### correct reported errors for third edition

 ... ... @@ -166,4 +166,12 @@ General: William Karstens for much help with the Topic on Coupled Pendula. 2016-Aug-20 Christelle Vincent Add slides that use handout option \ No newline at end of file Christelle Vincent Add slides that use handout option. 2016-Nov-23 Darij Grinberg Typo. David Guichard Many typos and suggestions. Mitch Rothstein Exercise error, and proof suggestion. James Moore Typo. Richard Satterfield Thought-provoking report about free variables. Christopher Thron Incorrect exercise in voting Topic. \ No newline at end of file
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 ... ... @@ -1074,13 +1074,13 @@ write it as a function of the rows We want a formula to determine whether an $\nbyn{n}$ matrix is nonsingular. We will not begin by stating such a formula. Instead we will begin by considering the function that such a formula calculates. We will define this function by its properties, then prove that the function with these properties exists and is unique, Instead we will begin by considering, for eacn~$n$, the function that such a formula calculates. We will define this function by a list of properties. We will then prove that a function with these properties exists and is unique, and also describe how to compute it. (Because we will eventually show that the function exists and is unique, from the start we will just say $$\det(T)$$' instead of (Because we will eventually prove this, from the start we will just say $$\det(T)$$' instead of if there is a unique determinant function then $$\det(T)$$'.) \begin{definition} \label{def:Det} ... ... @@ -1325,7 +1325,7 @@ quickly with the Gauss's Method procedure. The prior example illustrates an important point. Although we have not yet found a $\nbyn{4}$ determinant formula, if one exists then we know what value it gives to the matrix \Dash if there is a function with properties (1)-(4) then on the above if there is a function with properties (1)\,--\,(4) then on the above matrix the function must return $5$. \begin{lemma} \label{lm:DetFcnIsUnique} ... ...
 ... ... @@ -330,7 +330,7 @@ their associated cofactors. \tag*{($*$)}\end{equation*} Recall that a matrix with two identical rows has a zero determinant. Thus, weighing the cofactors by entries from weighting the cofactors by entries from row~$k$ with $k\neq i$ gives zero \begin{equation*} t_{i,1}\cdot T_{k,1}+t_{i,2}\cdot T_{k,2}+\dots+t_{i,n}\cdot T_{k,n}=0 ... ...
 ... ... @@ -2125,7 +2125,7 @@ We use matrices to do Gauss's Method in essentially the same way that we did it for systems of equations: where a row's \definend{leading entry}\index{echelon form!leading entry}% \index{leading!entry}. % \index{leading!entry} % is its first nonzero entry (if it has one), we perform row operations to arrive at \definend{matrix echelon form},\index{echelon form!matrix}% ... ... @@ -2420,11 +2420,10 @@ is a particular solution of the system. \end{example} Before the exercises, we will consider what we have accomplished and what we have yet to do. and what we will do in the remainder of the chapter. So far we have done the mechanics of Gauss's Method. We have not stopped to consider any of the interesting questions We have not stopped to consider any of the questions that arise, except for proving \nearbytheorem{th:GaussMethod}\Dash which justifies the method by showing that it gives ... ... @@ -2436,10 +2435,13 @@ a particular solution vector added to an unrestricted linear combination of some other vectors? We've noted that the solution sets we described in this way have infinitely many solutions so an answer to this question so answering this question would tell us about the size of solution sets. The following subsection shows that the answer is yes.'' This chapter's second section then uses that answer to describe the geometry of solution sets. Many questions arise from our observation that we can do Gauss's Method in Other questions arise from the observation that we can do Gauss's Method in more than one way (for instance, when swapping rows we may have a choice of more than one row). \nearbytheorem{th:GaussMethod} says that we must get the same solution set ... ... @@ -2450,18 +2452,12 @@ Must those be the same variables, that is, is it impossible to solve a problem one way to get $y$ and~$w$ free and solve it another way to get $y$ and~$z$ free? The third section of this chapter answers yes,'' that from any starting linear system, all derived echelon form versions have the same free variables. In the rest of this chapter we will answer these questions. The answer to each is yes'. In the next subsection we do the first one: we will prove that we can always describe solution sets in that way. Then, in this chapter's second section, we will use that understanding to describe the geometry of solution sets. In this chapter's final section, we will settle the questions about the parameters. When we are done, we will not only have a Thus, by the end of the chapter we will not only have a solid grounding in the practice of Gauss's Method but we will also have a solid grounding in the theory. We will know exactly what can and cannot happen in a reduction. ... ...
 ... ... @@ -36,8 +36,8 @@ from $-2\rho_1+\rho_2$ followed by $2\rho_2+\rho_1$ echelon form so the second one is extra work but it is nonetheless a legal row operation). The fact that echelon form is not unique raises questions. In this chapter's first section we noted that this raises questions. Will any two echelon form versions of a linear system have the same number of free variables? If yes, ... ... @@ -45,8 +45,8 @@ will the two have exactly the same free variables? In this section we will give a way to decide if one linear system can be derived from another by row operations. The answers to both questions, both yes', will follow from this. The answers to both questions, both yes,'' will follow. ... ... @@ -1794,22 +1794,44 @@ So $$B=C$$. % \end{proof} That result answers the two questions from this section's introduction:~do We have asked whether any two echelon form versions of a linear system have the same number of free variables, and if so are they exactly the same variables? With the prior result we can answer both questions yes.'' There is no linear system such that, say, we could apply Gauss's Method one way and get $y$ and $z$ free but apply it another way and get $y$ and $w$ free. Before the proof, recall the distinction between free variables and parameters. This system \begin{equation*} \begin{linsys}{3} x &+ &y & & &= &1 \\ & &y &+ &z &= &2 \end{linsys} \end{equation*} has one free variable,~$z$, because it is the only variable not leading a row. We have the habit of parametrizing using the free variable $y=2-z$, $x=-1+z$, but we could also parametrize using another variable, such as $z=2-y$, $x=1-y$. So the set of parameters is not unique, it is the set of free variables that is unique. \begin{corollary} If from a starting linear systems we derive by Gauss's Method two different echelon form systems, then the two have the same free variables. \end{corollary} \begin{proof} The prior result says that the reduced echelon form is unique. We get from any echelon form version to the reduced echelon form by eliminating up, so any echelon form version of a system has the same free variables as the reduced echelon form, and therefore uniqueness of reduced echelon form gives that the same variables are free in all echelon form version of a system. Thus both questions are answered yes.'' There is no linear system and no combination of row operations such that, say, we could solve the system one way and get $y$ and $z$ free but solve it another way and get $y$ and $w$ free. reduced echelon form version. \end{proof} We close with a recap. In Gauss's Method we start with a matrix and then ... ...
 ... ... @@ -64,7 +64,7 @@ quote any result from the book).} $\map{\composed{g}{h}}{\Re^3}{\matspace_{\nbyn{2}}}$ directly from the above definition. \item Represent $h$ and~$g$ with respect to the appropriate bases. \item Represent $\composed{g}{h}$ with resepct to the appropriate bases. \item Represent $\composed{g}{h}$ with respect to the appropriate bases. \item Check that the two matrices from the second part multiply to the matrix from the third part. \end{enumerate} ... ...
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 ... ... @@ -964,7 +964,7 @@ is more fruitful and more central to progress. is a linear transformation of that space. \begin{equation*} f\mapsto \frac{d^k}{dx^k}f+c_{k-1}\frac{d^{k-1}}{dx^{k-1}}f c_{k}\frac{d^k}{dx^k}f+c_{k-1}\frac{d^{k-1}}{dx^{k-1}}f +\dots+ c_1\frac{d}{dx}f+c_0f \end{equation*} ... ...
 ... ... @@ -997,8 +997,8 @@ function composition is possible. \text{dimension $$m$$ space} \tag{$*$} \end{equation*} Thus, matrix product combines an $\nbym{m}{r}$ matrix~$G$ with an $\nbym{r}{n}$ matrix~$F$ to yield the Thus, matrix product combines the $\nbym{m}{r}$ matrix~$G$ with the $\nbym{r}{n}$ matrix~$F$ to yield the $\nbym{m}{n}$ result~$GF$. Briefly: $\nbym{m}{r}\text{\ times\ }\nbym{r}{n}\text{\ equals\ }\nbym{m}{n}$. ... ... @@ -3077,7 +3077,7 @@ The \definend{elementary reduction matrices}% \index{matrix!elementary reduction}\index{elementary reduction matrix} (or just \definend{elementary matrices})\index{elementary matrix}% \index{matrix!elementary} result from applying a one Gaussian operation to an identity matrix. result from applying a single Gaussian operation to an identity matrix. \begin{enumerate} \item $$I\grstep{k\rho_i}M_i(k)$$ for $$k\neq 0$$ \item $$I\grstep{\rho_i\leftrightarrow\rho_j}P_{i,j}$$ for ... ...
 ... ... @@ -506,7 +506,7 @@ Square matrices of the same size have a defined product. \end{mat} \end{equation*} This reflects the fact that we can compose two functions from a space to itself $\map{f,g}{V}{V}$. functions from a space to itself $\map{g,h}{V}{V}$. \end{frame} ... ... @@ -571,10 +571,10 @@ Fix these bases. 0 &4 \end{mat}} \end{gather*} Suppose that $\map{f}{\Re^2}{\polyspace_2}$ and Suppose that $\map{h}{\Re^2}{\polyspace_2}$ and $\map{g}{\polyspace_2}{\matspace_{\nbyn{2}}}$ have these actions. \begin{equation*} \colvec{a \\ b}\mapsunder{f} ax^2+(a+b) \colvec{a \\ b}\mapsunder{h} ax^2+(a+b) \qquad px^2+qx+r\mapsunder{g} \begin{mat} ... ... @@ -584,7 +584,7 @@ $\map{g}{\polyspace_2}{\matspace_{\nbyn{2}}}$ have these actions. \end{equation*} Then the composition does this. \begin{equation*} \colvec{a \\ b}\mapsunder{f}ax^2+(a+b)\mapsunder{g} \colvec{a \\ b}\mapsunder{h}ax^2+(a+b)\mapsunder{g} \begin{mat} a &a \\ 0 &a+b ... ... @@ -592,7 +592,7 @@ Then the composition does this. \end{equation*} Here is the same statement in the other notation. \begin{equation*} \composed{g}{f}\,(\colvec{a \\ b})= \composed{g}{h}\,(\colvec{a \\ b})= \begin{mat} a &a \\ 0 &a+b ... ... @@ -601,19 +601,19 @@ Here is the same statement in the other notation. \end{frame} \begin{frame} So far in this example we have given the maps above the arrows, $f$, $g$, and $\composed{g}{f}$. and $\composed{g}{h}$. \centergraphic{../ch3.20} We next compute the matrices representing those maps, and we will finish by checking that the product of $H$ and~$G$ is the matrix finish by checking that the product of $G$ and~$H$ is the matrix representing $\composed{g}{h}$. \pause First, find $H=\rep{f}{B,D}$: compute the action of $h$ First, find $H=\rep{h}{B,D}$: compute the action of $h$ on the domain basis vectors, \begin{equation*} \colvec{1 \\ 1}\mapsunder{f}x^2+2 \colvec{1 \\ 1}\mapsunder{h}x^2+2 \quad \colvec{1 \\ -1}\mapsunder{f}x^2 \colvec{1 \\ -1}\mapsunder{h}x^2 \end{equation*} represent the results with respect to $D$, and make the matrix. \begin{equation*} ... ... @@ -662,7 +662,7 @@ and represent those with respect to its codomain basis. \end{equation*} \pause Next, $\composed{g}{f}$ has this action. Next, $\composed{g}{h}$ has this action. \begin{equation*} \colvec{1 \\ 1}\mapsto \begin{mat} ... ... @@ -1267,7 +1267,7 @@ can be expressed as this product; note that the order is right-to-left. \begin{frame} \ex To can bring this augmented matrix to echelon form with matrix multiplication You can bring this augmented matrix to echelon form with matrix multiplication. \begin{equation*} \begin{amat}{3} 1 &-1 &2 &4 \\ ... ... @@ -1277,7 +1277,7 @@ To can bring this augmented matrix to echelon form with matrix multiplication \end{equation*} \end{frame} \begin{frame} first perform $-2\rho_1+\rho_2$ via left multiplication by $C_{1,2}(-2)$. First perform $-2\rho_1+\rho_2$ via left multiplication by $C_{1,2}(-2)$. \begin{equation*} \begin{mat} 1 &0 &0 \\ ... ...
 %\documentclass{article} %\usepackage{verbatim} % %\begin{document} %\newcommand{\votepref}{\overrightarrow{#1}} %\newcommand{\votepreflist}{#1\,\votepref{DT} % +#2\,\votepref{TR}+#3\,\votepref{RS}} % \newcommand{\votepreflist}{\colvec{#1 \\ #2 \\ #3}} % \newcommand{\votecycle}{\psset{xunit=#4pt,yunit=#4pt,runit=#4pt} % \pspicture[.4](-7.25,-6)(7.25,5) %\psgrid(-1,-1)(1,1) % \SpecialCoor % \rput(3;90){\scriptsize $D$} % \psarc{->}(0,0){3}{10}{70} %-rd-> % \rput[lb](3.3;20){\scriptsize $#1$} % \rput[B](3;210){\scriptsize $T$} % \psarc{->}(0,0){3}{220}{320} %-tr-> % \rput[t](3.1;270){\scriptsize $#2$} % \rput[B](3;330){\scriptsize $R$} % \psarc{->}(0,0){3}{110}{175} %-dt-> % \rput[rb](3.3;160){\scriptsize $#3$} % \endpspicture} % %save it to the answer file. % \begin{Filesave}{bookans} % \newcommand{\votepreflist}{\colvec{#1 \\ #2 \\ #3}} % \newcommand{\votecycle}{\psset{xunit=#4pt,yunit=#4pt,runit=#4pt} % \pspicture[.4](-7.25,-6)(7.25,5) %\psgrid(-1,-1)(1,1) % \SpecialCoor % \rput(3;90){\scriptsize $D$} % \psarc{->}(0,0){3}{10}{70} %-rd-> % \rput[lb](3.3;20){\scriptsize $#1$} % \rput[B](3;210){\scriptsize $T$} % \psarc{->}(0,0){3}{220}{320} %-tr-> % \rput[t](3.1;270){\scriptsize $#2$} % \rput[B](3;330){\scriptsize $R$} % \psarc{->}(0,0){3}{110}{175} %-dt-> % \rput[rb](3.3;160){\scriptsize $#3$} % \endpspicture} % \end{Filesave} %\newcommand{\votinggraphic}{\hspace*{.8em}\mathord{\raisebox{-.2in}[.3in][.2in]{\includegraphics{voting.#1}}}\hspace*{.8em}} \topic{Voting Paradoxes} \index{voting paradox|(} %\emph{(Optional material from this chapter is discussed here, ... ... @@ -336,10 +295,22 @@ voting paradox can happen only when the tendencies toward cyclic preference reinforce each other. For the proof, assume that we have canceled opposite preference orders and we are left with one set of preference lists from each of the three rows. Consider the sum of these three (here, the numbers $a$, $b$, and $c$ could be positive, negative, or zero). opposite preference orders and are left with one set of preference lists for each of the three rows. Consider the first row's remaining preference lists. They could be from the first row's left or right (or neither, since the lists could have canceled exactly). We shall write \begin{equation*} \votinggraphic{27} \end{equation*} where $a$ is an integer that is positive if the remaining lists are on the left, where $a$ is negative if the lists are on the right, and zero if the cancellation was perfect. Similiarly we have integers $b$ and $c$ for the second and third rows, which can each be positive, negative, or zero. Then the election is determined by the sum of these three. \begin{equation*} \votinggraphic{27} %\votecycle{a}{a}{-a}{4} + ... ... @@ -348,23 +319,27 @@ Consider the sum of these three \votinggraphic{29} %\votecycle{c}{-c}{c}{4} =\hspace{.30in} \votinggraphic{30} %\votecycle{a-b+c}{a+b-c}{-a+b+c}{4} \hspace{.30in}\hbox{} \hspace{.30in}\hbox{} \end{equation*} A voting paradox occurs when the three numbers on the right, $a-b+c$ and $a+b-c$ and $-a+b+c$, are all nonnegative or all nonpositive. On the left, at least two of the three numbers $a$ and $b$ and $c$ are both nonnegative or both nonpositive. We can assume that they are $a$ and $b$. That makes four cases:~the cycle is nonnegative and $a$ and $b$ are nonnegative, the cycle is nonpositive and $a$ and $b$ are nonpositive, etc. We will do only the first case, since the second is similar and the other two are also easy. A voting paradox occurs when the three numbers in the total cycle on the right, % $-a+b+c$ and $a-b+c$ and $a+b-c$, are all nonnegative or all nonpositive. We will prove this occurs only when either all three of $a$, $b$, and~$c$ are nonnegative or all three are nonpositive. So assume that the cycle is nonnegative and that $a$ and $b$ are nonnegative. The conditions $0\leq a-b+c$ and $0\leq -a+b+c$ add to give that $0\leq 2c$, which implies that $c$ is also nonnegative, as desired. Suppose that the total cycle is nonnegative; the other case is similar. \begin{equation*} \begin{linsys}{3} -a &+ &b &+ &c &\geq &0 \\ a &- &b &+ &c &\geq &0 \\ a &+ &b &- &c &\geq &0 \end{linsys} \end{equation*} Adding the first two rows shows that $c\geq 0$. Adding the first and third rows gives $b\geq 0$. And the sum of the second and third shows $a\geq 0$. Thus if the total cycle is nonnegative then in each row the remaining preference lists are from the table's left. That ends the proof. This result says only that having all three spin in the same direction is a ... ... @@ -379,12 +354,14 @@ this work, showing that no voting system is entirely fair (for a reasonable definition of fair''). Some good introductory articles are \cite{Gardner70}, \cite{Gardner74}, \cite{Gardner80}, and \cite{NeimiRiker}. \cite{Taylor} is a readable recent book. \cite{Taylor} is a readable text. The long list of cases from recent American political history in \cite{GamingVote} shows these paradoxes are routinely manipulated in practice. This Topic is largely drawn from \cite{Zwicker}. (On the other hand, quite recent research shows that computing how to manipulate elections can in general be unfeasible, but this is beyond our scope.) This Topic is drawn from \cite{Zwicker}. \emph{(Author's Note:~I would like to thank Professor~Zwicker for his kind and illuminating discussions.)} ... ... @@ -491,7 +468,7 @@ for his kind and illuminating discussions.)} gives $c_3=2/3$, $c_2=-4/3$, and $c_1=1/3$. \end{answer} \item \label{exer:CancelPolSci} Do the cancellations of opposite preference orders Perform the cancellations of opposite preference orders for the Political Science class's mock election. Are all the remaining preferences from the left three rows of the table or from the right? ... ... @@ -622,15 +599,16 @@ for his kind and illuminating discussions.)} to get the same outcome. \end{answer} \item \label{exer:VoterCondNotSuff} The necessary condition that is proved above\Dash a The necessary condition that a voting paradox can happen only if all three preference lists remaining after cancellation have the same spin---is not also sufficient. after cancellation have the same spin is not also sufficient. \begin{exparts} \partsitem Continuing the positive cycle case considered in the proof, use the two inequalities $0\leq a-b+c$ and $0\leq -a+b+c$ to show that $\absval{a-b}\leq c$. \partsitem Also show that $c\leq a+b$, and hence that $\absval{a-b}\leq c\leq a+b$. % \partsitem Continuing the nonnegative total cycle case % considered in the proof, % use the two inequalities $0\leq a-b+c$ and $0\leq -a+b+c$ % to show that $\absval{a-b}\leq c$. % \partsitem Also show that $c\leq a+b$, and hence that % $\absval{a-b}\leq c\leq a+b$. \partsitem Give an example of a vote where there is a majority cycle, and addition of one more voter with the same spin causes the cycle to go away. ... ... @@ -641,17 +619,18 @@ for his kind and illuminating discussions.)} \end{exparts} \begin{answer} \begin{exparts} \partsitem We can rewrite the two as $-c\leq a-b$ and $-c\leq b-a$. Either $a-b$ or $b-a$ is nonpositive and so $-c\leq-\absval{a-b}$, as required. \partsitem This is immediate from the supposition that $0\leq a+b-c$. \partsitem A trivial example starts with the zero-voter election and adds any one voter. % \partsitem We can rewrite the two as $-c\leq a-b$ and $-c\leq b-a$. % Either $a-b$ or $b-a$ is nonpositive and so $-c\leq-\absval{a-b}$, % as required. % \partsitem This is immediate from the supposition that $0\leq a+b-c$. \partsitem A trivial example starts with the zero-voter election, which has a trivial majority cycle, and adds any one voter. A more interesting example takes the Political Science mock election and add two $T>D>R$ voters (they can be added one at a time, to satisfy the addition of one more voter'' criteria in the question). The additional voters have positive spin, which is the election and adds two $T>D>R$ voters (to satisfy the `addition of one more voter'' criteria in the question we can add them one at a time). The new voters have positive spin, which is the spin of the votes remaining after cancellation in the original mock election. This is the resulting table of voters and next to it is ... ... @@ -775,11 +754,14 @@ for his kind and illuminating discussions.)} % \endpspicture \tag*{}\end{equation*} The majority cycle has indeed disappeared. \partsitem Reverse the prior exercise. That is, add a voter to the result of the prior exercise that cancels the one who made the cycle disappear. \partsitem One such condition is that, after cancellation, all three be nonnegative or all three be nonpositive, and:~$\absval{c}<\absval{a+b}$ and $\absval{b}<\absval{a+c}$ and $\absval{a}<\absval{b+c}$. That follows from this diagram. That follows from this diagram. \begin{equation*} \votinggraphic{27} %\votecycle{a}{a}{-a}{4} + ... ...
 ... ... @@ -2175,7 +2175,7 @@ the inherited ones. =\colvec{rx \\ 0} \end{equation*} As in the prior example, to verify directly from the definition that this is a subspace we simply that it is a that this is a subspace we simply note that it is a subset and then check that it satisfies the conditions in definition of a vector space. For instance the two closure conditions are ... ...
 ... ... @@ -2712,7 +2712,7 @@ set. Lemma~One.III.\ref{le:EchFormNoLinCombo} says that no nonzero row of an echelon form matrix is a linear combination of the other rows. This result just restates that in this chapter's terminology. This result restates that using this chapter's terminology. % \end{proof} ... ...
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