Commit 35d2ab72 authored by Jim Hefferon's avatar Jim Hefferon

second Lightening Source hardcopy version

parent 4bfa2434
......@@ -129,4 +129,7 @@ General:
Mana Borwornpadungkitti Correction to exercise answer.
2014-Feb-21
Mana Borwornpadungkitti Number of corrections to homogeom.
\ No newline at end of file
Mana Borwornpadungkitti Number of corrections to homogeom.
2014-Apr-16
Fred Richman Number of suggestions and corrections, including terminology.
\ No newline at end of file
......@@ -22,8 +22,9 @@
% have to "rm *.aux" to get it to compile.
\includeonly{
covernew,%
\ifbool{hardcopybool}{titlepage}{symlist},
% symlist, % comment this out for printed version
titlepage, % uncomment for printed version
% titlepage, % uncomment for printed version
pref,%
gr1,gr2,gr3,cas,leontief,ppivot,network,%
vs1,vs2,vs3,fields,crystal,voting,dimen,%
......@@ -42,8 +43,9 @@ bib%
\begin{document}
\frontmatter
\pagenumbering{roman}
\ifbool{hardcopybool}{\include{titlepage}}{\include{covernew}\include{symlist}\clearemptydoublepage}
% \include{covernew}\include{symlist}\clearemptydoublepage % comment out for printed version
\include{titlepage} % uncomment for printed version (also remember to change the color of hyperlinks in bookjhconcrete.sty)
% \include{titlepage} % uncomment for printed version (also remember to change the color of hyperlinks in bookjhconcrete.sty)
\pagestyle{bookfront}
\setcounter{page}{1}\thispagestyle{empty}\include{pref} %
\pagestyle{booktoc}\clearemptydoublepage\tableofcontents\bigskip\par\noindent${}^*\!$Starred subsections are optional.\clearemptydoublepage %
......
......@@ -26927,7 +26927,7 @@ ans = 0.017398
\end{ans}
\subsection{Five.II.3: Eigenvalues and Eigenvectors}
\begin{ans}{Five.II.3.21}
\begin{ans}{Five.II.3.20}
\begin{exparts}
\partsitem This
\begin{equation*}
......@@ -26949,7 +26949,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.22}
\begin{ans}{Five.II.3.21}
\begin{exparts}
\partsitem The characteristic equation is \( (3-x)(-1-x)=0 \).
Its roots, the eigenvalues, are \( \lambda_1=3 \) and
......@@ -27062,7 +27062,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.23}
\begin{ans}{Five.II.3.22}
The characteristic equation
\begin{equation*}
0=
......@@ -27130,7 +27130,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.24}
\begin{ans}{Five.II.3.23}
The characteristic equation is
\begin{equation*}
0=
......@@ -27168,7 +27168,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.25}
\begin{ans}{Five.II.3.24}
\begin{exparts}
\partsitem The characteristic equation is
\begin{equation*}
......@@ -27371,7 +27371,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.26}
\begin{ans}{Five.II.3.25}
With respect to the natural basis $B=\sequence{1,x,x^2}$
the matrix representation is this.
\begin{equation*}
......@@ -27458,7 +27458,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.27}
\begin{ans}{Five.II.3.26}
$\lambda=1,
\begin{mat}
0 &0 \\
......@@ -27480,7 +27480,7 @@ ans = 0.017398
\end{mat}$
\end{ans}
\begin{ans}{Five.II.3.28}
\begin{ans}{Five.II.3.27}
Fix the natural basis $B=\sequence{1,x,x^2,x^3}$.
The map's action is $1\mapsto 0$, $x\mapsto 1$, $x^2\mapsto 2x$,
and $x^3\mapsto 3x^2$ and its representation is easy to compute.
......@@ -27529,13 +27529,13 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.29}
\begin{ans}{Five.II.3.28}
The determinant of the triangular matrix $T-xI$ is the product
down the diagonal, and so it factors into the product of
the terms $t_{i,i}-x$.
\end{ans}
\begin{ans}{Five.II.3.30}
\begin{ans}{Five.II.3.29}
Just expand the determinant of $T-xI$.
\begin{equation*}
\begin{vmatrix}
......@@ -27547,12 +27547,12 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.31}
\begin{ans}{Five.II.3.30}
Any two representations of that transformation are similar, and
similar matrices have the same characteristic polynomial.
\end{ans}
\begin{ans}{Five.II.3.32}
\begin{ans}{Five.II.3.31}
It is not true.
All of the eigenvalues of this matrix are $0$.
\begin{equation*}
......@@ -27563,7 +27563,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.33}
\begin{ans}{Five.II.3.32}
\begin{exparts}
\partsitem Use \( \lambda=1 \) and the identity map.
\partsitem Yes, use the transformation that multiplies all
......@@ -27571,12 +27571,12 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.34}
\begin{ans}{Five.II.3.33}
If $t(\vec{v})=\lambda\cdot\vec{v}$ then
$\vec{v}\mapsto\zero$ under the map $t-\lambda\cdot\identity$.
\end{ans}
\begin{ans}{Five.II.3.35}
\begin{ans}{Five.II.3.34}
The characteristic equation
\begin{equation*}
0=
......@@ -27591,7 +27591,7 @@ ans = 0.017398
(under the $a+b=c+d$ condition) is routine.
\end{ans}
\begin{ans}{Five.II.3.36}
\begin{ans}{Five.II.3.35}
Consider an eigenspace $V_{\lambda}$.
Any $\vec{w}\in V_{\lambda}$ is the image
$\vec{w}=\lambda\cdot\vec{v}$ of some $\vec{v}\in V_{\lambda}$
......@@ -27602,7 +27602,7 @@ ans = 0.017398
and so $1/\lambda$ is an eigenvalue of $t^{-1}$.
\end{ans}
\begin{ans}{Five.II.3.37}
\begin{ans}{Five.II.3.36}
\begin{exparts}
\partsitem We have
$(cT+dI)\vec{v}=cT\vec{v}+dI\vec{v}=c\lambda\vec{v}+d\vec{v}
......@@ -27613,7 +27613,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.38}
\begin{ans}{Five.II.3.37}
The scalar $\lambda$ is an eigenvalue if and only if the transformation
$t-\lambda \identity$ is singular.
A transformation is singular if and only if it is not an isomorphism
......@@ -27621,7 +27621,7 @@ ans = 0.017398
nonsingular).
\end{ans}
\begin{ans}{Five.II.3.39}
\begin{ans}{Five.II.3.38}
\begin{exparts}
\partsitem Where the eigenvalue $\lambda$ is associated with the
eigenvector $\vec{x}$ then
......@@ -27633,7 +27633,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\begin{ans}{Five.II.3.40}
\begin{ans}{Five.II.3.39}
No.
These are two same-sized, equal rank, matrices
with different eigenvalues.
......@@ -27650,12 +27650,12 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.41}
\begin{ans}{Five.II.3.40}
The characteristic polynomial has an odd power and so
has at least one real root.
\end{ans}
\begin{ans}{Five.II.3.42}
\begin{ans}{Five.II.3.41}
The characteristic polynomial $x^3+5x^2+6x$ has distinct roots
\( \lambda_1=0 \), \( \lambda_2=-2 \), and \( \lambda_3=-3 \).
Thus the matrix can be diagonalized into this form.
......@@ -27668,7 +27668,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\begin{ans}{Five.II.3.43}
\begin{ans}{Five.II.3.42}
We must show that it is one-to-one and onto, and that it respects the
operations of matrix addition and scalar multiplication.
......@@ -27687,7 +27687,7 @@ ans = 0.017398
similar:~$t_P(cT)=P(c\cdot T)P^{-1}=c\cdot (PTP^{-1})=c\cdot t_P(T)$.
\end{ans}
\begin{ans}{Five.II.3.44}
\begin{ans}{Five.II.3.43}
\answerasgiven %
If the argument of the characteristic function of \( A \) is set equal to
\( c \), adding the first \( (n-1) \) rows (columns) to the
......@@ -28542,9 +28542,7 @@ ans = 0.017398
\section{Jordan Form}
\subsection{Five.IV.1: Polynomials of Maps and Matrices}
\begin{ans}{Five.IV.1.13}
For each,
the minimal polynomial must have a leading coefficient of $1$
and \nearbytheorem{th:CayHam}, the Cayley-Hamilton Theorem, says that
The Cayley-Hamilton Theorem \nearbytheorem{th:CayHam} says that
the minimal polynomial must contain the same linear factors
as the characteristic polynomial, although possibly of lower degree
but not of zero degree.
......@@ -28552,8 +28550,6 @@ ans = 0.017398
\partsitem The possibilities are
$m_1(x)=x-3$, $m_2(x)=(x-3)^2$, $m_3(x)=(x-3)^3$,
and $m_4(x)=(x-3)^4$.
Note that the $8$ has been dropped because a minimal
polynomial must have a leading coefficient of one.
The first is a degree one polynomial, the second is degree two,
the third is degree three, and the fourth is degree four.
\partsitem The possibilities are $m_1(x)=(x+1)(x-4)$,
......@@ -2,6 +2,21 @@
%% bookjh.sty
% The INITIAL CODE part
% I want the booleans offered by this package
\usepackage{etoolbox}
% Are printing a hard copy or not?
\newbool{hardcopybool}
% set the default by uncommenting one or the other
% \booltrue{hardcopybool}
\boolfalse{hardcopybool}
% You can cause guidelines to be shown by invoking with
% pdflatex "\def\hardcopy{}\input{book}"
% See http://stackoverflow.com/a/1466610
\ifdefined\hardcopy
\booltrue{hardcopybool}
\fi
\ifbool{hardcopybool}{\typeout{!!! PRINTING HARD COPY}}{}
% The DECLARATION OF OPTIONS part
%\DeclareOption{singleanswerfile}{}
......@@ -876,8 +891,7 @@
% \ifhrefout %
\usepackage[
colorlinks=true,
% linkcolor=darkcolor,citecolor=darkcolor,filecolor=darkcolor,urlcolor=darkcolor, % usual colors for online version of book
linkcolor=black,citecolor=black,filecolor=black,urlcolor=black, % black links for printed version
\ifbool{hardcopybool}{linkcolor=black,citecolor=black,filecolor=black,urlcolor=black}{linkcolor=darkcolor,citecolor=darkcolor,filecolor=darkcolor,urlcolor=darkcolor}, % usual colors for online version of book is links are blue. but for hard copy links are black
naturalnames=false,
hypertexnames=false, % if true, index links works but a chapter 3 link to 2.6 will go to chapter 1's 2.6
hyperindex=false, % if true, page numbers are correct but links are to a page ten too early.
......
......@@ -5,6 +5,7 @@ verbatimtex
\documentclass{book}
\usepackage{bookjh}
\usepackage{linalgjh}
\usepackage{tabstackengine} \setstacktabbedgap{1pt}
\begin{document}
etex
......@@ -1007,7 +1008,7 @@ endfig;
% 28 is below. Why?
......@@ -1048,15 +1049,23 @@ beginfig(30) % equiv relation; two specific row-equiv mats
path p[]; partition;
label(btex {\small \ldots} etex,z13);
x14=.65[x0,x9]; y14=.7[y10,y9]; drawpoint(z14);
label.rt(btex {\tiny $\Bigl(\begin{smallmatrix}
% label.rt(btex {\tiny $\Bigl(\begin{smallmatrix}
% 1 &3 \\
% 2 &7
% \end{smallmatrix}\Bigr)$} etex,z14);
label.rt(btex {\tiny $\parenMatrixstack{
1 &3 \\
2 &7
\end{smallmatrix}\Bigr)$} etex,z14);
}$} etex,z14);
x15=.7[x0,x9]; y15=.35[y10,y9]; drawpoint(z15);
label.rt(btex {\tiny $\Bigl(\begin{smallmatrix}
% label.rt(btex {\tiny $\Bigl(\begin{smallmatrix}
% 1 &3 \\
% 0 &1
% \end{smallmatrix}\Bigr)$} etex,z15);
label.rt(btex {\tiny $\parenMatrixstack{
1 &3 \\
0 &1
\end{smallmatrix}\Bigr)$} etex,z15);
}$} etex,z15);
endfig;
......@@ -1084,10 +1093,14 @@ beginfig(31) % equiv relation; canonical reps shown, incl 2x2 identity
label(btex {\small $\star$} etex,z16);
x17=.28[x0,x1]; y17=.6[y0,y3]; % middle
label(btex {\small $\star$} etex,z17);
label.bot(btex {\tiny $\Bigl(\begin{smallmatrix}
% label.bot(btex {\tiny $\Bigl(\begin{smallmatrix}
% 1 &0 \\
% 0 &1
% \end{smallmatrix}\Bigr)$} etex,z17);
label.bot(btex {\tiny $\parenMatrixstack{
1 &0 \\
0 &1
\end{smallmatrix}\Bigr)$} etex,z17);
}$} etex,z17);
endfig;
......
......@@ -5,6 +5,7 @@ verbatimtex
\documentclass{book}
\usepackage{bookjh}
\usepackage{linalgjh}
\usepackage{tabstackengine} \setstacktabbedgap{1pt}
\begin{document}
etex
......@@ -972,20 +973,32 @@ beginfig(24) % equiv relation; all matrices partitioned by mat-equiv
draw partitionbox;
x10=.00[x0,x1]; y10=.2[y3,y0];
label.rt(btex {\tiny $\star\;\left(\begin{smallmatrix}
% label.rt(btex {\tiny $\star\;\left(\begin{smallmatrix}
% 0 &0 \\
% 0 &0
% \end{smallmatrix}\right)$} etex,z10);
label.rt(btex {\tiny $\star\;\parenMatrixstack{
0 &0 \\
0 &0
\end{smallmatrix}\right)$} etex,z10);
0 &0
}$} etex,z10);
x11=.60[x0,x1]; y11=.25[y3,y0];
label.rt(btex {\tiny $\star\;\left(\begin{smallmatrix}
% label.rt(btex {\tiny $\star\;\left(\begin{smallmatrix}
% 1 &0 \\
% 0 &0
% \end{smallmatrix}\right)$} etex,z11);
label.rt(btex {\tiny $\star\;\parenMatrixstack{
1 &0 \\
0 &0
\end{smallmatrix}\right)$} etex,z11);
0 &0
}$} etex,z11);
x12=.20[x0,x1]; y12=.75[y3,y0];
label.rt(btex {\tiny $\star\;\left(\begin{smallmatrix}
% label.rt(btex {\tiny $\star\;\left(\begin{smallmatrix}
% 1 &0 \\
% 0 &1
% \end{smallmatrix}\right)$} etex,z12);
label.rt(btex {\tiny $\star\;\parenMatrixstack{
1 &0 \\
0 &1
\end{smallmatrix}\right)$} etex,z12);
0 &1
}$} etex,z12);
path p[];
......
......@@ -117,7 +117,7 @@ beginfig(4) % equiv relation; row-equiv mats from ch1.27, further split
label(btex {\small \ldots} etex,z13);
x14=.8[x0,x9]; y14=.7[y10,y9]; drawpoint(z14);
label.rt(btex {\small $S$} etex,z14);
x15=.75[x0,x9]; y15=.35[y10,y9]; drawpoint(z15);
x15=.85[x0,x9]; y15=.3[y10,y9]; drawpoint(z15);
label.rt(btex {\small $T$} etex,z15);
pickup pencircle scaled line_width_light;
......
\documentclass[12pt]{article}
\documentclass{article}
\usepackage{covergraphic}
\usepackage{stackengine}
\renewcommand{\stackalignment}{c}
......@@ -25,14 +25,14 @@
\usepackage[papersize={16in,9.25in},margin=0in]{geometry}
\newcommand{\spinetext}[1]{\color{black}\fontsize{40pt}{8pt}{\fontfamily{ugq}\selectfont #1}}
\newcommand{\spineauthortext}[1]{\color{coverdarkcolor}\fontsize{14pt}{8pt}{\fontfamily{ugq}\selectfont #1}}
\newcommand{\spineauthortext}[2]{\color{coverdarkcolor}\fontsize{#2pt}{8pt}{\fontfamily{ugq}\selectfont{#1}}}
\newcommand{\spine}{%
\setlength{\unitlength}{1in}
\begin{picture}(1,9.5)
\setstackgap{S}{2.1ex}
\put(0.25,5.65){\Shortstack{{\spinetext{L}} {\spinetext{I}} {\spinetext{N}} {\spinetext{E}} {\spinetext{A}} {\spinetext{R}}}}
\put(0.235,5.65){\Shortstack{{\spinetext{L}} {\spinetext{I}} {\spinetext{N}} {\spinetext{E}} {\spinetext{A}} {\spinetext{R}}}}
\setstackgap{S}{0.9ex}
\put(0.375,1.15){\Shortstack{{\spineauthortext{H}} {\spineauthortext{e}} {\spineauthortext{f}} {\spineauthortext{f}} {\spineauthortext{e}} {\spineauthortext{r}} {\spineauthortext{o}} {\spineauthortext{n}}}}
\put(0.3925,1.15){\Shortstack{{\spineauthortext{H}{14}} {\spineauthortext{E}{14}} {\spineauthortext{F}{14}} {\spineauthortext{F}{14}} {\spineauthortext{E}{14}} {\spineauthortext{R}{14}} {\spineauthortext{O}{14}} {\spineauthortext{N}{14}}}} % tried fake small caps by making the H be 14 pt, others be 11
\end{picture}
}
......
......@@ -23,6 +23,6 @@
\fontfamily{ugq}\selectfont #1}}}
\put(-0.0,-6.9){\includegraphics{asy/shadow.pdf}}
\put(0,-6.5){\includegraphics{asy/axes.pdf}}
\put(2.32,-6.4){{\color{coverlightcolor} \texttt{http://joshua.smcvt.edu/linearalgebra}}}
\put(2.32,-6.4){{\color{coverlightcolor}\large \texttt{http://joshua.smcvt.edu/linearalgebra}}}
\end{picture}
}
No preview for this file type
......@@ -439,7 +439,7 @@ operations to set up back-substitution.
In each row of a system,
the first variable with a nonzero coefficient is the row's
\definend{leading variable}\index{echelon form!leading variable}%
\index{leading variable}. %
\index{leading!variable}. %
A system is in \definend{echelon form}\index{echelon form}
if each leading variable
is to the right of the leading variable in the row above it,
......@@ -1882,7 +1882,7 @@ Each number in the matrix is an
%</df:matrix>
\end{definition}
Matrices are usually named by upper case roman letters.
We usually denote a matrix with an upper case roman letters.
For instance,
\begin{equation*}
A=
......@@ -1906,10 +1906,25 @@ Note that the order of the subscripts matters:
$a_{1,2}\neq a_{2,1}$ since \( a_{1,2}=2.2 \).
% Matrices occur throughout this book.
We use
\( \matspace_{\nbym{n}{m}} \) to denote the set of all \( \nbym{n}{m} \)
We write
\( \matspace_{\nbym{n}{m}} \) for the set of all \( \nbym{n}{m} \)
matrices.
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}. %
is its first nonzero entry (if it has one),
we perform row operations to arrive at
\definend{matrix echelon form},\index{echelon form!matrix}%
\index{matrix!echelon form}%
where the leading entry in lower rows are to the right of those in
the rows above.
We switch to this notation because it lightens
the clerical load of Gauss's Method\Dash the copying of variables and the
writing of $+$'s and $=$'s.
\begin{example}
We can abbreviate this linear system
\begin{equation*}
......@@ -1927,13 +1942,10 @@ with this matrix.
1 &0 &2 &4
\end{amat}
\end{equation*}
The vertical bar simply reminds a reader of the difference between the
The vertical bar reminds a reader of the difference between the
coefficients on the system's left hand side and the constants on the right.
With a bar, this is an
\definend{augmented\/}\index{matrix!augmented}\index{augmented matrix} matrix.
In this notation
the clerical load of Gauss's Method\Dash the copying of variables, the
writing of $+$'s and $=$'s\Dash is lighter.
\begin{equation*}
\begin{amat}[r]{3}
1 &2 &0 &4 \\
......
......@@ -2057,20 +2057,6 @@ Some authors call these \definend{characteristic}%
\index{characteristic!vectors, values} values and vectors.
No authors call them ``peculiar.'')
\begin{remark}
This definition requires that the eigenvector be non-$\zero$.
Some authors allow $\zero$ as
an eigenvector for $\lambda$ as long as there are also
non-$\zero$ vectors associated with $\lambda$.
% Neither style of definition is clearly better; both involve small tradeoffs.
The key point is
to disallow the trivial case where $\lambda$ is such that
$t(\vec{v})=\lambda\vec{v}$ for only the single vector $\vec{v}=\zero$.
Also, note that the eigenvalue $\lambda$ could be~$0$.
The issue is whether $\vec{\zeta}$ equals $\zero$.
\end{remark}
\begin{example}
The projection map
\begin{equation*}
......@@ -2084,13 +2070,24 @@ has an eigenvalue of \( 1 \) associated with any eigenvector
\colvec{x \\ y \\ 0}
\end{equation*}
where \( x \) and \( y \) are scalars that are not both zero.
A number that is not an eigenvalue of $\pi$ is \( 2 \),
since no non-$\zero$ vector is doubled.
In contrast, a number that is not an eigenvalue of of this map is \( 2 \),
since assuming that $\pi$ doubles a vector leads to
the equations $x=2x$, $y=2y$, and $0=2z$, and thus
no non-$\zero$ vector is doubled.
\end{example}
% That shows why the `nonzero' appears in \nearbydefinition{def:Eigen}.
% Disallowing \( \zero \) as an eigenvector eliminates trivial eigenvalues.
% Note, however, that a matrix can have an eigenvalue of $\lambda=0$.
Note that
the definition requires that the eigenvector be non-$\zero$.
Some authors allow $\zero$ as
an eigenvector for $\lambda$ as long as there are also
non-$\zero$ vectors associated with $\lambda$.
The key point is
to disallow the trivial case where $\lambda$ is such that
$t(\vec{v})=\lambda\vec{v}$ for only the single vector $\vec{v}=\zero$.
Also, note that the eigenvalue $\lambda$ could be~$0$.
The issue is whether $\vec{\zeta}$ equals $\zero$.
\begin{example} \label{ex:NoEigenOnTrivSp}
The only transformation on the trivial space \( \set{\zero} \) is
......@@ -2116,6 +2113,9 @@ This map also has an eigenvalue of \( 0 \) associated with eigenvectors of
the form \( c-cx \) where \( c\neq 0 \).
\end{example}
The definition above is for maps.
We can give a matrix version.
\begin{definition} \label{df:EigenOfMatrix}
%<*df:EigenOfMatrix>
A square matrix \( T \) has a scalar
......@@ -2127,10 +2127,11 @@ A square matrix \( T \) has a scalar
\end{definition}
This extension of the definition for maps to a definition
for matrices is natural but we must be careful.
While the eigenvalues of a map are also the eigenvalues of matrices representing
that map, and so similar matrices have the same eigenvalues,
the eigenvectors can
for matrices is natural but there is a point on which we must take care.
The eigenvalues of a map are also the eigenvalues of matrices r
epresenting
that map, and so similar matrices have the same eigenvalues.
However, the eigenvectors can
differ\Dash similar matrices need not have the
same eigenvectors.
The next example explains.
......
......@@ -844,18 +844,18 @@ We might think to make something like
the map \( \map{t}{\C^5}{\C^5} \) with this action.
\begin{center}
\begin{minipage}{1in}
\setlength{\unitlength}{1pt}
$\begin{strings}{ccccc}
\setlength{\unitlength}{1pt}
\begin{picture}(4,15)(0,0)
\begin{picture}(4,15)
\put(0,14){$\vec{e}_1$}
\put(0,-12){$\vec{e}_2$}
\end{picture}
&\begin{picture}(8,10)(0,0)
&\begin{picture}(8,10)
\put(0,8){\rotatebox{-30}{$\mapsto$}}%
\put(0,-8){\rotatebox{30}{$\mapsto$}}
\end{picture}
&\vec{e}_3 &\mapsto &\zero \\[4ex]
\vec{e}_4&\mapsto &\vec{e}_5 &\mapsto &\zero
\vec{e}_4 &\mapsto &\vec{e}_5 &\mapsto &\zero
\end{strings}$
\end{minipage}
\hspace*{3em}
......
......@@ -570,16 +570,14 @@ and so \( m(x)=(x-1)(x-2)^2 \).
What are the possible minimal polynomials if a matrix has
the given characteristic polynomial?
\begin{exparts*}
\partsitem $8\cdot (x-3)^4$
\partsitem $(1/3)\cdot (x+1)^3(x-4)$
\partsitem $-1\cdot (x-2)^2(x-5)^2$
\partsitem \( 5\cdot(x+3)^2(x-1)(x-2)^2 \)
\partsitem $(x-3)^4$
\partsitem $(x+1)^3(x-4)$
\partsitem $(x-2)^2(x-5)^2$
\partsitem \( (x+3)^2(x-1)(x-2)^2 \)
\end{exparts*}
What is the degree of each possibility?
\begin{answer}
For each,
the minimal polynomial must have a leading coefficient of $1$
and \nearbytheorem{th:CayHam}, the Cayley-Hamilton Theorem, says that
\begin{answer}
The Cayley-Hamilton Theorem \nearbytheorem{th:CayHam} says that
the minimal polynomial must contain the same linear factors
as the characteristic polynomial, although possibly of lower degree
but not of zero degree.
......@@ -587,8 +585,6 @@ and so \( m(x)=(x-1)(x-2)^2 \).
\partsitem The possibilities are
$m_1(x)=x-3$, $m_2(x)=(x-3)^2$, $m_3(x)=(x-3)^3$,
and $m_4(x)=(x-3)^4$.
Note that the $8$ has been dropped because a minimal
polynomial must have a leading coefficient of one.
The first is a degree one polynomial, the second is degree two,
the third is degree three, and the fourth is degree four.
\partsitem The possibilities are $m_1(x)=(x+1)(x-4)$,
......
......@@ -386,7 +386,11 @@
\usepackage{calc} % needed for \widthof
\usepackage{transparent}
% \usepackage{tikz} % for transparency
\newcommand{\highlight}[1]{\makebox[\widthof{#1}]{\makebox[0pt]{\transparent{0.5}\colorbox{\highlightcolorname}{#1}}}}
% \newcommand{\highlight}[1]{\makebox[\widthof{#1}]{\makebox[0pt]{\transparent{0.5}\colorbox{\highlightcolorname}{#1}}}}
\newcommand{\highlight}[1]{{%
\setlength{\fboxsep}{1pt}%
\setlength{\fboxrule}{0.2pt}%
\framebox{#1}}}
% \newcommand{\highlight}[1]{\makebox[\widthof{#1}]{\begin{tikzpicture}[unit=1em]#1 \draw[fill=lightgray, ultra thin, lightgray, opacity=0.6] (0,0) rectangle (1,1); \end{tikzpicture}}}
\endinput
......@@ -1464,15 +1464,17 @@ The prior result shows that,
in passing from the definition of isomorphism to the more
general definition of homomorphism,
omitting the onto requirement doesn't make an essential difference.
Any homomorphism is onto a space, the range space.
Any homomorphism is onto some space, namely its range.
However, omitting the one-to-one condition does make an essential difference.
However, omitting the one-to-one condition does make a difference.
A homomorphism may have many elements
of the domain that map to one element of the codomain.
Below is a bean sketch of a many-to-one
map between sets.\appendrefs{many-to-one maps}\spacefactor=1000 %
It shows three elements of the codomain that are each the image of
many members of the domain.
(Rather than picture lots of individual $\mapsto$ arrows, each association
of many inputs with one output shows only one such arrow.)
\begin{center}
\includegraphics{ch3.5} % bean to bean; many to one
\end{center}
......
......@@ -984,7 +984,9 @@ entries as the column of the right-hand matrix, or else some entry
will be left without a matching entry from the other matrix.
Another aspect of the combinatorics of
matrix multiplication is that in the definition of the $i,j$ entry
matrix multiplication,
in the sum defining the $i,j$ entry,
is brought out here by the boxing the equal subscripts.
\begin{equation*} % \setlength{\fboxsep}{.15em}
p_{i,j}
=
......@@ -992,22 +994,8 @@ matrix multiplication is that in the definition of the $i,j$ entry
+g_{i,\text{\highlight{$2 $}}}h_{\text{\highlight{$2 $}},j}
+\dots+g_{i,\text{\highlight{$r $}}}h_{\text{\highlight{$r $}},j}
\end{equation*}
% \begin{equation*} \setlength{\fboxsep}{.15em}
% p_{i,j}
% =
% g_{i,\text{\colorbox{lightgray}{$\scriptsize 1 $}}}h_{\text{\colorbox{lightgray}{$\scriptsize 1 $}},j}
% +g_{i,\text{\colorbox{lightgray}{$\scriptsize 2 $}}}h_{\text{\colorbox{lightgray}{$\scriptsize 2 $}},j}
% +\dots+g_{i,\text{\colorbox{lightgray}{$\scriptsize r $}}}h_{\text{\colorbox{lightgray}{$\scriptsize r $}},j}
% \end{equation*}
% \begin{equation*}
% p_{i,j}
% =
% g_{i,\fbox{\scriptsize \( 1 \)}}h_{\fbox{\scriptsize \( 1 \)},j}
% +g_{i,\fbox{\scriptsize \( 2 \)}}h_{\fbox{\scriptsize \( 2 \)},j}
% +\dots+g_{i,\fbox{\scriptsize \( r \)}}h_{\fbox{\scriptsize \( r \)},j}
% \end{equation*}
the highlighted subscripts on the $g$'s are column indices while those on the
$h$'s indicate rows.
The highlighted subscripts on the $g$'s are column indices while those on the
$h$'s are for rows.
That is, the summation takes place over the columns of $G$
but over the rows of $H$\Dash
the definition treats left differently than right.
......@@ -2247,7 +2235,8 @@ are the ones with a single nonzero entry.
%<*df:UnitMatrix>
A matrix with all $0$'s except for a $1$ in the \( i,j \) entry
is an \( i,j \) \definend{unit}\index{matrix!unit}\index{unit matrix}
matrix.
matrix
(or \definend{matrix unit}).\index{matrix!unit}
%</df:UnitMatrix>
\end{definition}
......@@ -2868,6 +2857,8 @@ perform the combination operation \( -2\rho_2+\rho_3 \).
%<*df:ElementaryReductionMatrices>
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.
\begin{enumerate}
\item \( I\grstep{k\rho_i}M_i(k) \) for \( k\neq 0 \)
......
No preview for this file type
This diff is collapsed.
......@@ -271,7 +271,7 @@ and of Linear Algebra in particular.
\input{publicationdate}
\end{tabular}
\vspace{2ex}
\vspace{3ex}
\par\noindent\textit{Author's Note.}
Inventing a good exercise, one that enlightens as well as tests,
is a creative act and hard work.
......
......@@ -6,16 +6,14 @@
\newpage
\setlength{\unitlength}{1in}
\noindent\begin{picture}(0,0)
% \put(1,-2){\color{black}\fontsize{50pt}{8pt}{\fontfamily{ugq}\selectfont Linear Algebra}}
\put(0.75,-2){\centering{\begin{tabular}{c}
\titlepagetext{40}{Linear Algebra} \\[.3in]
\titlepagetext{20}{Jim Hef{}feron}
\titlepagetext{36}{Linear Algebra} \\[.35in]
\titlepagetext{18}{Jim Hef{}feron}
\end{tabular}}}
% \put(2.25,-7){\begin{tabular}{l}
% Edition: \input{publicationdate} \\
% Cover: colors by Gabriel S~Santiago \\
% See: \texttt{http://joshua.smcvt.edu/linearalgebra}
% \end{tabular}}
\put(2.25,-7){\begin{tabular}{l}
Edition: \input{publicationdate} \\
See: \texttt{http://joshua.smcvt.edu/linearalgebra}
\end{tabular}}
\end{picture}
\newpage
\input symlist
......@@ -2587,67 +2587,68 @@ a small number of members.)
\end{example}
\begin{example} \label{ex:SubspRThree}
These are the subspaces of \( \Re^3 \) that we now know of, the
The picture below shows the subspaces of \( \Re^3 \) that we now know of, the
trivial subspace, the lines through the origin,
the planes through the origin, and the whole space
(of course, the picture shows only a few of the infinitely many subspaces).
In the next section we will prove that $\Re^3$ has no other
type of subspaces, so in fact this picture shows them all.
We have described the subspaces
as spans of sets with a minimal number of members