{answer} \end{pro} \begin{pro} Prove that the dimension of the vector subspace of lower triangular $n\times n$ matrices is $\dfrac{n(n+1)}{2}$ and find a basis for this space. \begin{answer} As a basis we may take the $\dfrac{n(n+1)}{2}$ matrices ${\bf E}_{ij}\in \mat{n}{\BBF}$ for $1\leq i \leq j \leq n$. \end{answer} \end{pro} \begin{pro} Find a basis and the dimension of $$X = \span{\v{v_1} = \colvec{1\\ 1\\ 1 \\ 1}, \ \ \ \v{v_2} = \colvec{1\\ 1\\ 1\\ 0},\ \ \ \v{v_3} =\colvec{2\\ 2\\ 2\\ 1}}. $$ \begin{answer} $\dim{X} = 2$, as basis one may take $\{\v{v_1}, \v{v_2}\}$.\end{answer} \end{pro} \begin{pro} Find a basis and the dimension of $$X = \span{\v{v_1} = \colvec{1\\ 1\\ 1\\ 1}, \ \ \ \v{v_2} = \colvec{1\\ 1\\ 1\\ 0},\ \ \ \v{v_3} =\colvec{2\\ 2 \\ 2\\ 2}}. $$ \begin{answer} $\dim{X} = 3$, as basis one may take $\{\v{v_1}, \v{v_2}, \v{v_3}\}$.\end{answer} \end{pro} \begin{pro} Find a basis and the dimension of $$X = \span{\v{v_1} = \begin{bmatrix}1 & 0 \cr 0 & 1 \cr \end{bmatrix}, \ \ \ \v{v_2} = \begin{bmatrix} 1 & 0 & \cr 2 & 0 \cr \end{bmatrix},\ \ \ \v{v_3} =\begin{bmatrix}0 & 1 \cr 2 & 0 \cr \end{bmatrix}, \ \v{v_4} =\begin{bmatrix}1 & -1 \cr 0 & 0 \cr \end{bmatrix} }. $$ \begin{answer} $\dim{X} = 3$, as basis one may take $\{\v{v_1}, \v{v_2}, \v{v_3}\}$.\end{answer} \end{pro} \begin{pro} Prove that the set $$V = \left\{\begin{bmatrix} a& b & c \cr 0 & d & f \cr 0 & 0 & g \cr \end{bmatrix} \in\mat{3\times 3}{\BBR}: a+b+c=0, \quad a+d+g=0\right\} $$ is a vector space of $\mat{3\times 3}{\BBR}$ and find a basis for it and its dimension. \begin{answer} Let $\lambda\in\BBR$. Observe that $$\begin{bmatrix} a& b & c \cr 0 & d & f \cr 0 & 0 & g \cr \end{bmatrix}+ \lambda \begin{bmatrix} a'& b' & c' \cr 0 & d' & f' \cr 0 & 0 & g' \cr \end{bmatrix} =\begin{bmatrix} a + \lambda a'& b+ \lambda b' & c+ \lambda c' \cr 0 & d+ \lambda d' & f+ \lambda f' \cr 0 & 0 & g+ \lambda g' \cr \end{bmatrix} $$ and if $a+b+c = 0, \quad a+d+g=0,\quad a'+b'+c' = 0, \quad a'+d'+g'=0 $, then $$a+\lambda a' +b+\lambda b'+c+\lambda c' = (a+b+c) + \lambda (a'+b'+c')= 0+\lambda 0= 0, $$ and $$a+\lambda a' +d+\lambda d'+g+\lambda g' = (a+d+g) + \lambda (a'+d'+g')= 0+\lambda 0= 0, $$ proving that $V$ is a subspace. \bigskip Now, $a+b+c=0=a+d+g\implies a=-b-c, g=b+c-d$. Thus $$\begin{bmatrix} a& b & c \cr 0 & d & f \cr 0 & 0 & g \cr \end{bmatrix} = \begin{bmatrix} -b-c& b & c \cr 0 & d & f \cr 0 & 0 & b+c-d \cr \end{bmatrix}= b\begin{bmatrix} -1& 1 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} +c\begin{bmatrix} -1& 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix} +d\begin{bmatrix} 0& 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & -1 \cr \end{bmatrix} +f\begin{bmatrix} 0& 0 & 0 \cr 0 & 0 & 1 \cr 0 & 0 & 0 \cr \end{bmatrix}.$$ It is clear that these four matrices span $V$ and are linearly independent. Hence, $\dim V = 4$. \end{answer} \end{pro} \section{Coordinates} \begin{thm} Let $\{\v{v}_1, \v{v}_2, \ldots, \v{v}_n\}$ be a basis for a vector space $V$. Then any $\v{v}\in V$ has a unique representation $$\v{v} = a_1\v{v}_1 + a_2\v{v}_2 + \cdots + a_n\v{v}_n. $$ \label{thm:unique_rep_in_basis}\end{thm} \begin{pf} Let $$\v{v} = b_1\v{v}_1 + b_2\v{v}_2 + \cdots + b_n\v{v}_n $$be another representation of $\v{v}$. Then $$\v{0} = (a_1 - b_1)\v{v}_1 + (a_2 - b_2)\v{v}_2 + \cdots + (a_n - b_n)\v{v}_n. $$Since $\{\v{v}_1, \v{v}_2, \ldots, \v{v}_n\}$ forms a basis for $V$, they are a linearly independent family. Thus we must have $$ a_1 - b_1 = a_2 - b_2 = \cdots = a_n - b_n = 0_{\BBF }, $$ that is $$a_1 = b_1; a_2 = b_2; \cdots ;a_n = b_n, $$proving uniqueness. \end{pf} \begin{df} An {\em ordered basis} $\{\v{v}_1, \v{v}_2, \ldots, \v{v}_n\}$ of a vector space $V$ is a basis where the order of the $\v{v}_k$ has been fixed. Given an ordered basis $\{\v{v}_1, \v{v}_2, \ldots, \v{v}_n\}$ of a vector space $V$, Theorem \ref{thm:unique_rep_in_basis} ensures that there are unique $(a_1, a_2, \ldots, a_n)\in \BBF ^n$ such that $$\v{v} = a_1\v{v}_1 + a_2\v{v}_2 + \cdots + a_n\v{v}_n. $$ The $a_k$'s are called the {\em coordinates} of the vector $\v{v}$. \end{df} \begin{exa} The standard ordered basis for $\BBR^3$ is ${\mathscr S} = \{\v{i}, \v{j}, \v{k}\}$. The vector $\colvec{1\\ 2\\ 3}\in \BBR^3$ for example, has coordinates $(1, 2, 3)_{\mathscr S}$. If the order of the basis were changed to the ordered basis ${\mathscr S_1} = \{\v{i}, \v{k}, \v{j}\}$, then $\colvec{1\\ 2\\ 3}\in \BBR^3$ would have coordinates $(1, 3, 2)_{\mathscr S_1}$. \end{exa} \begin{rem} Usually, when we give a coordinate representation for a vector $\v{v}\in \BBR^n$, we assume that we are using the standard basis. \end{rem} \begin{exa} Consider the vector $\colvec{1\\ 2\\ 3}\in \BBR^3$ (given in standard representation). Since $$ \colvec{1\\ 2\\ 3} = -1\colvec{1\\ 0\\ 0} -1 \colvec{1\\ 1\\ 0} + 3\colvec{1\\ 1\\ 1}, $$under the ordered basis $\dis{{\mathscr B}_1 = \left\{\colvec{1\\ 0\\ 0}, \colvec{1\\ 1\\ 0}, \colvec{1\\ 1\\ 1}\right\}}$, $\colvec{1\\ 2\\ 3} $ has coordinates $(-1, -1, 3)_{\mathscr B_1}$. We write $$ \colvec{1\\ 2\\ 3} = \colvec{-1\\ -1\\ 3}_{\mathscr B_1}. $$ \end{exa} \begin{exa} The vectors of $$\mathscr{B}_1 = \left\{\colvec{1 \\ 1}, \colvec{1\\ 2}\right\}$$are non-parallel, and so form a basis for $\BBR^2$. So do the vectors $$\mathscr{B}_2 = \left\{\colvec{2 \\ 1}, \colvec{1\\ -1}\right\}.$$Find the coordinates of $\colvec{3\\ 4}_{\mathscr{B}_1}$ in the base $\mathscr{B}_2$. \end{exa} \begin{solu}We are seeking $x, y$ such that $$ 3\colvec{1\\ 1} + 4\colvec{1 \\ 2} = x\colvec{2 \\ 1} + y\colvec{1\\ -1} \implies \begin{bmatrix}1 & 1 \cr 1 & 2 \end{bmatrix}\colvec{3 \\ 4} = \begin{bmatrix}2 & 1 \cr 1 & -1 \cr \end{bmatrix}\colvec{x \\ y}_{\mathscr{B}_2} . $$ Thus $$\begin{array}{lll} \colvec{x\\ y}_{\mathscr{B}_2} & = & \begin{bmatrix}2 & 1 \cr 1 & -1 \cr \end{bmatrix}^{-1}\begin{bmatrix}1 & 1 \cr 1 & 2 \end{bmatrix}\colvec{3 \\ 4} \vspace{2mm}\\ & = & \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \cr \frac{1}{3} & -\frac{2}{3} \cr \end{bmatrix} \begin{bmatrix}1 & 1 \cr 1 & 2 \end{bmatrix}\colvec{3 \\ 4} \vspace{2mm} \\ & = & \begin{bmatrix}\frac{2}{3} & 1 \cr -\frac{1}{3} & -1 \cr \end{bmatrix}\colvec{3 \\ 4} \vspace{2mm}\\ & = & \colvec{6 \\ -5}_{\mathscr{B}_2}. \end{array}$$ Let us check by expressing both vectors in the standard basis of $\BBR^2$: $$\colvec{3 \\ 4}_{\mathscr{B}_1} = 3\colvec{1\\ 1} + 4\colvec{1 \\ 2} = \colvec{7\\ 11},$$ $$\colvec{6 \\ -5}_{\mathscr{B}_2} = 6\colvec{2\\ 1} - 5\colvec{1 \\ -1} = \colvec{7\\ 11}.$$ \end{solu} \bigskip In general let us consider bases $\mathscr{B}_1$ , $\mathscr{B}_2$ for the same vector space $V$. We want to convert $X_{\mathscr{B}_1}$ to $Y_{\mathscr{B}_2}$. We let $A$ be the matrix formed with the column vectors of $\mathscr{B}_1$ in the given order an $B$ be the matrix formed with the column vectors of $\mathscr{B}_2$ in the given order. Both $A$ and $B$ are invertible matrices since the $\mathscr{B}$'s are bases, in view of Theorem \ref{thm:base_matrix_is_invertible}. Then we must have $$AX _{\mathscr{B}_1} = BY_{\mathscr{B}_2} \implies Y_{\mathscr{B}_2} = B^{-1}AX_{\mathscr{B}_1}. $$Also, $$X _{\mathscr{B}_1} = A^{-1}BY_{\mathscr{B}_2} . $$ This prompts the following definition. \begin{df} Let ${\mathscr B_1} = \{\v{u}_1, \v{u}_2, \ldots, \v{u}_n\}$ and ${\mathscr B_2} = \{\v{v}_1, \v{v}_2, \ldots, \v{v}_n\}$ be two ordered bases for a vector space $V$. Let $A\in \mat{n\times n}{ \BBF }$ be the matrix having the $\v{u}$'s as its columns and let $B\in \mat{n\times n}{ \BBF }$ be the matrix having the $\v{v}$'s as its columns. The matrix $P = B^{-1}A$ is called the {\em transition matrix} from ${\mathscr B_1}$ to ${\mathscr B_2}$ and the matrix $P^{-1} = A^{-1}B$ is called the {\em transition matrix} from ${\mathscr B_2}$ to ${\mathscr B_1}$. \index{matrix!transition} \end{df} \begin{exa} Consider the bases of $\BBR^3$ $$ \mathscr{B}_1 = \left\{\colvec{1 \\ 1 \\ 1}, \colvec{1 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 0} \right\}, $$ $$ \mathscr{B}_2 = \left\{\colvec{1 \\ 1 \\ -1}, \colvec{1 \\ -1 \\ 0}, \colvec{2 \\ 0 \\ 0} \right\}. $$Find the transition matrix from ${\mathscr B_1}$ to ${\mathscr B_2}$ and also the transition matrix from ${\mathscr B_2}$ to ${\mathscr B_1}$. Also find the coordinates of $\colvec{1\\ 2 \\ 3}_{\mathscr{B}_1}$ in terms of $\mathscr{B}_2$. \end{exa} \begin{solu}Let $$A = \begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 0 \cr 1& 0 & 0 \cr \end{bmatrix}, \ \ B = \begin{bmatrix} 1 & 1 & 2 \cr 1 & -1 & 0 \cr -1 & 0 & 0 \cr \end{bmatrix}.$$ The transition matrix from ${\mathscr B_1}$ to ${\mathscr B_2}$ is $$\begin{array}{lll}P & = & B^{-1}A \vspace{2mm}\\ & = &\begin{bmatrix} 1 & 1 & 2 \cr 1 & -1 & 0 \cr -1 & 0 & 0 \cr \end{bmatrix}^{-1}\begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 0 \cr 1& 0 & 0 \cr \end{bmatrix}\vspace{2mm}\\ & = & \begin{bmatrix} 0 & 0 & -1 \cr 0 & -1 & -1 \cr \frac{1}{2} & \frac{1}{2} & 1 \cr \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \cr 1 & 1 & 0 \cr 1& 0 & 0 \cr \end{bmatrix} \vspace{2mm}\\ & = & \begin{bmatrix} -1 & 0 & 0 \cr -2 & -1 & -0 \cr 2 & 1 & \frac{1}{2} \end{bmatrix}. \end{array} $$ The transition matrix from ${\mathscr B_2}$ to ${\mathscr B_1}$ is $$P^{-1} = \begin{bmatrix} -1 & 0 & 0 \cr -2 & -1 & 0 \cr 2 & 1 & \frac{1}{2} \end{bmatrix}^{-1} = \begin{bmatrix} -1 & 0 & 0 \cr 2 & -1 & 0 \cr 0 & 2 & 2 \cr\end{bmatrix}. $$Now, $$ Y_{\mathscr{B}_2} = \begin{bmatrix} -1 & 0 & 0 \cr -2 & -1 & 0 \cr 2 & 1 & \frac{1}{2} \end{bmatrix}\colvec{1\\ 2 \\ 3}_{\mathscr{B}_1} = \colvec{-1 \\ -4 \\ \frac{11}{2}}_{\mathscr{B}_2}. $$ As a check, observe that in the standard basis for $\BBR^3$ $$ \colvec{1\\ 2 \\ 3}_{\mathscr{B}_1} = 1\colvec{1\\ 1\\ 1} + 2\colvec{1\\ 1\\ 0 } + 3\colvec{1\\ 0 \\ 0} = \colvec{6\\ 3 \\ 1}, $$ $$ \colvec{-1\\ -4 \\ \frac{11}{2}}_{\mathscr{B}_2} = -1\colvec{1\\ 1\\ -1} - 4\colvec{1\\ -1\\ 0 } + \frac{11}{2}\colvec{2\\ 0 \\ 0} = \colvec{6\\ 3 \\ 1}. $$ \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} \begin{enumerate} \item Prove that the following vectors are linearly independent in $\BBR^4$ $$\v{a}_1 = \begin{bmatrix} 1 \cr 1 \cr 1 \cr 1\cr \end{bmatrix}, \ \ \v{a}_2 =\begin{bmatrix} 1 \cr 1 \cr -1 \cr -1\cr \end{bmatrix}, \ \ \v{a}_3 =\begin{bmatrix} 1 \cr -1 \cr 1 \cr -1\cr \end{bmatrix}, \ \ \v{a}_4 =\begin{bmatrix} 1 \cr -1 \cr -1 \cr 1\cr \end{bmatrix}. $$ \item Find the coordinates of $\colvec{1 \\ 2 \\ 1 \\ 1}$ under the ordered basis $(\v{a}_1, \v{a}_2, \v{a}_3, \v{a}_4)$. \item Find the coordinates of $\colvec{1 \\ 2 \\ 1 \\ 1}$ under the ordered basis $(\v{a}_1, \v{a}_3, \v{a}_2, \v{a}_4).$ \end{enumerate} \begin{answer} \begin{enumerate}\item It is enough to prove that the matrix $$A = \begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix} $$ is invertible. But an easy computation shews that $$ A^2 = \begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix}^2 = 4{\bf I}_4, $$whence the inverse sought is $$ A^{-1} = \frac{1}{4}A = \frac{1}{4}\begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix} = \begin{bmatrix} 1/4 & 1/4 & 1/4 & 1/4 \cr 1/4 & 1/4 & -1/4 & -1/4 \cr 1/4 & -1/4 & 1/4 & -1/4 \cr 1/4 & -1/4 & -1/4 & 1/4 \cr\end{bmatrix}. $$ \item Since the $\v{a}_k$ are four linearly independent vectors in $\BBR^4$ and $\dim \BBR^4 = 4$, they form a basis for $\BBR^4$. Now, we want to solve $$A\colvec{x\\ y\\ z \\ w} = \begin{bmatrix} 1 & 1 & 1 & 1 \cr 1 & 1 & -1 & -1\cr 1 & -1 & 1 & -1\cr 1 & -1 & -1 & 1\cr\end{bmatrix} \colvec{x\\ y\\ z \\ w} = \colvec{1 \\ 2 \\ 1 \\ 1} $$ and so $$\colvec{x\\ y\\ z \\ w} = A^{-1}\colvec{1 \\ 2 \\ 1 \\ 1} = \begin{bmatrix} 1/4 & 1/4 & 1/4 & 1/4 \cr 1/4 & 1/4 & -1/4 & -1/4 \cr 1/4 & -1/4 & 1/4 & -1/4 \cr 1/4 & -1/4 & -1/4 & 1/4 \cr\end{bmatrix}\colvec{1 \\ 2 \\ 1 \\ 1} = \colvec{5/4 \\ 1/4 \\ -1/4 \\ -1/4}. $$ It follows that $$\colvec{1 \\ 2 \\ 1 \\ 1} = \frac{5}{4}\begin{bmatrix} 1 \cr 1 \cr 1 \cr 1\cr \end{bmatrix} + \frac{1}{4}\begin{bmatrix} 1 \cr 1 \cr -1 \cr -1\cr \end{bmatrix} - \frac{1}{4}\begin{bmatrix} 1 \cr -1 \cr 1 \cr -1\cr \end{bmatrix} - \frac{1}{4}\begin{bmatrix} 1 \cr -1 \cr -1 \cr 1\cr \end{bmatrix}. $$ The coordinates sought are $$ \left(\frac{5}{4}, \frac{1}{4}, -\frac{1}{4}, -\frac{1}{4}\right). $$ \item Since we have $$\colvec{1 \\ 2 \\ 1 \\ 1} = \frac{5}{4}\begin{bmatrix} 1 \cr 1 \cr 1 \cr 1\cr \end{bmatrix}- \frac{1}{4}\begin{bmatrix} 1 \cr -1 \cr 1 \cr -1\cr \end{bmatrix} + \frac{1}{4}\begin{bmatrix} 1 \cr 1 \cr -1 \cr -1\cr \end{bmatrix} - \frac{1}{4}\begin{bmatrix} 1 \cr -1 \cr -1 \cr 1\cr \end{bmatrix}, $$ the coordinates sought are $$ \left(\frac{5}{4}, -\frac{1}{4}, \frac{1}{4}, -\frac{1}{4}\right). $$ \end{enumerate} \end{answer} \end{pro} \begin{pro}Consider the matrix $$A(a) = \begin{bmatrix} a &1 &1&1\cr 0&1&0 & 1 \cr 1& 0& a &1 \cr 1&1&1&1 \cr \end{bmatrix}.$$ \begin{dingautolist}{202} \item Determine all $a$ for which $A(a)$ is not invertible. \item Find the inverse of $A(a)$ when $A(a)$ is invertible. \item Find the transition matrix from the basis $${\mathscr B}_1 = \colvec{1 \\ 1\\ 1\\ 1}, \colvec{1\\ 1\\ 1\\ 0}, \colvec{1\\ 1\\ 0 \\ 0}, \colvec{1\\ 0 \\ 0 \\ 0} $$to the basis $${\mathscr B}_2 = \colvec{a \\ 0\\ 1\\ 1}, \colvec{1\\ 1\\ 0\\ 1}, \colvec{1\\ 0\\ a\\ 1}, \colvec{1\\ 1 \\ 1 \\ 1}.$$ \end{dingautolist} \begin{answer}[1] $a = 1$, [2] $(A(a))^{-1} = \begin {bmatrix} \dfrac{1}{ a-1}&0&0&- \dfrac{1}{ a-1}\cr -1&1-a&-1&a+1\cr -\dfrac{1}{a-1}&-1&0&\dfrac{a}{a-1}\cr 1&a &1&-a-1\end {bmatrix}$ [3] $$ \begin{bmatrix} 0& \dfrac{1}{a-1} & \dfrac{1}{a-1}& \dfrac{1}{a-1}\cr 0&-a-1&-a& -1\cr 0&- -\dfrac{a}{a-1}&-\dfrac{a}{a-1}&-\dfrac{1}{ a-1}\cr 1&2+a&a+1&1\cr\end{bmatrix} $$ \end{answer} \end{pro} \end{multicols} \chapter{Linear Transformations} \section{Linear Transformations} \begin{df} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ and $\vecspace{W}{+}{\cdot}{ \BBF }$ be vector spaces over the same field $\BBF$. A {\em linear transformation} or {\em homomorphism} \index{linear transformation} \index{linear homomorphism} $$\fun{L}{\v{a}}{L(\v{ a})}{V}{W},$$is a function which is \begin{itemize} \item {\bf Linear:} $L(\v{a} + \v{b}) = L(\v{a}) + L(\v{b}),$ \item {\bf Homogeneous:} $L(\alpha\v{a}) = \alpha L(\v{a}),$ for $\alpha\in \BBF.$ \end{itemize} \end{df} \begin{rem} It is clear that the above two conditions can be summarised conveniently into $$L(\v{a} + \alpha\v{b}) = L(\v{a}) + \alpha L(\v{b}). $$ \end{rem} \begin{exa} Let $$\fun{L}{A}{\tr{A}}{\mat{n\times n}{\BBR}}{\BBR}. $$ Then $L$ is linear, for if $(A, B)\in\mat{n\times n}{\BBR}$, then $$L(A + \alpha B) = \tr{A + \alpha B} = \tr{A} + \alpha\tr{B} = L(A) + \alpha L(B). $$ \end{exa} \begin{exa} Let $$\fun{L}{A}{A^T}{\mat{n\times n}{\BBR}}{\mat{n\times n}{\BBR}}. $$ Then $L$ is linear, for if $(A, B)\in\mat{n\times n}{\BBR}$, then $$L(A + \alpha B) = (A + \alpha B)^T = A^T + \alpha B^T = L(A) + \alpha L(B). $$ \end{exa} \begin{exa} For a point $(x, y)\in\BBR^2$, its reflexion about the $y$-axis is $(-x, y)$. Prove that $$\fun{R}{(x, y)}{(-x, y)}{\BBR^2}{\BBR^2} $$is linear. \end{exa} \begin{solu}Let $(x_1, y_1)\in\BBR^2$, $(x_2, y_2)\in\BBR^2$, and $\alpha\in\BBR$. Then $$\begin{array}{lll} R((x_1, y_1) + \alpha (x_2, y_2) ) & = & R((x_1 + \alpha x_2, y_1 + \alpha y_2)) \\ & = & (-(x_1 + \alpha x_2), y_1 + \alpha y_2) \\ & = & (-x_1, y_1) + \alpha (-x_2, y_2) \\ & = & R((x_1, y_1)) + \alpha R((x_2, y_2)), \end{array}$$whence $R$ is linear. \end{solu} \begin{exa} Let $L:\BBR^2\rightarrow \BBR^4$ be a linear transformation with $$L\colvec{1 \\ 1} = \colvec{-1 \\ 1 \\ 2\\ 3}; \ \ \ \ L\colvec{-1 \\ 1} = \colvec{2 \\ 0 \\ 2\\ 3}. $$Find $L\colvec{5 \\ 3}.$ \end{exa} \begin{solu}Since $$ \colvec{5 \\ 3} = 4\colvec{1 \\ 1} - \colvec{-1 \\ 1}, $$ we have $$ L\colvec{5 \\ 3} = 4 L\colvec{1 \\ 1} - L\colvec{-1 \\ 1} = 4\colvec{-1 \\ 1 \\ 2\\ 3} - \colvec{2 \\ 0 \\ 2\\ 3} = \colvec{-6 \\ 4\\ 6\\ 9}. $$ \end{solu} \begin{thm}\label{thm:linear_takes_0_to_0}Let $\vecspace{V}{+}{\cdot}{ \BBF }$ and $\vecspace{W}{+}{\cdot}{ \BBF }$ be vector spaces over the same field $\BBF$, and let $L:V \rightarrow W$ be a linear transformation. Then \begin{itemize} \item $L(\v{0}_V) = \v{0}_W$. \item $\forall \v{x}\in V, L(-\v{x}) = - L(\v{x}).$ \end{itemize} \end{thm} \begin{pf} We have $$L(\v{0}_V) = L(\v{0}_V + \v{0}_V) = L(\v{0}_V) + L(\v{0}_V),$$ hence $$ L(\v{0}_V) - L(\v{0}_V) = L(\v{0}_V).$$ Since $$L(\v{0}_V) - L(\v{0}_V) = \v{0}_W,$$we obtain the first result. \bigskip Now $$ \v{0}_W = L(\v{0}_V) = L(\v{x}+ (-\v{x}))= L(\v{x}) + L(-\v{x}), $$from where the second result follows. \end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Consider $L:\BBR^3 \rightarrow \BBR^3$, $$L\colvec{x \\ y \\ z} = \colvec{x - y - z \\ x + y + z \\ z } .$$ Prove that $L$ is linear. \label{exa:linear_1}\begin{answer} Let $\alpha\in \BBR.$ Then $$\begin{array}{lll} L\colvec{x + \alpha a\\ y + \alpha b \\ z + \alpha c} & = & \colvec{(x + \alpha a) - (y + \alpha b) - (z + \alpha c) \\ (x + \alpha a) + (y + \alpha b) + (z + \alpha c) \\ z + \alpha c} \vspace{2mm} \\ & = & \colvec{x - y - z \\ x + y + z \\ z } + \alpha \colvec{a - b - c \\ a + b + c\\ c } \vspace{2mm} \\ & = & L\colvec{x \\ y \\ z} + \alpha L\colvec{a \\ b \\ c}, \\ \end{array}$$proving that $L$ is a linear transformation. \end{answer} \end{pro} \begin{pro} Let $A\in\gl{n}{\BBR}$ be a fixed matrix. Prove that $$\fun{L}{H}{-A^{-1}HA^{-1}}{\mat{n\times n}{\BBR}}{\mat{n\times n}{\BBR}}$$is a linear transformation. \begin{answer} $$\begin{array}{lll} L(H + \alpha H') & = & -A^{-1}(H + \alpha H')A^{-1}\\ & = & -A^{-1}HA^{-1} + \alpha(-A^{-1}H'A^{-1}) \\ & = & L(H) + \alpha L(H'), \\ \end{array}$$proving that $L$ is linear. \end{answer} \end{pro} \begin{pro} \index{convex set} Let $V$ be a vector space and let $S \subseteq V.$ The set $S$ is said to be {\em convex} if $\forall \alpha \in [0; 1], \forall {\bf x, y} \in S$, $(1 - \alpha){\bf x} + \alpha {\bf y} \in S$, that is, for any two points in $S$, the straight line joining them also belongs to $S$. Let $T: V \rightarrow W$ be a linear transformation from the vector space $V$ to the vector space $W$. Prove that $T$ maps convex sets into convex sets. \begin{answer} Let $S$ be convex and let $\v{a}, \v{b}\in T(S).$ We must prove that $\forall \alpha \in [0; 1], \ (1 - \alpha)\v{a} + \alpha\v{b} \in T(S).$ But since $\v{a}, \v{b}$ belong to $T(S)$, $\exists \v{x}\in S, \v{y}\in S$ with $T(\v{x}) = \v{a}, T(\v{y}) = \v{b}$. Since $S$ is convex, $(1 - \alpha)\v{x} + \alpha\v{y} \in S$. Thus $$T((1 - \alpha)\v{x} + \alpha\v{y}) \in T(S),$$ which means that $$(1 - \alpha)T(\v{x}) + \alpha T(\v{y}) \in T(S),$$that is, $$\ (1 - \alpha)\v{a} + \alpha\v{b} \in T(S),$$as we wished to show. \end{answer} \end{pro} \end{multicols} \section{Kernel and Image of a Linear Transformation} \begin{df}\index{linear transformation!kernel} \index{linear transformation!image} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ and $\vecspace{W}{+}{\cdot}{ \BBF }$ be vector spaces over the same field $\BBF$, and $$\fun{T}{\v{v}}{T(\v{v})}{V}{W} $$be a linear transformation. The {\em kernel} of $T$ is the set $$\ker{T} = \{\v{v}\in V: T(\v{v}) = \v{0}_W\}. $$ The {\em image} of $T$ is the set $$\im{T} = \{\v{w}\in w:\exists \v{v}\in V\ {\rm such\ that\ } T(\v{v}) = \v{w}\} = T(V). $$ \end{df} \begin{rem} Since $T(\v{0}_V) = \v{0}_W$ by Theorem \ref{thm:linear_takes_0_to_0}, we have $\v{0}_V\in\ker{T}$ and $\v{0}_W\in \im{T}$. \end{rem} \begin{thm} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ and $\vecspace{W}{+}{\cdot}{ \BBF }$ be vector spaces over the same field $\BBF$, and $$\fun{T}{\v{v}}{T(\v{v})}{V}{W} $$be a linear transformation. Then $\ker{T}$ is a vector subspace of $V$ and $\im{T}$ is a vector subspace of $W$. \end{thm} \begin{pf} Let $(\v{v}_1,\v{v}_2)\in (\ker{T})^2$ and $\alpha \in \BBF$. Then $T(\v{v}_1) = T(\v{v}_2) = \v{0}_V$. We must prove that $\v{v}_1 + \alpha \v{v}_2\in\ker{T}$, that is, that $T(\v{v}_1 + \alpha \v{v}_2) = \v{0}_W$. But $$T(\v{v}_1 + \alpha \v{v}_2) = T(\v{v}_1) + \alpha T(\v{v}_2) = \v{0}_V + \alpha \v{0}_V = \v{0}_V $$proving that $\ker{T}$ is a subspace of $V$. \bigskip Now, let $(\v{w}_1,\v{w}_2)\in (\im{T})^2$ and $\alpha \in \BBF$. Then $\exists (\v{v}_1,\v{v}_2)\in V^2$ such that $T(\v{v}_1) = \v{w}_1$ and $T(\v{v}_2) = \v{w}_2$. We must prove that $\v{w}_1 + \alpha \v{w}_2\in\im{T}$, that is, that $\exists \v{v}$ such that $T(\v{v}) = \v{w}_1 + \alpha \v{w}_2$. But $$\v{w}_1 + \alpha \v{w}_2 = T(\v{v}_1) + \alpha T(\v{v}_2) = T(\v{v}_1 + \alpha \v{v}_2), $$ and so we may take $\v{v} = \v{v}_1 + \alpha \v{v}_2$. This proves that $\im{T}$ is a subspace of $W$. \end{pf} \begin{thm} Let $\vecspace{V}{+}{\cdot}{ \BBF }$ and $\vecspace{W}{+}{\cdot}{ \BBF }$ be vector spaces over the same field $\BBF$, and $$\fun{T}{\v{v}}{T(\v{v})}{V}{W} $$be a linear transformation. Then $T$ is injective if and only if $\ker{T} = \v{0}_V$. \end{thm} \begin{pf} Assume that $T$ is injective. Then there is a unique $\v{x}\in V$ mapping to $\v{0}_W$: $$T(\v{x}) = \v{0}_W. $$ By Theorem \ref{thm:linear_takes_0_to_0}, $T(\v{0}_V) = \v{0}_W$, i.e., a linear transformation takes the zero vector of one space to the zero vector of the target space, and so we must have $\v{x} = \v{0}_V$. \bigskip Conversely, assume that $\ker{T} = \{\v{0}_V\}$, and that $T(\v{x}) = T(\v{y})$. We must prove that $\v{x} = \v{y}$. But $$\begin{array}{lll}T(\v{x}) = T(\v{y}) & \implies & T(\v{x}) - T(\v{y}) = \v{0}_W\\ & \implies & T(\v{x} - \v{y}) = \v{0}_W \\ & \implies & (\v{x} - \v{y})\in\ker{T}\\ & \implies & \v{x} - \v{y} = \v{0}_V \\ & \implies &\v{x} = \v{y}, \end{array}$$as we wanted to shew. \end{pf} \begin{thm}[Dimension Theorem] Let $\vecspace{V}{+}{\cdot}{ \BBF }$ and $\vecspace{W}{+}{\cdot}{ \BBF }$ be vector spaces of finite dimension over the same field $\BBF$, and $$\fun{T}{\v{v}}{T(\v{v})}{V}{W} $$be a linear transformation. Then $$\dim\ker{T} + \dim\im{T} = \dim V. $$ \label{thm:dimension_theorem}\end{thm} \begin{pf} Let $\{\v{v}_1, \v{v}_2, \ldots , \v{v}_k\}$ be a basis for $\ker{T}$. By virtue of Theorem \ref{thm:enlargement_of_basis}, we may extend this to a basis ${\mathscr A} = \{\v{v}_1, \v{v}_2, \ldots , \v{v}_k, \v{v}_{k +1}, \v{v}_{k +2}, \ldots , \v{v}_n\}$ of $V$. Here $n = \dim V$. We will now shew that ${\mathscr B}=\{T(\v{v}_{k +1}), T(\v{v}_{k +2}), \ldots , T(\v{v}_n)\}$ is a basis for $\im{T}$. We prove that (i) ${\mathscr B}$ spans $\im{T}$, and (ii) ${\mathscr B}$ is a linearly independent family. \bigskip Let $\v{w}\in \im{T}$. Then $\exists \v{v}\in V$ such that $T(\v{v}) = \v{w}$. Now since ${\mathscr A}$ is a basis for $V$ we can write $$ \v{v} = \sum _{i = 1} ^n \alpha_i\v{v}_i. $$Hence $$\v{w} = T(\v{v}) = \sum _{i = 1} ^n \alpha_iT(\v{v}_i) = \sum _{i = k + 1} ^n \alpha_iT(\v{v}_i), $$ since $T(\v{v}_i) = \v{0}_V$ for $1 \leq i \leq k$. Thus ${\mathscr B}$ spans $\im{T}$. \bigskip To prove the linear independence of the ${\mathscr B}$ assume that $$\v{0}_W = \sum _{i = k + 1} ^n \beta_iT(\v{v}_i) = T\left(\sum _{i = k + 1} ^n \beta_i\v{v}_i \right). $$ This means that $\sum _{i = k + 1} ^n \beta_i\v{v}_i\in\ker{T}$, which is impossible unless $\beta_{k + 1} = \beta_{k + 2} = \cdots = \beta_n = 0_{\BBF }$. \end{pf} \begin{cor} If $\dim V = \dim W < +\infty$, then $T$ is injective if and only if it is surjective. \end{cor} \begin{pf} Simply observe that if $T$ is injective then $\dim\ker{T} = 0$, and if $T$ is surjective $\im{T} =T(V) = W$ and $\im{T} = \dim{W}$. \end{pf} \begin{exa} Let $$\fun{L}{A}{A^T - A}{\mat{2\times 2}{\BBR}}{\mat{2\times 2}{\BBR}}. $$ Observe that $L$ is linear. Determine $\ker{L}$ and $\im{L}.$ \end{exa} \begin{solu}Put $A = \begin{bmatrix} a & b \cr c & d\end{bmatrix}$ and assume $L(A) = {\bf 0}_2$. Then $$ \begin{bmatrix} 0 & 0 \cr 0 & 0\end{bmatrix} =L(A) = \begin{bmatrix} a & c \cr b & d\end{bmatrix} - \begin{bmatrix} a & b \cr c & d\end{bmatrix} = (c - b)\begin{bmatrix} 0& 1\cr -1 & 0\end{bmatrix}. $$This means that $c = b$. Thus $$\ker{L} = \left\{\begin{bmatrix} a & b \cr b & d\end{bmatrix}: (a, b, d)\in \BBR^3\right\}, $$ $$\im{L} = \left\{\begin{bmatrix} 0 & k \cr -k & 0\end{bmatrix}: k\in \BBR\right\}. $$ This means that $\dim\ker{L} = 3$, and so $\dim\im{L} = 4 - 3 = 1$. \end{solu} \begin{exa} Consider the linear transformation $L:\mat{2\times 2}{\BBR} \rightarrow \BBR_3[X] $ given by $$L\begin{bmatrix} a & b \cr c & d \cr \end{bmatrix} = (a + b)X^2 + (a - b)X^3. $$ Determine $\ker{L}$ and $\im{L}$. \end{exa} \begin{solu}We have $$0 = L\begin{bmatrix} a & b \cr c & d \cr \end{bmatrix} = (a + b)X^2 + (a - b)X^3 \implies a+b = 0, \ a-b = 0, \implies a = b = 0. $$ Thus $$ \ker{L} = \left\{\begin{bmatrix} 0 & 0 \cr c & d \cr \end{bmatrix}: (c, d)\in\BBR^2 \right\}. $$ Thus $\dim\ker{L} = 2$ and hence $\dim\im{L} = 2$. Now $$(a + b)X^2 + (a - b)X^3 \implies a(X^2 + X^3) + b(X^2 - X^3). $$ Clearly $X^2 + X^3$, and $X^2 - X^3$ are linearly independent and span $\im{L}$. Thus $$\im{L} = \span{X^2 + X^3, X^2 - X^3}. $$ \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} In problem \ref{exa:linear_1} we saw that $L:\BBR^3 \rightarrow \BBR^3$, $$L\colvec{x \\ y \\ z} = \colvec{x - y - z \\ x + y + z \\ z } $$is linear. Determine $\ker{L}$ and $\im{L}$. \begin{answer} Assume $\colvec{x\\ y \\ z}\in\ker{L}$. Then $$L\colvec{x \\ y \\ z} = \colvec{0 \\ 0 \\ 0}, $$that is $$x - y - z = 0, $$ $$x + y + z = 0, $$ $$z = 0.$$ This implies that $x - y = 0$ and $x + y = 0$, and so $x = y = z = 0$. This means that $$\ker{L} = \left\{\colvec{0 \\ 0 \\ 0}\right\},$$and $L$ is injective. \bigskip By the Dimension Theorem \ref{thm:dimension_theorem}, $\dim\im{L} = \dim V - \dim\ker{L} = 3 - 0 = 3$, which means that $$\im{L} = \BBR^3$$ and $L$ is surjective. \end{answer} \end{pro} \begin{pro} Consider the function $L:\BBR^4 \to \BBR^2$ given by $$ L\colvec{x\\ y \\ z\\ w} = \colvec{x+y \\ x-y}. $$ \begin{enumerate} \item Prove that $L$ is linear. \item Determine $\ker{L}$. \item Determine $\im{L}$. \end{enumerate} \begin{answer} \noindent \begin{enumerate} \item If $a$ is any scalar, $$ L\left(\colvec{x\\ y \\ z\\ w}+a\colvec{x'\\ y' \\ z'\\ w'}\right)= L\colvec{x+ax'\\ y+ay'\\ z+az' \\ w+aw'} = \colvec{(x+ax')+(y+ay')\\ (x+ax')-(y+ay')}=\colvec{x+y\\ x-y}+a\colvec{x'+y'\\ x'-y'}= L\colvec{x\\ y \\ z\\ w}+aL\colvec{x'\\ y' \\ z'\\ w'}, $$ whence $L$ is linear. \item We have,$$ L\colvec{x\\ y \\ z\\ w} = \colvec{x+y \\ x-y}=\colvec{0\\ 0}\implies x=y, x=-y \implies x=y=0 \implies \ker{L}= \left\{\colvec{0\\ 0 \\ z\\ w}: z\in \BBR, w\in\BBR\right\}. $$ Thus $\dim\ker{L}=2$. In particular, the transformation is not injective. \item From the previous part, $\dim\im{L}=4-2=2$. Since $\im{L}\subseteq \BBR^2$ and $\dim \im{L}=2$, we must have $\im{L}=\BBR^2$. In particular, the transformation is surjective. \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $$\fun{L}{\v{a}}{L(\v{a})}{\BBR^3}{\BBR^4}$$ satisfy $$L\colvec{1\\ 0 \\ 0} = \colvec{1 \\ 0 \\ -1 \\ 0}; \ \ \ L\colvec{1\\ 1 \\ 0} = \colvec{2 \\ -1 \\ 0 \\ 0}; \ \ \ L\colvec{0\\ 0 \\ 1} = \colvec{1 \\ -1 \\ 1 \\ 0}.$$ Determine $\ker{L}$ and $\im{L}$. \begin{answer} Assume that $\colvec{a\\ b \\ c}\in\ker{T}$, $$\colvec{a\\ b \\ c} = (a - b)\colvec{1\\ 0 \\ 0} + b\colvec{1\\ 1 \\ 0} + c\colvec{0\\ 0 \\ 1}. $$ Then $$\begin{array}{lll}\colvec{0 \\ 0 \\ 0\\ 0 } & = & T\colvec{a\\ b \\ c}\\ & = & (a - b)T\colvec{1\\ 0 \\ 0} + bT\colvec{1\\ 1 \\ 0} + cT\colvec{0\\ 0 \\ 1}\vspace{2mm}\\ & = & (a - b)\colvec{1 \\ 0 \\ -1 \\ 0} + b\colvec{2 \\ -1 \\ 0 \\ 0} + c\colvec{1 \\ -1 \\ 1 \\ 0} \vspace{2mm}\\ & = & \colvec{a + b + c \\ -b - c \\ -a + b + c \\ 0}.\end{array}$$ It follows that $a = 0$ and $b = -c$. Thus $$ \ker{T} = \left\{c\colvec{0 \\ -1 \\ 1}: c\in\BBR\right\},$$and so $\dim\ker{T} = 1$. \bigskip By the Dimension Theorem \ref{thm:dimension_theorem}, $$\dim\im{T} = \dim V - \dim\ker{T} = 3 - 1 = 2.$$ We readily see that $$\colvec{2 \\ -1 \\ 0 \\ 0} = \colvec{1 \\ 0 \\ -1 \\ 0} + \colvec{1 \\ -1 \\ 1 \\ 0}, $$ and so $$\im{T} = \span{\colvec{1 \\ 0 \\ -1 \\ 0}, \colvec{1 \\ -1 \\ 1 \\ 0}}. $$ \end{answer} \end{pro} \begin{pro} It is easy to see that $L:\BBR^2 \rightarrow \BBR^3$,$$L\colvec{x \\ y} = \colvec{x + 2y \\ x + 2y \\ 0 }$$ is linear. Determine $\ker{L}$ and $\im{L}$. \begin{answer} Assume that $$L\colvec{x \\ y} = \colvec{x + 2y \\ x + 2y \\ 0 } = \colvec{0 \\ 0 \\ 0}. $$ Then $x = -2y $ and so $$\colvec{x \\ y} = y\colvec{-2 \\ 1} . $$This means that $\dim\ker{L} = 1$ and $\ker{L}$ is the line through the origin and $(-2, 1)$. Observe that $L$ is not injective. \bigskip By the Dimension Theorem \ref{thm:dimension_theorem}, $\dim\im{L} = \dim V - \dim\ker{L} = 2 - 1 = 1$. Assume that $\colvec{a \\ b\\ c}\in \im{L}$. Then $\exists (x, y)\in \BBR^2$ such that $$L\colvec{x \\ y} = \colvec{x + 2y \\ x + 2y \\ 0 } = \colvec{a \\ b \\ c}. $$ This means that $$\colvec{a \\ b \\ c} = \colvec{x + 2y \\ x + 2y \\ 0 } = (x + 2y)\colvec{1 \\ 1 \\ 0}. $$Observe that $L$ is not surjective. \end{answer} \end{pro} \begin{pro} It is easy to see that $L:\BBR^2 \rightarrow \BBR^3$,$$L\colvec{x \\ y} = \colvec{x - y \\ x + y \\ 0 }$$ is linear. Determine $\ker{T}$ and $\im{T}$. \begin{answer} Assume that $$L\colvec{x \\ y} = \colvec{x - y \\ x + y \\ 0 } = \colvec{0 \\ 0 \\ 0}. $$ Then $x + y = 0 = x - y $, that is, $x = y = 0$, meaning that $$\ker{L} = \left\{\colvec{0\\ 0}\right\},$$ and so $L$ is injective. \bigskip By the Dimension Theorem \ref{thm:dimension_theorem}, $\dim\im{L} = \dim V - \dim\ker{L} = 2 - 0 = 2$. Assume that $\colvec{a \\ b\\ c}\in \im{L}$. Then $\exists (x, y)\in \BBR^2$ such that $$L\colvec{x \\ y} = \colvec{x - y \\ x + y \\ 0 } = \colvec{a \\ b \\ c}. $$ This means that $$\colvec{a \\ b \\ c} = \colvec{x - y \\ x + y \\ 0 } = x\colvec{1 \\ 1 \\ 0} + y\colvec{-1 \\ 1 \\ 0} . $$ Since $$ \colvec{1 \\ 1 \\ 0},\colvec{-1 \\ 1 \\ 0} $$are linearly independent, they span a subspace of dimension $2$ in $\BBR^3$, that is, a plane containing the origin. Observe that $L$ is not surjective. \end{answer} \end{pro} \begin{pro} It is easy to see that $L:\BBR^3 \rightarrow \BBR^2$,$$L\colvec{x \\ y\\ z} = \colvec{x - y - z \\ y - 2z}$$ is linear. Determine $\ker{L}$ and $\im{L}$. \begin{answer} Assume that $$L\colvec{x \\ y\\ z} = \colvec{x - y - z \\ y - 2z} = \colvec{0 \\ 0}.$$Then $y = 2z; x = y + z = 3z$. This means that $\ker{L} = \left\{z\colvec{3 \\2 \\ 1}: z\in\BBR \right\}.$ Hence $\dim \ker{L} = 1$, and so $L$ is not injective. \bigskip Now, if $$L\colvec{x \\ y\\ z} = \colvec{x - y - z \\ y - 2z} = \colvec{a \\ b}.$$Then $$\colvec{a \\ b} = \colvec{x - y - z \\ y - 2z} = x\colvec{1 \\ 0} + y\colvec{-1 \\ 1} + z\colvec{-1 \\ -2}.$$Now, $$-3\colvec{1 \\ 0} - 2\colvec{-1 \\ 1} = \colvec{-1 \\ -2}$$and $$\colvec{1 \\ 0}, \colvec{-1 \\ 1}$$are linearly independent. Since $\dim\im{L} = 2$, we have $\im{L} = \BBR^2$, and so $L$ is surjective. \end{answer} \end{pro} \begin{pro} Let $$\fun{L}{A}{\tr{A}}{\mat{2\times 2}{\BBR}}{\BBR}.$$ Determine $\ker{L}$ and $\im{L}$. \begin{answer} Assume that $$ 0 = \tr{\begin{bmatrix} a & b \cr c & d \cr \end{bmatrix}} = a + d. $$ Then $a = -d$ and so, $$\begin{bmatrix} a & b \cr c & d \cr \end{bmatrix} = \begin{bmatrix} -d & b \cr c & d \cr \end{bmatrix} = d\begin{bmatrix} -1 & 0 \cr 0 & 1 \cr \end{bmatrix} + b\begin{bmatrix} 0 & 1 \cr 0 & 0 \cr \end{bmatrix} + c\begin{bmatrix} 0 & 0 \cr 1 & 0 \cr \end{bmatrix}, $$and so $\dim \ker{L} = 3.$ Thus $L$ is not injective. $L$ is surjective, however. For if $\alpha\in \BBR$, then $$\alpha = \tr{\begin{bmatrix} \alpha & 0 \cr 0 & 0 \cr \end{bmatrix}}. $$ \end{answer} \end{pro} \begin{pro} \begin{enumerate} \item Demonstrate that $$ \fun{L}{A}{A^T + A}{\mat{2\times 2}{\BBR}}{\mat{2\times 2}{\BBR}}$$ is a linear transformation. \item Find a basis for $\ker{L}$ and find $\dim\ker{L}$ \item Find a basis for $\im{L}$ and find $\dim\im{L}$. \end{enumerate} \begin{answer} \begin{enumerate} \item Let $(A,B)^2\in\mat{2\times 2}{\BBR}, \alpha \in \BBR$. Then $$\begin{array}{lll}L(A + \alpha B) & = & (A + \alpha B)^T + (A + \alpha B)\\ & = & A^T + B^T + A + \alpha B \\ & = & A^T + A + \alpha B^T + \alpha B \\ & = & L(A) + \alpha L(B), \end{array} $$ proving that $L$ is linear. \item Assume that $A = \begin{bmatrix}a & b \cr c & d\cr \end{bmatrix}\in \ker{L}$. Then $$ \begin{bmatrix}0 & 0 \cr 0 & 0\cr \end{bmatrix}= L(A) = \begin{bmatrix}a & b \cr c & d\cr \end{bmatrix} + \begin{bmatrix}a & c \cr b & d\cr \end{bmatrix} = \begin{bmatrix}2a & b + c \cr b + c & 2d\cr \end{bmatrix}, $$ whence $a = d = 0$ and $b = -c$. Hence $$\ker{L} = \span{\begin{bmatrix}0 & -1 \cr 1 & 0\cr \end{bmatrix}}, $$ and so $\dim \ker{L} = 1$. \item By the Dimension Theorem, $\dim \im{L} = 4 -1 = 3$. As above, $$\begin{array}{lll}L(A) & = & \begin{bmatrix}2a & b + c \cr b + c & 2d\cr \end{bmatrix}\\ & = & a\begin{bmatrix}2 & 0 \cr 0 & 0\cr \end{bmatrix} + (b + c)\begin{bmatrix}0 & 1 \cr 1 & 0\cr \end{bmatrix} + d\begin{bmatrix}0 & 0 \cr 0 & 2\cr \end{bmatrix}, \end{array} $$from where $$\im{L} = \span{\begin{bmatrix}2 & 0 \cr 0 & 0\cr \end{bmatrix},\begin{bmatrix}0 & 1 \cr 1 & 0\cr \end{bmatrix},\begin{bmatrix}0 & 0 \cr 0 & 2\cr \end{bmatrix}}. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $V$ be an $n$-dimensional vector space, where the characteristic of the underlying field is different from $2$. A linear transformation $T:V\rightarrow V$ is {\em idempotent} if $T^2 = T$. Prove that if $T$ is idempotent, then \begin{dingautolist}{202} \item $I - T$ is idempotent ($I$ is the identity function). \item $I+T$ is invertible. \\ \item $\ker{T} = \im{I-T}$ \\ \end{dingautolist} \begin{answer} \begin{dingautolist}{202} \item Observe that $$ (I-T)^2 = I -2T+T^2 = I-2T+T = I-T, $$proving the result. \item The inverse is $I-\frac{1}{2}T$, for $$ (I + T)(I-\frac{1}{2}T) = I+T-\frac{1}{2}T - \frac{1}{2}T^2 = I+T-\frac{1}{2}T - \frac{1}{2}T = I, $$proving the claim. \item We have $$\begin{array}{lll}\v{x}\in \ker{T} & \iff & \v{x} - T(\v{x}) \in \ker{T} \\ & \iff & I(\v{x}) - T(\v{x}) \in \ker{T} \\ & \iff & (I-T)(\v{x}) \in \ker{T} \\ & \iff & \v{x} \in \im{I-T}. \\ \end{array} $$ \end{dingautolist} \end{answer} \end{pro} \end{multicols} \section{Matrix Representation} Let $V, W$ be two vector spaces over the same field $\BBF$. Assume that $\dim V = m$ and $\{\v{v}_i\}_{i\in [1;m]}$ is an ordered basis for $V$, and that $\dim W = n$ and ${\mathscr A} = \{\v{w}_i\}_{i\in [1;n]}$ an ordered basis for $W$. Then $$ \begin{array}{ccccc}L(\v{v}_1) & = & a_{11}\v{w}_1 + a_{21}\v{w}_2 + \cdots + a_{n1}\v{w}_n & = & \colvec{a_{11} \\ a_{21} \\ \vdots \\ a_{n1}}_{\mathscr A} \\ L(\v{v}_2) & = & a_{12}\v{w}_1 + a_{22}\v{w}_2 + \cdots + a_{n2}\v{w}_n & = & \colvec{a_{12} \\ a_{22} \\ \vdots \\ a_{n2}}_{\mathscr A} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ L(\v{v}_m) & = & a_{1m}\v{w}_1 + a_{2m}\v{w}_2 + \cdots + a_{nm}\v{w}_n & = & \colvec{a_{1n} \\ a_{2n} \\ \vdots \\ a_{nm}}_{\mathscr A}\\ \end{array}. $$ \begin{df} The $n\times m$ matrix $$ M_L = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \cr a_{21} & a_{12} & \cdots & a_{2n} \cr \vdots & \vdots & \vdots & \vdots \cr a_{n1} & a_{n2} & \cdots & a_{nm} \cr \end{bmatrix} $$formed by the column vectors above is called the {\em matrix representation of the linear map $L$ with respect to the bases $\{\v{v}_i\}_{i\in [1;m]}, \{\v{w}_i\}_{i\in [1;n]}$.} \end{df} \begin{exa} Consider $L:\BBR^3 \rightarrow \BBR^3$, $$L\colvec{x \\ y \\ z} = \colvec{x - y - z \\ x + y + z \\ z } .$$ Clearly $L$ is a linear transformation. \begin{enumerate} \item Find the matrix corresponding to $L$ under the standard ordered basis. \item Find the matrix corresponding to $L$ under the ordered basis $\colvec{1 \\ 0 \\ 0}, \colvec{1 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 1},$ for both the domain and the image of $L$. \end{enumerate} \end{exa} \begin{solu} \begin{enumerate} \item The matrix will be a $3\times 3$ matrix. We have $L\colvec{ 1 \\ 0 \\ 0 } = \colvec{ 1 \\ 1 \\ 0 }$, $L\colvec{ 0 \\ 1 \\ 0 } = \colvec{ -1 \\ 1 \\ 0}$, and $L\colvec{0 \\ 0\\ 1} = \colvec{-1 \\ 1 \\ 1}$, whence the desired matrix is $$\begin{bmatrix} 1 & -1 & -1 \cr 1 & 1 & 1 \cr 0 & 0 & 1 \cr\end{bmatrix}.$$ \item Call this basis ${\mathscr A}$. We have $$L\colvec{ 1 \\ 0 \\ 0 } = \colvec{ 1 \\ 1 \\ 0 } = 0 \colvec{ 1 \\ 0 \\ 0 } + 1\colvec{ 1 \\ 1 \\ 0 } + 0\colvec{ 1 \\ 0 \\ 1 } = \colvec{ 0 \\ 1 \\ 0 }_{\mathscr A},$$ $$L\colvec{ 1\\ 1 \\ 0 } = \colvec{ 0 \\ 2 \\ 0} = -2\colvec{ 1 \\ 0 \\ 0 } + 2\colvec{ 1 \\ 1 \\ 0 } + 0\colvec{ 1 \\ 0 \\ 1 } = \colvec{ -2 \\ 2 \\ 0 }_{\mathscr A},$$ and $$L\colvec{1 \\ 0\\ 1} = \colvec{0 \\ 2 \\ 1} = -3\colvec{ 1 \\ 0 \\ 0 } + 2\colvec{ 1 \\ 1 \\ 0 }+ 1\colvec{ 1 \\ 0 \\ 1} = \colvec{ -3 \\ 2 \\ 1 }_{\mathscr A},$$ whence the desired matrix is $$\begin{bmatrix} 0 & -2 & -3 \cr 1 & 2 & 2 \cr 0 & 0 & 1 \cr\end{bmatrix}.$$ \end{enumerate} \end{solu} \begin{exa} Let $\BBR_n[x]$ denote the set of polynomials with real coefficients with degree at most $n$. \begin{enumerate} \item Prove that $$\fun{L}{p(x)}{p''(x)}{\BBR_3[x]}{\BBR_1[x]} $$is a linear transformation. Here $p''(x)$ denotes the second derivative of $p(x)$ with respect to $x$. \item Find the matrix of $L$ using the ordered bases $\{1, x, x^2, x^3\}$ for $\BBR_3[x]$ and $\{1, x\}$ for $\BBR_2[x]$. \item Find the matrix of $L$ using the ordered bases $\{1, x, x^2, x^3\}$ for $\BBR_3[x]$ and $\{1, x + 2\}$ for $\BBR_1[x]$. \item Find a basis for $\ker{L}$ and find $\dim\ker{L}$. \item Find a basis for $\im{L}$ and find $\dim\im{L}$. \end{enumerate}\end{exa} \begin{solu} \begin{enumerate} \item Let $(p(x), q(x))\in(\BBR_3[x])^2$ and $\alpha\in\BBR$. Then $$L(p(x) + \alpha q(x)) = (p(x) + \alpha q(x))'' = p''(x) + \alpha q''(x) = L(p(x)) + \alpha L(q(x)),$$ whence $L$ is linear. \item We have $$\begin{array}{ccccccccc}L(1) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} 1 & = & 0 & = & 0(1) + 0(x) & = & \colvec{0\\ 0}, \vspace{2mm}\\ L(x) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} x & = & 0 & = & 0(1) + 0(x) & = & \colvec{0\\ 0}, \vspace{2mm}\\ L(x^2) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} x^2 & = & 2 & = & 2(1) + 0(x) & = & \colvec{2\\ 0},\vspace{2mm} \\ L(x^3) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} x^3& = & 6x & = & 0(1) + 6(x) & = & \colvec{0\\ 6}, \end{array}$$whence the matrix representation of $L$ under the standard basis is $$\begin{bmatrix} 0 & 0 & 2 & 0 \cr 0 & 0 & 0 & 6 \cr \end{bmatrix}. $$ \item We have $$\begin{array}{ccccccccc}L(1) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} 1 & = & 0 & = & 0(1) + 0(x + 2) & = & \colvec{0\\ 0}, \vspace{2mm}\\ L(x) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} x & = & 0 & = & 0(1) + 0(x + 2) & = & \colvec{0\\ 0}, \vspace{2mm}\\ L(x^2) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} x^2 & = & 2 & = & 2(1) + 0(x + 2) & = & \colvec{2\\ 0}, \vspace{2mm}\\ L(x^3) & = & \dfrac{{\rm d}^2}{{\rm d}x^2} x^3& = & 6x & = & -12(1) + 6(x + 2) & = & \colvec{-12\\ 6}, \end{array}$$whence the matrix representation of $L$ under the standard basis is $$\begin{bmatrix} 0 & 0 & 2 & -12 \cr 0 & 0 & 0 & 6 \cr \end{bmatrix}. $$ \item Assume that $p(x) = a + bx + cx^2 + dx^3\in\ker{L}$. Then $$ 0= L(p(x)) = 2c + 6dx, \ \ \ \forall x\in \BBR.$$ This means that $c = d = 0$. Thus $a, b$ are free and $$\ker{L} = \{a + bx: (a, b)\in\BBR^2\}. $$Hence $\dim\ker{L} = 2$. \item By the Dimension Theorem, $\dim\im{L} = 4 - 2 = 2$. Put $q(x) = \alpha + \beta x + \gamma x^2 + \delta x^3$. Then $$L(q(x)) = 2\gamma + 6\delta (x) = (2\gamma)(1) + (6\delta)(x). $$Clearly $\{1, x\}$ are linearly independent and span $\im{L}$. Hence $$\im{L} = \span{1, x} = \BBR_1[x].$$ \end{enumerate} \end{solu} \begin{exa}\noindent \begin{enumerate} \item A linear transformation $T:\BBR^3 \rightarrow \BBR^3 $ is such that $$T(\v{i}) = \colvec{2 \\ 1\\ 1}; \ \ \ T(\v{j}) = \colvec{3 \\ 0\\ -1}.$$It is known that $$\im{T} = \span{T(\v{i}), T(\v{j})} $$and that $$\ker{T} = \span{\colvec{1\\ 2 \\ -1}}. $$ Argue that there must be $\lambda$ and $\mu $ such that $$T(\v{k}) = \lambda T(\v{i}) + \mu T(\v{j}).$$ \item Find $\lambda$ and $\mu $, and hence, the matrix representing $T$ under the standard ordered basis. \end{enumerate} \end{exa} \begin{solu}\begin{enumerate} \item Since $T(\v{k}) \in \im{T}$ and $\im{T}$ is generated by $T(\v{i}) $ and $T(\v{k}) $ there must be $(\lambda, \mu)\in\BBR^2$ with $$ T(\v{k}) = \lambda T(\v{i}) + \mu T(\v{j}) = \lambda \colvec{2\\ 1 \\ 1} + \mu\colvec{3 \\ 0 \\ -1} = \colvec{2\lambda + 3\mu \\ \lambda \\ \lambda - \mu}. $$ \item The matrix of $T$ is $$\begin{bmatrix}T(\v{i}) & T(\v{j}) & T(\v{k})\cr \end{bmatrix} = \begin{bmatrix} 2 & 3 & 2\lambda + 3\mu \cr 1 & 0 & \lambda \cr 1 & -1 & \lambda - \mu\end{bmatrix}. $$ Since $\colvec{1\\2\\ -1}\in\ker{T}$ we must have $$\begin{bmatrix} 2 & 3 & 2\lambda + 3\mu \cr 1 & 0 & \lambda \cr 1 & -1 & \lambda - \mu\end{bmatrix}\colvec{1\\2\\ -1} = \colvec{0\\0 \\ 0}. $$Solving the resulting system of linear equations we obtain $\lambda = 1, \mu = 2$. The required matrix is thus $$\begin{bmatrix} 2 & 3 & 8 \cr 1 & 0 & 1\cr 1 & -1 & -1\end{bmatrix}.$$ \end{enumerate} \end{solu} \begin{rem} If the linear mapping $L:V \rightarrow W$, $\dim V = n, \dim W = m$ has matrix representation $A\in\mat{m\times n}{ \BBF }$, then $\dim \im{L} = \rank{A}$. \end{rem} \section*{\psframebox{Homework}} \begin{pro} Let $T:\BBR ^4\rightarrow \BBR^3$ be a linear transformations such that $$ T\colvec{1 \\ 1\\ 1\\ 1} = \colvec{0 \\ 0 \\ 1}, \qquad T\colvec{1 \\ 0\\ 1\\ 0} = \colvec{1 \\ 1 \\ -1}, \qquad T\colvec{1 \\ 1\\ 1\\ 0} = \colvec{0 \\ 0 \\ -1}, \qquad T\colvec{-1 \\ -2\\ 0\\ 0} = \colvec{1 \\ 1 \\ 1}. $$Find the matrix of $T$ with respect to the canonical bases. Find the dimensions and describe $\ker{T}$ and $\im{T}$. \begin{answer} Observe that $$\colvec{a\\ b\\ c\\ d} = d\colvec{1\\ 1 \\ 1\\ 1} + (2a-c-b)\colvec{1 \\ 0\\ 1\\ 0} +(-d-2a+2c+b)\colvec{1 \\ 1\\ 1\\ 0}+(-a+c)\colvec{-1 \\ -2\\ 0\\ 0}. $$ Hence $$\begin{array}{lll}T\colvec{a\\ b\\ c\\ d} & = & dT\colvec{1\\ 1 \\ 1\\ 1} + (2a-c-b)T\colvec{1 \\ 0\\ 1\\ 0} +(-d-2a+2c+b)T\colvec{1 \\ 1\\ 1\\ 0}+(-a+c)T\colvec{-1 \\ -2\\ 0\\ 0}\\ & = & d\colvec{0 \\ 0\\ 1} + (2a-c-b)\colvec{1 \\ 1\\ -1} +(-d-2a+2c+b)\colvec{0\\ 0\\ -1}+(-a+c)\colvec{1 \\ 1\\ 1}\\ & = & \colvec{a-b\\ a-b \\ -a+2d}. \end{array} $$ This gives $$ T\colvec{1\\ 0 \\ 0 \\ 0} = \colvec{1\\ 1\\ -1}, \quad T\colvec{0\\ 1 \\ 0 \\ 0} = \colvec{-1\\ -1\\ 0}, \quad T\colvec{0\\ 0 \\ 1 \\ 0} = \colvec{0\\ 0\\ 0}, \quad T\colvec{0\\ 0 \\ 0 \\ 1} = \colvec{0\\ 0\\ 2}. $$ The required matrix is therefore $$\begin{bmatrix} 1 & -1 & 0 & 0 \cr 1 & -1 & 0 & 0 \cr -1 & 0 & 0 & 2 \cr \end{bmatrix}. $$ This matrix has rank $2$, and so $\dim\im{T}=2$. We can use $\left\{\colvec{1\\ 1\\ -1}, \colvec{-1\\ -1\\ 0}\right\}$ as a basis for $\im{T}$. Thus by the dimension theorem $\dim\ker{T}=2$. If $\colvec{0\\ 0\\ 0}=T\colvec{a\\ b \\ c \\ d} = \colvec{a-b\\ a-b\\ -a+2d}$, Hence the vectors in $\ker{T}$ have the form $\colvec{2d\\ 2d\\ c \\ d}$ and hence we may take $\left\{\colvec{2\\ 2\\ 0 \\ 1}, \colvec{0\\ 0\\ 1 \\ 0}\right\}$ as a basis for $\ker{T}$. \end{answer} \end{pro} \begin{pro} \begin{enumerate} \item A linear transformation $T:\BBR^3 \rightarrow \BBR^3$ has as image the plane with equation $x + y + z = 0$ and as kernel the line $x = y = z$. If $$T\colvec{1\\ 1\\ 2} = \colvec{a\\ 0\\ 1}, \ \ \ T\colvec{2\\ 1\\ 1} = \colvec{3\\ b\\ -5},\ \ \ T\colvec{1\\ 2\\ 1} = \colvec{-1\\ 2\\ c}.$$Find $a, b, c$. \item Find the matrix representation of $T$ under the standard basis. \end{enumerate}\begin{answer}\begin{enumerate} \item Since the image of $T$ is the plane $x + y + z = 0$, we must have $$a + 0 + 1 = 0 \implies a = -1, $$ $$3 + b - 5 = 0 \implies b = 2, $$ $$-1 + 2 + c = 0 \implies c = -1. $$ \item Observe that $\colvec{1\\ 1\\ 1}\in\ker{T}$ and so $$ T\colvec{1\\ 1\\ 1} = \colvec{0\\ 0\\ 0}. $$ Thus $$ T\colvec{1\\ 0\\ 0} = T\colvec{2\\ 1\\ 1} - T\colvec{1\\ 1\\ 1} = \colvec{3\\ 2\\ -5}, $$ $$ T\colvec{0\\ 1\\ 0} = T\colvec{1\\ 2\\ 1} - T\colvec{1\\ 1\\ 1} = \colvec{-1\\ 2\\ -1}, $$ $$ T\colvec{0\\ 0\\ 1} = T\colvec{1\\ 1\\ 2} - T\colvec{1\\ 1\\ 1} = \colvec{-1\\ 0\\ 1}. $$ The required matrix is therefore $$\begin{bmatrix}3 & -1 & -1 \cr 2 & 2 & 0 \cr -5 & -1 & 1 \cr \end{bmatrix}. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro} \begin{enumerate} \item Prove that $T:\BBR^2 \rightarrow \BBR^3$ $$T\colvec{x \\ y} = \colvec{x + y \\ x - y \\ 2x + 3y} $$ is a linear transformation. \item Find a basis for $\ker{T}$ and find $\dim\ker{T}$ \item Find a basis for $\im{T}$ and find $\dim\im{T}$. \item Find the matrix of $T$ under the ordered bases ${\mathscr A} = \left\{\colvec{1 \\ 2}, \colvec{1\\ 3}\right\}$ of $\BBR^2$ and ${\mathscr B} = \left\{\colvec{1 \\ 1\\ 1}, \colvec{1\\ 0\\ -1}, \colvec{0\\ 1 \\ 0}\right\}$ of $\BBR^3$. \end{enumerate}\begin{answer} \begin{enumerate} \item Let $\alpha\in\BBR$. We have $$ \begin{array}{lll} T\left(\colvec{x \\ y}+ \alpha \colvec{u \\ v}\right) & = & T\left(\colvec{x + \alpha u \\ y + \alpha v}\right) \\ & = & \colvec{x + \alpha u + y + \alpha v \\ x + \alpha u - y - \alpha v \\ 2(x + \alpha u) + 3(y + \alpha v)} \\ & = & \colvec{x + y \\ x - y \\ 2x + 3y} + \alpha\colvec{u + v \\ u - v \\ 2u + 3v} \\ & = &T\left(\colvec{x \\ y}\right) + \alpha T\left( \colvec{u \\ v}\right), \end{array}$$ proving that $T$ is linear. \item We have $$T\left(\colvec{x \\ y}\right) = \colvec{0\\ 0 \\ 0} \iff \colvec{x + y \\ x - y \\ 2x + 3y} = \colvec{0\\ 0 \\ 0} \iff x = y = 0,$$$\dim\ker{T} = 0$, and whence $T$ is injective. \item By the Dimension Theorem, $\dim \im{T} = 2 - 0 = 2$. Now, since $$T\left(\colvec{x \\ y}\right) = \colvec{x + y \\ x - y \\ 2x + 3y} = x \colvec{1\\ 1\\ 2} + y\colvec{1 \\ -1 \\ 3}, $$ whence $$\im{T} = \span{\colvec{1\\ 1\\ 2}, \colvec{1 \\ -1 \\ 3}}. $$ \item We have $$ T\left(\colvec{1 \\ 2}\right) = \colvec{3\\ -1\\ 8} = \frac{11}{2}\colvec{1 \\ 1\\ 1} - \frac{5}{2}\colvec{1 \\ 0\\ -1} -\frac{13}{2}\colvec{0 \\ 1\\ 0} = \colvec{11/2 \\ -5/2\\ -13/2}_{\mathscr B}, $$ and $$ T\left(\colvec{1 \\ 3}\right) = \colvec{4\\ -2\\ 11} = \frac{15}{2}\colvec{1 \\ 1\\ 1} - \frac{7}{2}\colvec{1 \\ 0\\ -1} -\frac{19}{2}\colvec{0 \\ 1\\ 0} = \colvec{15/2 \\ -7/2\\ -19/2}_{\mathscr B}. $$ The required matrix is $$\begin{bmatrix}11/2 & 15/2 \cr -5/2 & -7/2 \cr -13/2 & -19/2 \cr \end{bmatrix}_{\mathscr B}. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $$\fun{L}{\v{a}}{L(\v{a})}{\BBR^3}{\BBR^2},$$ where $$L\colvec{x \\ y \\ z} = \colvec{x + 2y \\ 3x - z}.$$ Clearly $L$ is linear. Find a matrix representation for $L$ if \begin{enumerate} \item The bases for both $\BBR^3$ and $\BBR^2$ are both the standard ordered bases. \item The ordered basis for $\BBR^3$ is $\dis{\colvec{1 \\ 0 \\ 0 }, \colvec{1 \\ 1 \\ 0 }, \colvec{1 \\ 1 \\ 1 }}$ and $\BBR^2$ has the standard ordered basis . \item The ordered basis for $\BBR^3$ is $\dis{\colvec{1 \\ 0 \\ 0 }, \colvec{1 \\ 1 \\ 0 }, \colvec{1 \\ 1 \\ 1 }}$ and the ordered basis for $\BBR^2$ is $\dis{{\mathscr A} = \left\{ \colvec{1 \\ 0}, \colvec{1 \\ 1}\right\}}$. \end{enumerate} \begin{answer} The matrix will be a $2\times 3$ matrix. In each case, we find the action of $L$ on the basis elements of $\BBR^3$ and express the result in the given basis for $\BBR^3$. \begin{enumerate} \item We have $$L\left(\colvec{1 \\ 0 \\ 0 }\right) = \colvec{1 \\ 3}, L\left(\colvec{0 \\ 1 \\ 0 }\right) = \colvec{2 \\ 0}, L\left(\colvec{0 \\ 0 \\ 1 }\right) = \colvec{0 \\ -1}. $$The required matrix is $$\begin{bmatrix} 1 & 2 & 0 \cr 3 & 0 & -1 \cr \end{bmatrix}.$$ \item We have $$L\left(\colvec{1 \\ 0 \\ 0 }\right) = \colvec{1 \\ 3}, L\left(\colvec{1 \\ 1 \\ 0 }\right) = \colvec{3 \\ 3}, L\left(\colvec{1 \\ 1 \\ 1 }\right) = \colvec{3 \\ 2}. $$The required matrix is $$\begin{bmatrix} 1 & 3 & 3 \cr 3 & 3 & 2 \cr \end{bmatrix}.$$ \item We have $$L\left(\colvec{1 \\ 0 \\ 0 }\right) = \colvec{1 \\ 3} = -2\colvec{1 \\ 0} + 3\colvec{1 \\ 1} = \colvec{-2 \\ 3}_{\mathscr A} ,$$ $$L\left(\colvec{1 \\ 1 \\ 0 }\right) = \colvec{3 \\ 3} = 0\colvec{1 \\ 0} + 3\colvec{1 \\ 1} = \colvec{0 \\ 3}_{\mathscr A}, $$ $$ L\left(\colvec{1 \\ 1 \\ 1 }\right) = \colvec{3 \\ 2} = 1\colvec{1 \\ 0} + 2\colvec{1 \\ 1} = \colvec{1 \\ 2}_{\mathscr A}. $$The required matrix is $$\begin{bmatrix} -2 & 0 & 1 \cr 3 & 3 & 2 \cr \end{bmatrix}.$$ \end{enumerate} \end{answer} \end{pro} \begin{pro} A linear transformation $T:\BBR^2 \rightarrow \BBR^2$ satisfies $\ker{T} = \im{T}$, and $T\colvec{1\\ 1} = \colvec{2 \\ 3}$. Find the matrix representing $T$ under the standard ordered basis. \begin{answer} Observe that $\colvec{2 \\ 3}\in \im{T} = \ker{T}$ and so $$T\colvec{2 \\ 3} = \colvec{0 \\ 0}. $$Now $$T\colvec{1 \\ 0} = T\left( 3\colvec{1 \\ 1} - \colvec{2 \\ 3}\right) = 3T\colvec{1 \\ 1} - T\colvec{2 \\ 3} = \colvec{6 \\ 9},$$ and $$T\colvec{0 \\ 1} = T\left( \colvec{2 \\ 3} - 2\colvec{1 \\ 1} \right) = T\colvec{2 \\ 3}- 2T\colvec{1 \\ 1} = \colvec{-4 \\ -6}.$$ The required matrix is thus $$\begin{bmatrix}6 & -4 \cr 9 & -6 \cr \end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Find the matrix representation for the linear map $$\fun{L}{A}{\tr{A}}{\mat{2\times 2}{\BBR}}{\BBR}, $$under the standard basis $${\mathscr A} = \left\{\begin{bmatrix} 1 & 0 \cr 0 & 0 \cr \end{bmatrix}, \begin{bmatrix} 0 & 1 \cr 0 & 0 \cr \end{bmatrix}, \begin{bmatrix} 0 & 0 \cr 1 & 0 \cr \end{bmatrix}, \begin{bmatrix} 0 & 0 \cr 0 & 1 \cr \end{bmatrix} \right\}$$for $\mat{2\times 2}{\BBR}$. \begin{answer} The matrix will be a $1\times 4$ matrix. We have $$\tr{\begin{bmatrix} 1 & 0 \cr 0 & 0 \cr \end{bmatrix}} = 1, $$ $$\tr{\begin{bmatrix} 0 & 1 \cr 0 & 0 \cr \end{bmatrix}} = 0, $$ $$\tr{\begin{bmatrix} 0 & 0 \cr 1 & 0 \cr \end{bmatrix}} = 0, $$ $$\tr{\begin{bmatrix} 0 & 0 \cr 0 & 1 \cr \end{bmatrix}} = 1. $$ Thus $$ M_L = (1 \ \ 0 \ \ 0 \ \ 1). $$ \end{answer} \end{pro} \begin{pro} Let $A\in\mat{n\times p}{\BBR}$, $B\in\mat{p\times q}{\BBR}$, and $C\in\mat{q\times r}{\BBR}$, be such that $\rank{B} = \rank{AB}$. Shew that $$\rank{BC} = \rank{ABC}.$$ \begin{answer} First observe that $\ker{B} \subseteq \ker{AB}$ since $\forall X \in\mat{q\times 1}{\BBR}$, $$BX = 0 \implies (AB)X = A(BX) = 0.$$Now $$\begin{array}{lll}\dim\ker{B} & = & q - \dim\im{B} \\ & = & q - \rank{B} \\ & = & q - \rank{AB} \\ & = & q - \dim\im{AB} \\ & = & \dim \ker{AB}. \end{array}$$ Thus $\ker B = \ker{AB}.$ Similarly, we can demonstrate that $\ker{ABC} = \ker{BC}.$ Thus $$ \begin{array}{lll}\rank{ABC} & = & \dim\im{ABC} \\ & = & r - \dim\ker{ABC} \\ & = & r - \dim\ker{BC} \\ & = & \dim\im{BC} \\ & = & \rank{BC}.\end{array}$$ \end{answer} \end{pro} \chapter{Determinants} \section{Permutations} \begin{df} Let $S$ be a finite set with $n \geq 1$ elements. A {\em permutation} is a bijective function $\tau : S \rightarrow S$. It is easy to see that there are $n!$ permutations from $S$ onto itself. \index{permutation} \end{df} Since we are mostly concerned with the {\em action} that $\tau$ exerts on $S$ rather than with the particular names of the elements of $S$, we will take $S$ to be the set $S = \{1, 2, 3, \ldots , n \}$. We indicate a permutation $\tau$ by means of the following convenient diagram $$\tau = \begin{bmatrix} 1 & 2 & \cdots & n \cr \tau (1) & \tau (2) & \cdots & \tau (n) \end{bmatrix}.$$ \begin{df} The notation $S_n$ will denote the set of all permutations on $\{1, 2, 3, \ldots , n \}$. Under this notation, the composition of two permutations $(\tau, \sigma)\in S_n ^2$ is $$ \begin{array}{lll}\tau\circ \sigma & = & \begin{bmatrix} 1 & 2 & \cdots & n \cr \tau (1) & \tau (2) & \cdots & \tau (n) \end{bmatrix} \circ\begin{bmatrix} 1 & 2 & \cdots & n \cr \sigma (1) & \sigma (2) & \cdots & \sigma (n) \end{bmatrix} \\ & = & \begin{bmatrix} 1 & 2 & \cdots & n \cr (\tau \circ \sigma) (1) & (\tau \circ \sigma) (2) & \cdots & (\tau \circ \sigma) (n) \end{bmatrix}\end{array}. $$ The $k$-fold composition of $\tau$ is $$ \underbrace{\tau \circ \cdots \circ \tau}_{k\ {\rm compositions}} = \tau^k. $$ \end{df} \begin{rem} We usually do away with the $\circ$ and write $\tau\circ\sigma$ simply as $\tau\sigma$. This ``product of permutations'' is thus simply function composition. \end{rem} Given a permutation $\tau: S \rightarrow S$, since $\tau$ is bijective, $$\tau ^{-1}: S \rightarrow S$$ exists and is also a permutation. In fact if $$\tau = \begin{bmatrix} 1 & 2 & \cdots & n \\ \tau (1) & \tau (2) & \cdots & \tau (n) \end{bmatrix},$$ then $$\tau^{-1} = \begin{bmatrix}\tau (1) & \tau (2) & \cdots & \tau (n) \\ 1 & 2 & \cdots & n \\ \end{bmatrix}.$$ \vspace{2cm} \begin{figure}[htb] $$\psset{unit=2pc}\psline[linewidth=2pt](-1, 0)(1, 0)(0, 1.4142135623730950488016887242097)(-1,0) \psline[linestyle=dotted](0,1.4142135623730950488016887242097)(0,0) \uput[u](0, 1.4142135623730950488016887242097){1} \psline[linestyle=dotted](-1,0)(.5, 0.70710678118654752440084436210485) \uput[l](-1, 0){2} \psline[linestyle=dotted](1,0)(-.5, 0.70710678118654752440084436210485) \uput[r](1,0){3} $$\vspace{1cm}\caption{$S_3$ are rotations and reflexions.} \label{fig:S_3} \end{figure} \begin{exa} The set $S_3$ has $3! = 6$ elements, which are given below. \begin{enumerate} \item $\idefun: \{1, 2, 3\} \rightarrow \{1, 2, 3\}$ with $$\idefun = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix}.$$ \item $\tau_1: \{1, 2, 3\} \rightarrow \{1, 2, 3\} $ with $$\tau_1 = \begin{bmatrix}1 & 2 & 3 \\ 1 & 3 & 2 \end{bmatrix}.$$ \item $\tau_2: \{1, 2, 3\} \rightarrow \{1, 2, 3\} $ with $$\tau_2 = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{bmatrix}.$$ \item $\tau_3: \{1, 2, 3\} \rightarrow \{1, 2, 3\} $ with $$\tau_3 = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix}.$$ \item $\sigma_1: \{1, 2, 3\} \rightarrow \{1, 2, 3\} $ with $$\sigma_1 = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}.$$ \item $\sigma_2: \{1, 2, 3\} \rightarrow \{1, 2, 3\} $ with $$\sigma_2 = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix}.$$ \end{enumerate} \label{ex:S_3}\end{exa} \begin{exa} The compositions $\tau_1 \circ \sigma_1$ and $\sigma_1\circ \tau_1$ can be found as follows. $$ \tau_1 \circ \sigma_1 = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{bmatrix}\circ \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix}1 & 2 & 3 \\ 3 & 2 & 1 \end{bmatrix} = \tau_2 .$$ (We read from right to left $1 \rightarrow 2 \rightarrow 3$ (``1 goes to 2, 2 goes to 3, so 1 goes to 3''), etc. Similarly $$ \sigma_1 \circ \tau_1 = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}\circ\begin{bmatrix}1 & 2 & 3 \\ 1 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix} = \tau_3.$$Observe in particular that $\sigma_1 \circ \tau_1 \neq \tau_1 \circ \sigma_1 .$ Finding all the other products we deduce the following ``multiplication table'' (where the ``multiplication'' operation is really composition of functions). $$\begin{array}{|c||c|c|c|c|c|c|} \hline \circ & \idefun & \tau_1 & \tau_2 & \tau_3 & \sigma_1 & \sigma_2\\ \hline \hline \idefun & \idefun & \tau_1 & \tau_2 & \tau_3 & \sigma_1 & \sigma_2 \\ \hline \tau_1 & \tau_1 & \idefun & \sigma_1 &\sigma_2 & \tau_2 & \tau_3 \\ \hline \tau_2 & \tau_2 & \sigma_2 & \idefun & \sigma_1 & \tau_3&\tau_1 \\ \hline \tau_3 & \tau_3 & \sigma_1& \sigma_2 & \idefun &\tau_1 & \tau_2 \\ \hline \sigma_2 & \sigma_2 & \tau_2& \tau_3& \tau_1& \idefun & \sigma_1 \\ \hline \sigma_1 & \sigma_1 & \tau_3 & \tau_1 & \tau_2& \sigma_2 &\idefun \\ \hline \end{array}$$ \label{ex:tau}\end{exa} The permutations in example \ref{ex:S_3} can be conveniently interpreted as follows. Consider an equilateral triangle with vertices labelled $1$, $2$ and $3$, as in figure \ref{fig:S_3}. Each $\tau_a$ is a reflexion (``flipping'') about the line joining the vertex $a$ with the midpoint of the side opposite $a$. For example $\tau_1$ fixes $1$ and flips $2$ and $3$. Observe that two successive flips return the vertices to their original position and so $(\forall a\in\{1, 2, 3\})(\tau_a ^2 = \idefun)$. Similarly, $\sigma_1$ is a rotation of the vertices by an angle of $120^\circ$. Three successive rotations restore the vertices to their original position and so $\sigma_1 ^3 = \idefun$. \begin{exa} To find $\tau_1^{-1}$ take the representation of $\tau_1$ and exchange the rows: $$ \tau_1^{-1} = \begin{bmatrix}1 & 3 & 2 \\ 1 & 2 & 3 \\ \end{bmatrix}.$$ This is more naturally written as $$ \tau_1^{-1} = \begin{bmatrix}1 & 2 & 3 \\ 1 & 3 & 2 \\ \end{bmatrix}.$$ Observe that $\tau_1^{-1} = \tau _1$. \end{exa} \begin{exa} To find $\sigma_1^{-1}$ take the representation of $\sigma_1$ and exchange the rows: $$ \sigma_1^{-1} = \begin{bmatrix}2 & 3 & 1 \\ 1 & 2 & 3 \\ \end{bmatrix}.$$ This is more naturally written as $$ \sigma_1^{-1} = \begin{bmatrix}1 & 2 & 3 \\ 3 & 1 & 2 \\ \end{bmatrix}.$$ Observe that $\sigma_1^{-1} = \sigma_2$. \end{exa} \section{Cycle Notation} We now present a shorthand notation for permutations by introducing the idea of a {\em cycle}. Consider in $S_9$ the permutation $$\tau = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 1 & 3 & 6 & 9 & 7 & 8 & 4 & 5 \end{bmatrix}.$$ We start with 1. Since 1 goes to 2 and 2 goes back to 1, we write $(12)$. Now we continue with 3. Since 3 goes to 3, we write $(3)$. We continue with 4. As 4 goes 6, 6 goes to 7, 7 goes 8, and 8 goes back to 4, we write $(4678)$. We consider now $5$ which goes to 9 and 9 goes back to 5, so we write $(59)$. We have written $\tau$ as a product of disjoint cycles $$\tau = (12)(3)(4678)(59).$$This prompts the following definition. \begin{df} Let $l\geq 1$ and let $i_1,\ldots,i_l\in\{1,2,\ldots n\}$ be distinct. We write $(i_1\ i_2\ \ldots\ i_l)$ for the element $\sigma\in S_n$ such that $\sigma(i_r)=i_{r+1}, \ \ 1\leq r1$: $\exists a$ such that $\forall k$: $i_k = j_{k+a \mod l}$. Two cycles $(i_1,\ldots,i_l)$ and $(j_1,\ldots,j_m)$ are disjoint if $\{i_1, \ldots , i_l\}\cap\{j_1, \ldots , j_m\}=\varnothing$. Disjoint cycles commute and if $\tau = \sigma_1\sigma_2 \cdots \sigma_t$ is the product of disjoint cycles of length $l_1, l_2, \ldots , l_t$ respectively, then $\tau$ has order $$ \lcm{l_1, l_2, \ldots , l_t}.$$ \end{rem} \begin{exa} A cycle decomposition for $\alpha\in S_9,$ $$\alpha = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 8 & 7 & 6 & 2 & 3 & 4 & 5 & 9 \end{bmatrix}$$ is $$(285)(3746).$$ The order of $\alpha$ is $\lcm{3, 4} = 12$. \label{ex:alpha}\end{exa} \begin{exa} The cycle decomposition $\beta = (123)(567)$ in $S_9$ arises from the permutation $$\beta = \begin{bmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 2 & 3 & 1 & 4 & 6 & 7 & 5 & 8 & 9 \\ \end{bmatrix}.$$ Its order is $\lcm{3,3} = 3$. \label{ex:beta}\end{exa} \begin{exa} Find a shuffle of a deck of $13$ cards that requires $42$ repeats to return the cards to their original order. \end{exa} \begin{solu}Here is one (of many possible ones). Observe that $7 + 6 = 13$ and $7\times 6 = 42$. We take the permutation $$(1\ 2\ 3\ 4\ 5\ 6\ 7)(8\ 9\ 10\ 11\ 12\ 13)$$which has order 42. This corresponds to the following shuffle: For $$i\in \{1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12\},$$ take the $i$th card to the ($i + 1$)th place, take the $7$th card to the first position and the 13th card to the $8$th position. Query: Of all possible shuffles of 13 cards, which one takes the longest to restitute the cards to their original position? \end{solu} \begin{exa} Let a shuffle of a deck of 10 cards be made as follows: The top card is put at the bottom, the deck is cut in half, the bottom half is placed on top of the top half, and then the resulting bottom card is put on top. How many times must this shuffle be repeated to get the cards in the initial order? Explain. \end{exa} \begin{solu}Putting the top card at the bottom corresponds to $$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\cr 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 1 \cr \end{bmatrix}. $$ Cutting this new arrangement in half and putting the lower half on top corresponds to $$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\cr 7 & 8 & 9 & 10 & 1 & 2 & 3 & 4 & 5 & 6 \cr \end{bmatrix}. $$ Putting the bottom card of this new arrangement on top corresponds to $$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\cr 6 & 7 & 8 & 9 & 10 & 1 & 2 & 3 & 4 & 5 \cr \end{bmatrix} = (1\ 6)(2\ 7)(3\ 8)(4\ 9)(5\ 10). $$The order of this permutation is ${\rm lcm} (2, 2, 2, 2, 2) = 2$, so in 2 shuffles the cards are restored to their original position. \end{solu} The above examples illustrate the general case, given in the following theorem. \begin{thm} Every permutation in $S_n$ can be written as a product of disjoint cycles.\label{thm:permu_is_prod_cycles} \end{thm} \begin{pf} Let $\tau \in S_n, a_1\in\{1, 2, \ldots, n\}$. Put $\tau ^k (a_1) = a_{k + 1}, k \geq 0$. Let $a_1, a_2, \ldots , a_s$ be the longest chain with no repeats. Then we have $\tau (a_s) = a_1$. If the $\{a_1, a_2, \ldots , a_s\}$ exhaust $\{1, 2, \ldots, n\}$ then we have $\tau = (a_1\ a_2\ \ldots\ a_s)$. If not, there exist $b_1\in \{1, 2, \ldots, n\} \setminus\{a_1, a_2, \ldots , a_s\}$. Again, we find the longest chain of distinct $b_1, b_2, \ldots, b_t$ such that $\tau (b_k) = b_{k + 1}, k = 1, \ldots, t -1$ and $\tau (b_t) = b_1.$ If the $\{a_1, a_2, \ldots , a_s, b_1, b_2, \ldots, b_t\}$ exhaust all the $\{1, 2, \ldots, n\}$ we have $\tau = (a_1\ a_2\ \ldots\ a_s)( b_1\ b_2\ \ldots\ b_t)$. If not we continue the process and find $$\tau = (a_1\ a_2\ \ldots\ a_s)( b_1\ b_2\ \ldots\ b_t)(c_1\ldots )\ldots . $$This process stops because we have only $n$ elements. \end{pf} \begin{df} A {\em transposition} is a cycle of length $2$.\footnote{A cycle of length $2$ should more appropriately be called a {\em bicycle}.} \end{df} \begin{exa} The cycle $(23468)$ can be written as a product of transpositions as follows $$(23468) = (28)(26)(24)(23).$$ Notice that this decomposition as the product of transpositions is not unique. Another decomposition is $$(23468) = (23)(34)(46)(68).$$ \end{exa} \begin{lem} Every permutation is the product of transpositions. \label{lem:permu_is_prod_bicycles}\end{lem} \begin{pf} It is enough to observe that $$ (a_1\ a_2\ \ldots\ a_s) = (a_1\ a_s)(a_1\ a_{s - 1}) \cdots (a_1\ a_2) $$and appeal to Theorem \ref{thm:permu_is_prod_cycles}. \end{pf} Let $\sigma\in S_n$ and let $(i, j)\in \{1, 2, \ldots, n\}^2, \ \ i\neq j$. Since $\sigma$ is a permutation, $\exists (a, b)\in \{1, 2, \ldots, n\}^2, \ \ a\neq b$, such that $\sigma (j) - \sigma (i) = b - a$. This means that $$ \left|\prod_{1\leq i\sigma(j)\}$, i.e., ${\bf I}(\sigma)$ is the number of inversions that $\sigma$ effects to the identity permutation $\idefun$.\end{rem} \begin{exa} The transposition $(1\ 2)$ has one inversion. \end{exa} \begin{lem} For any transposition $(k\ l)$ we have $\sgn{(k\ l)} = -1$. \end{lem} \begin{pf} Let $\tau$ be transposition that exchanges $k$ and $l$, and assume that $k 1$ be an odd integer. Recall that the binomial coefficients $\binom{n}{k}$ satisfy $\binom{n}{n} = \binom{n}{0} = 1$ and that for $1 \leq k \leq n$, $$\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}. $$Prove that $$ \det\begin{bmatrix} 1 & \binom{n}{1} & \binom{n}{2} & \cdots & \binom{n}{n - 1} & 1 \cr 1 & 1 & \binom{n}{1} & \cdots & \binom{n}{n - 2} & \binom{n}{n - 1} \cr \binom{n}{n - 1} & 1 & 1 & \cdots & \binom{n}{n - 3} & \binom{n}{n - 2} \cr \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \cr \binom{n}{1} & \binom{n}{2} & \binom{n}{3} & \cdots & 1 & 1 \cr \end{bmatrix} = (1 + (-1)^n)^n.$$ \begin{answer} Recall that $\binom{n}{k} = \binom{n}{n - k}$, $$ \sum _{k = 0} ^n \binom{n}{k} = 2^n $$ and$$ \sum _{k = 0} ^n (-1)^{k}\binom{n}{k} = 0, \ \ \ \ {\rm if}\ \ n > 0. $$ Assume that $n$ is odd. Observe that then there are $n + 1$ (an even number) of columns and that on the same row, $\binom{n}{k}$ is on a column of opposite parity to that of $\binom{n}{n - k}$. By performing $C_1 - C_2 + C_3 - C_4 + \cdots + C_n - C_{n + 1} \rightarrow C_1$, the first column becomes all $0$'s, whence the determinant if $0$ if $n$ is odd. \end{answer} \end{pro} \begin{pro} Let $A\in\gl{n}{ \BBF }$, $n>1$. Prove that $\det (\adj{A}) = (\det A)^{n-1}$. \end{pro} \begin{pro} Let $(A, B, S)\in(\gl{n}{ \BBF })^3$. Prove that\begin{dingautolist}{202} \item $ \adj{\adj{A}} = (\det A)^{n-2}A$. \item $\adj{AB} = \adj{A}\adj{B}$. \item $\adj{SAS^{-1}} = S(\adj{ A})S^{-1}$. \end{dingautolist}\end{pro} \begin{pro} If $A\in\gl{2}{\BBF}$, , and let $k$ be a positive integer. Prove that $\det (\underbrace{\mathrm{adj} \cdots \mathrm{adj}}_{k}(A)) = \det A$. \end{pro} \begin{pro} Find the determinant $$\det \begin{bmatrix} (b+c)^2 & ab & ac \cr ab & (a+c)^2 & bc\cr ac & bc & (a+b)^2 \cr \end{bmatrix} $$ {\em by hand, making explicit all your calculations.} \begin{answer} I will prove that $$\det \begin{bmatrix} (b+c)^2 & ab & ac \cr ab & (a+c)^2 & bc\cr ac & bc & (a+b)^2 \cr \end{bmatrix} = 2abc(a+b+c)^3. $$ Using permissible row and column operations, $$\begin{array}{lll} \det \begin{bmatrix} (b+c)^2 & ab & ac \cr ab & (a+c)^2 & bc\cr ac & bc & (a+b)^2 \cr \end{bmatrix} & = & \det \begin{bmatrix} b^2+2bc+c^2 & ab & ac \cr ab & a^2+2ca+c^2 & bc\cr ac & bc & a^2+2ab+b^2 \cr \end{bmatrix}\\ & = \grstep{C_1+C_2+C_3\to C_1} & \det \begin{bmatrix} b^2+2bc+c^2+ab+ac & ab & ac \cr ab+a^2+2ca+c^2+bc & a^2+2ca+c^2 & bc\cr ac+bc+a^2+2ab+b^2 & bc & a^2+2ab+b^2 \cr \end{bmatrix}\\ & = & \det \begin{bmatrix} (b+c)(a+b+c) & ab & ac \cr (a+c)(a+b+c) & a^2+2ca+c^2 & bc\cr (a+b)(a+b+c) & bc & a^2+2ab+b^2 \cr \end{bmatrix}\\ \end{array}$$ Pulling out a factor, the above equals $$ (a+b+c)\det \begin{bmatrix} b+c & ab & ac \cr a+c & a^2+2ca+c^2 & bc\cr a+b & bc & a^2+2ab+b^2 \cr \end{bmatrix} $$ and performing $R_1+R_2+R_3\to R_1$, this is $$ (a+b+c)\det \begin{bmatrix} 2a+2b+2c & ab+a^2+2ca+c^2+bc & ac+bc+ a^2+2ab+b^2 \cr a+c & a^2+2ca+c^2 & bc\cr a+b & bc & a^2+2ab+b^2 \cr \end{bmatrix}$$ Factoring this is $$ (a+b+c)\det \begin{bmatrix} 2(a+b+c) & (a+c)(a+b+c) & (a+b)(a+b+c) \cr a+c & a^2+2ca+c^2 & bc\cr a+b & bc & a^2+2ab+b^2 \cr \end{bmatrix},$$which in turn is $$(a+b+c)^2\det \begin{bmatrix} 2 & a+c & a+b \cr a+c & a^2+2ca+c^2 & bc\cr a+b & bc & a^2+2ab+b^2 \cr \end{bmatrix}$$ Performing $C_2-(a+c)C_1\to C_2$ and $C_3-(a+b)C_1\to C_3$ we obtain $$(a+b+c)^2\det \begin{bmatrix} 2 & -a-c & -a-b \cr a+c & 0 & -a^2-ab-ac\cr a+b &-a^2- ab-ac & 0 \cr \end{bmatrix}$$ This last matrix we will expand by the second column, obtaining that the original determinant is thus $$ (a+b+c)^2\left((a+c)\det\begin{bmatrix}a+c & -a^2-ab-ac \cr a+b & 0 \end{bmatrix} +(a^2+ab+ac)\det\begin{bmatrix} 2 & -a-b \cr a+c & -a^2-ab-ac \end{bmatrix}\right) $$ This simplifies to $$\begin{array}{lll} (a+b+c)^2\left((a+c)(a+b)(a^2+ab+ac\right.)\\ \qquad \left.+(a^2+ab+ac)(-a^2-ab-ac+bc)\right) & = & a(a+b+c)^3((a+c)(a+b)-a^2-ab-ac+bc)\\ & = & 2abc(a+b+c)^3,\end{array} $$ as claimed. \end{answer} \end{pro} \begin{pro} The matrix $$ \begin{bmatrix} a & b & c & d \cr d & a & b & c \cr c & d & a & b \cr b & c & d & a\cr \end{bmatrix} $$ is known as a {\em circulant matrix}. Prove that its determinant is $(a+b+c+d)(a-b+c-d)((a-c)^2+(b-d)^2)$. \begin{answer} We have \begin{eqnarray*}\det \begin{bmatrix} a & b & c & d \cr d & a & b & c \cr c & d & a & b \cr b & c & d & a\cr \end{bmatrix} & \grstep[=]{R_1+R_2+R_3+R_4\to R_1} & \det \begin{bmatrix} a+b+c+d & a+b+c+d & a+b+c+d & a+b+c+d \cr d & a & b & c \cr c & d & a & b \cr b & c & d & a\cr \end{bmatrix}\\ & = & (a+b+c+d)\det \begin{bmatrix} 1 & 1 & 1 & 1 \cr d & a & b & c \cr c & d & a & b \cr b & c & d & a\cr \end{bmatrix}\\ & \grstep[=]{C_4-C_3+C_2-C_1\to C_4} & (a+b+c+d)\begin{bmatrix} 1 & 1 & 1 & 0\cr d & a & b & c-b+a-d \cr c & d & a & b-a+d-c \cr b & c & d & a-d+c-b\cr \end{bmatrix}\\ & = & (a+b+c+d)(a-b+c-d)\begin{bmatrix} 1 & 1 & 1 & 0\cr d & a & b & 1\cr c & d & a & -1 \cr b & c & d & 1\cr \end{bmatrix}\\ & \grstep[=]{R_2+R_3\to R_2,\ R_4+R_3\to R_4} & (a+b+c+d)(a-b+c-d)\begin{bmatrix} 1 & 1 & 1 & 0\cr d+c & a+d & b+a & 0\cr c & d & a & -1 \cr b+c & c+d & a+d & 0\cr \end{bmatrix}\\ &= & (a+b+c+d)(a-b+c-d)\begin{bmatrix} 1 & 1 & 1 \cr d+c & a+d & b+a \cr b+c & c+d & a+d \cr \end{bmatrix}\\ & \grstep[=]{C_1-C_3\to C_1, \ C_2-C_3\to C_2} & (a+b+c+d)(a-b+c-d)\begin{bmatrix} 0 & 0 & 1 \cr d+c-b-a & d-b & b+a \cr b+c-a-d & c-a & a+d \cr \end{bmatrix}\\ & = & (a+b+c+d)(a-b+c-d)\begin{bmatrix} d+c-b-a & d-b \cr b+c-a-d & c-a \cr \end{bmatrix}\\ & = & (a+b+c+d)(a-b+c-d) (d+c-b-a)(c-a)-(d-b)(b+c-a-d) \\ & = & (a+b+c+d)(a-b+c-d)\\ & & \qquad ((c-a)(c-a)+(c-a)(d-b)-(d-b)(c-a)-(d-b)(b-d)) \\ & = & (a+b+c+d)(a-b+c-d)((a-c)^2+(b-d)^2). \end{eqnarray*} \begin{rem} Since $$ (a-c)^2+(b-d)^2=(a-c+i(b-d))(a-c-i(b-d)), $$ the above determinant is then $$ (a+b+c+d)(a-b+c-d)(a+ib-c-id)(a-ib-c+id). $$ Generalisations of this determinant are possible using roots of unity. \end{rem} \end{answer} \end{pro} \section{Determinants and Linear Systems} \begin{thm}\label{thm:systems_determinants_invertibility}Let $A\in\mat{n\times n}{ \BBF }$. The following are all equivalent \begin{dingautolist}{202} \item \label{thm:sys_det_1} $\det A \neq 0_{\BBF }$. \item \label{thm:sys_det_2}$A$ is invertible. \item\label{thm:sys_det_3} There exists a unique solution $X\in\mat{n\times 1}{ \BBF }$ to the equation $AX = Y$. \item \label{thm:sys_det_4}If $AX = {\bf 0}_{n\times 1}$ then $X = {\bf 0}_{n\times 1}$. \end{dingautolist}\end{thm} \begin{pf} We prove the implications in sequence: \\ $\ref{thm:sys_det_1} \implies \ref{thm:sys_det_2}$: follows from Corollary \ref{cor:inverse_via_adjoint}\\ $\ref{thm:sys_det_2} \implies \ref{thm:sys_det_3}$: If $A$ is invertible and $AX = Y$ then $X = A^{-1}Y$ is the unique solution of this equation.\\ $\ref{thm:sys_det_3} \implies \ref{thm:sys_det_4}$: follows by putting $Y = {\bf 0}_{n\times 1}$\\ $\ref{thm:sys_det_4} \implies \ref{thm:sys_det_1}$: Let $R$ be the row echelon form of $A$. Since $RX = {\bf 0}_{n\times 1}$ has only $X={\bf 0}_{n\times 1}$ as a solution, every entry on the diagonal of $R$ must be non-zero, $R$ must be triangular, and hence $\det R \neq 0_{\BBF }$. Since $A = PR$ where $P$ is an invertible $n\times n$ matrix, we deduce that $\det A = \det P\det R \neq 0_{\BBF }$.\\ \end{pf} The contrapositive form of the implications \ref{thm:sys_det_1} and \ref{thm:sys_det_4} will be used later. Here it is for future reference. \begin{cor}\label{cor:0determinant_non_null_kernel} Let $A\in\mat{n\times n}{ \BBF }$. If there is $X\neq {\bf 0}_{n\times 1}$ such that $AX = {\bf 0}_{n\times 1}$ then $\det A = 0_{\BBF }$. \end{cor} \section*{\psframebox{Homework}} \begin{pro} For which $a$ is the matrix $\begin{bmatrix}-1 & 1 & 1 \cr 1 & a & 1 \cr 1 & 1 & a \cr \end{bmatrix}$ singular (non-invertible)? \end{pro} \chapter{Eigenvalues and Eigenvectors} \section{Similar Matrices} \begin{df} We say that $A\in\mat{n\times n}{ \BBF }$ is {\em similar} to $B\in\mat{n\times n}{ \BBF }$ if there exist a matrix $P\in\gl{n}{ \BBF }$ such that $$B = PAP^{-1}. $$ \index{matrix!similarity} \end{df} \begin{thm} Similarity is an equivalence relation. \end{thm} \begin{pf} Let $A\in\mat{n\times n}{ \BBF }$. Then $A = {\bf I}_nA{\bf I}_n ^{-1}$, so similarity is reflexive. If $B = PAP^{-1} $ ($P\in\gl{n}{ \BBF }$ ) then $A = P^{-1}BP$ so similarity is symmetric. Finally, if $B = PAP^{-1}$ and $C = QBQ^{-1} $ ($P\in\gl{n}{ \BBF }$ , $Q\in\gl{n}{ \BBF }$) then $C = QPAP^{-1}Q^{-1} = QPA(QP)^{-1}$ and so similarity is transitive. \end{pf} Since similarity is an equivalence relation, it partitions the set of $n\times n$ matrices into equivalence classes by Theorem \ref{thm:equiv_relation_yields_partition}. \begin{df} A matrix is said to be {\em diagonalisable} if it is similar to a diagonal matrix. \end{df} Suppose that $$ A = \begin{bmatrix}\lambda_1 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n \end{bmatrix}.$$Then if $K$ is a positive integer $$A^K = \begin{bmatrix}\lambda_1 ^K & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 ^K& 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n ^K \end{bmatrix}. $$In particular, if $B$ is similar to $A$ then $$B^K = \underbrace{(PAP^{-1})(PAP^{-1})\cdots (PAP^{-1})}_{K\ \mathrm{factors}} = PA^KP^{-1} = P\begin{bmatrix}\lambda_1 ^K & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 ^K& 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n ^K \end{bmatrix}P^{-1}, $$so we have a simpler way of computing $B^K$. Our task will now be to establish when a particular square matrix is diagonalisable. \section{Eigenvalues and Eigenvectors} Let $A\in\mat{n\times n}{ \BBF }$ be a square diagonalisable matrix. Then there exist $P\in\gl{n}{ \BBF }$ and a diagonal matrix $D\in\mat{n\times n}{ \BBF }$ such that $P^{-1}AP = D$, whence $AP = DP$. Put $$D = \begin{bmatrix}\lambda_1 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n \end{bmatrix}, \ \ \ P = [P_1; P_2; \cdots ; P_n], $$where the $P_k$ are the columns of $P$. Then $$AP=DP \implies [AP_1; AP_2; \cdots ; AP_n] = [\lambda_1P_1; \lambda_2P_2; \cdots ; \lambda_nP_n], $$ from where it follows that $AP_k = \lambda_kP_k$. This motivates the following definition. \begin{df} Let $V$ be a finite-dimensional vector space over a field $\BBF$ and let $T:V \rightarrow V$ be a linear transformation. A scalar $\lambda \in \BBF$ is called an {\em eigenvalue} of $T$ if there is a $\v{v}\neq \v{0}$ (called an {\em eigenvector}) such that $T(\v{v}) = \lambda\v{v}$. \index{eigenvalue} \index{eigenvector} \end{df} \begin{exa} Shew that if $\lambda$ is an eigenvalue of $T:V \rightarrow V$, then $\lambda ^k$ is an eigenvalue of $T^k:V \rightarrow V$, for $k\in\BBN\setminus \{0\} $. \end{exa}\begin{solu}Assume that $T(\v{v}) = \lambda\v{v}$. Then $$T^2(\v{v}) = TT(\v{v}) = T(\lambda\v{v}) = \lambda T(\v{v}) = \lambda(\lambda\v{v}) = \lambda^2\v{v}.$$Continuing the iterations we obtain $T^k(\v{v}) = \lambda^k \v{v}$, which is what we want. \end{solu} \begin{thm} Let $A\in\mat{n\times n}{ \BBF }$ be the matrix representation of $T:V \rightarrow V$. Then $\lambda\in \BBF$ is an eigenvalue of $T$ if an only if $\det (\lambda {\bf I}_n - A) = 0_{\BBF }.$ \end{thm} \begin{pf} $\lambda$ is an eigenvalue of $A$ $\iff$ there is $\v{v} \neq \v{0}$ such that $A\v{v} = \lambda\v{v}$ $\iff$ $\lambda \v{v} - A\v{v} = \v{0}$ $\iff$ $\lambda {\bf I}_n \v{v} - A\v{v}= \v{0}$ $\iff$ $\det (\lambda {\bf I}_n - A) = 0_{\BBF }$ by Corollary \ref{cor:0determinant_non_null_kernel}.\end{pf} \begin{df} The equation $$\det (\lambda {\bf I}_n - A) = 0_{\BBF } $$is called the {\em characteristic equation of $A$} or {\em secular equation of $A$}. The polynomial $p(\lambda) = \det (\lambda {\bf I}_n - A)$ is the characteristic polynomial of $A$. \index{characteristic equation} \end{df} \begin{exa} Let $\dis{A = \begin{bmatrix} 1 & 1 & 0 & 0 \cr 1 & 1 & 0 & 0 \cr 0 & 0 & 1 & 1 \cr 0 & 0 & 1 & 1 \cr \end{bmatrix}}$. Find \begin{dingautolist}{202} \item The characteristic polynomial of $A$. \item The eigenvalues of $A$. \item The corresponding eigenvectors. \end{dingautolist} \end{exa} \begin{solu}We have \begin{dingautolist}{202} \item $$\begin{array}{lll} \det (\lambda {\bf I}_4 - A) & = & \det\begin{bmatrix} \lambda -1 & -1 & 0 & 0 \cr -1 & \lambda -1 & 0 & 0 \cr 0 & 0 & \lambda -1 & -1 \cr 0 & 0 & -1 & \lambda -1 \cr\end{bmatrix} \\ & = & (\lambda -1)\det \begin{bmatrix}\lambda - 1 & 0 & 0 \cr 0 & \lambda -1 & -1 \cr 0 & -1 & \lambda -1 \cr \end{bmatrix} + \det\begin{bmatrix} -1 & 0 & 0 \cr 0 & \lambda -1 & -1 \cr 0 & -1 & \lambda -1\end{bmatrix}\\ & = & (\lambda -1)((\lambda -1)((\lambda -1)^2-1)) + (-((\lambda -1)^2 -1)) \\ & = & (\lambda -1)((\lambda -1)(\lambda -2)(\lambda)) - (\lambda -2)(\lambda) \\ & = & (\lambda -2)(\lambda)((\lambda -1)^2-1) \\ & = & (\lambda -2)^2(\lambda)^2 \\ \end{array}$$ \item The eigenvalues are clearly $\lambda = 0$ and $\lambda = 2$. \item If $\lambda = 0$, then $$\begin{array}{lll}0 {\bf I}_4 - A & = & \begin{bmatrix} -1 & -1 & 0 & 0 \cr -1 & -1 & 0 & 0 \cr 0 & 0 & -1 & -1 \cr 0 & 0 & -1 & -1 \cr \end{bmatrix}.\\ \end{array}$$This matrix has row-echelon form $$\begin{bmatrix} -1 & -1 & 0 & 0 \cr 0 & 0 & -1 & -1 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr\end{bmatrix}, $$and if $$\begin{bmatrix} -1 & -1 & 0 & 0 \cr 0 & 0 & -1 & -1 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr\end{bmatrix}\colvec{a\\ b \\ c\\ d} = \colvec{0 \\ 0\\ 0 \\ 0}, $$ then $c = -d$ and $a = -b$ Thus the general solution of the system $(0 {\bf I}_4 - A)X = {\bf 0}_{n\times 1}$ is $$\colvec{a\\ b\\ c\\ d} = a\colvec{1 \\ -1 \\ 0 \\ 0} + c\colvec{0 \\ 0 \\ 1 \\ -1}. $$ If $\lambda = 2$, then $$\begin{array}{lll}2 {\bf I}_4 - A & = & \begin{bmatrix} 1 & -1 & 0 & 0 \cr -1 & 1 & 0 & 0 \cr 0 & 0 & 1 & -1 \cr 0 & 0 & -1 & 1 \cr \end{bmatrix}.\\ \end{array}$$This matrix has row-echelon form $$\begin{bmatrix} -1 & 1 & 0 & 0 \cr 0 & 0 & 1 & -1 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr\end{bmatrix}, $$and if $$\begin{bmatrix} 1 & -1 & 0 & 0 \cr 0 & 0 & -1 & 1 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr\end{bmatrix}\colvec{a\\ b \\ c\\ d} = \colvec{0 \\ 0\\ 0 \\ 0}, $$ then $c = d$ and $a = b$ Thus the general solution of the system $(2 {\bf I}_4 - A)X = {\bf 0}_{n\times 1}$ is $$\colvec{a\\ b\\ c\\ d} = a\colvec{1 \\ 1 \\ 0 \\ 0} + c\colvec{0 \\ 0 \\ 1 \\ 1}. $$ Thus for $\lambda = 0$ we have the eigenvectors $$\colvec{1 \\ -1 \\ 0 \\ 0}, \colvec{0 \\ 0 \\ 1 \\ -1} $$and for $\lambda = 2$ we have the eigenvectors $$\colvec{1 \\ 1 \\ 0 \\ 0}, \colvec{0 \\ 0 \\ 1 \\ 1}. $$ \end{dingautolist} \end{solu} \begin{thm} If $\lambda = 0_{\BBF }$ is an eigenvalue of $A$, then $A$ is non-invertible. \end{thm} \begin{pf} Put $p(\lambda) = \det(\lambda {\bf I}_n - A)$. Then $p(0_{\BBF }) = \det(-A) = (-1)^n\det A$ is the constant term of the characteristic polynomial. If $\lambda = 0_{\BBF }$ is an eigenvalue then $$ p(0_{\BBF }) = 0_{\BBF } \implies \det A = 0_{\BBF }, $$and hence $A$ is non-invertible by Theorem \ref{thm:systems_determinants_invertibility}. \end{pf} \begin{thm} Similar matrices have the same characteristic polynomial. \end{thm} \begin{pf} We have $$\begin{array}{lll}\det (\lambda {\bf I}_n -SAS^{-1} ) & = & \det (\lambda S{\bf I}_nS^{-1} -SAS^{-1} ) \\ & = & \det S(\lambda {\bf I}_n-A)S^{-1} \\ & = &(\det S)(\det(\lambda {\bf I}_n-A))(\det S^{-1}) \\ & = & (\det S)(\det(\lambda {\bf I}_n-A))\left(\dfrac{1}{\det S}\right) \\ & = & \det(\lambda {\bf I}_n-A), \end{array} $$from where the result follows.\end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Find the eigenvalues and eigenvectors of $\dis{A = \begin{bmatrix} 1 & -1 \cr -1 & 1 \cr \end{bmatrix}}$ \begin{answer}We have $$\det (\lambda {\bf I}_2 - A) = \det \begin{bmatrix} \lambda -1 & 1 \cr 1 & \lambda - 1 \cr \end{bmatrix} = (\lambda -1)^2 - 1 = \lambda (\lambda -2), $$whence the eigenvalues are $0$ and $2$. For $\lambda = 0$ we have $$ 0 {\bf I}_2 - A = \begin{bmatrix} -1 & 1 \cr 1 & -1 \cr \end{bmatrix}.$$This has row-echelon form $$\begin{bmatrix} 1 & -1 \cr 0 & 0 \cr \end{bmatrix}.$$ If $$\begin{bmatrix} 1 & -1 \cr 0 & 0 \cr \end{bmatrix}\colvec{a\\ b} = \colvec{0 \\ 0}$$then $a = b$. Thus $$\colvec{a\\ b} = a\colvec{1 \\ 1} $$ and we can take $\dis{\colvec{1\\ 1}}$ as the eigenvector corresponding to $\lambda = 0$. Similarly, for $\lambda = 2,$ $$ 2 {\bf I}_2 - A = \begin{bmatrix} 1 & 3 \cr 1 & 3 \cr \end{bmatrix},$$which has row-echelon form $$\begin{bmatrix} 1 & 3\cr 0 & 0 \cr \end{bmatrix}.$$ If $$\begin{bmatrix} 1 & 3 \cr 0 & 0 \cr \end{bmatrix}\colvec{a\\ b} = \colvec{0 \\ 0}$$then $a =-3b$. Thus $$\colvec{a\\ b} = a\colvec{1 \\ -3} $$ and we can take $\dis{\colvec{1\\ -3}}$ as the eigenvector corresponding to $\lambda = 2$. \end{answer} \end{pro} \begin{pro} Let $A$ be a $2\times 2$ matrix over some some field $\BBF$. Prove that the characteristic polynomial of $A$ is $$ \lambda ^2 - (\tr{A})\lambda + \det A. $$ \end{pro} \begin{pro} A matrix $A\in\mat{2\times 2}{\BBR}$ satisfies $\tr{A}=-1$ and $\det{A}=-6$. Find the value of $\det ({\bf I}_2+A)$. \end{pro} \begin{pro} A $2\times 2$ matrix $A$ with real entries has characteristic polynomial $p(\lambda ) = \lambda ^2 + 2\lambda -1$. Find the value of $\det (2{\bf I}_2 + A)$. \end{pro} \begin{pro} Let $\dis{A = \begin{bmatrix}0 & 2 & -1 \cr 2 & 3 & -2 \cr -1 & -2 & 0 \cr\end{bmatrix}}$. Find \begin{dingautolist}{202} \item The characteristic polynomial of $A$. \item The eigenvalues of $A$. \item The corresponding eigenvectors. \end{dingautolist} \begin{answer} \begin{dingautolist}{202} \item We have $$\begin{array}{lll}\det (\lambda {\bf I}_3 - A) & = & \det\begin{bmatrix}\lambda & -2 & 1 \cr -2 & \lambda- 3 & 2 \cr 1 & 2 & \lambda \cr\end{bmatrix} \\ & = & \lambda\det\begin{bmatrix} \lambda - 3 & 2 \cr 2 & \lambda \cr \end{bmatrix} + 2\det\begin{bmatrix} -2 & 2 \cr 1 & \lambda \cr \end{bmatrix} + \det\begin{bmatrix} -2 & \lambda - 3\cr 1 & 2\cr \end{bmatrix}\\ & = & \lambda (\lambda ^2 -3\lambda - 4) + 2(-2\lambda - 2) + (-\lambda - 1)\\ & = & \lambda (\lambda - 4)(\lambda + 1) - 5(\lambda + 1) \\ & = & (\lambda^2 - 4\lambda - 5)(\lambda + 1) \\ & = & (\lambda + 1)^2(\lambda - 5)\end{array}$$ \item The eigenvalues are $-1, -1, 5$. \item If $\lambda = -1$, $$\begin{array}{lll}(-{\bf I}_3 - A) = \begin{bmatrix}-1 & -2 & 1 \cr -2 & -4 & 2 \cr 1 & 2 & -1 \cr \end{bmatrix}\colvec{a \\ b\\ c} = \colvec{0 \\ 0 \\ 0} & \iff & a = -2b + c\\ & \iff & \colvec{a\\ b\\ c} = b\colvec{-2 \\ 1 \\ 0} + c\colvec{1 \\ 0 \\ 1}. \end{array}$$We may take as eigenvectors $\colvec{-2 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 1}$, which are clearly linearly independent. \bigskip If $\lambda = 5$, $$\begin{array}{lll} (5{\bf I}_3 - A) = \begin{bmatrix}5 & -2 & 1 \cr -2 & 2 & 2 \cr 1 & 2 & 5 \cr \end{bmatrix}\colvec{a \\ b\\ c} = \colvec{0 \\ 0 \\ 0} & \iff & a = -c, b = -2c,\\ & \iff & \colvec{a\\ b\\ c} = c\colvec{-1 \\ -2 \\ 1}. \end{array}$$We may take as eigenvector $\colvec{1 \\ 2 \\ -1}$. \end{dingautolist} \end{answer} \end{pro} \begin{pro} Describe all matrices $A\in\mat{2\times 2}{\BBR}$ having eigenvalues $1$ and $-1$. \begin{answer}The characteristic polynomial of $A$ must be $\lambda ^2-1$, which means that $\tr{A}=0$ and $\det A = -1$. Hence $A$ must be of the form $\begin{bmatrix}a & c \cr b & -a \end{bmatrix}$, with $-a^2-bc=-1$, that is, $a^2+bc=1$. \end{answer} \end{pro} \begin{pro} Let $A\in \mat{n\times n}{\BBR}$. Demonstrate that $A$ has the same characteristic polynomial as its transpose. \begin{answer} We must shew that $\det (\lambda {\bf I}_n - A)=\det (\lambda {\bf I}_n - A)^T$. Now, recall that the determinant of a square matrix is the same as the determinant of its transpose. Hence $$ \det (\lambda {\bf I}_n - A) = \det ((\lambda {\bf I}_n - A)^T) = \det (\lambda {\bf I}_n ^T - A^T) = \det (\lambda {\bf I}_n - A^T), $$ as we needed to shew. \end{answer} \end{pro} \end{multicols} \section{Diagonalisability} In this section we find conditions for diagonalisability. \begin{thm} Let $\{\v{v}_1,\v{v}_2, \ldots, \v{v}_k \}\subset V$ be the eigenvectors corresponding to the {\em different} eigenvalues $\{\lambda_1, \lambda_2, \ldots, \lambda_k\}$ (in that order). Then these eigenvectors are linearly independent. \end{thm} \begin{pf} Let $T:V\rightarrow V$ be the underlying linear transformation. We proceed by induction. For $k = 1$ the result is clear. Assume that every set of $k-1$ eigenvectors that correspond to $k-1$ distinct eigenvalues is linearly independent and let the eigenvalues $\lambda_1, \lambda_2, \ldots , \lambda_{k-1}$ have corresponding eigenvectors $\v{v}_1, \v{v}_2, \ldots , \v{v}_{k-1}$. Let $\lambda$ be a eigenvalue different from the $\lambda_1, \lambda_2, \ldots , \lambda_{k-1}$ and let its corresponding eigenvector be $\v{v}$. If $\v{v}$ were linearly dependent of the $\v{v}_1, \v{v}_2, \ldots , \v{v}_{k-1}$, we would have \begin{equation}\label{pf:distinct_eigen1}x\v{v} + x_1\v{v}_1 + x_2\v{v}_2 + \cdots + x_{k-1}\v{v}_{k-1} = \v{0}. \end{equation}Now $$T(x\v{v} + x_1\v{v}_1 + x_2\v{v}_2 + \cdots + x_{k-1}\v{v}_{k-1}) = T(\v{0}) = \v{0}, $$by Theorem \ref{thm:linear_takes_0_to_0}. This implies that \begin{equation}\label{pf:distinct_eigen2}x\lambda\v{v} + x_1\lambda_1\v{v}_1 + x_2\lambda_2\v{v}_2 + \cdots + x_{k-1}\lambda_{k-1}\v{v}_{k-1} = \v{0}.\end{equation} From \ref{pf:distinct_eigen2} take away $\lambda$ times \ref{pf:distinct_eigen1}, obtaining \begin{equation}\label{pf:distinct_eigen3} x_1(\lambda_1-\lambda)\v{v}_1 + x_2(\lambda_2\v{v}_2 + \cdots + x_{k-1}(\lambda_{k-1}-\lambda)\v{v}_{k-1} = \v{0} \end{equation} Since $\lambda - \lambda_i \neq 0_{\BBF }$ \ref{pf:distinct_eigen3} is saying that the eigenvectors $\v{v}_1, \v{v}_2, \ldots , \v{v}_{k-1}$ are linearly dependent, a contradiction. Thus the claim follows for $k$ distinct eigenvalues and the result is proven by induction. \end{pf} \begin{thm}A matrix $A\in\mat{n\times n}{ \BBF }$ is diagonalisable if and only if it possesses $n$ linearly independent eigenvectors. \end{thm} \begin{pf} Assume first that $A$ is diagonalisable, so there exists $P\in\gl{n}{ \BBF }$ and $$D = \begin{bmatrix}\lambda_1 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n \end{bmatrix} $$ such that $$P^{-1}AP = \begin{bmatrix}\lambda_1 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n \end{bmatrix}. $$ Then $$[AP_1; AP_2; \cdots ; AP_n] = AP = P\begin{bmatrix}\lambda_1 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n \end{bmatrix} = [\lambda_1P_1; \lambda_2P_2; \cdots ; \lambda_nP_n], $$where the $P_k$ are the columns of $P$. Since $P$ is invertible, the $P_k$ are linearly independent by virtue of Theorems \ref{thm:lin_ind_columns} and \ref{thm:systems_determinants_invertibility}. \bigskip Conversely, suppose now that $\v{v}_1, \ldots , \v{v}_n$ are $n$ linearly independent eigenvectors, with corresponding eigenvalues $\lambda_1, \lambda_2, \ldots , \lambda_n$. Put $$P = [\v{v}_1; \ldots ; \v{v}_n], \ \ \ D = \begin{bmatrix}\lambda_1 & 0 & 0 & \cdots & 0 \cr 0 & \lambda_2 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr 0 & 0 & 0 & \cdots & \lambda_n \end{bmatrix}. $$Since $A\v{v}_i = \lambda_i\v{v}_i$ we see that $AP = PD$. Again $ P$ is invertible by Theorems \ref{thm:lin_ind_columns} and \ref{thm:systems_determinants_invertibility} since the $\v{v}_k$ are linearly independent. Left multiplying by $P^{-1}$ we deduce $P^{-1}AP = D$, from where $A$ is diagonalisable. \end{pf} \begin{exa} \label{exa:diagonalisable3x3}Shew that the following matrix is diagonalisable: $$\begin{bmatrix}1 & - 1 & -1 \cr 1 & 3 & 1 \cr -3 & 1 & -1 \cr\end{bmatrix}$$and find a diagonal matrix $D$ and an invertible matrix $P$ such that $$A = PDP^{-1}.$$ \end{exa} \begin{solu} Verify that the characteristic polynomial of $A$ is $$\lambda^3-3\lambda^2-4\lambda+12 = (\lambda - 2)(\lambda + 2)(\lambda - 3).$$ The eigenvector for $\lambda = -2$ is $$\colvec{1 \\ -1 \\ 4}.$$ The eigenvector for $\lambda = 2$ is $$\colvec{-1 \\ 0 \\ 1}.$$ The eigenvector for $\lambda = 3$ is $$\colvec{-1 \\ 1 \\ 1}.$$ We may take $$D = \begin{bmatrix}-2 & 0 & 0 \cr 0 & 2 & 0 \cr 0 & 0 & 3 \cr\end{bmatrix}, \ \ P = \begin{bmatrix}1 & -1 & 4 \cr -1 & 0 & 1 &\cr -1 & 1 & 1 \cr\end{bmatrix}.$$ We also find $$P^{-1} = \begin{bmatrix}\frac{1}{5} & -1 & \frac{1}{5} \cr 0 & -1 & 1 \cr \frac{1}{5} & 0 & \frac{1}{5} \cr \end{bmatrix}.$$ \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro}Let $A$ be a $2\times 2$ matrix with eigenvalues $1$ and $-2$ and corresponding eigenvectors $\dis{\begin{bmatrix}1 \cr 0\cr\end{bmatrix}}$ and $\dis{\begin{bmatrix}1 \cr -1\cr\end{bmatrix}}$, respectively. Determine $A^{10}.$ \begin{answer} Put $$D = \begin{bmatrix}1 & 0 \cr 0 & -2\end{bmatrix}, \ \ , P = \begin{bmatrix}1 & 0 \cr 1 & -1 \cr \end{bmatrix}.$$ We find $$P^{-1} = \begin{bmatrix}1 & 1 \cr 0 & -1\end{bmatrix}.$$ Since $A = PDP^{-1}$ $$A^{10} = PD^{10}P^{-1} = \begin{bmatrix}1 & 0 \cr 1 & -1 \cr \end{bmatrix}\begin{bmatrix}1 & 0 \cr 0 & 1024\end{bmatrix}\begin{bmatrix}1 & 1 \cr 0 & -1\end{bmatrix} = \begin{bmatrix}1 & -1023 \cr 0 & 1024\end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Consider the matrix $A=\begin{bmatrix} 9 & -4 \cr 20 & -9 \cr \end{bmatrix}$. \begin{enumerate} \item Find the characteristic polynomial of $A$. \item Find the eigenvalues of $A$. \item Find the eigenvectors of $A$. \item If $A^{20} = \begin{bmatrix}a & b \cr c & d \cr \end{bmatrix}$, find $a+d$. \end{enumerate} \begin{answer} \noindent \begin{enumerate} \item $A$ has characteristic polynomial $\det\begin{bmatrix} \lambda -9 & 4 \cr -20 & \lambda+9 \cr \end{bmatrix}=(\lambda -9)(\lambda +9)+80=\lambda ^2-1=(\lambda -1)(\lambda +1)$. \item $(\lambda -1)(\lambda +1)=0\implies \lambda \in \{-1,1\}$. \item For $\lambda =-1$ we have $$ \begin{bmatrix} 9 & -4 \cr 20 & -9 \cr \end{bmatrix}\colvec{a\\ b}=-1\colvec{a\\ b}\implies 10a=4b \implies a=\dfrac{2b}{5}, $$ so we can take $\colvec{2\\ 5}$ as an eigenvector. \bigskip For $\lambda =1$ we have $$ \begin{bmatrix} 9 & -4 \cr 20 & -9 \cr \end{bmatrix}\colvec{a\\ b}=1\colvec{a\\ b}\implies 8a=4b \implies a=\dfrac{b}{2}, $$ so we can take $\colvec{1\\ 2}$ as an eigenvector. \item We can do this problem in at least three ways. The quickest is perhaps the following. \bigskip Recall that a $2\times 2$ matrix has characteristic polynomial $\lambda ^2 - (\tr{A})\lambda+\det A$. Since $A$ has eigenvalues $-1$ and $1$, $A^{20}$ has eigenvalues $1^{20}=1$ and $(-1)^{20}=1$, i.e., the sole of $A^{20}$ is $1$ and so $A^{20}$ has characteristic polynomial $(\lambda -1)^2=\lambda ^2 -2\lambda + 1$. This means that $-\tr{A^{20}}=-2$ and so $\tr{A^{20}}=2$. \bigskip The direct way would be to argue that $$\begin{array}{lll} A^{20} & = & \begin{bmatrix} 2 & 1 \cr 5 & 2\end{bmatrix}\begin{bmatrix} -1 & 0 \cr 0 & 1\end{bmatrix}^{20} \begin{bmatrix} 2 & 1 \cr 5 & 2\end{bmatrix} ^{-1} \\ & = &\begin{bmatrix} 2 & 1 \cr 5 & 2\end{bmatrix}\begin{bmatrix} 1 & 0 \cr 0 & 1\end{bmatrix} \begin{bmatrix} 2 & 1 \cr 5 & 2\end{bmatrix}^{-1}\\ & = & \begin{bmatrix} 1 & 0 \cr 0 & 1\end{bmatrix}, \end{array}$$and so $a+d=2$. One may also use the fact that $\tr{XY}=\tr{YX}$ and hence $$\tr{A^{20}}=\tr{PD^{20}P^{-1}}=\tr{PP^{-1}D^{20}}=\tr{D^{20}}=2. $$ \end{enumerate} \end{answer} \end{pro} \begin{pro} Let $A\in\mat{3\times 3}{\BBR}$ have characteristic polynomial $$(\lambda + 1)^2(\lambda - 3).$$ One of the eigenvalues has two eigenvectors $\begin{bmatrix}1 \cr 0 \cr 0\cr\end{bmatrix}$ and $\begin{bmatrix}1 \cr 1\cr 0 \cr \end{bmatrix}$. The other eigenvalue has corresponding eigenvector $\colvec{1 \\ 1 \\ 1}$. Determine $A$.\begin{answer} Put $$D = \begin{bmatrix}-1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 3 \cr \end{bmatrix}, \ \ \ X = \begin{bmatrix}1 & 1 & 1 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix}.$$ Then we know that $A = XDX^{-1}$ and so we need to find $X^{-1}$. But this is readily obtained by performing $R_1 - R_2 \rightarrow R_1$ and $R_2 - R_3 \rightarrow R_3$ in the augmented matrix $$\begin{bmat}{ccc|ccc}1 & 1 & 1 & 1 & 0 & 0 \cr 0 & 1 & 1 & 0 & 1 & 0 \cr 0 & 0 & 1 & 0 & 0 & 1 \cr\end{bmat}, $$getting $$X^{-1} = \begin{bmatrix}1 & -1 & 0 \cr 0 & 1 & -1 \cr 0 & 0 & 1 \cr \end{bmatrix}.$$ Thus $$\begin{array}{lll} A & = & XDX^{-1} \\ & = & \begin{bmatrix}1 & 1 & 1 \cr 0 & 1 & 1 \cr 0 & 0 & 1 \cr \end{bmatrix} \begin{bmatrix}-1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 3 \cr \end{bmatrix}\begin{bmatrix}1 & -1 & 0 \cr 0 & 1 & -1 \cr 0 & 0 & 1 \cr \end{bmatrix}\\ & = &\begin{bmatrix}-1 & 0 & 4 \cr 0 & -1 & 4 \cr 0 & 0 & 3 \cr \end{bmatrix}. \end{array}$$ \end{answer} \end{pro} \begin{pro} Let $$A = \begin{bmatrix} 0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr 1 & 0 & 0 & 0 \cr\end{bmatrix}.$$ \begin{enumerate} \item Find $\det A$. \item Find $A^{-1}$. \item Find $\rank{A - {\bf I}_4}$. \item Find $\det (A - {\bf I}_4)$. \item Find the characteristic polynomial of $A$. \item Find the eigenvalues of $A$. \item Find the eigenvectors of $A$. \item Find $A^{10}$. \end{enumerate} \begin{answer} The determinant is $1$, $A = A^{-1}$, and the characteristic polynomial is $(\lambda ^2 - 1)^2$. \end{answer} \end{pro} \begin{pro} Consider the matrix $$A = \begin{bmatrix} 1 & a & 1 \cr 0 & 1 & b \cr 0 & 0 & c \cr \end{bmatrix}. $$ \begin{dingautolist}{202} \item Find the characteristic polynomial of $A$. \item Explain whether $A$ is diagonalisable when $a=0, c=1$. \item Explain whether $A$ is diagonalisable when $a\neq 0, c=1$. \item Explain whether $A$ is diagonalisable when $ c\neq 1$. \end{dingautolist} \end{pro} \begin{pro} Find a closed formula for $A^n$, if $$ A = \begin{bmatrix} -7 & -6 \cr 12 & 10 \cr \end{bmatrix} .$$ \begin{answer} We find $$ \det (\lambda {\bf I}_2-A)= \det \begin{bmatrix} \lambda +7 & 6 \cr -12 & \lambda -10 \cr \end{bmatrix} = \lambda ^2-3\lambda +2=(\lambda-1)(\lambda-2). $$ A short calculation shews that the eigenvalue $\lambda =2$ has eigenvector $\colvec{2\\ -3}$ and that the eigenvalue $\lambda =1$ has eigenvector $\colvec{3\\ -4}$. Thus we may form $$D = \begin{bmatrix} 2 & 0 \cr 0 & 1\cr \end{bmatrix}, \quad P =\begin{bmatrix} 2 & -3 \cr 3 & -4\cr \end{bmatrix} , \quad P^{-1}=\begin{bmatrix} 4 & -3 \cr 3 & 1\cr \end{bmatrix}.$$ This gives $$ A=PDP^{-1} \implies A^n = PD^nP^{-1} = \begin{bmatrix} 2 & -3 \cr 3 & -4\cr \end{bmatrix}\begin{bmatrix} 2^n & 0 \cr 0 & 1\cr \end{bmatrix}\begin{bmatrix} 4 & -3 \cr 3 & 1\cr \end{bmatrix} =\begin{bmatrix} -8\cdot 2^n+9 & -6\cdot 2^n+6 \cr 12\cdot 2^n-12 & 9\cdot 2^n-8\end{bmatrix}. $$ \end{answer} \end{pro} \begin{pro} Let $U\in\mat{n\times n}{\BBR}$ be a square matrix all whose entries are equal to $1$. \begin{enumerate} \item Demonstrate that $U^2 = nU$. \item Find $\det U$.\item Prove that $\det(\lambda {\bf I}_n - U) = \lambda ^{n - 1}(\lambda - n).$ \item Shew that $\dim\ker U = n - 1$. \item Shew that $$U = P\begin{bmatrix}n & 0 & \cdots & 0 \cr 0 & 0 & \cdots & 0 \cr \vdots & \vdots & \vdots & \vdots \cr 0 & 0 & \cdots & 0 \cr\end{bmatrix} P^{-1}, $$where $$P = \begin{bmatrix}1 & 1 & 0 & \cdots & 0 & 0 \cr 1 & 0 & 1 & \cdots & 0 & 0 \cr 1 & 0 & 0 & \ddots & \vdots & \vdots \cr \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \cr 1 & 0 & 0 & \cdots & 0 & 1 \cr 1 & -1 & -1 & \cdots & -1 & -1 \cr\end{bmatrix}.$$ \end{enumerate} \end{pro} \end{multicols} \section{Theorem of Cayley and Hamilton } \begin{thm}[Cayley-Hamilton] A matrix $A\in\mat{n}{\BBF}$ satisfies its characteristic polynomial. \end{thm} \begin{pf} Put $B = \lambda {\bf I}_n - A$. We can write $$\det B = \det (\lambda {\bf I}_n - A) = \lambda ^n + b_1\lambda ^{n - 1} + b_2\lambda ^{n - 2}+ \cdots + b_n,$$as $\det (\lambda {\bf I}_n - A)$ is a polynomial of degree $n$. \bigskip Since $\adj{B}$ is a matrix obtained by using $(n-1)\times (n-1)$ determinants from $B$, we may write $$\adj{B} = \lambda ^{n-1}B_{n-1} + \lambda ^{n-2}B_{n-2} + \cdots + B_0. $$ Hence $$\det (\lambda {\bf I}_n - A){\bf I}_n = (B)(\adj{B}) = (\lambda {\bf I}_n - A)(\adj{B}),$$ from where $$\lambda ^n{\bf I}_n + b_1{\bf I}_n\lambda ^{n - 1} + b_2{\bf I}_n\lambda ^{n - 2}+ \cdots + b_n{\bf I}_n = (\lambda {\bf I}_n - A)( \lambda ^{n-1}B_{n-1} + \lambda ^{n-2}B_{n-2} + \cdots + B_0). $$ By equating coefficients we deduce $$\begin{array}{lll} {\bf I}_n & = & B_{n-1}\\ b_1{\bf I}_n & = & -AB_{n-1}+B_{n-2}\\ b_2{\bf I}_n & = & -AB_{n-2}+B_{n-3}\\ & \vdots & \\ b_{n-1}{\bf I}_n & = & -AB_1 + B_0 \\ b_{n}{\bf I}_n & = & -AB_0. \\ \end{array} $$ Multiply now the $k$-th row by $A^{n-k}$ (the first row appearing is really the $0$-th row): $$\begin{array}{lll} A^{n} & = & A^{n}B_{n-1}\\ b_1A^{n-1} & = & -A^{n}B_{n-1}+A^{n-1}B_{n-2}\\ b_2A^{n-2} & = & -A^{n-1}B_{n-2}+A^{n-2}B_{n-3}\\ & \vdots & \\ b_{n-1}A & = & -A^2B_1 + AB_0 \\ b_{n}{\bf I}_n & = & -AB_0. \\ \end{array} $$ Add all the rows and through telescopic cancellation obtain $$A^n + b_1A^{n-1} + \cdots + b_{n-1}A + b_n{\bf I}_n= {\bf 0}_n, $$ from where the theorem follows. \end{pf} \begin{exa} From example \ref{exa:diagonalisable3x3} the matrix $$\begin{bmatrix}1 & - 1 & -1 \cr 1 & 3 & 1 \cr -3 & 1 & -1 \cr\end{bmatrix}$$ has characteristic polynomial $$(\lambda - 3)(\lambda - 2)(\lambda + 2)= \lambda^3-3\lambda^2-4\lambda+12,$$hence the inverse of this matrix can be obtained by observing that $$A^3 - 3A^2 - 4A + 12{\bf I}_3 = {\bf 0}_3 \implies A^{-1} = -\dfrac{1}{12}\left(A^2 - 3A - 4{\bf I}_3\right) = \left[ \begin {array}{ccc} 1/3&1/6&-1/6\\\noalign{\medskip}1/6&1/3&1/6\\\noalign{\medskip}-5/6&-1/6&-1/3\end {array} \right] . $$ \end{exa} \section*{\psframebox{Homework}} \begin{pro} A $3\times 3$ matrix $A$ has characteristic polynomial $\lambda (\lambda - 1)(\lambda + 2)$. What is the characteristic polynomial of $A^2$? \begin{answer} The eigenvalues of $A$ are $0$, $1$, and $-2$. Those of $A^2$ are $0$, $1$, and $4$. Hence, the characteristic polynomial of $A^2$ is $\lambda (\lambda - 1)(\lambda - 4)$. \end{answer} \end{pro} \chapter{Linear Algebra and Geometry} \section{Points and Bi-points in \protect$\BBR^2\protect$} $\BBR^2$ is the set of all points $A = \colpoint{a_1\\ a_2}$ with real number coordinates on the plane, as in figure \ref{fig:pointinr2}. We use the notation $\zeropoint = \colpoint{0\\ 0}$ to denote the {\em origin}. \vspace{2cm} \begin{figure}[h] $$\psset{unit=2pc}\psline[linewidth=2pt, linecolor=red]{->}(-3,0)(3, 0)\uput[r](3,0){x} \psline[linewidth=2pt, linecolor=blue]{->}(0,-3)(0, 3)\uput[u](0,3){y} \psdots[dotstyle=*,dotscale=1.5](0,0)(1.5,1.5)\uput[ur](1.5, 1.5){A = \colpoint{a_1\\ a_2}} \uput[ur](0, 0){\zeropoint} \psline[linestyle=dashed](0,1.5)(1.5, 1.5)\psline[linestyle=dashed](1.5, 0)(1.5, 1.5) $$\vspace{2cm}\footnotesize\hangcaption{Rectangular coordinates in $\BBR^2$.} \label{fig:pointinr2}\end{figure} \bigskip Given $A = \colpoint{a_1\\ a_2}\in\BBR^2$ and $B = \colpoint{b_1\\ b_2}\in\BBR^2$ we define their addition as \begin{equation} A + B = \colpoint{a_1\\ a_2} + \colpoint{b_1\\ b_2} = \colpoint{a_1 + b_1\\ a_2 + b_2}. \label{eq:addition_in_r2}\end{equation} Similarly, we define the scalar multiplication of a point of $\BBR^2$ by the scalar $\alpha\in\BBR$ as \begin{equation} \alpha A= \alpha \colpoint{a_1\\ a_2} = \colpoint{\alpha a_1\\ \alpha a_2}. \label{eq:multiplication_in_r2}\end{equation} \begin{rem} Throughout this chapter, unless otherwise noted, we will use the convention that a point $A\in \BBR^2$ will have its coordinates named after its letter, thus $$ A = \colpoint{a_1\\ a_2}.$$ \end{rem} \begin{df} Consider the points $A\in\BBR^2, B\in \BBR^2$. By the {\em bi-point} starting at $A$ and ending at $B$, denoted by $\bipoint{A}{B}$, we mean the directed line segment from $A$ to $B$. We define $$\bipoint{A}{A} = \zeropoint = \colpoint{0\\ 0}.$$ \index{bi-point} \end{df} \begin{rem} The bi-point $\bipoint{A}{B}$ can be thus interpreted as an arrow starting at $A$ and finishing, with the arrow tip, at $B$. We say that $A$ is the {\em tail} of the bi-point $\bipoint{A}{B}$ and that $B$ is its {\em head}. Some authors use the terminology ``{\em fixed vector}'' instead of ``bi-point.'' \end{rem} \begin{df} Let $A \neq B$ be points on the plane and let $L$ be the line passing through $A$ and $B$. The {\em direction} of the bi-point $\bipoint{A}{B}$ is the direction of the line $L$, that is, the angle $\theta \in \left]-\frac{\pi}{2}; \frac{\pi}{2}\right]$ that the line $L$ makes with the horizontal. See figure \ref{fig:bipointdirection}. \end{df}\begin{df} Let $A, B$ lie on line $L$, and let $ C, D$ lie on line $L'$. If $L||L'$ then we say that $\bipoint{A}{B}$ has the same direction as $\bipoint{C}{D}$. We say that the bi-points $\bipoint{A}{B}$ and $\bipoint{C}{D}$ have the {\em same sense} if they have the same direction and if both their heads lie on the same half-plane made by the line joining their tails. They have {\em opposite sense } if they have the same direction and if both their heads lie on alternative half-planes made by the line joining their tails. See figures \ref{fig:bipointsense1} and \ref{fig:bipointsense2} . \end{df} \vspace{2cm} \begin{figure}[h] \hfill \begin{minipage}{3cm}$$\psset{unit=.7pc} \psline(-5, -5)(5, 5)\psline(-6, -3)(6, -3) \psdots[dotstyle=*, dotscale=1.5](-3, -3) \psarc{->}(-3, -3){2}{0}{45}\uput[r](-2.5,-2.5){\theta} \psline[linewidth=2pt, linecolor=red]{*->}(-1, -1)(2, 2) \uput[r](-1,-1){A}\uput[r](2, 2 ){B}$$\vspace{2cm}\scriptsize\hangcaption{Direction of a bi-point}\label{fig:bipointdirection} \end{minipage} \hfill \begin{minipage}{3cm}$$\psset{unit=.7pc} \psline[linewidth=2pt, linecolor=red]{*->}(-1, -1)(2, 2) \psline[linewidth=2pt, linecolor=blue]{*->}(2, -3)(5, 0) \uput[r](-.7,-1){A}\uput[r](2, 2 ){B}\uput[r](2,-3){C}\uput[r](5, 0 ){D}\psline(-1, -1)(2,-3)$$\vspace{2cm}\scriptsize\hangcaption{Bi-points with the same sense. }\label{fig:bipointsense1}\end{minipage} \hfill \begin{minipage}{3cm}$$\psset{unit=.7pc} \psline[linewidth=2pt, linecolor=red]{*->}(-1, -1)(2, 2)\psline[linewidth=2pt, linecolor=blue]{*->}(2, -3)(-1, -6) \uput[r](-.7,-1){A}\uput[r](2, 2 ){B}\uput[r](2,-3){C}\uput[r](-1, -6 ){D}\psline(2, -3)(-1, -1)$$\vspace{2cm}\scriptsize\hangcaption{Bi-points with opposite sense. }\label{fig:bipointsense2}\end{minipage}\hfill \end{figure} \begin{rem} Bi-point $\bipoint{B}{A}$ has the opposite sense of $\bipoint{A}{B}$ and so we write $$\bipoint{B}{A} = -\bipoint{A}{B}.$$ \end{rem} \begin{df}\index{norm!of a bi-point}\index{norm!unit} Let $A \neq B$. The {\em Euclidean length or norm} of bi-point $\bipoint{A}{B}$ is simply the distance between $A$ and $B$ and it is denoted by $$\norm{\bipoint{A}{B}} = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2.}$$ We define $$\norm{\bipoint{A}{A}}= \norm{\zeropoint} = 0.$$ A bi-point is said to have {\em unit length} if it has norm $1$. \end{df} \begin{rem} A bi-point is completely determined by three things: (i) its norm, (ii) its direction, and (iii) its sense. \end{rem} \begin{df}[Chasles' Rule] Two bi-points are said to be {\em contiguous} if one has as tail the head of the other. In such case we define the sum of contiguous bi-points $\bipoint{A}{B}$ and $\bipoint{B}{C}$ by {\em Chasles' Rule} \index{Chasles' Rule} $$\bipoint{A}{B} + \bipoint{B}{C} = \bipoint{A}{C}.$$ See figure \ref{fig:addbipoints}.\end{df} \begin{df}[Scalar Multiplication of Bi-points] Let $\lambda \in \BBR \setminus \{0\}$ and $A \neq B$. We define $$0\bipoint{A}{B} = \zeropoint$$ and $$\lambda\bipoint{A}{A} = \zeropoint .$$ We define $\lambda\bipoint{A}{B}$ as follows. \begin{enumerate} \item $\lambda\bipoint{A}{B}$ has the direction of $\bipoint{A}{B}$. \item $\lambda\bipoint{A}{B}$ has the sense of $\bipoint{A}{B}$ if $\lambda > 0$ and sense opposite $\bipoint{A}{B}$ if $\lambda < 0$. \item $\lambda\bipoint{A}{B}$ has norm $|\lambda|\norm{\bipoint{A}{B}}$ which is a contraction of $\bipoint{A}{B}$ if $0 < |\lambda| < 1$ or a dilatation of $\bipoint{A}{B}$ if $|\lambda| > 1$. \end{enumerate} See figure \ref{fig:scalarmul_1} for some examples. \end{df} \vspace{2cm} \begin{figure}[h]\begin{minipage}{6cm} $$\psset{unit=.5pc} \rput(-2,0){\psline[linewidth=2pt, linecolor=red]{->}(-3,3)(3, 5) \psline[linewidth=2pt, linecolor=blue]{->}(3,5)(9, 2) \uput[l](-3, 3 ){A}\uput[u](3,5){B}\uput[r](9, 2 ){C}\psline[linewidth=2pt, linecolor=green]{->}(-3,3)(9, 2)} $$\vspace{1cm}\footnotesize\hangcaption{Chasles' Rule.}\label{fig:addbipoints} \end{minipage}\hfill\begin{minipage}{6cm}$$\psset{unit=.5pc} \psline[linewidth=2pt, linecolor=red]{->}(-3,3)(3, 5)\uput[r](3, 5.5){\bipoint{A}{B}} \psline[linewidth=2pt, linecolor=green]{->}(0,0)(3, 1)\uput[d](5, .5){\frac{1}{2}\bipoint{A}{B}} \psline[linewidth=2pt, linecolor=blue]{->}(6, 10)(-6,6)\uput[u](6, 10){-2\bipoint{A}{B}} $$\vspace{1cm}\footnotesize\hangcaption{Scalar multiplication of bi-points.}\label{fig:scalarmul_1} \end{minipage}\end{figure} \section{Vectors in $\BBR^2$} \begin{df}[Midpoint] \index{midpoint} Let $A, B$ be points in $\BBR^2$. We define the {\em midpoint} of the bi-point $\bipoint{A}{B}$ as $$\frac{A + B}{2} = \colpoint{\frac{a_1 + b_1}{2}\\ \frac{a_2 + b_2}{2}}.$$ \end{df} \begin{df}[Equipollence] \index{equipollence} Two bi-points $\bipoint{X}{Y}$ and $\bipoint{A}{B}$ are said to be {\em equipollent} written $\bipoint{X}{Y} \sim \bipoint{A}{B}$ if the midpoints of the bi-points $\bipoint{X}{B}$ and $\bipoint{Y}{A}$ coincide, that is, $$ \bipoint{X}{Y} \sim \bipoint{A}{B} \Leftrightarrow \frac{{X} + B}{2} = \frac{{Y} + A}{2}.$$ See figure \ref{fig:equibi-points1}.\label{df:equipollence}\end{df} Geometrically, equipollence means that the quadrilateral ${XYBA}$ is a parallelogram. Thus the bi-points $\bipoint{X}{Y}$ and $\bipoint{A}{B}$ have the same norm, sense, and direction. \vspace{2cm} \begin{figure}[h] $$\psset{unit=1pc} \psline[linewidth=2pt, linecolor=red]{->}(-3,3)(3, 5) \psline[linewidth=2pt, linecolor=blue]{->}(3,0)(9, 2) \psline[linestyle=dashed](-3,3)(9, 2) \rput(1.5, 2.75){||}\rput(6, 2.25){||} \psline[linestyle=dashed](3,5)(3,0)\rput(3, 3.75){-}\rput(3, 1.25){-} \uput[l](-3, 3){{X}} \uput[u](3, 5){{Y}}\uput[d](3, 0){A}\uput[r](9, 2){B} $$\caption{Equipollent bi-points.}\label{fig:equibi-points1} \end{figure} \begin{lem}\label{lem:condition_for_equipollence} Two bi-points $\bipoint{X}{Y}$ and $\bipoint{A}{B}$ are equipollent if and only if $$\colpoint{y_1 - x_1\\ y_2 - x_2} = \colpoint{b_1 - a_1\\ b_2 - a_2}.$$ \end{lem} \begin{pf} This is immediate, since $$\begin{array}{lll} \bipoint{X}{Y} \sim \bipoint{A}{B} & \iff & \colpoint{\frac{a_1 + y_1}{2}\vspace{2mm}\\ \frac{a_2 + y_2}{2}} = \colpoint{\frac{b_1 + x_1}{2}\\ \frac{b_2 + x_2}{2}} \vspace{2mm}\\ & \iff & \colpoint{y_1 - x_1\\ y_2 - x_2} = \colpoint{b_1 - a_1\\ b_2 - a_2}, \end{array}$$as desired. \end{pf} \begin{rem} From Lemma \ref{lem:condition_for_equipollence}, equipollent bi-points have the same norm, the same direction, and the same sense. \end{rem} \begin{thm} \label{thm:equipollence_equiv_rel} Equipollence is an equivalence relation. \end{thm} \begin{pf} Write $\bipoint{X}{Y} \sim \bipoint{A}{B}$ if $\bipoint{X}{Y}$ if equipollent to $\bipoint{A}{B}$. Now $\bipoint{X}{Y} \sim \bipoint{X}{Y}$ since $\colpoint{y_1 - x_1\\ y_2 - x_2} = \colpoint{y_1 - x_1\\ y_2 - x_2}$ and so the relation is reflexive. Also $$ \begin{array}{lll} \bipoint{X}{Y} \sim \bipoint{A}{B} &\iff & \colpoint{y_1 - x_1\\ y_2 - x_2} = \colpoint{b_1 - a_1\\ b_2 - a_2} \vspace{2mm}\\ & \iff & \colpoint{b_1 - a_1\\ b_2 - a_2} = \colpoint{y_1 - x_1\\ y_2 - x_2} \\ & \iff & \bipoint{A}{B} \sim \bipoint{X}{Y}, \end{array}$$and the relation is symmetric. Finally $$ \begin{array}{lll} \bipoint{X}{Y} \sim \bipoint{A}{B} \wedge \bipoint{A}{B} \sim \bipoint{U}{V} &\iff & \colpoint{y_1 - x_1\\ y_2 - x_2} = \colpoint{b_1 - a_1\\ b_2 - a_2} \vspace{2mm} \\ & & \qquad \wedge \colpoint{b_1 - a_1\\ b_2 - a_2} = \colpoint{v_1 - u_1\\ v_2 - au_2} \vspace{2mm}\\ & \iff & \colpoint{y_1 - x_1\\ y_2 - x_2} = \colpoint{v_1 - u_1\\ v_2 - u_2} \vspace{2mm} \\ & \iff & \bipoint{X}{Y} \sim \bipoint{U}{V}, \end{array}$$ and the relation is transitive. \end{pf} \begin{df}[Vectors on the Plane] \index{vector!on the plane} The equivalence class in which the bi-point $\bipoint{X}{Y}$ falls is called the {\em vector} (or {\em free vector}) from ${X}$ to ${Y}$, and is denoted by $\vect{XY}$. Thus we write $$\bipoint{X}{Y} \in \vect{XY} = \colvec{y_1 - x_1 \\ y_2 - x_2}. $$If we desire to talk about a vector without mentioning a bi-point representative, we write, say, $\v{v}$, thus denoting vectors with boldface lowercase letters. If it is necessary to mention the coordinates of $\v{v}$ we will write $$\v{v} = \colvec{v_1 \\ v_2 }. $$ \label{df:disvect}\end{df} \begin{rem} For any point $X$ on the plane, we have $\vect{XX} = \v{0}$, the {\em zero vector}. If $\bipoint{X}{Y}\in \v{v}$ then $\bipoint{Y}{X}\in -\v{v}$. \end{rem} \begin{df}[Position Vector] \index{vector!position} For any particular point $P = \colpoint{p_1\\ p_2}\in \BBR^2$ we may form the vector $\vect{OP} = \colvec{p_1 \\ p_2}$. We call $\vect{OP}$ the {\em position vector} of $P$ and we use boldface lowercase letters to denote the equality $\vect{OP} = \v{p}$. \end{df} \begin{exa} The vector into which the bi-point with tail at $A = \colpoint{-1\\ 2}$ and head at $B = \colpoint{3\\ 4}$ falls is $$\vect{AB} = \colvec{3 - (-1)\\ 4 - 2} = \colvec{4 \\ 2}. $$ \end{exa} \begin{exa} The bi-points $\bipoint{A}{B}$ and $\bipoint{X}{Y}$ with $$A = \colpoint{-1\\ 2}, B = \colpoint{3\\ 4},$$ $$X = \colpoint{3\\ 7}, Y = \colpoint{7\\ 9}$$ represent the same vector $$\vect{AB} = \colvec{3 - (-1)\\ 4 - 2} = \colvec{4 \\ 2} = \colvec{7-3 \\ 9-7} = \vect{XY}. $$ In fact, if $S = \colpoint{-1 + n\\ 2 + m}, T = \colpoint{3 + n \\ 4 + m}$ then the infinite number of bi-points $\bipoint{S}{T}$ are representatives of of the vectors $\vect{AB} = \vect{XY} = \vect{ST}$. \end{exa} \index{vector!sum!on the plane}Given two vectors $\v{u}$, $\v{v}$ we define their sum $\v{u} + \v{v}$ as follows. Find a bi-point representative $\vect{AB}\in\v{u}$ and a contiguous bi-point representative $\vect{BC}\in\v{v}$. Then by Chasles' Rule $$ \v{u} + \v{v} = \vect{AB} + \vect{BC} = \vect{AC}. $$ Again, by virtue of Chasles' Rule we then have \begin{equation} \vect{AB} = \vect{AO} + \vect{OB} = -\vect{OA} + \vect{OB} = \v{b} - \v{a} \label{eq:difference_pos_vectors} \end{equation} \bigskip \index{scalar multiplication!of a plane vector}Similarly we define scalar multiplication of a vector by scaling one of its bi-point representatives.\index{norm!of a vector}We define the norm of a vector $\v{v}\in\BBR^2$ to be the norm of any of its bi-point representatives. \bigskip Componentwise we may see that given vectors $\v{u} = \colvec{u_1 \\ u_2}$, $\v{v} = \colvec{v_1 \\ v_2}$, and a scalar $\lambda\in\BBR$ then their sum and scalar multiplication take the form $$ \v{u} + \v{v}= \colvec{u_1 \\ u_2} + \colvec{v_1 \\ v_2}, \ \ \ \ \lambda\v{u} = \colvec{\lambda u_1 \\ \lambda u_2}. $$ \vspace{2cm} \begin{figure}[h]\begin{minipage}{6cm} $$\psset{unit=.5pc} \rput(-2,0){\psline[linewidth=2pt, linecolor=red]{->}(-3,3)(3, 5) \psline[linewidth=2pt, linecolor=blue]{->}(3,5)(9, 2) \uput[l](-3, 3 ){A}\uput[u](3,5){B}\uput[r](9, 2 ){C}\uput[u](5.5, 3.5){\v{v}} \uput[l](0,4.5){\v{u}} \uput[l](3.5,1){\v{u} + \v{v}}\psline[linewidth=2pt, linecolor=green]{->}(-3,3)(9, 2)} $$\vspace{1cm}\footnotesize\hangcaption{Addition of Vectors.}\label{fig:addvectors} \end{minipage}\hfill\begin{minipage}{6cm}$$\psset{unit=.5pc} \psline[linewidth=2pt, linecolor=red]{->}(-3,3)(3, 5)\uput[r](3, 5.5){\v{u}} \psline[linewidth=2pt, linecolor=green]{->}(0,0)(3, 1)\uput[d](5, .5){\frac{1}{2}\v{u}} \psline[linewidth=2pt, linecolor=blue]{->}(6, 10)(-6,6)\uput[u](6, 10){-2\v{u}} $$\vspace{1cm}\footnotesize\hangcaption{Scalar multiplication of vectors.}\label{fig:scalarmultvectors} \end{minipage}\end{figure} \begin{exa} Diagonals are drawn in a rectangle $ABCD$. If $\vect{AB} = \v{x}$ and $\vect{AC} = \v{y}$, then $\vect{BC} = \v{y} -\v{x}$, $\vect{CD} = -\v{x}$, $\vect{DA} = \v{x} - \v{y}$, and $\vect{BD} = \v{y} - 2\v{x}$. \end{exa} \begin{df}[Parallel Vectors] Two vectors $\v{u}$ and $\v{v}$ are said to be {\em parallel} if there is a scalar $\lambda$ such that $\v{u}=\lambda\v{v}$. If $\v{u}$ is parallel to $\v{v}$ we write $\v{u}||\v{v}$. We denote by $\BBR\v{v} = \{\alpha \v{v}: \alpha \in \BBR \}$, the set of all vectors parallel to $\v{v}$. \end{df} \begin{rem} $\v{0}$ is parallel to every vector. \end{rem} \begin{df}If $\v{u} = \colvec{u_1 \\ u_2}$, then we define its {\em norm} as $\norm{\v{u}} = \sqrt{u_1 ^2 + u_2 ^2}$. The distance between two vectors $\v{u}$ and $\v{v}$ is ${\bf d}\langle \v{u}, \v{v} \rangle = \norm{\v{u}- \v{v}}$. \end{df} \begin{exa} Let $a\in\BBR$, $a>0$ and let $\v{v} \neq \v{0}$. Find a vector with norm $a$ and parallel to $\v{v}$. \end{exa} \begin{solu}Observe that $\dfrac{\v{v}}{\norm{\v{v}}}$ has norm $1$ as $$\left|\left|\dfrac{\v{v}}{\norm{\v{v}}}\right|\right| = \dfrac{\norm{\v{v}}}{\norm{\v{v}}} = 1.$$ Hence the vector $a\dfrac{\v{v}}{\norm{\v{v}}}$ has norm $a$ and it is in the direction of $\v{v}$. One may also take $-a\dfrac{\v{v}}{\norm{\v{v}}}$. \end{solu} \begin{exa} If ${M}$ is the midpoint of the bi-point $\bipoint{X}{Y}$ then $\vect{XM}= \vect{MY}$ from where $\vect{XM}= \frac{1}{2}\vect{XY}$. Moreover, if ${T}$ is any point, by Chasles' Rule $$\begin{array}{lll}\vect{TX} + \vect{TY} & = & \vect{TM} + \vect{MX} + \vect{TM} + \vect{MY}\\ & = & 2\vect{TM} - \vect{XM}+ \vect{MY} \\ & = & 2\vect{TM}. \\ \end{array} $$ \label{exa:propertyofmidpoint}\end{exa} \begin{exa} Let $\triangle {ABC}$ be a triangle on the plane. Prove that the line joining the midpoints of two sides of the triangle is parallel to the third side and measures half its length. \label{exa:medtriag_1}\end{exa} \begin{solu}Let the midpoints of $\bipoint{A}{B}$ and $\bipoint{A}{C}$ be ${M_C}$ and ${M_B}$, respectively. We shew that $\vect{BC} = 2\vect{M_CM_B}$. We have $2\vect{AM_C} = \vect{AB}$ and $2\vect{AM_B} = \vect{AC}$. Thus $$\begin{array}{lll}\vect{BC} & = & \vect{BA} + \vect{AC} \\ & = & -\vect{AB} + \vect{AC} \\ & = & -2\vect{AM_C} + 2\vect{AM_B} \\ & = & 2\vect{M_CA} + 2\vect{AM_B} \\ & = & 2(\vect{M_CA} + \vect{AM_B}) \\ & = & 2\vect{M_CM_B}, \end{array}$$ as we wanted to shew. \end{solu} \begin{exa} In $\triangle {ABC}$, let ${M_C}$ be the midpoint of side ${AB}$. Shew that $$\vect{CM_C} = \frac{1}{2}\left(\vect{CA} + \vect{CB}\right).$$ \label{exa:medtriag_2}\end{exa} \begin{solu}Since $\vect{AM_C} = \vect{M_CB}$, we have $$\begin{array}{lll} \vect{CA} + \vect{CB} & = & \vect{CM_C} + \vect{M_CA} + \vect{CM_C} + \vect{M_CB} \\ & = & 2\vect{CM_C} - \vect{AM_C} + \vect{M_CB}\\ & = & 2\vect{CM_C}, \end{array} $$which yields the desired result. \end{solu} \begin{thm}[Section Formula] Let $APB$ be a straight line and $\lambda$ and $\mu$ be real numbers such that $$\dfrac{\norm{\bipoint{A}{P}}}{\norm{\bipoint{P}{B}}} = \dfrac{\lambda}{\mu}. $$With $\v{a} = \vect{OA}$, $\v{b} = \vect{OB}$, and $\v{p} = \vect{OP}$, then \begin{equation}\v{p} =\dfrac{\lambda\v{b} + \mu\v{a}}{\lambda + \mu}.\label{eq:section_formula}\index{section formula}\end{equation} \end{thm} \begin{pf} Using Chasles' Rule for vectors, $$\vect{AB} = \vect{AO} + \vect{OB} = -\v{a} + \v{b}, $$ $$\vect{AP} = \vect{AO} + \vect{OP} = -\v{a} + \v{p}. $$Also, using Chasles' Rule for bi-points, $$\bipoint{A}{P}\mu = \lambda (\bipoint{P}{B}) = \lambda (\bipoint{P}{A} +\bipoint{A}{B}) = \lambda (-\bipoint{A}{P} +\bipoint{A}{B}), $$whence $$\bipoint{A}{P} = \dfrac{\lambda}{\lambda + \mu}\bipoint{A}{B} \implies \vect{AP} = \dfrac{\lambda}{\lambda + \mu}\vect{AB} \implies \v{p} - \v{a} = \dfrac{\lambda}{\lambda + \mu}(\v{b} -\v{a}). $$ On combining these formul\ae \ $$(\lambda + \mu)(\v{p} - \v{a}) = \lambda (\v{b} - \v{a}) \implies (\lambda + \mu)\v{p} = \lambda\v{b} + \mu\v{a}, $$from where the result follows. \end{pf} \vspace{1cm} \begin{figure}[h] \begin{minipage}{4cm} $$\psset{unit=2pc} \pscircle[linewidth=1pt](0,0){1} \psline[linewidth=1.5pt]{o->}(0,0)(0,1) \psline[linewidth=1.5pt]{o->}(0,0)(0.866, -0.5) \psline[linewidth=1.5pt]{o->}(0,0)(-0.866,-0.5) \uput[u](0,1){\v{a}} \uput[dl](-0.866, -0.5){\v{b}} \uput[dr](0.866, -0.5){\v{c}} $$\vspace{1cm}\footnotesize\hangcaption{[A]. Problem \ref{pro:sum_of_vectors1}.} \label{fig:sum_of_vectors1} \end{minipage} \hfill \begin{minipage}{4cm} $$\psset{unit=2pc} \pscircle[linewidth=1pt](0,0){1} \psline[linewidth=1.5pt]{<-o}(0,0)(0,1) \psline[linewidth=1.5pt]{<-o}(0,0)(0.866, -0.5) \psline[linewidth=1.5pt]{<-o}(0,0)(-0.866, -0.5) \uput[u](0,1){\v{a}} \uput[dr](0.866, -0.5){\v{b}} \uput[dl](-0.866, -0.5){\v{c}} $$\vspace{1cm}\footnotesize\hangcaption{[B]. Problem \ref{pro:sum_of_vectors1}.} \label{fig:sum_of_vectors2} \end{minipage}\hfill \begin{minipage}{4cm} $$\psset{unit=2pc} \pscircle[linewidth=1pt](0,0){1} \psline[linewidth=1.5pt]{o->}(0,0)(1,0) \psline[linewidth=1.5pt]{o->}(0,0)(-1,0) \psline[linewidth=1.5pt]{o->}(0,0)(0,1) \psline[linewidth=1.5pt]{o->}(0,0)(0,-1) \uput[u](0,1){\v{a}} \uput[d](0, -1){\v{b}} \uput[l](-1, 0){\v{c}} \uput[r](1, 0){\v{d}} $$\vspace{1cm}\footnotesize\hangcaption{[C]. Problem \ref{pro:sum_of_vectors1}.} \label{fig:sum_of_vectors3} \end{minipage} \end{figure} \vspace{1cm} \begin{figure}[h]\begin{minipage}{4cm} $$\psset{unit=2pc} \pscircle[linewidth=1pt](0,0){1} \psline[linewidth=1.5pt]{<-o}(0,0)(1,0) \psline[linewidth=1.5pt]{<-o}(0,0)(-1,0) \psline[linewidth=1.5pt]{<-o}(0,0)(0,1) \psline[linewidth=1.5pt]{<-o}(0,0)(0,-1) \uput[u](0,1){\v{a}} \uput[d](0, -1){\v{b}} \uput[l](-1, 0){\v{c}} \uput[r](1, 0){\v{d}} $$\vspace{1cm}\footnotesize\hangcaption{[D]. Problem \ref{pro:sum_of_vectors1}.} \label{fig:sum_of_vectors4} \end{minipage}\hfill \begin{minipage}{4cm} $$\psset{unit=2pc} \pscircle[linewidth=1pt](0,0){1} \psline[linewidth=1.5pt]{o->}(0,0)(1,0) \psline[linewidth=1.5pt]{<-o}(0,0)(-1,0) \psline[linewidth=1.5pt]{o->}(0,0)(0,1) \psline[linewidth=1.5pt]{o->}(0,0)(0,-1) \uput[u](0,1){\v{a}} \uput[d](0, -1){\v{b}} \uput[l](-1, 0){\v{c}} \uput[r](1, 0){\v{d}} $$\vspace{1cm}\footnotesize\hangcaption{[E]. Problem \ref{pro:sum_of_vectors1}.} \label{fig:sum_of_vectors5} \end{minipage}\hfill \begin{minipage}{4cm} $$\psset{unit=2pc} \pscircle[linewidth=1pt](0,0){1} \psline[linewidth=1.5pt]{o->}(0,0)(0.5,-0.866) \psline[linewidth=1.5pt]{<-o}(0,0)(-0.5,-0.866) \psline[linewidth=1.5pt]{o->}(0,0)(0.5,0.866) \psline[linewidth=1.5pt]{<-o}(0,0)(-0.5,0.866) \psline[linewidth=1.5pt]{<-o}(0,0)(1,0) \psline[linewidth=1.5pt]{o->}(0,0)(-1,0) \uput[l](-1,0){\v{a}} \uput[r](1, 0){ \v{d}} \uput[ul](-0.5, 0.866){\v{b}} \uput[ur](0.5, 0.866){\v{c}} \uput[dl](-0.5, -0.866){\v{f}} \uput[dr](0.5, -0.866){\v{e}} $$ \vspace{1cm} \footnotesize\hangcaption{[F]. Problem \ref{pro:sum_of_vectors1}.} \label{fig:sum_of_vectors6} \end{minipage} \end{figure} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Let $a$ be a real number. Find the distance between $\colvec{1\\ a}$ and $\colvec{1 - a \\ 1}$. \begin{answer} $\sqrt{2a^2-2a + 1}$ \end{answer} \end{pro} \begin{pro} Find all scalars $\lambda$ for which $\norm{\lambda\v{v}} = \frac{1}{2}$, where $\v{v} = \colvec{1\\ -1}$. \begin{answer} $\norm{\lambda\v{v}} = \frac{1}{2} \implies \sqrt{(\lambda)^2 + (-\lambda)^2} = \frac{1}{2} \implies 2\lambda^2 = \frac{1}{4} \implies \lambda = \pm \dfrac{1}{\sqrt{8}}$. \end{answer} \end{pro} \begin{pro} Given a pentagon $ABCDE$, find $\vect{AB} + \vect{BC}+\vect{CD}+\vect{DE}+\vect{EA}$. \begin{answer} $\v{0}$ \end{answer} \end{pro} \begin{pro} For which values of $a$ will the vectors $$\v{a} = \colvec{a + 1\\ a^2 -1}, \ \ \ \ \v{b} = \colvec{2a + 5 \\ a^2 - 4a + 3}$$ will be parallel? \begin{answer} $a = \pm 1$ or $a = -8$.\end{answer} \end{pro} \begin{pro} In $\triangle ABC$ let the midpoints of $\bipoint{A}{B}$ and $\bipoint{A}{C}$ be ${M_C}$ and ${M_B}$, respectively. Put $\vect{M_CB} = \v{x}$, $\vect{M_BC} = \v{y}$, and $\vect{CA} = \v{z}$. Express [A] $\vect{AB} + \vect{BC} + \vect{M_CM_B}$, [B] $\vect{AM_C} + \vect{M_CM_B} + \vect{M_BC}$, [C] $\vect{AC} + \vect{M_CA} - \vect{BM_B}$ in terms of $\v{x}$, $\v{y}$, and $\v{z}$. \begin{answer} [A] $2(\v{x} + \v{y}) - \frac{1}{2}\v{z} $, [B] $\v{x} + \v{y} - \frac{1}{2}\v{z}$, [C] $-(\v{x}+\v{y} + \v{z})$ \end{answer} \end{pro} \begin{pro} \label{pro:sum_of_vectors1} A circle is divided into three, four equal, or six equal parts (figures \ref{fig:sum_of_vectors1} through \ref{fig:sum_of_vectors6}). Find the sum of the vectors. Assume that the divisions start or stop at the centre of the circle, as suggested in the figures. \begin{answer} [A]. $\v{0}$, [B]. $\v{0}$, [C]. $\v{0}$, [D]. $\v{0}$, [E]. $2\v{c} (=2\v{d})$ \end{answer} \end{pro} \begin{pro} Diagonals are drawn in a square (figures \ref{fig:sum_of_vectors7} through \ref{fig:sum_of_vectors9}). Find the vectorial sum $\v{a} + \v{b} + \v{c}$. Assume that the diagonals either start, stop, or pass through the centre of the square, as suggested by the figures. \vspace{1cm} \begin{figure}[htb] \hfill\begin{minipage}{4cm} $$\psset{unit=2pc} \pspolygon[linewidth=.6pt](-1,-1)(1,-1)(1,1)(-1,1) \psline[linewidth=1.5pt]{o->}(-1,-1)(-1,1) \uput[u](-1,1){\v{a}} \psline[linewidth=1.5pt]{o->}(-1,-1)(1,1) \uput[ur](1,1){\v{b}} \psline[linewidth=1.5pt]{o->}(1,1)(1,-1) \uput[d](1,-1){\v{c}} $$\vspace{1cm}\footnotesize\hangcaption{[G].} \label{fig:sum_of_vectors7} \end{minipage} \begin{minipage}{4cm} $$\psset{unit=2pc} \pspolygon[linewidth=.6pt](-1,-1)(1,-1)(1,1)(-1,1) \psline[linewidth=1.5pt]{<-o}(-1,1)(0,0) \uput[l](-1,1){\v{a}} \psline[linewidth=1.5pt]{o->}(0,0)(1,1) \uput[r](1,1){\v{b}} \psline[linewidth=1.5pt]{o->}(1,1)(1,-1) \uput[d](1,-1){\v{c}} $$\vspace{1cm}\footnotesize\hangcaption{[H].} \label{fig:sum_of_vectors8} \end{minipage}\hfill \begin{minipage}{4cm} $$\psset{unit=2pc} \pspolygon[linewidth=.6pt](-1,-1)(1,-1)(1,1)(-1,1) \psline[linewidth=1.5pt]{o->}(-1,1)(1,1) \uput[r](1,1){\v{a}} \psline[linewidth=1.5pt]{o->}(1,1)(0,0) \uput[l](-1,1){\v{b}} \psline[linewidth=1.5pt]{o->}(1,-1)(0,0) \uput[d](1,-1){\v{c}} $$\vspace{1cm}\footnotesize\hangcaption{[I].} \label{fig:sum_of_vectors9} \end{minipage}\hfill \end{figure} \begin{answer} [F]. $\v{0}$, [G]. $\v{b}$, [H]. $2\v{0}$, [I]. $\v{0}$. \end{answer} \end{pro} \begin{pro} Prove that the mid-points of the sides of a skew quadrilateral form the vertices of a parallelogram. \begin{answer} Let the skew quadrilateral be $ABCD$ and let $P, Q, R, S$ be the midpoints of $\bipoint{A}{B}$, $\bipoint{B}{C}$, $\bipoint{C}{D}$, $\bipoint{D}{A}$, respectively. Put $\v{x} = \v{OX}$, where $X\in \{A, B, C, D, P, Q, R, S\}$. Using the Section Formula \ref{eq:section_formula} we have $$\v{p} = \dfrac{\v{a} + \v{b}}{2}, \ \ \ \v{q} = \dfrac{\v{b} + \v{c}}{2}, \ \ \ \v{r} = \dfrac{\v{c} + \v{d}}{2}, \ \ \ \v{s} = \dfrac{\v{d} + \v{a}}{2}. $$This gives $$\v{p} - \v{q} = \dfrac{\v{a} - \v{c}}{2},\ \ \ \v{s} - \v{r} = \dfrac{\v{a} - \v{c}}{2}. $$This means that $\vect{QP} = \vect{RS}$ and so $PQRS$ is a parallelogram since one pair of sides are equal and parallel. \end{answer} \end{pro} \begin{pro} ${ABCD}$ is a parallelogram. ${E}$ is the midpoint of $\bipoint{B}{C}$ and ${F}$ is the midpoint of $\bipoint{D}{C}$. Prove that $$\vect{AC} + \vect{BD} = 2\vect{BC}.$$ \begin{answer} We have $2\vect{BC} = \vect{BE} + \vect{EC}$. By Chasles' Rule $\vect{AC} = \vect{AE} + \vect{EC}$, and $\vect{BD} = \vect{BE} + \vect{ED}$. We deduce that $$ \vect{AC} + \vect{BD} = \vect{AE} + \vect{EC} + \vect{BE} + \vect{ED} = \vect{AD} + \vect{BC}.$$ But since ${ABCD}$ is a parallelogram, $\vect{AD} = \vect{BC}$. Hence $$ \vect{AC} + \vect{BD} = \vect{AD} + \vect{BC} = 2\vect{BC}.$$ \end{answer} \end{pro} \begin{pro} Let $A, B$ be two points on the plane. Construct two points ${I}$ and ${J}$ such that $$\vect{IA} = -3\vect{IB}, \ \ \ \vect{JA} = -\frac{1}{3}\vect{JB}, $$and then demonstrate that for any arbitrary point ${M}$ on the plane $$\vect{MA} + 3\vect{MB} = 4\vect{MI} $$ and $$3\vect{MA} + \vect{MB} = 4\vect{MJ}. $$ \begin{answer} We have $\vect{IA} = -3\vect{IB} \iff \vect{IA} = -3(\vect{IA} + \vect{AB}) = -3\vect{IA} - 3\vect{AB}$. Thus we deduce $$\begin{array}{lll} \vect{IA} + 3\vect{IA} = -3\vect{AB} & \iff & 4\vect{IA} = -3\vect{AB} \\ & \iff & 4\vect{AI} = 3\vect{AB} \\ & \iff & \vect{AI} = \frac{3}{4}\vect{AB}. \end{array} $$Similarly $$\begin{array}{lll} \vect{JA} = -\frac{1}{3}\vect{JB} & \iff & 3\vect{JA} = -\vect{JB} \\ & \iff & 3\vect{JA} = -\vect{JA} - \vect{AB} \\ & \iff & 4\vect{JA} = -\vect{AB} \\ & \iff & \vect{AJ} = \frac{1}{4}\vect{AB} \\ . \end{array} $$ Thus we take ${I}$ such that $\vect{AI} = \frac{3}{4}\vect{AB}$ and ${J}$ such that $\vect{AJ} = \frac{1}{4}\vect{AB}$. \bigskip Now $$\begin{array}{lll} \vect{MA} + 3\vect{MB} & = & \vect{MI} + \vect{IA} + 3\vect{IB} \\ & = & 4\vect{MI} + \vect{IA} + 3\vect{IB} \\ & = & 4\vect{MI}, \end{array} $$ and $$\begin{array}{lll} 3\vect{MA} + \vect{MB} & = & 3\vect{MJ} + 3\vect{JA} + \vect{MJ} + \vect{JB} \\ & = & 4\vect{MJ} + 3\vect{JA} + \vect{JB} \\ & = & 4\vect{MJ}. \end{array} $$ \end{answer} \end{pro} \begin{pro} You find an ancient treasure map in your great-grandfather's sea-chest. The sketch indicates that from the gallows you should walk to the oak tree, turn right $90^\circ$ and walk a like distance, putting and $x$ at the point where you stop; then go back to the gallows, walk to the pine tree, turn left $90^\circ$, walk the same distance, mark point $Y$. Then you will find the treasure at the midpoint of the segment $\overline{XY}$. So you charter a sailing vessel and go to the remote south-seas island. On arrival, you readily locate the oak and pine trees, but unfortunately, the gallows was struck by lightning, burned to dust and dispersed to the winds. No trace of it remains. What do you do? \begin{answer} Let $\v{G}$, $\v{O}$ and $\v{P}$ denote vectors from an arbitrary origin to the gallows, oak, and pine, respectively. The conditions of the problem define $\v{X}$ and $\v{Y}$, thought of similarly as vectors from the origin, by $\v{X} = \v{O} + R(\v{O} -\v{G})$, $\v{Y} = \v{P} - R(\v{P} -\v{G})$, where $R$ is the $90^\circ$ rotation to the right, a linear transformation on vectors in the plane; the fact that $-R$ is $90^\circ$ leftward rotation has been used in writing $Y$. Anyway, then $$ \dfrac{\v{X} + \v{Y}}{2} = \dfrac{\v{O} + \v{P}}{2} + \dfrac{R(\v{O} - \v{P})}{2}$$is independent of the position of the gallows. This gives a simple algorithm for treasure-finding: take $\v{P}$ as the (hitherto) arbitrary origin, then the treasure is at $\dfrac{\v{O} + R(\v{O})}{2}$. \end{answer} \end{pro} \end{multicols} \section{Dot Product in \protect$\BBR^2\protect$} \begin{df} Let $(\v{a}, \v{b})\in (\BBR^2)^2$. The {\em dot product} $\dotprod{a}{b}$ of $\v{a}$ and $\v{b}$ is defined by $$\dotprod{a}{b} = \begin{bmatrix} a_1 \cr a_2 \cr\end{bmatrix} \bulletproduct \begin{bmatrix} b_1 \cr b_2 \end{bmatrix} = a_1b_1 + a_2b_2.$$\end{df} The following properties of the dot product are easy to deduce from the definition. \begin{enumerate} \item[DP1] {\bf Bilinearity} \begin{equation}(\v{x} + \v{y})\bulletproduct \v{z} = \dotprod{x}{z} + \dotprod{y}{z}, \ \ \ \v{x}\bulletproduct (\v{y} + \v{z}) = \dotprod{x}{y} + \dotprod{x}{z}\label{dp:bilinearity}\end{equation} \item[DP2] {\bf Scalar Homogeneity} \begin{equation}(\alpha\v{ x})\bulletproduct \v{ y} = \v{ x}\bulletproduct (\alpha\v{ y}) = \alpha(\dotprod{\bf x}{\bf y}), \ \alpha \in \BBR. \label{dp:scalar_homogeneity}\end{equation} \item[DP3] {\bf Commutativity} \begin{equation}\dotprod{x}{y} = \dotprod{\bf y}{\bf x}\label{dp:commutativity}\end{equation} \item[DP4] \begin{equation}\dotprod{x}{x} \ \ \geq \ \ 0 \label{dp:dot_itself_is_positive}\end{equation} \item[DP5] \begin{equation}\dotprod{x}{x} = 0 \Leftrightarrow \v{ x} = \v{{\bf 0}}\label{dp:when_zero_is_zero}\end{equation} \item[DP6] \begin{equation}\norm{\bf x} = \sqrt{\dotprod{x}{x}} \label{dp:dot_and_norm}\end{equation} \end{enumerate} \begin{exa} If we put $$\v{{i}} = \colvec{ 1 \\ 0}, \ \v{{j}} = \colvec{0 \\ 1},$$then we can write any vector $\v{a} = \begin{bmatrix} a_1 \cr a_2 \cr \end{bmatrix}$ as a sum $$\v{a} = a_1\v{i} + a_2\v{j}.$$ The vectors $$\v{{i}} = \begin{bmatrix} 1 \cr 0 \cr\end{bmatrix}, \ \v{{j}} = \begin{bmatrix} 0 \cr 1 \cr \end{bmatrix},$$ satisfy $\v{i}\bulletproduct\v{j} = 0,$ and $\norm{\v{i}} = \norm{\v{j}} = 1$. \label{exa:ij}\end{exa} \begin{df}\index{vectors!angle between} Given vectors $\v{a}$ and $\v{b}$, we define the angle between them, denoted by $\anglebetween{a}{b}$, as the angle between any two contiguous bi-point representatives of $\v{a}$ and $\v{b}$. \end{df} \begin{thm} $$\v{{\bf a}}\bulletproduct\v{ b} = ||\v{ a}||||\v{ b}||\cos\anglebetween{a}{b}.$$ \label{thm:cosanglebetween}\end{thm} \begin{pf} Using Al-Kashi's Law of Cosines on the length of the vectors, we have $$\begin{array}{l}||\v{ b} - \v{ a}||^2 = ||\v{ a}||^2 + ||\v{ b}||^2 - 2||\v{ a}||||\v{ b}||\cos\anglebetween{a}{b}\\ \Leftrightarrow (\v{ b} - \v{ a})\bulletproduct (\v{ b} -\v{ a}) = ||\v{ a}||^2 + ||\v{ b}||^2 - 2||\v{ a}||||\v{ b}||\cos\anglebetween{a}{b}\\ \Leftrightarrow \v{ b}\bulletproduct\v{ b} - 2\v{ a}\bulletproduct\v{ b} + \v{ a}\bulletproduct\v{ a} = ||\v{ a}||^2 + ||\v{ b}||^2 - 2||\v{ a}||||\v{ b}||\cos\anglebetween{a}{b}\\ \Leftrightarrow ||\v{{\bf b}}||^2 - 2\v{{\bf a}}\bulletproduct\v{{\bf b}} + ||\v{{\bf b}}||^2 = ||\v{ a}||^2 + ||\v{ b}||^2 - 2||\v{ a}||||\v{ b}||\cos\anglebetween{a}{b}\\ \Leftrightarrow \v{ a}\bulletproduct\v{ b} = ||\v{ a}||||\v{ b}||\cos\anglebetween{a}{b} , \end{array}$$as we wanted to shew. \end{pf} Putting $\anglebetween{a}{b} = \frac{\pi}{2}$ in Theorem \ref{thm:cosanglebetween} we obtain the following corollary. \begin{cor} Two vectors in $\BBR^2$ are perpendicular if and only if their dot product is $0$. \end{cor} \vspace{2cm} \begin{figure}[htb] $$\psset{unit=1pc} \psline[linewidth=2pt, linecolor=red]{->}(-3,3)(3, 5) \psline[linewidth=2pt, linecolor=blue]{->}(3,5)(9, 2) \uput[l](-2.5, 4 ){\v{a}}\uput[u](6.5,4){\v{b} -\v{a}}\uput[r](4, 1 ){\v{b}}\psline[linewidth=2pt, linecolor=green]{->}(-3,3)(9, 2)\ $$\footnotesize\hangcaption{Theorem \ref{thm:cosanglebetween}.}\label{fig:dotproduct} \end{figure} \begin{df} Two vectors are said to be {\em orthogonal} if they are perpendicular. If $\v{a}$ is orthogonal to $\v{b}$, we write $\v{a} \perp \v{b}.$ \index{orthogonal} \end{df} \begin{df} If $\v{a} \perp \v{b}$ and $\norm{\v{a}}= \norm{\v{b}} = 1$ we say that $\v{a}$ and $\v{b}$ are {\em orthonormal}. \index{orthonormal} \end{df} \begin{rem} It follows that the vector $\v{0}$ is simultaneously parallel and perpendicular to any vector! \end{rem} \begin{df}\index{orthogonal space} Let $\v{a} \in \BBR^2$ be fixed. Then the {\em orthogonal space} to $\v{a}$ is defined and denoted by $$\v{a}^{\perp} = \{\v{x}\in \BBR^2: \v{x} \perp \v{ a}\}.$$ \end{df} Since $|\cos \theta| \leq\ 1$ we also have \begin{cor}[Cauchy-Bunyakovsky-Schwarz Inequality]\index{inequality!Cauchy-Bunyakovsky-Schwarz!in R2} $$\left|\dotprod{a}{b}\right| \leq \norm{\v{a}}\norm{\v{b}}.$$ \end{cor} \begin{cor}[Triangle Inequality] \index{inequality!triangle!in R2} $$\norm{\v{a} + \v{b}} \leq \norm{\v{a}} + \norm{\v{b}}.$$ \end{cor} \begin{pf} $$\begin{array}{lll} ||\v{a} + \v{b}||^2 & = & (\v{a} + \v{b})\bulletproduct (\v{a} + \v{b}) \\ & = & \v{a}\bulletproduct\v{a} + 2\v{a}\bulletproduct\v{b} + \v{b}\bulletproduct\v{b} \\ & \leq & ||\v{a}||^2 + 2||\v{a}||||\v{b}|| + ||\v{b}||^2 \\ & = & (||\v{a}|| + ||\v{b}||)^2, \end{array}$$from where the desired result follows. \end{pf} \begin{cor}[Pythagorean Theorem] If $\v{a} \perp \v{b}$ then $$||\v{a} + \v{b}||^2 = \norm{\v{a}}^2 + \norm{\v{b}}^2. $$\label{cor:pythagoras} \index{theorem!Pythagoras'} \end{cor} \begin{pf} Since $\dotprod{a}{b} = 0$, we have $$\begin{array}{lll} ||\v{a} + \v{b}||^2 & = & (\v{a} + \v{b})\bulletproduct (\v{a} + \v{b}) \\ & = & \v{a}\bulletproduct\v{a} + 2\v{a}\bulletproduct\v{b} + \v{b}\bulletproduct\v{b} \\ & = & \v{a}\bulletproduct\v{a} + 0 + \v{b}\bulletproduct\v{b} \\ & = & ||\v{a}||^2 + ||\v{b}||^2, \end{array}$$from where the desired result follows. \end{pf} \begin{df} The {\em projection} of $\v{t}$ onto $\v{v}$ (or the $\v{v}$-component of $\v{t}$) is the vector $$ \proj{t}{v} = (\cos \anglebetween{t}{v})\norm{\v{t}}\frac{1}{\norm{\v{v}}}\v{v},$$where $\anglebetween{v}{t}\in [0; \pi]$ is the convex angle between $\v{v}$ and $\v{t}$ read in the positive sense.\index{projection of a vector} \end{df} \begin{rem} Given two vectors $\v{t}$ and vector $\v{v} \neq \v{0}$, find bi-point representatives of them having a common tail and join them together at their tails. The projection of $\v{t}$ onto $\v{v}$ is the ``shadow'' of $\v{t}$ in the direction of $\v{v}$. To obtain $\proj{t}{v}$ we prolong $\v{v}$ if necessary and drop a perpendicular line to it from the head of $\v{t}$. The projection is the portion between the common tails of the vectors and the point where this perpendicular meets $\v{t}$. See figure \ref{fig:vectorprojections}. \end{rem} \vspace{1cm} \begin{figure}[htb] \begin{minipage}[t]{6cm}$$ \psset{unit=1pc} \psline[linewidth=2pt, linecolor=blue]{*->}(0, 0)(4, 0)\psline[linewidth=2pt, linecolor=red]{*->}(0, 0)(2, 3) \psline[linestyle=dashed](2, 3)(2, 0)\psdots[dotstyle=*, dotscale=2](0,0) $$ \end{minipage} \hfill \begin{minipage}[t]{6cm}$$ \psset{unit=1pc} \psline[linewidth=2pt, linecolor=blue]{*->}(0, 0)(4, 0)\psline[linewidth=2pt, linecolor=red]{*->}(0, 0)(-2, 3) \psline[linestyle=dashed](-2, 3)(-2, 0)\psdots[dotstyle=*, dotscale=2](0,0) \psline[linestyle=dashed]{->}(0,0)(-2,0) $$ \end{minipage} \vspace{1cm}\footnotesize\caption{Vector Projections.}\label{fig:vectorprojections} \end{figure} \begin{cor} Let $\v{a} \neq \v{0}$. Then $$\proj{x}{a} = \cos\anglebetween{x}{a}\norm{\v{x}}\frac{1}{\norm{\v{a}}} \v{a} = \frac{\dotprod{x}{a}}{\norm{\v{a}}^2}\v{a}.$$ \end{cor} \begin{thm} Let $\v{a}\in \BBR^2 \setminus \{\v{0}\}$. Then any $\v{x}\in \BBR^2$ can be decomposed as $$\v{x} = \v{u} + \v{v},$$where $\v{u} \in \BBR\v{a}$ and $\v{v} \in \v{a}^\perp$. \label{thm:orthodecomp}\end{thm} \begin{pf} We know that $\proj{x}{a}$ is parallel to $\v{ a}$, so we take $\v{u} = \proj{x}{a}$. This means that we must then take $\v{v} = \v{x} - \proj{x}{a}$. We must demonstrate that $\v{v}$ is indeed perpendicular to $\v{a}$. But this is clear, as \renewcommand{\arraystretch}{1.7} $$ \begin{array}{lll}\dotprod{a}{v} & = & \dotprod{a}{x} - \v{a}\bulletproduct \proj{x}{a} \\ & = & \dotprod{a}{x} - \v{a}\bulletproduct \frac{\dotprod{x}{a}}{\norm{\v{a}}^2}\v{a} \\ & = & \dotprod{a}{x} - \dotprod{x}{a}\\ & = & 0, \end{array}$$\renewcommand{\arraystretch}{1}completing the proof. \end{pf} \begin{cor} Let $\v{v} \perp \v{w}$ be non-zero vectors in $\BBR^2$. Then any vector $\v{a} \in \BBR^2$ has a unique representation as a linear combination of $\v{v}, \v{w}$ , $$\v{a} = s\v{v} + t\v{w}, \ \ (s, t)\in \BBR^2.$$ \label{cor:orthodecomp_1}\end{cor} \begin{pf} By Theorem \ref{thm:orthodecomp}, there exists a decomposition $$\v{a} = s\v{v} + s'\v{v}',$$where $\v{v}'$ is orthogonal to $\v{v}$. But then $\v{v}' || \v{w}$ and hence there exists $\alpha\in \BBR$ with $\v{v}' = \alpha \v{w}$. Taking $t = s'\alpha$ we achieve the decomposition $$\v{a} = s\v{v} + t\v{w}.$$ To prove uniqueness, assume $$s\v{v} + t\v{w} = \v{a} = p\v{v} + q\v{w}.$$ Then $(s - p)\v{v} = (q - t)\v{w}.$ We must have $s = p$ and $q = t$ since otherwise $\v{v}$ would be parallel to $\v{w}$. This completes the proof. \end{pf} \begin{cor} Let $\v{p},\v{q}$ be non-zero, non-parallel vectors in $\BBR^2$. Then any vector $\v{a} \in \BBR^2$ has a unique representation as a linear combination of $\v{p}, \v{q}$ , $$\v{a} = l\v{p} + m\v{q}, \ \ (l, m)\in \BBR^2.$$ \label{cor:orthodecomp_2}\end{cor} \begin{pf} Consider $\v{z} = \v{q} - \proj{q}{p}$. Clearly $\v{p} \perp \v{z}$ and so by Corollary \ref{cor:orthodecomp_1}, there exists unique $(s, t)\in \BBR^2$ such that $$\begin{array}{lll}\v{a} & = & s\v{p} + t\v{z} \\ & = & s\v{p} - t\proj{q}{p} + t\v{q} \\ & = & \left(s - t\frac{\dotprod{q}{p}}{\norm{\v{p}}^2}\right)\v{p} + t\v{q}, \end{array} $$establishing the result upon choosing $l = s - t\frac{\dotprod{q}{p}}{\norm{\v{p}}^2}$ and $m = t$. \end{pf} \begin{exa} Let $\v{p} = \colvec{1 \\ 1}$, $\v{q} = \colvec{1 \\ 2}$. Write $\v{p}$ as the sum of two vectors, one parallel to $\v{q}$ and the other perpendicular to $\v{q}$. \end{exa} \begin{solu}We use Theorem \ref{thm:orthodecomp}. We know that $\proj{p}{q}$ is parallel to $\v{q}$, and we find $$\proj{p}{q} = \dfrac{\dotprod{p}{q}}{\norm{\v{q}}^2} \v{q} = \dfrac{3}{5}\v{q} = \colvec{\frac{3}{5} \\ \frac{6}{5}}.$$ We also compute $$ \v{p} - \proj{p}{q} = \colvec{1-\frac{3}{5} \\ 1 - \frac{6}{5}} = \colvec{\frac{2}{5} \\ -\frac{1}{5}}. $$Observe that $$ \colvec{\frac{3}{5} \\ \frac{6}{5}} \bulletproduct \colvec{\frac{2}{5} \\ -\frac{1}{5}} = \frac{6}{25} - \frac{6}{25} = 0,$$ and the desired decomposition is $$\colvec{1\\ 1} = \colvec{\frac{3}{5} \\ \frac{6}{5}} + \colvec{\frac{2}{5} \\ -\frac{1}{5}}. $$ \end{solu} \vspace{2cm} \begin{figure}[h] \centering\begin{minipage}{4cm}\psset{unit=1pc} \pstTriangle[PosAngleA=-90, PosAngleB=180, PosAngleC=0] (4,1){A}(1,3){B}(5,5){C} \pstCircleABC[CodeFig=true, CodeFigColor=blue, linecolor=red]{A}{B}{C}{H} \vspace{.3cm}\footnotesize\hangcaption{Orthocentre.} \end{minipage} \end{figure} \begin{exa}\label{exa:orthocentre} Prove that the altitudes of a triangle $\triangle {ABC}$ on the plane are concurrent. This point is called the {\em orthocentre} of the triangle. \end{exa} \begin{solu}Put $\v{a} = \vect{OA}, \v{b} = \vect{OB}, \v{c} = \vect{OC}$. First observe that for any $\v{x}$, we have, upon expanding, \begin{equation}(\v{x} - \v{a})\bulletproduct (\v{b} - \v{c}) + (\v{x} - \v{b})\bulletproduct(\v{c} - \v{a}) + (\v{x} - \v{c})\bulletproduct(\v{a} - \v{b}) = 0. \label{eq:rel_altitudes_1}\end{equation} Let $H$ be the point of intersection of the altitude from $A$ and the altitude from $B.$ Then \begin{equation}0 = \vect{AH}\bulletproduct\vect{CB} = (\vect{OH} - \vect{OA})\bulletproduct (\vect{OB} - \vect{OC}) = (\vect{OH} - \v{a})\bulletproduct(\v{b} - \v{c}), \label{eq:rel_altitudes_2}\end{equation}and \begin{equation}0 = \vect{BH}\bulletproduct\vect{AC} = (\vect{OH} - \vect{OB})\bulletproduct (\vect{OC} - \vect{OA}) = (\vect{OH} - \v{b})\bulletproduct(\v{c} - \v{a}). \label{eq:rel_altitudes_3}\end{equation} Putting $\v{x} = \vect{OH}$ in (\ref{eq:rel_altitudes_1}) and subtracting from it (\ref{eq:rel_altitudes_2}) and (\ref{eq:rel_altitudes_3}), we gather that $$0 = (\vect{OH} - \v{c})\bulletproduct (\v{a} - \v{b}) = \vect{CH}\bulletproduct\vect{AB}, $$ which gives the result. \end{solu} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Determine the value of $a$ so that $\colvec{a\\ 1-a}$ be perpendicular to $ \colvec{1 \\ -1}.$ \begin{answer} $a = \frac{1}{2}$\end{answer} \end{pro} \begin{pro} Demonstrate that $$(\v{b} + \v{c} = \v{0}) \wedge (\norm{\v{a}} = \norm{\v{b}}) \iff (\v{a} - \v{b})\bulletproduct (\v{a} - \v{c}) = 0.$$ \end{pro} \begin{pro} Let $\v{p} = \colvec{4 \\ 5}$, $\v{r} = \colvec{-1 \\ 1}$, $\v{s} = \colvec{2 \\ 1}$. Write $\v{p}$ as the sum of two vectors, one parallel to $\v{r}$ and the other parallel to $\v{s}$. \begin{answer} $$\v{p} = \colvec{4\\ 5} = 2\colvec{-1\\ 1} + 3\colvec{2 \\ 1} = 2\v{r} + 3\v{s}. $$ \end{answer} \end{pro} \begin{pro} Prove that $$ \norm{\v{a}}^2 = (\dotprod{a}{i})^2 + (\dotprod{a}{j})^2. $$ \begin{answer} Since $a_1 = \dotprod{a}{i}, a_2 = \dotprod{a}{j},$ we may write $$ \v{a} = (\dotprod{a}{i})\v{i} + (\dotprod{a}{j})\v{j} $$from where the assertion follows. \end{answer} \end{pro} \begin{pro} Let $\v{a} \neq \v{0} \neq \v{b}$ be vectors in $\BBR^2$ such that $\dotprod{a}{b} = 0$. Prove that $$\alpha\v{a} + \beta\v{b} = \v{0} \implies \alpha = \beta = 0. $$ \begin{answer} $$\begin{array}{lll} \alpha\v{a} + \beta\v{b} = \v{0} & \implies & \v{a}\bulletproduct (\alpha\v{a} + \beta\v{b}) = \v{a}\bulletproduct\v{0} \\ & \implies & \alpha (\v{a}\bulletproduct \v{a}) = 0\\ & \implies & \alpha \norm{\v{a}}^2 = 0. \end{array}$$Since $\v{a} \neq \v{0}$, we must have $\norm{\v{a}} \neq 0$ and thus $\alpha = 0.$ But if $\alpha = 0$ then $$\begin{array}{lll}\alpha\v{a} + \beta\v{b} = \v{0} & \implies & \beta\v{b} = \v{0}\\ & \implies & \beta = 0, \\ \end{array} $$since $\v{b} \neq \v{0}$. \end{answer} \end{pro} \begin{pro} Let $(\v{x},\ \v{y}) \in (\BBR^2)^2 $ with $||\v{x}|| = \frac{3}{2}||\v{y}||$. Shew that $2\v{x} + 3\v{y}$ and $2\v{x} - 3\v{y}$ are perpendicular. \begin{answer} We must shew that $$(2\v{x} + 3\v{y})\bulletproduct( 2\v{x} - 3\v{y}) = 0.$$ But $$(2\v{x} + 3\v{y})\bulletproduct(2\v{x} - 3\v{y}) = 4||\v{x}||^2 - 9||\v{y}||^2 = 4(\frac{9}{4}||\v{y}||^2) - 9||\v{y}||^2 = 0.$$ \end{answer} \end{pro} \begin{pro} Let $\v{a}, \v{b}$ be fixed vectors in $\BBR^2$. Prove that if $$\forall \v{v}\in\BBR^2, \v{v}\bulletproduct\v{a} = \v{v}\bulletproduct\v{b},$$then $\v{a} = \v{b}$. \begin{answer} We have $\forall \v{v}\in\BBR^2, \v{v}\bulletproduct(\v{a} - \v{b}) = 0.$ In particular, choosing $\v{v} = \v{a} - \v{b}$, we gather $$(\v{a} - \v{b})\bulletproduct (\v{a} - \v{b}) = ||\v{a} - \v{b}||^2 = 0.$$But the norm of a vector is $0$ if and only if the vector is the $\v{0}$ vector. Therefore $\v{a} - \v{b} = \v{0}$, i.e., $\v{a} = \v{b}$. \end{answer} \end{pro} \begin{pro} Let $(\v{a}, \v{b})\in (\BBR^2)^2$. Prove that $$ \norm{\v{a} + \v{b}}^2 + \norm{\v{a} - \v{b}}^2 = 2 \norm{\v{a}}^2 + 2 \norm{\v{b}}^2 . $$ \begin{answer} We have $$ \begin{array}{lll} \norm{\v{a} \pm \v{b}}^2 & = & (\v{a} \pm \v{b}) \bulletproduct (\v{a} \pm \v{b}) \\ & = & \dotprod{a}{a} \pm 2\dotprod{a}{b} + \dotprod{b}{b} \\ & = & \norm{\v{a}}^2 \pm 2\dotprod{a}{b}+ \norm{\v{b}}^2,\end{array} $$ whence the result follows. \end{answer} \end{pro} \begin{pro} Let $\v{u}, \v{v}$ be vectors in $\BBR^2$. Prove the {\em polarisation identity:} $$\v{u}\bullet\v{v} = \frac{1}{4}\left(||\v{u} + \v{v}||^2 - ||\v{u} - \v{v}||^2 \right).$$ \begin{answer} We have $$\begin{array}{lll} ||\v{u} + \v{v}||^2 - ||\v{u} - \v{v}||^2 & = & (\v{u} + \v{v})\bulletproduct (\v{u} + \v{v}) - (\v{u} - \v{v})\bulletproduct (\v{u} - \v{v}) \\ &=& \dotprod{u}{u} + 2\dotprod{u}{v} + \dotprod{v}{v} - (\dotprod{u}{u} - 2\dotprod{u}{v} + \dotprod{v}{v}) \\ &=& 4\dotprod{u}{v}, \end{array}$$giving the result. \end{answer} \end{pro} \begin{pro} Let $\v{x}, \v{a}$ be non-zero vectors in $\BBR^2$. Prove that $${\rm proj}\begin{array}{c}\proj{a}{x}\\ \v{a}\end{array} = \alpha \v{a},$$with $0 \leq \alpha \leq 1.$ \begin{answer} By definition \renewcommand{\arraystretch}{2.5} $${\everymath{\displaystyle} \begin{array}{lll} {\rm proj}\begin{array}{c}\proj{a}{x}\\ \v{a}\end{array} & = & \frac{\proj{a}{x}\bulletproduct \v{a}}{\norm{\v{a}}^2}\v{a}\\ & = & \frac{ \frac{\dotprod{a}{x}}{\norm{\v{x}}^2}\v{x}\bulletproduct \v{ a} }{\norm{\v{a}}^2}\v{a} \\ & = & \frac{(\dotprod{a}{x})^2}{\norm{\bf x}^2\norm{\v{a}}^2} \v{a}, \\ \end{array}}$$ \renewcommand{\arraystretch}{1} Since $\dis{ 0 \leq \frac{(\dotprod{a}{x})^2}{\norm{\v{x}}^2\norm{\v{a}}^2} \leq 1}$ by the CBS Inequality, the result follows. \end{answer} \end{pro} \begin{pro} Let $(\lambda , \v{ a})\in\BBR\times \BBR^2$ be fixed. Solve the equation $$\dotprod{a}{x} = \lambda $$for $\v{x}\in\BBR^2$. \begin{answer} Clearly, if $\v{ a} = \v{{\bf 0}}$ and $\lambda \neq 0$ then there are no solutions. If both $\v{ a} = \v{{\bf 0}}$ and $\lambda = 0$, then the solution set is the whole space $\BBR^2$. So assume that $\v{ a} \neq \v{{\bf 0}}$. By Theorem \ref{thm:orthodecomp}, we may write $\v{ x} = \v{u} + \v{v}$ with $\proj{\bf x}{\bf a} = \v{u}||\v{ a}$ and $\v{v}\perp \v{ a}$. Thus there are infinitely many solutions, each of the form $$\v{ x} = \v{u} + \v{v} = \frac{\dotprod{x}{a}}{\norm{\bf a}^2}\v{a} + \v{v} = \frac{\lambda}{\norm{\bf a}^2}\v{ a} + \v{v},$$where $\v{v}\in \v{a}^{\perp}$. \end{answer} \end{pro} \end{multicols} \section{Lines on the Plane} \begin{df} Three points $A$, $B$, and $C$ are {\em collinear} if they lie on the same line. \end{df} It is clear that the points $A$, $B$, and $C$ are collinear if and only if $\vect{AB}$ is parallel to $\vect{AC}$. Thus we have the following definition. \begin{df} The parametric equation with parameter $t\in \BBR$ of the straight line passing through the point $P = \colpoint{p_1 \\ p_2}$ in the direction of the vector $\v{v} \neq \v{0}$ is $$\colvec{x-p_1 \\ y-p_2} = t \v{v}. $$ If $\v{r} = \colvec{x\\ y}$, then the equation of the line can be written in the form \begin{equation} \v{r} - \v{p} = t\v{v}. \label{eq:equation_line_R2}\end{equation} The {\em Cartesian equation of a line} is an equation of the form $ax + by = c$, where $a^2 + b^2 \neq 0$. We write $(AB)$ for the line passing through the points $A$ and $B$. \end{df} \begin{thm}\label{thm:alternative_eq_line_R2} Let $\v{v} \neq \v{0}$ and let $\v{n} \perp \v{v}$. An alternative form for the equation of the line $\v{r} - \v{p} = t\v{v}$ is $$ (\v{r} - \v{p})\bulletproduct \v{n} = 0. $$ Moreover, the vector $\colvec{a\\ b}$ is perpendicular to the line with Cartesian equation $ax + by = c$. \end{thm} \begin{pf} The first part follows at once by observing that $\dotprod{v}{n} = 0$ and taking dot products to both sides of \ref{eq:equation_line_R2}. For the second part observe that at least one of $a$ and $b$ is $\neq 0$. First assume that $a \neq 0$. Then we can put $y = t$ and $x = -\frac{b}{a}t + \frac{c}{a}$ and the parametric equation of this line is $$\colvec{x - \frac{c}{a}\\ y} = t\colvec{-\frac{b}{a} \\ 1}, $$ and we have $$\colvec{-\frac{b}{a} \\ 1} \bulletproduct \colvec{a\\ b} = -\frac{b}{a}\cdot a + b = 0. $$Similarly if $b \neq 0$ we can put $x = t$ and $y = -\frac{a}{b}t + \frac{c}{b}$ and the parametric equation of this line is $$\colvec{x \\ y - \frac{c}{b}} = t\colvec{1 \\ -\frac{a}{b}}, $$ and we have $$\colvec{1\\ -\frac{a}{b}} \bulletproduct \colvec{a\\ b} = a-\frac{a}{b}\cdot b = 0, $$proving the theorem in this case. \end{pf} \begin{rem} The vector $\colvec{\frac{a}{\sqrt{a^2 + b^2}} \\ \frac{b}{\sqrt{a^2 + b^2}}}$ has norm $1$ and is orthogonal to the line $ax + by = c$. \end{rem} \begin{exa} The equation of the line passing through $A = \colpoint{2\\ 3}$ and in the direction of $\v{v} = \colvec{-4 \\ 5}$ is $$\colvec{x -2 \\ y-3} = \lambda \colvec{-4 \\ 5}.$$ \end{exa} \begin{exa} Find the equation of the line passing through $A = \colpoint{-1\\ 1}$ and $B = \colpoint{-2\\ 3}$. \end{exa} \begin{solu}The direction of this line is that of $$\vect{AB} = \colvec{-2 - (-1) \\ 3 - 1} = \colvec{-1 \\ 2}.$$The equation is thus $$\colvec{x+1\\ y-1} = \lambda\colvec{-1 \\ 2}, \ \lambda \in\BBR .$$ \end{solu} \begin{exa} Suppose that $(m, b)\in \BBR^2$. Write the Cartesian equation of the line $y = mx + b$ in parametric form. \end{exa}\begin{solu}Here is a way. Put $x = t$. Then $y = mt + b$ and so the desired parametric form is $$\colvec{x \\ y - b} = t\colvec{1\\ m }. $$ \end{solu} \begin{exa} Let $(m_1, m_2, b_1, b_2)\in\BBR^4, m_1m_2 \neq 0.$ Consider the lines $L_1: y = m_1x + b_1$ and $L_2: y = m_2x + b_2$. By translating this problem in the language of vectors in $\BBR^2$, shew that $L_1 \perp L_2$ if and only if $m_1m_2 = -1.$ \end{exa} \begin{solu}The parametric equations of the lines are $$L_1: \ \colvec{x \\ y-b_1}= s\colvec{1 \\ m_1},\ \ \ L_2: \ \colvec{x \\ y-b_2} = t\colvec{1 \\ m_2}.$$ Put $\v{v} = \colvec{1\\ m_1}$ and $\v{w} = \colvec{1\\ m_2}$. Since the lines are perpendicular we must have $\dotprod{v}{w} = 0$, which yields $$0 = \dotprod{v}{w} = 1(1) + m_1(m_2) \implies m_1m_2 = -1.$$ \end{solu} \begin{thm}[Distance Between a Point and a Line] \label{thm:distance_point_lineR2}\index{distance!between a point and a line!on the plane} Let $(\v{r} - \v{a})\bulletproduct\v{n} = 0$ be a line passing through the point $A$ and perpendicular to vector $\v{n}$. If $B$ is not a point on the line, then the distance from $B$ to the line is $$\frac{\left|(\v{a} - \v{b})\bulletproduct\v{n}\right|}{\norm{\v{n}}}.$$ If the line has Cartesian equation $ax + by = c$, then this distance is $$ \dfrac{|ab_1 + bb_2 - c|}{\sqrt{a^2 + b^2}}. $$ \end{thm} \begin{pf} Let $R_0$ be the point on the line that is nearest to $B$. Then $\vect{BR_0} = \v{r_0} - \v{b}$ is orthogonal to the line, and the distance we seek is $$ ||\proj{\v{r_0} - \v{b}}{n}|| = \left|\left| \frac{(\v{r_0} - \v{b})\bulletproduct\v{n}}{\norm{\v{n}}^2}\v{n}\right|\right| = \frac{|(\v{r_0} - \v{b})\bulletproduct\v{n}|}{\norm{\v{n}}}.$$ Since $R_0$ is on the line, $\dotprod{r_0}{n} = \dotprod{a}{n},$ and so $$ ||\proj{\v{r_0} - \v{b}}{n}|| = \frac{|\dotprod{r_0}{n} - \dotprod{b}{n}|}{\norm{\v{n}}|} = \frac{|\dotprod{a}{n} - \dotprod{b}{n}|}{\norm{\v{n}}} = \frac{|(\v{a} - \v{b})\bulletproduct\v{n}|}{\norm{\v{n}}},$$ as we wanted to shew. \bigskip If the line has Cartesian equation $ax + by = c$, then at least one of $a$ and $b$ is $\neq 0$. Let us suppose $a \neq 0$, as the argument when $a = 0$ and $b\neq 0$ is similar. Then $ax + by = c$ is equivalent to $$\left(\colvec{x\\ y} - \colvec{\frac{c}{a} \\ 0}\right)\bulletproduct \colvec{a\\ b} =0. $$We use the result obtained above with $\v{a} = \colvec{\frac{c}{a} \\ 0}$, $\v{n} = \colvec{a \\ b}$, and $B = \colpoint{b_1 \\ b_2}$. Then $\norm{\v{n}} = \sqrt{a^2 + b^2}$ and $$|(\v{a} - \v{b})\bulletproduct\v{n}| = \left|\colvec{\frac{c}{a} - b_1 \\ -b_2} \bulletproduct \colvec{a\\ b}\right| = |c - ab_1 - bb_2| = |ab_1 + bb_2 - c|,$$giving the result. \end{pf} \begin{exa} Recall that the medians of $\triangle {ABC}$ are lines joining the vertices of $\triangle {ABC}$ with the midpoints of the side opposite the vertex. Prove that the medians of a triangle are concurrent, that is, that they pass through a common point. \end{exa} \begin{rem} This point of concurrency is called, alternatively, the {\em isobarycentre}, {\em centroid}, or {\em centre of gravity} of the triangle. \end{rem} \begin{solu}Let ${M_A}$, ${M_B}$, and ${M_C}$ denote the midpoints of the lines opposite $A$, $B$, and $C$, respectively. The equation of the line passing through $A$ and in the direction of $\vect{AM_A}$ is (with $\v{r} = \colvec{x\\ y}$) $$\v{r} = \vect{OA} + r\vect{AM_A}.$$Similarly, the equation of the line passing through $B$ and in the direction of $\vect{BM_B}$ is $$\v{r} = \vect{OB} + s\vect{BM_A}.$$These two lines must intersect at a point ${G}$ inside the triangle. We will shew that $\vect{GC}$ is parallel to $\vect{CM_C}$, which means that the three points $G, C, M_C$ are collinear. Now, $\exists (r_0, s_0) \in \BBR^2$ such that $$\vect{OA} + r_0\vect{AM_A} = \vect{OG} = \vect{OB} + s_0\vect{BM_B},$$ that is $$r_0\vect{AM_A} - s_0\vect{BM_B} = \vect{OB} - \vect{OA},$$ or $$r_0(\vect{AB} + \vect{BM_A}) - s_0(\vect{BA}+ \vect{AM_B}) = \vect{AB}.$$ Since ${M_A}$ and ${M_B}$ are the midpoints of $\bipoint{B}{C}$ and $\bipoint{C}{A}$ respectively, we have $2\vect{BM_A} = \vect{BC}$ and $2\vect{AM_B} = \vect{AC} = \vect{AB} + \vect{BC}$. The relationship becomes $$r_0(\vect{AB} + \frac{1}{2}\vect{BC}) - s_0(-\vect{AB} + \frac{1}{2}\vect{AB} + \frac{1}{2}\vect{BC}) = \vect{AB},$$ $$ (r_0 + \frac{s_0}{2} - 1)\vect{AB} = (-\frac{r_0}{2} + \frac{s_0}{2})\vect{BC}.$$ We must have $$r_0 + \frac{s_0}{2} - 1 = 0,$$ $$-\frac{r_0}{2} + \frac{s_0}{2} = 0,$$since otherwise the vectors $\vect{AB}$ and $\vect{BC}$ would be parallel, and the triangle would be degenerate. Solving, we find $s_0 = r_0 = \frac{2}{3}$. Thus we have $\vect{OA} + \frac{2}{3}\vect{AM_A} = \vect{OG}$, or $\vect{AG} = \frac{2}{3}\vect{AM_A}$, and similarly, $\vect{BG} = \frac{2}{3}\vect{BM_B}$. \bigskip From $\vect{AG} = \frac{2}{3}\vect{AM_A}$, we deduce $\vect{AG} = 2\vect{GM_A}$. Since ${M_A}$ is the midpoint of $\bipoint{B}{C}$, we have $\vect{GB}+ \vect{GC} = 2\vect{GM_A} = \vect{AG}$, which is equivalent to $$\vect{GA} + \vect{GB}+ \vect{GC} = \v{0}.$$ As ${M_C}$ is the midpoint of $\bipoint{A}{B}$ we have $\vect{GA} + \vect{GB}= 2\vect{GM_C}$. Thus $$\v{0} = \vect{GA} + \vect{GB}+ \vect{GC} = 2\vect{GM_C} + \vect{GC}.$$ This means that $\vect{GC} = -2\vect{GM_C}$, that is, that they are parallel, and so the points ${G}$, $C$ and ${M_C}$ all lie on the same line. This achieves the desired result. \end{solu} \begin{rem} The centroid of $\triangle {ABC}$ satisfies thus $$\vect{GA} + \vect{GB}+ \vect{GC} = \v{0}, $$and divides the medians on the ratio $2:1$, reckoning from a vertex. \end{rem} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Find the angle between the lines $2x - y = 1$ and $x - 3y = 1$. \begin{answer} Since $\v{a} = \colvec{2 \\ -1}$ is normal to $2x - y = 1$ and $\v{b} = \colvec{1\\ -3}$ is normal to $x - 3y = 1$, the desired angle can be obtained by finding the angle between the normal vectors: $$\anglebetween{a}{b} = \arccos \dfrac{\dotprod{a}{b}}{\norm{\v{a}}\norm{\v{b}}}= \arccos \dfrac{5}{\sqrt{5} \cdot \sqrt{10}} = \arccos \dfrac{1}{\sqrt{2}} = \dfrac{\pi}{4}. $$ \end{answer} \end{pro} \begin{pro} Find the equation of the line passing through $\colpoint{1\\ -1}$ and in a direction perpendicular to $\colvec{2\\ 1}$. \begin{answer} $2(x-1) + (y + 1) = 0$ or $2x + y = 1$. \end{answer} \end{pro} \begin{pro} $\triangle {ABC}$ has centroid ${G}$, and $\triangle A'B'C'$ satisfies $$\vect{AA'} + \vect{BB'} + \vect{CC'} = \v{0}.$$Prove that ${G}$ is also the centroid of $\triangle A'B'C'$. \begin{answer} By Chasles' Rule $\vect{AA'} = \vect{AG} + \vect{GA'} $, $\vect{BB'} = \vect{BG} + \vect{GB'} $, and $\vect{CC'} = \vect{CG} + \vect{GC'} $. Thus $$\begin{array}{lll} \v{0} & = & \vect{AA'} + \vect{BB'} + \vect{CC'} \\ & =& \vect{AG} + \vect{GA'} + \vect{BG} + \vect{GB'} + \vect{CG} + \vect{GC'} \\ & = & -(\vect{GA} + \vect{GB}+ \vect{GC}) + (\vect{GA'} + \vect{GB'} + \vect{GC'}) \\ & = & \vect{GA'} + \vect{GB'} + \vect{GC'}, \end{array} $$whence the result. \end{answer} \end{pro} \begin{pro} Let ${ABCD}$ be a trapezoid, with bases $\bipoint{A}{B}$ and $\bipoint{C}{D}$. The lines $(AC)$ and $(BD)$ meet at ${E}$ and the lines $(AD)$ and $(BC)$ meet at $F$. Prove that the line $({EF})$ passes through the midpoints of $\bipoint{A}{B}$ and $\bipoint{C}{D}$ by proving the following steps. \begin{dingautolist}{202} \item Let ${I}$ be the midpoint of $\bipoint{A}{B}$ and let ${J}$ be the point of intersection of the lines $({FI})$ and $({DC})$. Prove that ${J}$ is the midpoint of $\bipoint{C}{D}$. Deduce that ${F}, {I}, {J}$ are collinear. \item Prove that ${E}, {I}, {J}$ are collinear. \end{dingautolist} \begin{answer} We have: \begin{dingautolist}{202}\item The points ${F}, A, D$ are collinear, and so $\vect{FA}$ is parallel to $\vect{FD}$, meaning that there is $k\in \BBR\setminus \{0\}$ such that $\vect{FA} = k\vect{FD}$. Since the lines $(AB)$ and $(DC)$ are parallel, we obtain through Thales' Theorem that $\vect{FI} = k\vect{FJ}$ and $\vect{FB} = k\vect{FC}$. This gives $$ \vect{FA} - \vect{FI} = k(\vect{FD} - \vect{FJ}) \implies \vect{IA} = k\vect{JD}. $$ Similarly $$ \vect{FB} - \vect{FI} = k(\vect{FC} - \vect{FJ}) \implies \vect{IB} = k\vect{JC}. $$Since ${I}$ is the midpoint of $\bipoint{A}{B}$, $\vect{IA} + \vect{IB} = \v{0}$, and thus $k(\vect{JC} + \vect{JD}) = \v{0}$. Since $k \neq 0,$ we have $\vect{JC} + \vect{JD} = \v{0}$, meaning that ${J}$ is the midpoint of $\bipoint{C}{D}$. Therefore the midpoints of $\bipoint{A}{B}$ and $\bipoint{C}{D}$ are aligned with ${F}$. \item Let ${J'}$ be the intersection of the lines $({EI})$ and $({DC})$. Let us prove that ${J'} = {J}$. \bigskip Since the points ${E}, A, C$ are collinear, there is $l\neq 0$ such that $\vect{EA}= l\vect{EC}$. Since the lines $({ab})$ and $({DC})$ are parallel, we obtain via Thales' Theorem that $\vect{EI} = l\vect{EJ'}$ and $\vect{EB} = l\vect{ED}$. These equalities give $$\vect{EA}-\vect{EI} = l(\vect{EC}- \vect{EJ'}) \implies \vect{IA} = l\vect{J'C},$$ $$\vect{EB} -\vect{EI} = l(\vect{ED} - \vect{EJ'}) \implies \vect{IB} = l\vect{J'D}.$$ Since ${I}$ is the midpoint of $\bipoint{A}{B}$, $\vect{IA} + \vect{IB} = \v{0}$, and thus $l(\vect{J'C} + \vect{J'D}) = \v{0}$. Since $l \neq 0$, we deduce $\vect{J'C} + \vect{J'D} = \v{0}$, that is, ${J'}$ is the midpoint of $\bipoint{C}{D}$, and so ${J'} = {J}$. \end{dingautolist} \end{answer} \end{pro} \begin{pro} Let ${ABCD}$ be a parallelogram.\begin{dingautolist}{202} \item Let ${E}$ and ${F }$ be points such that $$\vect{AE} = \frac{1}{4} \vect{AC} \ \ \ \ {\rm and}\ \ \ \vect{AF} = \frac{3}{4}\vect{AC}. $$Demonstrate that the lines $({BE})$ and $({DF})$ are parallel. \item Let ${I}$ be the midpoint of $\bipoint{A}{D}$ and ${J}$ be the midpoint of $\bipoint{B}{C}$. Demonstrate that the lines $({AB})$ and $({IJ})$ are parallel. What type of quadrilateral is ${IEJF}$? \end{dingautolist} \begin{answer} We have: \begin{dingautolist}{202} \item By Chasles' Rule $$\begin{array}{lll}\vect{AE} = \frac{1}{4} \vect{AC} & \iff & \vect{AB} + \vect{BE} = \frac{1}{4} \vect{AC} \end{array},$$ and $$\begin{array}{lll}\vect{AF} = \frac{3}{4} \vect{AC} & \iff & \vect{AD} + \vect{DF} = \frac{3}{4} \vect{AC} \end{array}.$$Adding, and observing that since ${ABCD}$ is a parallelogram, $ \vect{AB} = \vect{CD}$, $$\begin{array}{lll}\vect{AB} + \vect{BE} + \vect{AD} + \vect{DF} = \vect{AC} & \iff & \vect{BE} + \vect{DF} = \vect{AC} - \vect{AB} - \vect{AD} \\ & \iff & \vect{BE} + \vect{DF} = \vect{AD} + \vect{DC} - \vect{AB} - \vect{AD} \\ & \iff & \vect{BE} = -\vect {DF}. \end{array}. $$ The last equality shews that the lines $({BE})$ and $({DF})$ are parallel. \item Observe that $\vect{BJ} = \frac{1}{2}\vect{BC} = \frac{1}{2}\vect{AD} = \vect{AI} = -\vect{IA}$ . Hence $$\vect{IJ} = \vect{IA} + \vect{AB} + \vect{BJ} = \vect{AB}, $$ proving that the lines $({AB})$ and $({IJ})$ are parallel. \bigskip Observe that $$\vect{IE} = \vect{IA} + \vect{AE} = \frac{1}{2}\vect{DA} + \frac{1}{4}\vect{AC} = \frac{1}{2}\vect{CB} + \vect{FC} = \vect{CJ} + \vect{FC} = \vect{FC} + \vect{CJ} = \vect{FJ}, $$whence ${IEJF}$ is a parallelogram. \end{dingautolist} \end{answer} \end{pro} \begin{pro} ${ABCD}$ is a parallelogram; point ${I}$ is the midpoint of $\bipoint{A}{B}$. Point $E$ is defined by the relation $\vect{IE} = \frac{1}{3} \vect{ID}$. Prove that $$ \vect{AE} = \frac{1}{3}\left(\vect{AB} + \vect{AD}\right)$$ and prove that the points $A, C, E$ are collinear. \begin{answer} Since $\vect{IE} = \frac{1}{3} \vect{ID}$ and $\bipoint{I}{D}$ is a median of $\triangle ABD$, $E$ is the centre of gravity of $\triangle ABD$. Let ${M}$ be the midpoint of $\bipoint{B}{D}$, and observe that ${M}$ is the centre of the parallelogram, and so $2\vect{AM} = \vect{AB} + \vect{AD}$. Thus $$ \vect{AE} = \frac{2}{3}\vect{AM} = \frac{1}{3}(2\vect{AM}) = \frac{1}{3}(\vect{AB} + \vect{AD}) . $$To shew that $A, C, E$ are collinear it is enough to notice that $\vect{AE} = \frac{1}{3}\vect{AC}$. \end{answer} \end{pro} \begin{pro}\label{pro:barycentre_three_points} Put $\vect{OA} = \v{a}$, $\vect{OB} = \v{b}$, $\vect{OC} = \v{c}$. Prove that $A, B, C$ are collinear if and only if there exist real numbers $\alpha, \beta, \gamma$, not all zero, such that $$\alpha\v{a} + \beta\v{b} + \gamma\v{c} = \v{0}, \ \ \ \ \alpha + \beta + \gamma = 0. $$ \begin{answer} Suppose $A, B, C$ are collinear and that $ \dfrac{\norm{\bipoint{A}{B}}}{\norm{\bipoint{B}{C}}} = \dfrac{\lambda}{\mu}$. Then by the Section Formula \ref{eq:section_formula}, $$ \v{b} = \dfrac{\lambda\v{c} + \mu\v{a}}{\lambda + \mu}, $$ whence $\mu \v{a} - (\lambda + \mu)\v{b} + \lambda \v{c} = \v{0}$ and clearly $\mu - (\lambda + \mu) + \lambda = 0$. Thus we may take $\alpha = \mu$, $\beta = \lambda + \mu$, and $\gamma = \lambda$. \bigskip Conversely, suppose that $$\alpha\v{a} + \beta\v{b} + \gamma\v{c} = \v{0}, \ \ \ \ \alpha + \beta + \gamma = 0 $$ for some real numbers $\alpha, \beta, \gamma$, not all zero. Assume without loss of generality that $\gamma \neq 0$. Otherwise we simply change the roles of $\gamma$, and $\alpha$ and $\beta$ Then $\gamma = -(\alpha + \beta) \neq 0$. Hence $$\alpha \v{a} + \beta\v{b} = (\alpha + \beta)\v{c} \implies \v{c} = \dfrac{\alpha \v{a} + \beta\v{b}}{\alpha + \beta}, $$and thus $\bipoint{O}{C}$ divides $\bipoint{A}{B}$ into the ratio $\dfrac{\beta}{\alpha}$, and therefore, $A, B, C$ are collinear. \end{answer} \end{pro} \begin{pro} Prove Desargues' Theorem:\index{theorem!Desargues} If $\triangle ABC$ and $\triangle A'B'C'$ (not necessarily in the same plane) are so positioned that $(AA')$, $(BB')$, $(CC')$ all pass through the same point $V$ and if $(BC)$ and $(B'C')$ meet at $L$, $(CA)$ and $(C'A')$ meet at $M$, and $(AB)$ and $(A'B')$ meet at $N$, then $L, M, N$ are collinear. \begin{answer} Put $\vect{OX} = \v{x}$ for $X\in \{A, A', B, B', C, C', L, M, N, V\}$. Using problem \ref{pro:barycentre_three_points} we deduce \begin{equation} \v{v} + \alpha \v{a} + \alpha'\v{a'} = \v{0}, \ \ \ 1 + \alpha + \alpha' = 0, \label{eq:desargues1} \end{equation} \begin{equation} \v{v} + \beta \v{a} + \beta'\v{a'} = \v{0}, \ \ \ 1 + \beta + \beta' = 0, \label{eq:desargues2} \end{equation} \begin{equation} \v{v} + \gamma \v{a} + \gamma'\v{a'} = \v{0}, \ \ \ 1 + \gamma + \gamma' = 0. \label{eq:desargues3} \end{equation} From \ref{eq:desargues2}, \ref{eq:desargues3}, and the Section Formula \ref{eq:section_formula} we find $$\dfrac{\beta\v{b} - \gamma\v{c}}{\beta - \gamma} = \dfrac{\beta'\v{b'} - \gamma'\v{c'}}{\beta' - \gamma'}= \v{l}, $$whence $(\beta - \gamma)\v{l} = \beta\v{b} - \gamma\v{c}$. In a similar fashion, we deduce $$(\gamma - \alpha)\v{m} = \gamma \v{c} - \alpha\v{a}, $$ $$(\alpha - \beta)\v{n} = \alpha \v{a} - \beta\v{b}. $$This gives $$(\beta - \gamma)\v{l} + (\gamma - \alpha)\v{m} + (\alpha - \beta)\v{n} = \v{0}, $$ $$ (\beta - \gamma) + (\gamma - \alpha) + (\alpha - \beta) = 0, $$and appealing to problem \ref{pro:barycentre_three_points} once again, we deduce that $L, M, N$ are collinear. \end{answer} \end{pro} \end{multicols} \section{Vectors in \protect$\BBR^3\protect$} \index{vector!in space} We now extend the notions studied for $\BBR^2$ to $\BBR^3$. The rectangular coordinate form of a vector in $\BBR^3$ is $$\v{a} = \colvec{a_1 \\ a_2 \\ a_3}.$$In particular, if $$\v{{i}} = \colvec{ 1 \\ 0 \\ 0 }, \ \v{{j}} = \colvec{0 \\ 1 \\ 0}, \v{{k}} = \colvec{0 \\ 0 \\ 1}$$then we can write any vector $\v{a} = \colvec{ a_1 \\ a_2 \\ a_3}$ as a sum $$\v{a} = a_1\v{i} + a_2\v{j} + a_3\v{k}.$$ Given $\v{a} = \colvec{a_1 \\ a_2 \\ a_3}$ and $\v{b} = \colvec{b_1 \\ b_2 \\ b_3}$, their dot product is $$\dotprod{a}{b} = a_1b_1 + a_2b_2 + a_3b_3,$$ and $$\norm{\v{a}} = \sqrt{a_1 ^2 + a_2 ^2 + a_3 ^2}.$$ We also have $$\v{i}\bulletproduct\v{j} = \v{j}\bulletproduct\v{k} = \v{k}\bulletproduct\v{i} = 0,$$and $$\norm{\v{i}} = \norm{\v{j}} = \norm{\v{k}} = 1.$$ \begin{df} A system of unit vectors $\v{i}, \v{j}, \v{k}$ is {\em right-handed} if the shortest-route rotation which brings $\v{i}$ to coincide with $\v{j}$ is performed in a counter-clockwise manner. It is {\em left-handed} if the rotation is done in a clockwise manner. \index{right-handed} \index{left-handed} \end{df} To study points in space we must first agree on the orientation that we will give our coordinate system. We will use, unless otherwise noted, a right-handed orientation, as in figure \ref{fig:righthanded}. \vspace{1cm} \begin{figure}[htb] \begin{minipage}{6cm} $$\psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{->}(0,0)(3, 0)\uput[r](3,0){\v{j}} \psline[linewidth=1.5pt, linecolor=blue]{->}(0,0)(0, 3)\uput[u](0,3){\v{k}} \psline[linewidth=1.5pt, linecolor=green]{->}(0,0)(-2.213, -2.213)\uput[d](-2.213,-2.213){\v{i}} $$\vspace{1cm}\hangcaption{Right-handed system.} \label{fig:righthanded}\end{minipage} \hfill \begin{minipage}{6cm} $$\psset{unit=1pc} \psline[linewidth=1.5pt, linecolor=red]{->}(0,0)(-3, 0)\uput[l](-3,0){\v{j}} \psline[linewidth=1.5pt, linecolor=blue]{->}(0,0)(0, 3)\uput[u](0,3){\v{k}} \psline[linewidth=1.5pt, linecolor=green]{->}(0,0)(-2.213, -2.213)\uput[d](-2.213,-2.213){\v{i}} $$\vspace{1cm}\hangcaption{Left-handed system.}\end{minipage}\end{figure} \begin{rem} The analogues of the Cauchy-Bunyakovsky-Schwarz and the Triangle Inequality also hold in $\BBR^3$. \end{rem} \bigskip We now define the (standard) cross (wedge) product in $\BBR^3$ as a product satisfying the following properties. \begin{df} Let $(\v{x}, \v{y}, \v{z}, \alpha) \in \BBR^3\times\BBR^3\times\BBR^3\times\BBR.$ The wedge product $\cross: \BBR^3 \times \BBR^3 \rightarrow \BBR^3$ is a closed binary operation satisfying \begin{enumerate} \item[CP1] {\bf Anti-commutativity:} \begin{equation}\v{ x} \cross\v{ y} = -(\v{ y} \cross\v{ x})\label{cp:anti_commutativity}\end{equation} \item[CP2] {\bf Bilinearity:} \begin{equation}(\v{ x} + \v{z}) \cross\v{ y} = \v{ x} \cross\v{ y} + \v{z} \cross\v{ y}, \ \ \ \v{ x} \cross ( \v{z} + \v{ y}) = \v{ x} \cross\v{z} + \v{ x} \cross\v{ y}\label{cp:bilinearity}\end{equation} \item[CP3] {\bf Scalar homogeneity:} \begin{equation}(\alpha\v{ x}) \cross\v{ y} = \v{ x} \cross (\alpha \v{ y}) = \alpha(\v{ x} \cross\v{ y})\label{cp:scalar_homogeneity}\end{equation} \item[CP4] \begin{equation}\v{ x} \cross\v{ x} = \v{{\bf 0}}\label{cp:cross_itself_is_0}\end{equation} \item[CP5] {\bf Right-hand Rule:} \begin{equation}\crossprod{i}{j} = \v{k}, \ \ \ \crossprod{j}{k} = \v{i}, \ \ \ \crossprod{k}{i} = \v{j} \label{cp:right_hand_rule}\end{equation} \end{enumerate} \end{df} \begin{thm} Let $\v{x} = \colvec{x_1 \\ x_2 \\ x_3}$ and $\v{y} = \colvec{y_1 \\ y_2 \\ y_3}$ be vectors in $\BBR^3$. Then $$\crossprod{x}{y} = (x_2y_3 - x_3y_2 )\v{i} + (x_3y_1 - x_1y_3 )\v{j} + (x_1y_2 - x_2y_1 )\v{k}.$$ \end{thm} \begin{pf} Since $\crossprod{i}{i} =\crossprod{j}{j} = \crossprod{k}{k} = \v{0} $ we have $$\begin{array}{lll} (x_1\v{i} + x_2\v{j} + x_3\v{k}) \cross (y_1\v{i} + y_2\v{j} + y_3\v{k}) & = & x_1y_2\crossprod{i}{j} + x_1y_3\crossprod{i}{k} \\ & & \qquad + x_2y_1\crossprod{j}{i} + x_2y_3\crossprod{j}{k} \\ & & \qquad + x_3y_1\crossprod{k}{i} + x_3y_2\crossprod{k}{j} \\ & = & x_1y_2\v{k} - x_1y_3\v{j}- x_2y_1\v{k}\\ & & \qquad + x_2y_3\v{i} +x_3y_1\v{j} - x_3y_2\v{i}, \end{array}$$from where the theorem follows. \end{pf} \begin{exa} Find $$ \colvec{1 \\ 0 \\ -3} \cross \colvec{0 \\ 1 \\ 2}.$$ \label{exa:wedge1}\end{exa} \begin{solu}We have $$\begin{array}{lll} (\v{i} - 3\v{k}) \cross (\v{j} + 2\v{k}) & = & \crossprod{i}{j} + 2\crossprod{i}{k} - 3\crossprod{k}{j} - 6\crossprod{k}{k} \\ & = & \v{k} - 2 \v{j} - 3 \v{i} + 6\v{0} \\ & = & -3\v{i} - 2\v{j} + \v{k}. \end{array}$$ Hence $$ \colvec{1 \\ 0 \\ -3} \cross \colvec{0 \\ 1 \\ 2} = \colvec{-3 \\ -2 \\ 1}.$$ \end{solu} \begin{thm} The cross product vector $\v{x} \cross\v{y}$ is simultaneously perpendicular to $\v{x}$ and $ \v{y}$. \end{thm} \begin{pf} We will only check the first assertion, the second verification is analogous. $$\begin{array}{lll} \v{ x}\bulletproduct(\crossprod{ x}{\bf y}) & = & (x_1\v{i} + x_2 \v{j} + x_3\v{k})\bulletproduct ((x_2y_3 - x_3y_2 )\v{i} \\ & & \qquad + (x_3y_1 - x_1y_3 )\v{j} + (x_1y_2 - x_2y_1 )\v{k} ) \\ & = & x_1x_2y_3 - x_1x_3y_2 + x_2x_3y_1 - x_2x_1y_3 + x_3x_1y_2 - x_3x_2y_1 \\ & = & 0, \end{array}$$ completing the proof. \end{pf} \begin{thm}\label{thm:semi_associative_cross_prod} $\v{ a}\cross (\crossprod{ b}{c}) = (\dotprod{\bf a}{c})\v{ b} - (\dotprod{\bf a}{\bf b})\v{c}.$ \end{thm} \begin{pf} $$\begin{array}{lll} \v{ a}\cross (\crossprod{ b}{c}) & = & (a_1\v{i} + a_2 \v{j} + a_3\v{k})\cross ((b_2c_3 - b_3c_2 )\v{i} + \\ & & \qquad + (b_3c_1 - b_1c_3)\v{j} + (b_1c_2 - b_2c_1)\v{k} ) \\ & = & a_1(b_3c_1 - b_1c_3)\v{k} - a_1(b_1c_2 - b_2c_1)\v{j} \\ & & \qquad - a_2(b_2c_3 - b_3c_2 )\v{k}+ a_2(b_1c_2 - b_2c_1)\v{i} \\ & & \qquad + a_3(b_2c_3 - b_3c_2 )\v{j} - a_3(b_3c_1 - b_1c_3)\v{i} \\ & = & (a_1c_1 + a_2c_2 + a_3c_3)(b_1\v{i} + b_2\v{j} + b_3\v{i}) \\ & & \qquad +(-a_1b_1 - a_2b_2 - a_3b_3)(c_1\v{i} + c_2\v{j} + c_3\v{i})\\ & = & (\dotprod{\bf a}{c})\v{ b} - (\dotprod{\bf a}{\bf b})\v{c}, \\ \end{array}$$ completing the proof. \end{pf} \begin{thm}[Jacobi's Identity] $$\v{a}\cross (\crossprod{b}{c}) + \v{ b}\cross (\crossprod{c}{a}) + \v{c}\cross (\crossprod{ a}{b}) = \v{0}.$$ \end{thm} \begin{pf}From Theorem \ref{thm:semi_associative_cross_prod} we have $$\v{ a}\cross (\crossprod{ b}{c}) = (\dotprod{\bf a}{c})\v{ b} - (\dotprod{\bf a}{\bf b})\v{c}, $$ $$\v{ b}\cross (\crossprod{ c}{a}) = (\dotprod{ b}{a})\v{ c} - (\dotprod{ b}{ c})\v{a}, $$ $$\v{ c}\cross (\crossprod{ a}{b}) = (\dotprod{c}{b})\v{ a} - (\dotprod{c}{a})\v{b}, $$ and adding yields the result. \end{pf} \begin{thm} Let $\anglebetween{x}{y}\in [0;\pi[$ be the convex angle between two vectors $\v{ x}$ and $\v{ y}$. Then $$||\v{ x}\cross\v{ y}|| = ||\v{ x}||||\v{ y}||\sin\anglebetween{x}{y} .$$ \label{thm:sineanglebetweenc}\end{thm} \begin{pf}We have $$\begin{array}{lll} ||\v{ x}\cross\v{ y}||^2 & = & (x_2y_3 - x_3y_2 )^2 + (x_3y_1 - x_1y_3 )^2 + (x_1y_2 - x_2y_1 )^2 \\ & = & x_2 ^2 y_3^2 - 2x_2y_3x_3y_2 + x_3 ^2 y_2 ^2 + x_3 ^2 y_1 ^2 - 2x_3y_1x_1y_3 + \\ & & \qquad + x_1 ^2y_3 ^2 + x_1 ^2 y_2 ^2 - 2x_1y_2x_2y_1 + x_2 ^2y_1 ^2 \\ & = & (x_1 ^2 + x_2 ^2 + x_3 ^2)(y_1 ^2 + y_2 ^2 + y_3 ^2) - (x_1y_1 + x_2y_2 + x_3y_3)^2 \\ & = & ||\v{ x}||^2||\v{{\bf y}}||^2 - (\v{{\bf x}}\bulletproduct \v{{\bf y}})^2 \\ & = & ||\v{ x}||^2||\v{{\bf y}}||^2 - ||\v{ x}||^2||\v{{\bf y}}||^2\cos^2\anglebetween{x}{y} \\ & = & ||\v{ x}||^2||\v{{\bf y}}||^2\sin^2\anglebetween{x}{y}, \end{array} $$whence the theorem follows. The Theorem is illustrated in Figure \ref{fig:sineanglebetweenc}. Geometrically it means that the area of the parallelogram generated by joining $\v{x}$ and $\v{y}$ at their heads is $\norm{\crossprod{x}{y}}$. \end{pf} \vspace{1cm} \begin{figure}[htb] $$\psset{unit=4pc} \psset{viewpoint=1 1.5 1} \ThreeDput[normal=1 0 0](0,0,0){ %%This draws the vector (1,0,0) \psline[linewidth=2pt]{->}(0,0)(0,1) \uput[u](0,1){\crossprod{x}{y}} %%This draws the vector (0,1,0) \psline[linewidth=2pt]{->}(0,0)(1,0) \uput[u](1,0){\v{y}} } %% \ThreeDput[normal=0 1 0](0,0,0){ %%This draws the vector (1,0,0) \psline[linewidth=2pt]{->}(0,0)(-1,0) \uput[u](-1,0){\v{x}}} \ThreeDput[normal=0 0 1](0,0,0){ %%This draws the grid \psframe[fillstyle=solid, fillcolor=green](0,0)(1,1) } $$ \vspace{1cm} \footnotesize\hangcaption{Theorem \ref{thm:sineanglebetweenc}.}\label{fig:sineanglebetweenc} \end{figure} The following corollaries are now obvious. \begin{cor}\label{cor:a_cross_parallel_is_0} Two non-zero vectors $\v{ x}, \v{ y}$ satisfy $\v{ x}\cross \v{ y} = \v{{\bf 0}}$ if and only if they are parallel. \end{cor} \begin{cor}[Lagrange's Identity] $$||\crossprod{ x}{\bf y}||^2 = \norm{\bf x}^2\norm{\bf y}^2 - (\dotprod{\bf x}{\bf y})^2.$$ \end{cor} \begin{exa} Let $\v{ x}\in \BBR^3, \norm{\bf x} = 1$. Find $$||\crossprod{ x}{i}||^2 + ||\crossprod{ x}{j}||^2 + ||\crossprod{ x}{k}||^2.$$ \end{exa} \begin{solu}By Lagrange's Identity, $$||\crossprod{ x}{i}||^2 = \norm{\v{x}}^2\norm{\v{i}}^2 - (\dotprod{x}{i})^2 = 1 - (\dotprod{\bf x}{i})^2,$$ $$||\crossprod{ x}{k}||^2 = \norm{\v{x}}^2\norm{\v{j}}^2 - (\dotprod{x}{j})^2 = 1 - (\dotprod{\bf x}{j})^2,$$ $$||\crossprod{ x}{j}||^2 = \norm{\v{x}}^2\norm{\v{k}}^2 - (\dotprod{x}{k})^2 = 1 - (\dotprod{x}{k})^2,$$and since $(\dotprod{x}{i})^2 + (\dotprod{x}{j})^2 + (\dotprod{x}{k})^2 = \norm{\v{x}}^2 = 1$, the desired sum equals $3 - 1 = 2$. \end{solu} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Consider a tetrahedron $ABCS$. [A] Find $\vect{AB} + \vect{BC} + \vect{CS}$. [B]~Find $\vect{AC} + \vect{CS} + \vect{SA} + \vect{AB}$. \begin{answer} [A] $\vect{AS}$, [B] $\vect{AB}$. \end{answer} \end{pro} \begin{pro} Find a vector simultaneously perpendicular to $\colvec{1 \\ 1\\ 1}$ and $\colvec{1 \\ 1\\ 0}$ and having norm $3$.\begin{answer} Put $$\v{a} = \colvec{1 \\ 1\\ 1} \cross \colvec{1 \\ 1\\ 0} = (\v{i} + \v{j} + \v{k}) \cross (\v{i} + \v{j}) = \v{j} - \v{i} = \colvec{-1 \\ 1 \\ 0} . $$ Then either $$ \frac{3\v{a}}{\norm{\v{a}}} = \frac{3\v{a}}{\sqrt{2}} = \colvec{-\frac{3}{\sqrt{2}} \\ \frac{3}{\sqrt{2}} \\ 0}, $$ or $$ -\frac{3\v{a}}{\norm{\v{a}}} = \colvec{\frac{3}{\sqrt{2}} \\ -\frac{3}{\sqrt{2}} \\ 0}$$will satisfy the requirements. \end{answer} \end{pro} \begin{pro} Find the area of the triangle whose vertices are at $P = \colpoint{0\\ 0 \\ 1}$, $Q = \colpoint{0\\ 1 \\ 0}$, and $R = \colpoint{1\\ 0 \\ 0}$. \begin{answer} The desired area is $$\norm{\vect{PQ} \cross \vect{PR}} = \norm{\colvec{0 \\ 1 \\ -1} \cross \colvec{1 \\ 0 \\ -1}} = \norm{\colvec{-1\\ -1 \\ -1}} = \sqrt{3}.$$ \end{answer} \end{pro} \begin{pro} Prove or disprove! The cross product is associative. \begin{answer} It is not associative, since $\v{i}\cross (\crossprod{i}{j}) = \crossprod{i}{k} = -\v{j}$ but $(\crossprod{i}{i})\cross\v{j} = \crossprod{0}{j} = \v{{\bf 0}}$. \end{answer} \end{pro} \begin{pro} Prove that $\crossprod{x}{x} = \v{0}$ follows from the anti-commutativity of the cross product. \begin{answer} We have $\crossprod{x}{x} = -\crossprod{x}{x}$ by letting $\v{y} = \v{x}$ in \ref{cp:anti_commutativity}. Thus $2\crossprod{x}{x} = \v{0}$ and hence $\crossprod{x}{x} = \v{0}$. \end{answer} \end{pro} \begin{pro} Expand the product $(\v{a} - \v{b})\cross (\v{a} + \v{b})$. \begin{answer} $2\crossprod{a}{b}$ \end{answer} \end{pro} \begin{pro} The vectors $\v{a}, \v{b}$ are constant vectors. Solve the equation $\v{a}\cross (\crossprod{x}{b}) = \v{b} \cross (\crossprod{x}{a})$. \begin{answer} $$\v{a}\cross (\crossprod{x}{b}) = \v{b} \cross (\crossprod{x}{a}) \iff (\dotprod{a}{b})\v{x} - (\dotprod{a}{x})\v{b} = (\dotprod{b}{a})\v{x} - (\dotprod{b}{x})\v{a} \iff \dotprod{a}{x} = \dotprod{b}{x} = 0. $$The answer is thus $\{\v{x}: \v{x}\in \BBR\crossprod{a}{b}\}$. \end{answer} \end{pro} \begin{pro} The vectors $\v{a}, \v{b}, \v{c}$ are constant vectors. Solve the system of equations $$2\v{x} + \crossprod{y}{a} = \v{b}, \ \ \ 3\v{y} + \crossprod{x}{a} = \v{c}, $$ \begin{answer} $$\v{x} = \dfrac{(\dotprod{a}{b})\v{a} + 6\v{b} + 2\crossprod{a}{c}}{12 + 2\norm{\v{a}}^2}$$ $$\v{y} = \dfrac{(\dotprod{a}{c})\v{a} + 6\v{c} + 3\crossprod{a}{b}}{18 + 3\norm{\v{a}}^2}$$ \end{answer} \end{pro} \begin{pro} Prove that there do not exist three unit vectors in $\BBR^3$ such that the angle between any two of them be $> \dfrac{2\pi}{3}$. \begin{answer} Assume contrariwise that $\v{a}$, $\v{b}$, $\v{c}$ are three unit vectors in $\BBR^3$ such that the angle between any two of them is $> \dfrac{2\pi}{3}$. Then $\dotprod{a}{b} < -\dfrac{1}{2}$, $\dotprod{b}{c} < -\dfrac{1}{2}$, and $\dotprod{c}{a} < -\dfrac{1}{2}$. Thus $$\begin{array}{lll} \norm{\v{a} + \v{b} + \v{c}}^2 & = & \norm{\v{a}}^2 + \norm{\v{b}}^2 + \norm{\v{c}}^2\\ & & \qquad + 2\dotprod{a}{b} + 2\dotprod{b}{c} + 2\dotprod{c}{a} \\ & < & 1 + 1 + 1 -1-1-1 \\ & = & 0, \end{array} $$which is impossible, since a norm of vectors is always $\geq 0$. \end{answer} \end{pro} \begin{pro} Let $\v{a}\in \BBR^3$ be a fixed vector. Demonstrate that $$ X = \{\v{x}\in\BBR^3: \crossprod{a}{x} = \v{0}\}$$is a subspace of $\BBR^3$. \begin{answer} Take $(\v{u}, \v{v})\in X^2$ and $\alpha\in \BBR$. Then $$\v{a}\cross (\v{u}+\alpha \v{v}) = \crossprod{a}{u} + \alpha \crossprod{a}{v} = \v{0} + \alpha\v{0} =\v{0}, $$proving that $X$ is a vector subspace of $\BBR^n$. \end{answer} \end{pro} \begin{pro} Let $(\v{a}, \v{b})\in (\BBR^3)^2$ and assume that $\dotprod{\bf a}{\bf b} = 0$ and that $\v{a}$ and $\v{b}$ are linearly independent. Prove that $\v{a}, \v{b}, \crossprod{a}{b}$ are linearly independent. \label{exa:3_vectors_in_r3}\begin{answer} Since $\v{a}, \v{b}$ are linearly independent, none of them is $\v{0}$. Assume that there are $(\alpha, \beta, \gamma)\in\BBR^3$ such that \begin{equation}\label{eq:3_vectors_in_r3}\alpha\v{a} + \beta\v{b} + \gamma\crossprod{a}{b} = \v{0}. \end{equation} Since $\v{a}\bulletproduct(\crossprod{a}{b}) = 0$, taking the dot product of \ref{eq:3_vectors_in_r3} with $\v{a}$ yields $\alpha \norm{\v{a}}^2 = 0$, which means that $\alpha = 0$, since $\norm{\v{a}} \neq 0$. Similarly, we take the dot product with $\v{b}$ and $\crossprod{a}{b}$ obtaining respectively, $\beta = 0$ and $\gamma = 0$. This establishes linear independence. \end{answer} \end{pro} \begin{pro} Let $(\v{ a}, \v{ b})\in \BBR^3 \times \BBR^3$ be fixed. Solve the equation $$\crossprod{ a}{\bf x} = \v{b},$$for $\v{ x}$. \begin{answer} Since $\v{ a} \perp \crossprod{ a}{\bf x} = \v{ b}$, there are no solutions if $\dotprod{\bf a}{\bf b} \neq 0.$ Neither are there solutions if $\v{ a} = \v{{\bf 0}}$ and $\v{ b} \neq \v{{\bf 0}}$. If both $\v{ a} = \v{ b} = \v{{\bf 0}}$, then the solution set is the whole of $\BBR^3.$ Assume thus that $\dotprod{\bf a}{\bf b} = 0$ and that $\v{ a}$ and $\v{ b}$ are linearly independent. Then $\v{a}, \v{b}, \crossprod{a}{b}$ are linearly independent, and so they constitute a basis for $\BBR^3$. Any $\v{x} \in \BBR^3$ can be written in the form $$\v{x} = \alpha \v{a} + \beta\v{b} + \gamma\crossprod{a}{b}.$$ We then have $$\begin{array}{lll}\v{b} & = & \crossprod{a}{x} \\ & = & \beta\crossprod{a}{b} + \gamma \v{ a}\cross (\crossprod{a}{b}) \\ & = & \beta\crossprod{a}{b} + \gamma ((\dotprod{a}{b})\v{a} - (\dotprod{a}{a})\v{b}). \\ & = & \beta\crossprod{a}{b} - \gamma (\dotprod{a}{a}\v{b}) \\ & = & \beta\crossprod{a}{b} -\gamma\norm{\v{a}}^2\v{b}, \end{array}$$ from where $$\beta\crossprod{ a}{\bf b} + (-\gamma\norm{\v{a}}^2 - 1) \v{b} = \v{0},$$ which means that $\beta = 0$ and $\gamma = -\frac{1}{\norm{\v{a}}^2}$, since $\v{a}, \v{b}, \crossprod{a}{b}$ are linearly independent. Thus $$\v{x} = \alpha\v{a} - \frac{1}{\norm{\v{a}}^2}\crossprod{a}{b}$$in this last case. \end{answer} \end{pro} \begin{pro} Let $\v{h}, \v{k}$ be fixed vectors in $\BBR^3$. Prove that $$\fun{L}{(\v{x}, \v{y})}{\crossprod{x}{k} + \crossprod{h}{y}}{\BBR^3\times \BBR^3}{\BBR^3}$$is a linear transformation. \begin{answer} Let $\v{x}, \v{y}, \v{x}', \v{y}'$ be vectors in $\BBR^3$ and let $\alpha\in\BBR$ be a scalar. Then $$\begin{array}{lll} L((\v{x}, \v{y}) + \alpha (\v{x}',\v{y}')) & = & L(\v{x} + \alpha \v{x}', \v{y} + \alpha\v{y}') \\ & = & (\v{x} + \alpha \v{x}')\cross \v{k} + \v{h}\cross (\v{y} + \alpha\v{y}') \\ & = & \crossprod{x}{k} + \alpha\v{x}'\cross \v{k} + \crossprod{h}{y} + \v{h} \cross \alpha\v{y}' \\ & = & L(\v{x}, \v{y}) + \alpha L(\v{x}', \v{y}') \end{array}$$ \end{answer} \end{pro} \end{multicols} \section{Planes and Lines in $\BBR^3$} \begin{df} If bi-point representatives of a family of vectors in $\BBR^3$ lie on the same plane, we will say that the vectors are {\em coplanar} or parallel to the plane. \end{df} \begin{lem} Let $\v{v}, \v{w}$ in $\BBR^3$ be non-parallel vectors. Then every vector $\v{u}$ of the form $$\v{u} = a\v{v} + b\v{w},$$ ($(a, b) \in \BBR^2$ arbitrary) is coplanar with both $\v{v}$ and $\v{w}$. Conversely, any vector $\v{t}$ coplanar with both $\v{v}$ and $\v{w}$ can be uniquely expressed in the form $$\v{t} = p\v{v} + q\v{w}.$$ \label{lem:coplanarvectors}\end{lem} \begin{pf} This follows at once from Corollary \ref{cor:orthodecomp_2}, since the operations occur on a plane, which can be identified with $\BBR^2$. \end{pf} \bigskip A plane is determined by three non-collinear points. Suppose that $A$, $B$, and $C$ are non-collinear points on the same plane and that $R = \colpoint{x\\ y\\ z}$ is another arbitrary point on this plane. Since $A$, $B$, and $C$ are non-collinear, $\vect{AB}$ and $\vect{AC}$, which are coplanar, are non-parallel. Since $\vect{AR}$ also lies on the plane, we have by Lemma \ref{lem:coplanarvectors}, that there exist real numbers $p, q$ with $$\vect{AR} = p\vect{AB} + q\vect{AC}.$$By Chasles' Rule, $$\vect{OR} = \vect{OA} + p(\vect{OB} - \vect{OA}) + q(\vect{OC} - \vect{OA}),$$is the equation of a plane containing the three non-collinear points $A$, $B$, and $C$. By letting $\v{r} = \vect{OR}$, $\v{a} = \vect{OA}$, etc., we deduce that $$ \v{r}- \v{a} = p(\v{b} - \v{a}) + q(\v{c} - \v{a}).$$ Thus we have the following definition. \begin{df} The {\em parametric equation} of a plane containing the point $A$, and parallel to the vectors $\v{u}$ and $\v{v}$ is given by $$ \v{r} -\v{a} = p\v{u} + q\v{v}.$$ Componentwise this takes the form $$\begin{array}{c}x - a_1 = pu_1 + qv_1, \\ y - a_2 = pu_2 + qv_2, \\ z - a_3 = pu_3 + qv_3. \end{array} $$ The {\em Cartesian} equation of a plane is an equation of the form $ax + by + cz = d$ with $(a, b, c, d) \in \BBR^4$ and $a^2 + b^2 + c^2 \neq 0$. \end{df} \begin{exa}\label{exa:eqn_of_plane1} Find both the parametric equation and the Cartesian equation of the plane parallel to the vectors $\colvec{1 \\ 1 \\ 1}$ and $\colvec{1 \\ 1\\ 0}$ and passing through the point $\colpoint{0 \\ -1\\ 2}$. \end{exa} \begin{solu}The desired parametric equation is $$ \colvec{x \\ y + 1 \\ z-2} = s\colvec{1\\ 1\\ 1} + t\colvec{1\\ 1\\ 0}. $$ This gives $s = z-2$, $t = y+1 - s = y + 1 -z+2 = y-z + 3$ and $x = s + t = z-2 + y-z+3 = y+1$. Hence the Cartesian equation is $x-y =1$. \end{solu} \begin{thm}\label{thm:alternative_eq_planeR3} Let $\v{u}$ and $\v{v}$ be non-parallel vectors and let $\v{r} - \v{a} = p\v{u} + q\v{v}$ be the equation of the plane containing $A$ an parallel to the vectors $\v{u}$ and $\v{v}$. If $\v{n}$ is simultaneously perpendicular to $\v{u}$ and $\v{v}$ then $$(\v{r} - \v{a})\bulletproduct \v{n} = 0. $$ Moreover, the vector $\colvec{a\\ b \\ c}$ is normal to the plane with Cartesian equation $ax + by + cz = d$. \end{thm} \begin{pf} The first part is clear, as $\dotprod{u}{n} = 0 = \dotprod{v}{n}$. For the second part, recall that at least one of $a, b, c$ is non-zero. Let us assume $a \neq 0$. The argument is similar if one of the other letters is non-zero and $a = 0$. In this case we can see that $$ x =\frac{d}{a} -\frac{b}{a}y - \frac{c}{a}z. $$ Put $y = s$ and $z = t$. Then $$\colvec{x - \frac{d}{a}\\ y \\ z} = s\colvec{-\frac{b}{a} \\ 1 \\ 0} + t\colvec{-\frac{c}{a} \\ 0 \\ 1} $$ is a parametric equation for the plane. \end{pf} \begin{exa} Find once again, by appealing to Theorem \ref{thm:alternative_eq_planeR3}, the Cartesian equation of the plane parallel to the vectors $\colvec{1 \\ 1 \\ 1}$ and $\colvec{1 \\ 1\\ 0}$ and passing through the point $\colpoint{0 \\ -1\\ 2}$. \end{exa} \begin{solu}The vector $\colvec{1 \\ 1 \\ 1}\cross \colvec{1 \\ 1\\ 0} = \colvec{-1\\ 1 \\ 0}$ is normal to the plane. The plane has thus equation $$ \colvec{x \\ y + 1 \\ z-2}\bulletproduct \colvec{-1\\ 1 \\ 0} = 0 \implies -x + y + 1 = 0 \implies x-y= 1, $$as obtained before. \end{solu} \begin{thm}[Distance Between a Point and a Plane] \label{thm:distance_point_planeR3}\index{distance!between a point and a plane} Let $(\v{r} - \v{a})\bulletproduct\v{n} = 0$ be a plane passing through the point $A$ and perpendicular to vector $\v{n}$. If $B$ is not a point on the plane, then the distance from $B$ to the plane is $$\frac{\left|(\v{a} - \v{b})\bulletproduct\v{n}\right|}{\norm{\v{n}}}.$$ \end{thm} \begin{pf} Let $R_0$ be the point on the plane that is nearest to $B$. Then $\vect{BR_0} = \v{r_0} - \v{b}$ is orthogonal to the plane, and the distance we seek is $$ ||\proj{\v{r_0} - \v{b}}{n}|| = \left|\left| \frac{(\v{r_0} - \v{b})\bulletproduct\v{n}}{\norm{\v{n}}^2}\v{n}\right|\right| = \frac{|(\v{r_0} - \v{b})\bulletproduct\v{n}|}{\norm{\v{n}}}.$$ Since $R_0$ is on the plane, $\dotprod{r_0}{n} = \dotprod{a}{n},$ and so $$ ||\proj{\v{r_0} - \v{b}}{n}|| = \frac{|\dotprod{r_0}{n} - \dotprod{b}{n}|}{\norm{\v{n}}|} = \frac{|\dotprod{a}{n} - \dotprod{b}{n}|}{\norm{\v{n}}} = \frac{|(\v{a} - \v{b})\bulletproduct\v{n}|}{\norm{\v{n}}},$$ as we wanted to shew. \end{pf} \begin{rem} Given three planes in space, they may (i) be parallel (which allows for some of them to coincide), (ii) two may be parallel and the third intersect each of the other two at a line, (iii) intersect at a line, (iv) intersect at a point. \end{rem} \begin{df}The equation of a line passing through $A\in\BBR^3$ in the direction of $\v{v} \neq \v{0}$ is given by $$ \v{r} - \v{a} = t\v{v}, \ \ t\in\BBR. $$ \end{df} \begin{thm}Put $\vect{OA} = \v{a}$, $\vect{OB} = \v{b}$, and $\vect{OC} = \v{c}$. Points $(A, B, C)\in (\BBR^3)^3$ are collinear if and only if $$\crossprod{a}{b} + \crossprod{b}{c} + \crossprod{c}{a} = \v{0}.$$ \label{thm:coll_in_r3}\end{thm} \begin{pf} If the points $A, B, C$ are collinear, then $\vect{AB}$ is parallel to $\vect{AC}$ and by Corollary \ref{cor:a_cross_parallel_is_0}, we must have $$(\v{c} - \v{a})\cross (\v{b} - \v{a}) = \v{0}.$$ Rearranging, gives $$\crossprod{c}{b} - \crossprod{c}{a} - \crossprod{a}{b} = \v{0}.$$Further rearranging completes the proof. \end{pf} \begin{thm}[Distance Between a Point and a Line] \label{thm:distance_point_lineR3}\index{distance!between a point and a line!in space} Let $L: \v{r} = \v{a} + \lambda \v{v}, \ \ \v{v} \neq \v{0},$ be a line and let $B$ be a point not on $L$. Then the distance from $B$ to $L$ is given by $$ \frac{||(\v{a} - \v{b})\cross\v{v}||}{\norm{\v{v}}}.$$ \end{thm} \begin{pf} If $R_0$---with position vector $\v{r_0}$---is the point on $L$ that is at shortest distance from $B$ then $\vect{BR_0}$ is perpendicular to the line, and so $$||\vect{BR_0}\cross \v{v}|| = ||\vect{BR_0}||\norm{\v{v}}\sin\frac{\pi}{2} = ||\vect{BR_0}||\norm{\v{v}}.$$ The distance we must compute is $\norm{\vect{BR_0}} = ||\v{r_0} - \v{b}||$, which is then given by $$||\v{r_0} - \v{b}|| = \frac{||\vect{BR_0}\cross \v{v}||}{\norm{\v{v}}} = \frac{||(\v{r_0} - \v{b}) \cross \v{v}||}{\norm{\v{v}}}.$$ Now, since $R_0$ is on the line $\exists t_0\in\BBR$ such that $\v{r_0} = \v{a} + t_0\v{v}$. Hence $$(\v{r_0} - \v{b}) \cross \v{v} = (\v{a} - \v{b})\cross\v{v},$$ giving $$||\v{r_0} - \v{b}|| = \frac{||(\v{a} - \v{b})\cross\v{v}||}{\norm{\v{v}}},$$ proving the theorem. \end{pf} \begin{rem} Given two lines in space, one of the following three situations might arise: (i) the lines intersect at a point, (ii) the lines are parallel, (iii) the lines are skew (one over the other, without intersecting). \end{rem} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro} Find the equation of the plane passing through the points $(a, 0, a)$, $(-a, 1, 0)$, and $(0, 1, 2a)$ in $\BBR^3$. \begin{answer} The vectors $$\colvec{a - (-a) \\ 0 - 1 \\ a - 0} = \colvec{2a \\ -1 \\ a} $$ and $$\colvec{0 - (-a) \\ 1 - 1 \\ 2a - 0} = \colvec{a \\ 0 \\ 2a} $$are coplanar. A vector normal to the plane is $$ \colvec{2a \\ -1 \\ a} \cross \colvec{a \\ 0 \\ 2a} = \colvec{-2a\\ -3a^2 \\ a}. $$ The equation of the plane is thus given by $$ \colvec{-2a\\ -3a^2 \\ a}\bulletproduct \colvec{x - a\\ y - 0 \\ z - a} = 0, $$ that is, $$2ax +3a^2y - az = a^2. $$ \end{answer} \end{pro} \begin{pro} Find the equation of plane containing the point $(1, 1, 1)$ and perpendicular to the line $x = 1 + t, y = -2t, z = 1 - t$. \begin{answer} The vectorial form of the equation of the line is $$\v{r} = \colvec{1 \\ 0 \\ 1} + t\colvec{1 \\ -2 \\ -1}.$$ Since the line follows the direction of $\colvec{1 \\ -2 \\ -1}$, this means that $\colvec{1 \\ -2 \\ -1}$ is normal to the plane, and thus the equation of the desired plane is $$(x - 1) - 2(y - 1) - (z - 1) = 0.$$ \end{answer} \end{pro} \begin{pro} Find the equation of plane containing the point $(1, -1, -1)$ and containing the line $x = 2y = 3z$. \begin{answer} Observe that $(0, 0, 0)$ (as $0 = 2(0) = 3(0)$) is on the line, and hence on the plane. Thus the vector $$\colvec{1 - 0 \\ -1 - 0 \\ -1 - 0} = \colvec{1 \\ -1 \\ -1}$$lies on the plane. Now, if $x = 2y = 3z = t$, then $x = t, y = t/2, z = t/3$. Hence, the vectorial form of the equation of the line is $$\v{r} = \colvec{0 \\ 0 \\ 0} + t\colvec{1 \\ 1/2 \\ 1/3} = t\colvec{1 \\ 1/2 \\ 1/3}.$$ This means that $\colvec{1 \\ 1/2 \\ 1/3}$ also lies on the plane, and thus $$\colvec{1 \\ -1 \\ -1} \cross \colvec{1 \\ 1/2 \\ 1/3} = \colvec{1/6\\ -4/3\\ 3/2}$$is normal to the plane. The desired equation is thus $$\frac{1}{6}x - \frac{4}{3}y + \frac{3}{2}z = 0.$$ \end{answer} \end{pro} \begin{pro} Find the equation of the plane perpendicular to the line $ax = by = cz,\ \ \ abc \neq 0$ and passing through the point $(1, 1, 1)$ in $\BBR^3$. \begin{answer} Put $ax = by = cz = t$, so $x = t/a; y = t/b; z = t/c$. The parametric equation of the line is $$\colvec{x \\ y \\ z} = t\colvec{1/a \\ 1/b \\ 1/c}, \ \ \ t\in \BBR. $$ Thus the vector $\colvec{1/a \\ 1/b \\ 1/c}$ is perpendicular to the plane. Therefore, the equation of the plane is $$\colvec{1/a \\ 1/b \\ 1/c}\bulletproduct \colvec{x - 1 \\ y - 1 \\ z - 1} = \colvec{0 \\ 0\\ 0}, $$ or $$ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}. $$ We may also write this as $$bcx + cay + abz = ab + bc + ca. $$ \end{answer} \end{pro} \begin{pro} Find the equation of the line perpendicular to the plane $ax + a^2y + a^3z = 0,\ \ a \neq 0$ and passing through the point $(0, 0, 1)$. \begin{answer} A vector normal to the plane is $ \colvec{a \\ a^2 \\ a^2}$. The line sought has the same direction as this vector, thus the equation of the line is $$ \colvec{x \\ y \\ z} = \colvec{0\\ 0 \\ 1} + t\colvec{a \\ a^2 \\ a^2}, \ \ \ t\in\BBR. $$ \end{answer} \end{pro} \begin{pro} The two planes $$ x - y - z = 1,\ \ \ \ \ \ x - z = -1, $$intersect at a line. Write the equation of this line in the form $$\colvec{x\\ y \\ z} = \v{a} + t\v{v}, \ \ t\in\BBR . $$ \begin{answer} We have $$ x - z - y = 1 \implies -1 - y = 1 \implies y = -2. $$ Hence if $z = t$, $$ \colvec{x\\ y \\ z } = \colvec{t - 1\\ -2 \\ t} = \colvec{-1 \\ -2 \\ 0} + t\colvec{1 \\ 0 \\ 1}. $$ \end{answer} \end{pro} \begin{pro} Find the equation of the plane passing through the points $\colpoint{1\\ 0\\ -1}$, \ \ $\colpoint{2\\ 1\\ 1}$ and parallel to the line $ \colvec{x\\ y \\ z } = \colvec{-1 \\ -2 \\ 0} + t\colvec{1 \\ 0 \\ 1}. $. \begin{answer} The vector $$\colvec{2-1\\ 1 - 0 \\ 1 - (-1)} = \colvec{1\\ 1 \\ 2} $$lies on the plane. The vector $$\colvec{1 \\ 0 \\ 1} \cross \colvec{1 \\ 1\\ 2} = \colvec{1\\ 1 \\ -1} $$ is normal to the plane. Hence the equation of the plane is $$\colvec{1 \\ 1 \\ -1}\bulletproduct \colvec{x -1 \\ y \\ z + 1} = 0 \implies x+y-z=2. $$ \end{answer} \end{pro} \begin{pro} Points ${\bf a, b, c}$ in $\BBR^3$ are collinear and it is known that $\crossprod{a}{c} = \v{i} - 2\v{j}$ and $\crossprod{a}{b} = 2\v{k} - 3\v{i}$. Find $\crossprod{b}{c}$. \begin{answer} We have $\crossprod{c}{a} = -\v{i} + 2\v{j}$ and $\crossprod{a}{b} = 2\v{k} - 3\v{i}$. By Theorem \ref{thm:coll_in_r3}, we have $$\crossprod{b}{c} = -\crossprod{a}{b} - \crossprod{c}{a} = -2\v{k} + 3\v{i} +\v{i} - 2\v{j} = 4\v{i} - 2\v{j} - 2\v{k}.$$ \end{answer} \end{pro} \begin{pro} Find the equation of the plane which is equidistant of the points $\colpoint{3\\ 2\\ 1}$ and $\colpoint{1\\ -1\\ 1}$. \begin{answer} $4x + 6y =1 $ \end{answer} \end{pro} \begin{pro}[{\red\bf Putnam Exam, 1980}] Let $S$ be the solid in three-dimensional space consisting of all points $(x, y, z)$ satisfying the following system of six conditions: $$x \geq 0, \ \ \ y \geq 0, \ \ \ z \geq 0,$$ $$x + y + z \leq 11,$$ $$2x + 4y + 3z \leq 36,$$ $$2x + 3z \leq 24.$$Determine the number of vertices and the number of edges of $S$. \begin{answer} There are $7$ vertices ($V_0 = (0, 0, 0), V_1 = (11, 0, 0)$, $V_2 = (0, 9, 0), V_3 = (0, 0, 8)$, $V_4 = (0, 3, 8)$, $V_5 = (9, 0, 2)$, $V_6 = (4, 7, 0)$) and $11$ edges ($V_0V_1$, $V_0V_2$, $V_0V_3$, $V_1V_5$, $V_1V_6$, $V_2V_4$, $V_3V_4$, $V_3V_5$, $V_4V_5$, and $V_4V_6$). \end{answer} \end{pro} \end{multicols} \section{$\BBR^n$} \index{vector!n-dimensional} As a generalisation of $\BBR^2$ and $\BBR^3$ we define $\BBR^n$ as the set of $n$-tuples $$\left\{ \colvec{x_1\\x_2 \\ \vdots \\ x_n}: x_i\in\BBR \right\}.$$ The dot product of two vectors in $\BBR^n$ is defined as $$\dotprod{x}{y} = \colvec{x_1\\x_2 \\ \vdots \\ x_n}\bulletproduct \colvec{y_1\\y_2 \\ \vdots \\ y_n} = x_1y_1 + x_2y_2 + \cdots + x_ny_n.$$ The norm of a vector in $\BBR^n$ is given by $$ \norm{\v{x}} = \sqrt{\dotprod{x}{x}}. $$ As in the case of $\BBR^2$ and $\BBR^3$ we have \begin{thm}[Cauchy-Bunyakovsky-Schwarz Inequality] \index{inequality!Cauchy-Bunyakovsky-Schwarz!in Rn}Given $(\v{x}, \v{y})\in (\BBR^n)^2$ the following inequality holds $$ |\dotprod{x}{y}| \leq \norm{\v{x}}\norm{\v{y}}. $$ \end{thm} \begin{pf} Put $\dis{a = \sum _{k = 1} ^n x_k ^2}$, $\dis{b = \sum _{k = 1} ^n x_ky_k }$, and $\dis{c = \sum _{k = 1} ^n y_k ^2}$. Consider $$f(t) = \sum _{k = 1} ^n (tx_k - y_k)^2 = t^2\sum _{k = 1} ^n x_k ^2 - 2t \sum _{k = 1} ^n x_ky_k + \sum _{k = 1} ^n y_k ^2 = at^2 + bt + c. $$ This is a quadratic polynomial which is non-negative for all real $t$, so it must have complex roots. Its discriminant $b^2 - 4ac$ must be non-positive, from where we gather $$4\left(\sum _{k = 1} ^n x_ky_k\right)^2 \leq 4\left(\sum _{k = 1} ^n x_k ^2 \right)\left(\sum _{k = 1} ^n y_k ^2 \right). $$ This gives $$ |\dotprod{x}{y}|^2 \leq \norm{\v{x}}^2\norm{\v{y}}^2 $$from where we deduce the result. \end{pf} \begin{exa} Assume that $a_k, b_k, c_k, k = 1, \ldots, n$, are positive real numbers. Shew that $$\left(\sum _{k = 1} ^n a_kb_kc_k\right)^{4} \leq \left(\sum _{k = 1} ^n a_k ^4\right)\left(\sum _{k = 1} ^n b_k ^4\right) \left(\sum _{k = 1} ^n c_k ^2\right)^{2}.$$\end{exa} \begin{solu}Using CBS on $\sum _{k = 1} ^n (a_kb_k)c_k$ once we obtain $$\sum _{k = 1} ^n a_kb_kc_k \leq \left(\sum _{k = 1} ^n a_k ^2b_k ^2\right)^{1/2} \left(\sum _{k = 1} ^n c_k ^2\right)^{1/2}. $$Using CBS again on $\left(\sum _{k = 1} ^n a_k ^2b_k ^2\right)^{1/2}$ we obtain $$ \begin{array}{lll} \sum _{k = 1} ^n a_kb_kc_k & \leq & \left(\sum _{k = 1} ^n a_k ^2 b_k ^2\right)^{1/2} \left(\sum _{k = 1} ^n c_k ^2\right)^{1/2} \\ & \leq & \left(\sum _{k = 1} ^n a_k ^4\right)^{1/4} \left(\sum _{k = 1} ^n b_k ^4\right)^{1/4} \left(\sum _{k = 1} ^n c_k ^2\right)^{1/2}, \\ \end{array} $$which gives the required inequality. \end{solu} \begin{thm}[Triangle Inequality] \index{inequality!triangle!in Rn}Given $(\v{x}, \v{y})\in (\BBR^n)^2$ the following inequality holds $$ \norm{\v{x} + \v{y}}\leq \norm{\v{x}} + \norm{\v{y}}. $$ \end{thm} \begin{pf}We have $$\begin{array}{lll} ||\v{a} + \v{b}||^2 & = & (\v{a} + \v{b})\bulletproduct (\v{a} + \v{b}) \\ & = & \v{a}\bulletproduct\v{a} + 2\v{a}\bulletproduct\v{b} + \v{b}\bulletproduct\v{b} \\ & \leq & ||\v{a}||^2 + 2||\v{a}||||\v{b}|| + ||\v{b}||^2 \\ & = & (||\v{a}|| + ||\v{b}||)^2, \end{array}$$from where the desired result follows. \end{pf} We now consider a generalisation of the Euclidean norm. Given $p > 1$ and $\v{x}\in \BBR^n$ we put \begin{equation} \norm{\v{x}}_p = \left(\sum _{k = 1} ^n |x_k| ^p\right)^{1/p} \label{eq:p_norm}\end{equation} Clearly \begin{equation} \norm{\v{x}}_p \geq 0 \end{equation} \begin{equation} \norm{\v{x}}_p = 0 \Leftrightarrow \v{x} = \v{0} \end{equation} \begin{equation} \norm{\alpha\v{x}}_p = |\alpha|\norm{\v{x}}_p, \ \ \alpha \in \BBR\end{equation}We now prove analogues of the Cauchy-Bunyakovsky-Schwarz and the Triangle Inequality for $\norm{\cdot}_p$. For this we need the following lemma. \begin{lem}[Young's Inequality] Let $p > 1$ and put $\dfrac{1}{p} + \dfrac{1}{q} = 1$. Then for $(a, b)\in ([0;+\infty[)^2$ we have $$ ab \leq \frac{a^p}{p} + \frac{b^q}{q}. $$ \end{lem} \begin{pf} Let $0 < k < 1$, and consider the function $$\fun{f}{x}{x^k - k(x - 1)}{[0;+\infty[}{\BBR}. $$ Then $0 = f'(x) = kx^{k - 1} - k \Leftrightarrow x = 1$. Since $f''(x) = k(k - 1)x^{k - 2} < 0$ for $0 < k < 1, x \geq 0$, $x = 1$ is a maximum point. Hence $f(x) \leq f(1)$ for $x \geq 0$, that is $x^k \leq 1 + k(x - 1)$. Letting $k = \dfrac{1}{p}$ and $x = \dfrac{a^p}{b^q}$ we deduce $$ \frac{a}{b^{q/p}} \leq 1 + \frac{1}{p}\left(\frac{a^p}{b^q} - 1\right). $$ Rearranging gives $$ ab \leq b^{1 + p/q} + \frac{a^pb^{1 + p/q - p}}{p} - \frac{b^{1 + p/q}}{p} $$ from where we obtain the inequality. \end{pf} The promised generalisation of the Cauchy-Bunyakovsky-Schwarz Inequality is given in the following theorem. \begin{thm}[H\"{o}lder Inequality] \index{inequality!H\"{o}lder} Given $(\v{x}, \v{y})\in (\BBR^n)^2$ the following inequality holds $$ |\dotprod{x}{y}| \leq \norm{\v{x}}_p\norm{\v{y}}_q. $$ \end{thm} \begin{pf} If $\norm{\v{x}}_p = 0$ or $\norm{\v{y}}_q = 0$ there is nothing to prove, so assume otherwise. From the Young Inequality we have $$ \frac{|x_k|}{\norm{\v{x}}_p}\frac{|y_k|}{{\norm{\v{y}}_q}} \leq \frac{|x_k| ^p}{{\norm{\v{x}}_p}^pp} + \frac{|y_k| ^q}{{\norm{\v{y}}_q}^qq}. $$Adding, we deduce $$ \begin{array}{lll}\sum _{k = 1} ^n \dfrac{|x_k|}{\norm{\v{x}}_p}\dfrac{|y_k|}{{\norm{\v{y}}_q}} & \leq& \dfrac{1}{{\norm{\v{x}}_p}^pp}\sum _{k = 1} ^n |x_k| ^p + \dfrac{1}{{\norm{\v{y}}_q}^qq}\sum _{k = 1} ^n |y_k| ^q \\ & = & \dfrac{{\norm{\v{x}}_p}^p}{{\norm{\v{x}}_p}^pp} + \dfrac{{\norm{\v{y}}_q}^q}{{\norm{\v{y}}_q}^qq}\\ & = & \dfrac{1}{p} + \dfrac{1}{q}\\ & = & 1. \end{array} $$ This gives $$ \sum _{k = 1} ^n |x_ky_k| \leq \norm{\v{x}}_p\norm{\v{y}}_q. $$The result follows by observing that $$ \left| \sum _{k = 1} ^n x_ky_k \right|\leq\sum _{k = 1} ^n |x_ky_k| \leq \norm{\v{x}}_p\norm{\v{y}}_q. $$ \end{pf} As a generalisation of the Triangle Inequality we have \begin{thm}[Minkowski Inequality] Let $p \in ]1; +\infty[$. Given $(\v{x}, \v{y})\in (\BBR^n)^2$ the following inequality holds $$ \norm{\v{x} + \v{y}}_p \leq \norm{\v{x}}_p + \norm{\v{y}}_p. $$ \label{thm:minkowski_inequality}\index{inequality!Minkowski}\end{thm} \begin{pf}From the triangle inequality for real numbers \ref{ineq:triangle_real_numbers} $$|x_k + y_k|^p = |x_k + y_k||x_k + y_k|^{p - 1} \leq \left(|x_k| + |y_k|\right)|x_k + y_k|^{p - 1}.$$ Adding \begin{equation} \sum _{k = 1} ^n |x_k + y_k|^p \leq \sum _{k = 1} ^n |x_k||x_k + y_k|^{p - 1} + \sum _{k = 1} ^n |y_k||x_k + y_k|^{p - 1}.\label{eq:minkowski_1}\end{equation} By the H\"{o}lder Inequality \begin{equation}\begin{array}{lll} \sum _{k = 1} ^n |x_k||x_k + y_k|^{p - 1} & \leq & \left(\sum _{k = 1} ^n |x_k|^p\right)^{1/p}\left(\sum _{k = 1} ^n|x_k + y_k|^{(p - 1)q}\right)^{1/q} \\ & = & \left(\sum _{k = 1} ^n |x_k|^p\right)^{1/p}\left(\sum _{k = 1} ^n|x_k + y_k|^{p}\right)^{1/q}\\ & = & \norm{\v{x}}_p\norm{\v{x} + \v{y}}_p ^{p/q} \end{array}\label{eq:minkowski_2}\end{equation} In the same manner we deduce \begin{equation}\sum _{k = 1} ^n |y_k||x_k + y_k|^{p - 1} \leq \norm{\v{y}}_p\norm{\v{x} + \v{y}}_p ^{p/q}\label{eq:minkowski_3}.\end{equation} Hence (\ref{eq:minkowski_1}) gives $$ \norm{\v{x} + \v{y}}_p ^p = \sum _{k = 1} ^n |x_k + y_k|^p \leq \norm{\v{x}}_p\norm{\v{x} + \v{y}}_p ^{p/q} + \norm{\v{y}}_p\norm{\v{x} + \v{y}}_p ^{p/q}, $$from where we deduce the result. \end{pf} \section*{\psframebox{Homework}} \begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900 \begin{pro}Prove Lagrange's identity: $$ \begin{array}{lll}\left(\sum_{1\leq j\leq n} a_jb_j \right)^2 & = & \left(\sum_{1\leq j\leq n}a_j^2\right) \left(\sum_{1\leq j\leq n}b_j^2\right) \\ & &\qquad - \sum_{1\leq k < j\leq n}(a_kb_j-a_jb_k)^2 \end{array}$$ and then deduce the CBS Inequality in $\BBR^n$. \end{pro} \begin{pro}Let $\v{a}_i\in\BBR^n$ for $1 \leq i \leq n$ be unit vectors with $\sum _{i=1} ^n \v{a_i} = \v{0}$. Prove that $\sum_{1\leq i 0$. Use the CBS Inequality to shew that $$ \left(\sum _{k = 1} ^n a_k ^2\right)\left(\sum _{k = 1} ^n \frac{1}{a_k ^2}\right) \geq n^2. $$ \begin{answer} Observe that $\sum _{k = 1} ^n 1 = n$. Then we have $$ n^2 = \left(\sum _{k = 1} ^n 1\right)^2 =\left(\sum _{k = 1} ^n (a_k)\left(\frac{1}{a_k}\right)\right)^2 \leq \left(\sum _{k = 1} ^n a_k ^2\right)\left(\sum _{k = 1} ^n \frac{1}{a_k ^2}\right),$$ giving the result. \end{answer} \end{pro} \begin{pro} Let $\v{a}\in \BBR^n$ be a fixed vector. Demonstrate that $$ X = \{\v{x}\in\BBR^n: \dotprod{a}{x} = 0\}$$is a subspace of $\BBR^n$. \begin{answer} Take $(\v{u}, \v{v})\in X^2$ and $\alpha\in \BBR$. Then $$\v{a}\bulletproduct (\v{u}+\alpha \v{v}) = \dotprod{a}{u} + \alpha \dotprod{a}{v} = 0 + 0 = 0, $$proving that $X$ is a vector subspace of $\BBR^n$. \end{answer} \end{pro} \begin{pro} Let $\v{a_i}\in\BBR^n$, $1 \leq i \leq k$ ($k \leq n$) be $k$ non-zero vectors such that $\dotprod{a_i}{a_j} = 0$ for $i\neq j$. Prove that these $k$ vectors are linearly independent. \begin{answer} Assume that $$ \lambda_1\v{a_1} + \cdots + \lambda_k\v{a_k} = \v{0}. $$Taking the dot product with $\v{a_j}$ and using the fact that $\dotprod{a_i}{a_j} = 0$ for $i\neq j$ we obtain $$ 0 = \dotprod{0}{a_j} = \lambda_j\v{a_j}\bulletproduct \v{a_j} = \lambda_j \norm{a_j}^2. $$ Since $\v{a_j} \neq \v{0} \implies \norm{a_j}^2 \neq 0$, we must have $\lambda_j = 0$. Thus the only linear combination giving the zero vector is the trivial linear combination, which proves that the vectors are linearly independent. \end{answer} \end{pro} \begin{pro} Let $a_k \geq 0, 1 \leq k \leq n$ be arbitrary. Prove that $$ \left(\sum _{k = 1} ^n a_k\right)^2\leq \frac{n(n + 1)(2n + 1)}{6}\sum _{k = 1} ^n \frac{{a_k ^2}}{k^2}.$$ \begin{answer} This follows at once from the CBS Inequality by putting $$\v{v} = \colvec{\frac{a_1}{1}\\ \frac{a_2}{2}\\ \ldots \\ \frac{a_n}{n}}, \ \ \v{u} = \colvec{1\\ 2\\ \ldots\\ n} $$and noticing that $$\sum _{k = 1} ^n k^2 = \frac{n(n + 1)(2n + 1)}{6}.$$ \end{answer} \end{pro} \end{multicols} \Closesolutionfile{linearans} \appendix \renewcommand{\chaptername}{Appendix} \chapter{Answers and Hints}\addcontentsline{toc}{section}{Answers and Hints} \markright{Answers and Hints} {\small\input{linearansC1}} \begin{thebibliography}{9} \bibitem[Cul]{Cul} CULLEN, C., {\bf Matrices and Linear Transformations}, 2nd ed., New York: Dover Publications, 1990. \bibitem[Del]{Del} DELODE, C., {\bf G\'{e}om\'{e}trie Affine et Euclidienne}, Paris: Dunod, 2000. \bibitem[Fad]{Fad} FADD\'{E}EV, D., SOMINSKY, I., {\bf Recueil d'Exercises d'Alg\`{e}bre Sup\'{e}rieure}, 7th ed., Paris: Ellipses, 1994. \bibitem[Hau]{Hau} HAUSNER, M., {\bf A Vector Space Approach to Geometry}, New York: Dover, 1998. \bibitem[Lan]{Lan} LANG, S., {\bf Introduction to Linear Algebra}, 2nd ed., New York: Springer-Verlag, 1986. \bibitem[Pro]{Pro} PROSKURYAKOV, I. V., {\bf Problems in Linear Algebra}, Moscow: Mir Publishers, 1978. \bibitem[Riv]{Riv} RIVAUD, J., {\bf Ejercicios de \'{a}lgebra}, Madrid: Aguilar, 1968. \bibitem[Tru]{Tru} TRUFFAULT, B., {\bf G\'{e}om\'{e}trie \'{E}l\'{e}mentaire: Cours et exercises}, Paris: Ellipses, 2001. \end{thebibliography} \lhead{}\chead{}\rhead{} {\tiny\chapter*{\rlap{GNU Free Documentation License}} \phantomsection % so hyperref creates bookmarks \addcontentsline{toc}{chapter}{GNU Free Documentation License} %\label{label_fdl} \begin{center} Version 1.2, November 2002 Copyright \copyright{} 2000,2001,2002 Free Software Foundation, Inc. \bigskip 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA \bigskip Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. \end{center} \begin{center} {\bf\large Preamble} \end{center} The purpose of this License is to make a manual, textbook, or other functional and useful document ``free'' in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or noncommercially. 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