Commit 332cba0a authored by Jim Hefferon's avatar Jim Hefferon

milestone; up to date with bug reports on book

parent 07e3f571
......@@ -101,4 +101,10 @@ General:
DG formatting of some links in jhanswers not right
2012-Oct-20
Ken Shrirriff addition to introduction of determinant function.
\ No newline at end of file
Ken Shrirriff addition to introduction of determinant function.
2013-Jan-01
Itamar Stein hole in proof of Jordan form
David Guichard several corrections throughout text
Troy J. Jasso hyperref'ed links in index wrong (I couldn't figure out
how to correct; dropped links)
\ No newline at end of file
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......@@ -34,12 +34,13 @@ bib%
% \overfullrule=5pt % if a draft
\begin{document}
\frontmatter
\pagenumbering{roman}
\frontmatter \include{covernew}\include{symlist}
\include{covernew}\include{symlist}
\pagestyle{bookfront}\clearemptydoublepage\setcounter{page}{1}\thispagestyle{empty}\include{pref} %
\pagestyle{booktoc}\clearemptydoublepage\tableofcontents\bigskip\par\noindent${}^*\!$Starred subsections are optional.\clearemptydoublepage %
\pagenumbering{arabic}
\mainmatter
\pagenumbering{arabic}
\pagestyle{bookbody}
\include{gr1}
\include{gr2}
......
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......@@ -845,14 +845,18 @@
% The hrefout package allows me to specify on the command line if I want
% to have hyperreffed output.
% Note: I am unable to figure out how to get the index page links to point to
% the correct page. Hence I turned the links off.
% \ifhrefout %
\usepackage[
colorlinks=true,
linkcolor=darkcolor,citecolor=darkcolor,filecolor=darkcolor,urlcolor=darkcolor,
naturalnames=true, % if true, toc Three.IV.2 link no work
hypertexnames=false, %
% hyperindex=false, % if true, pages are ten off
plainpages=false,pdfpagelabels, % HO says on ctt 2001-dec
hyperindex=false, % if true, pages are correct but links are to a page ten too early.
plainpages=false, % These three from http://stackoverflow.com/a/1455672/238366
pdfpagelabels,
pagebackref, % HO says on ctt 2001-dec, to make page links in index point to the same-numbered page in mainmatter
pdfpagemode=UseNone,pdfstartview=FitH,
pdftitle={Linear Algebra},
pdfauthor={Jim Hefferon, Saint Michael's College, Vermont USA,
......@@ -869,7 +873,7 @@
\rule{5.7in}{0em}
\fontsize{260pt}{8pt}
{\fontfamily{qzc}\itshape\selectfont \contour{darkcolor}{\textcolor{lightcolor}{Linear}}}
} \\[0.1in]
} \\[0.05in]
% } \\[-2.15in]
% \noindent\makebox[0pt][c]{\rule{8.5in}{0pt}%
% \makebox[0pt][l]{\contourlength{0.6pt} %
......
......@@ -3787,7 +3787,7 @@ We must show this is the negative of $d(T)$.
\sgn(\phi)
\tag{*}
\end{equation*}
We will show that each term in ($*$) is associated with a term in $d(t)$,
We will show that each term in ($*$) is associated with a term in $d(T)$,
and that the two terms are negatives of each other.
Consider the matrix from the multilinear expansion
of $d(\hat{T})$ giving the term
......
......@@ -4,7 +4,7 @@
\topic{Speed of Calculating Determinants}
The permutation expansion formula for computing determinants is useful
for proving theorems, but the method of using row operations
is a much better for finding the determinants of a large matrix.
is much better for finding the determinants of a large matrix.
We can make this statement
precise by considering, as computer algorithm designers do,
the number of arithmetic operations that each method uses.
......
......@@ -3771,7 +3771,7 @@ general solution.
\medskip
%<*table:KindsSolutionSets>
\begin{center}
\begin{center} % \small
\begin{tabular}{r@{}c}
&\hspace*{2.5em}\begin{tabular}{c}
\textit{number of solutions of the} \\[-.5ex]
......
......@@ -60,8 +60,6 @@ be square and of the same size.
similarity is an equivalence relation.
\begin{example}
%One way to produce similar matrices is to start with an arbitrary nonsingular
%matrix $P$, and an arbitrary $T$.
Calculation with these two,
\begin{equation*}
P=
......@@ -80,8 +78,8 @@ gives that $S$ is similar to this matrix.
\begin{equation*}
T=
\begin{mat}[r]
0 &-1 \\
1 &1
12 &-19 \\
7 &-11
\end{mat}
\end{equation*}
\end{example}
......
......@@ -779,6 +779,17 @@ A \definend{\( t \)-string basis}\index{basis!string}\index{string!basis}
is a basis that is a concatenation of \( t \)-strings.
\end{definition}
\begin{example}
Consider differentiation \( \map{d/dx}{\polyspace_2}{\polyspace_2} \).
The sequence
\( \sequence{x^2, 2x, 2, 0} \)
is a \( d/dx \)-string of length~\( 4 \).
The sequence
\( \sequence{x^2, 2x, 2} \)
is a \( d/dx \)-string of length~\( 3 \)
that is a basis for \( \polyspace_2\).
\end{example}
Note that the strings cannot form a basis under concatenation
if they are not disjoint because a basis cannot have a repeated vector.
......@@ -796,19 +807,21 @@ length equal to the index of nilpotency of $t$.
\end{lemma}
\begin{proof}
Suppose not.
Those strings cannot be longer; if the index is
\( k \) then \( t^k \) sends any vector\Dash including those starting the
string\Dash to \( \zero \).
So suppose instead that there is a transformation $t$ of index~$k$
on some space, such that the space has a $t$-string basis where
all of the strings are shorter than length \( k \).
Assume the space has a basis of $t$-strings and that $t$'s index of
nilpotency is $k$.
We cannot have that the longest string in that basis is longer than
$t$'s index of nilpotency:
\( t^k \) sends any vector, including the vector starting the longest
string, to \( \zero \).
So suppose instead that the space has a $t$-string basis~$B$ where
all of the strings are shorter than length~\( k \).
Because $t$ has index~$k$, there is a vector \( \vec{v} \)
such that \( t^{k-1}(\vec{v})\neq\zero \).
Represent $\vec{v}$ as a linear combination of basis elements
Represent $\vec{v}$ as a linear combination of elements from~$B$
and apply \( t^{k-1} \).
We are supposing that \( t^{k-1} \) sends each basis element to \( \zero \)
but that it does not send \( \vec{v} \) to \( \zero \).
We are supposing that \( t^{k-1} \) maps each element of~$B$ to \( \zero \),
and therefore each summand in the linear combination to $\zero$,
but also that it does not map \( \vec{v} \) to \( \zero \).
That is impossible.
\end{proof}
......@@ -819,9 +832,10 @@ nilpotent map has an associated string basis.
% all zeros except for blocks of subdiagonal ones, is immediate,
%as in \nearbyexample{ThirdNilMap}.
Looking for a counterexample, a nilpotent map without an associated
string basis that is disjoint, will suggest the idea for the proof.
Consider the map \( \map{t}{\C^5}{\C^5} \) with this action.
We first see main idea of the argument by considering an example.
If we want to construct a counterexample, a nilpotent map without an associated
disjoint string basis, we would something like
the map \( \map{t}{\C^5}{\C^5} \) with this action.
\begin{center}
\begin{minipage}{1in}
$\begin{strings}{ccccc}
......@@ -848,20 +862,19 @@ Consider the map \( \map{t}{\C^5}{\C^5} \) with this action.
0 &0 &0 &1 &0
\end{mat}$
\end{center}
These three
strings aren't disjoint.
The first two strings $\vec{e}_1\mapsto\vec{e}_3\mapsto\zero$ and
$\vec{e}_1\mapsto\vec{e}_3\mapsto\zero$ overlap, even after omitting $\zero$.
But that doesn't mean that there is no
$t$-string basis;
it only means that \( \stdbasis_5 \) is not one.
To find a basis we first find
the number and lengths of its strings.
The action on the basis shows that this map is nilpotent but this is not
a disjoint string basis, because the first two of the
three strings aren't disjoint.
However, the fact that this basis isn't disjoint doesn't mean there is no
disjoint string basis.
To produce a such a basis for this map
we will first find the number and lengths of its strings.
Since $t$'s index of nilpotency is two,
\nearbylemma{le:LongestTowerIsIndex} says that
at least one string in the basis has length two.
Thus the map must act on a string basis in one of these two ways.
There are five basis elements so if there is a disjoint string basis then
the map must act on it in one of these two ways.
\begin{equation*}
\begin{strings}{ccccc}
\vec{\beta}_1 &\mapsto &\vec{\beta}_2 &\mapsto &\zero \\
......@@ -882,23 +895,26 @@ null space of dimension three since that's how many basis vectors are
mapped to zero.
A transformation with the right-hand action has a null space of
dimension four.
Using the matrix representation above, calculation of $t$'s null space
Thus, using the matrix representation above, we can determine which of the
two possible shapes is right by calculating $t$'s null space.
\begin{equation*}
\nullspace{t}=
\set{\colvec{x \\ -x \\ z \\ 0 \\ r}\suchthat x,z,r\in\C }
\end{equation*}
shows that it is three-dimensional,
meaning that we want the left-hand action.
It is three-dimensional,
meaning that this~\( t \) has the left-hand action, since in the
right-hand action the number of basis vectors mapped to zero
is~\( 4 \).
To produce a string basis, first
To produce a string basis first
pick \( \vec{\beta}_2 \) and \( \vec{\beta}_4 \) from
\( \rangespace{t}\intersection\nullspace{t} \)
\( \rangespace{t}\intersection\nullspace{t} \).
\begin{equation*}
\vec{\beta}_2=\colvec[r]{0 \\ 0 \\ 1 \\ 0 \\ 0}\qquad
\vec{\beta}_4=\colvec[r]{0 \\ 0 \\ 0 \\ 0 \\ 1}
\end{equation*}
(other choices are possible, just be sure that
\( \set{\vec{\beta}_2,\vec{\beta}_4} \) is linearly independent).
(Other choices are possible, just be sure that the set
\( \set{\vec{\beta}_2,\vec{\beta}_4} \) is linearly independent.)
For \( \vec{\beta}_5 \) pick a vector from \( \nullspace{t} \)
that is not in the span of \( \set{ \vec{\beta}_2,\vec{\beta}_4 } \).
\begin{equation*}
......@@ -911,8 +927,10 @@ Finally, take \( \vec{\beta}_1 \) and \( \vec{\beta}_3 \) such that
\vec{\beta}_1=\colvec[r]{0 \\ 1 \\ 0 \\ 0 \\ 0}\qquad
\vec{\beta}_3=\colvec[r]{0 \\ 0 \\ 0 \\ 1 \\ 0}
\end{equation*}
Now, with respect to \( B=\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_5} \),
the matrix of $t$ is as desired.
Therefore, we have a string basis
\( B=\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_5} \)
and with respect to that basis
the matrix of $t$ has blocks of subdiagonal~$1$'s.
\begin{equation*}
\rep{t}{B,B}=
\begin{pmat}{rr|rr|r}
......@@ -931,8 +949,8 @@ While the basis is not unique, the number
and the length of the strings is determined by \( t \).
\end{theorem}
This illustrates the argument below, which describes three kinds of
basis vectors (we shown them in squares if they are in the
This illustrates the proof, which describes three kinds of
basis vectors (shown in squares if they are in the
null space and in circles if they are not).
\begin{equation*}
\begin{strings}{ccccccccccccccccccc}
......
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......@@ -59,8 +59,8 @@
ItalicFont = sfoti10.pfb,
SmallCapsFont = sfocc10]{sform10.pfb}
\setsansfont{Computer Modern Sans Serif Demi Condensed}
\setmonofont[Path=/usr/local/texlive/texmf-local/fonts/type1/bh/luxi/,
Scale=0.87]{ul9r8a.pfb} % Luxi Mono Regular
% \setmonofont[Path=/usr/local/texlive/texmf-local/fonts/type1/bh/luxi/,
% Scale=0.87]{ul9r8a.pfb} % Luxi Mono Regular
% \setmonofont[Path=/usr/local/texlive/texmf-local/fonts/type1/bh/luxi/,
% Scale=0.87]{ul9r8a.pfb} % Luxi Mono Regular
% \setmonofont[Path=fonts/tmp/,
......
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......@@ -147,7 +147,7 @@ S(vector(QQ, [0, 1]))
\end{sageoutput}
\noindent
Another way to tell that the map is not one-to-one is to look at the
nullspace.
null space.
\begin{sageoutput}[d,0,2]
S_symbolic(a, b) = [a+2*b, a+2*b]
S = linear_transformation(QQ^2, QQ^2, S_symbolic)
......
......@@ -459,7 +459,7 @@ there are \( k \) or fewer summands, that is, whenever
$n=1$, or $n=2$, \ldots, or $n=k$.
Consider the $k+1$-summand case.
Use the first half of~(1)
to breaking the sum along the final `$+$'.
to break the sum along the final `$+$'.
\begin{equation*}
f(c_1\vec{v}_1+\dots+c_k\vec{v}_k+c_{k+1}\vec{v}_{k+1})
=f(c_1\vec{v}_1+\dots+c_k\vec{v}_k)+f(c_{k+1}\vec{v}_{k+1})
......
......@@ -12,7 +12,7 @@ Experience shows that these maps are
tremendously useful.
For one thing we shall see in the second subsection below
that while isomorphisms describe how spaces are the same,
we can think of these maps as describe how spaces are alike.
we can think of these maps as describing how spaces are alike.
......@@ -2059,7 +2059,7 @@ point and the map is an isomorphism between the domain and the range.
it is the image, under $h$, of $0\in\polyspace_3$.
\partsitem The polynomial $7\in\polyspace_3$ is not in the null space
because $h(7)=7x$ is not the zero polynomial in $\polyspace_4$.
The polynomial $x^3\in\polyspace_4$ is not in the range space
The polynomial $7\in\polyspace_4$ is not in the range space
because there is no member of the domain that when multiplied
by $x$ gives the constant polynomial $p(x)=7$.
\partsitem The polynomial $12x-0.5x^3\in\polyspace_3$ is not in the
......@@ -2127,7 +2127,7 @@ point and the map is an isomorphism between the domain and the range.
\end{equation*}
The range space
\begin{equation*}
\rangespace{h}
\rangespace{h}=
\set{a+d\suchthat a,b,c,d\in\Re}
\end{equation*}
is all of $\Re$ (we can get any real number by
......@@ -2182,22 +2182,22 @@ point and the map is an isomorphism between the domain and the range.
\begin{answer}
Because
\begin{equation*}
\frac{d}{dx}\,(a_0+a_1x+\dots+a_nx^n)
=a_1+2a_2x+3a_3x^2+\dots+na_nx^{n-1}
\frac{d}{dx}\,(a_0+a_1x+\cdots+a_nx^n)
=a_1+2a_2x+3a_3x^2+\cdots+na_nx^{n-1}
\end{equation*}
we have this.
\begin{align*}
\nullspace{\frac{d}{dx}}
&=\set{a_0+\dots+a_nx^n\suchthat a_1+2a_2x+\dots+na_nx^{n-1}
=0+0x+\dots+0x^{n-1}} \\
&=\set{a_0+\dots+a_nx^n\suchthat
&=\set{a_0+\cdots+a_nx^n\suchthat a_1+2a_2x+\cdots+na_nx^{n-1}
=0+0x+\cdots+0x^{n-1}} \\
&=\set{a_0+\cdots+a_nx^n\suchthat
\text{$a_1=0$, and $a_2=0$, \ldots, $a_n=0$}} \\
&=\set{a_0+0x+0x^2+\dots+0x^n\suchthat a_0\in\Re}
&=\set{a_0+0x+0x^2+\cdots+0x^n\suchthat a_0\in\Re}
\end{align*}
In the same way,
\begin{equation*}
\nullspace{\frac{d^k}{dx^k}}
=\set{a_0+a_1x+\dots+a_nx^n\suchthat a_0,\dots,a_{k-1}\in\Re}
=\set{a_0+a_1x+\cdots+a_nx^n\suchthat a_0,\ldots,a_{k-1}\in\Re}
\end{equation*}
for \( k\leq n \).
\end{answer}
......
......@@ -980,11 +980,11 @@ for any matrix there is an associated linear map.
The resulting matrix
\begin{equation*}
\renewcommand{\arraystretch}{1.2}
\rep{\int}{B,B}
\rep{\text{slide}_{-1}}{B,B}
=\begin{mat}
1 &1 &1 &1 &\ldots &1 \\
0 &1 &2 &3 &\ldots &\binom{n}{2} \\
0 &0 &1 &3 &\ldots &\binom{n}{3} \\
0 &1 &2 &3 &\ldots &\binom{n}{1} \\
0 &0 &1 &3 &\ldots &\binom{n}{2} \\
&\vdots \\
0 &0 &0 & &\ldots &1
\end{mat}
......@@ -2491,10 +2491,10 @@ And, we shall see how to find the matrix that represents a map's inverse.
Denote the given basis of
$\polyspace_2$
by $B$.
Then application of the linear map is represented by matrix-vector
Application of the linear map is represented by matrix-vector
multiplication.
Thus, the first vector in $\stdbasis_3$ maps to the element
of $\polyspace_2$ represented with respect to $B$ by
Thus the first vector in $\stdbasis_3$ maps to the element
of $\polyspace_2$ represented with respect to~$B$ by
\begin{equation*}
\begin{mat}[r]
1 &3 &0 \\
......@@ -2531,10 +2531,10 @@ And, we shall see how to find the matrix that represents a map's inverse.
We can thus decide if $1+2x$ is in the range of the map by
looking for scalars $c_1$, $c_2$, and $c_3$ such that
\begin{equation*}
c_1\cdot(1)+c_2\cdot(1+x^2)+c_3\cdot(x)=1+2x
c_1\cdot(1+x)+c_2\cdot(4+x^2)+c_3\cdot(x)=1+2x
\end{equation*}
and obviously $c_1=1$, $c_2=0$, and $c_3=1$ suffice.
Thus it is in the range, and in fact it is the image of
Thus $1+2x$ is in the range, since it is the image of
this vector.
\begin{equation*}
1\cdot\colvec[r]{1 \\ 0 \\ 0}+0\cdot\colvec[r]{0 \\ 1 \\ 0}
......@@ -2542,9 +2542,12 @@ And, we shall see how to find the matrix that represents a map's inverse.
\end{equation*}
\textit{Comment.}
A more slick argument is to note that the matrix is nonsingular,
so it has rank~$3$, so the codomain has dimension~$4$,
and thus every polynomial is the image of some vector.
A slicker argument is to note that the matrix is nonsingular,
so it has rank~$3$, so the range has dimension~$3$,
and since the codomain has dimension~$3$ the map is onto.
Thus every polynomial is the image of some vector and in
particular~$1+2x$
is the image of a vector in the domain.
\end{answer}
\item
\nearbyexample{ex:NonSMatHasNonSMap} gives a matrix that is
......
......@@ -557,7 +557,7 @@ no matter what domain and codomain bases we use.
\index{symmetric matrix} if each \( i,j \) entry equals
the \( j,i \) entry, that is, if the matrix equals its transpose.
\begin{exparts}
\partsitem Prove that for any $H$,
\partsitem Prove that for any square~$H$,
the matrix \( H+\trans{H} \) is symmetric.
Does every symmetric matrix have this form?
\partsitem Prove that the set of \( \nbyn{n} \) symmetric matrices is
......@@ -945,8 +945,8 @@ function composition is possible.
\;\stackrel{g}{\longrightarrow}\;
\text{dimension \( m \) space}
\end{equation*}
So the matrix product has an $\nbym{m}{r}$ matrix~$G$
times an $\nbym{r}{n}$ matrix~$F$ to get an
Thus, matrix product combines an $\nbym{m}{r}$ matrix~$G$
with an $\nbym{r}{n}$ matrix~$F$ to yield the
$\nbym{m}{n}$ result~$GF$.
Briefly,
`$\nbym{m}{r}\text{\ times\ }\nbym{r}{n}\text{\ equals\ }\nbym{m}{n}$'.
......@@ -1415,7 +1415,7 @@ will explore this operation more in the next subsection.
with respect to \( B,B \) where \( B \) is the natural basis
\( \sequence{1,x,\ldots,x^n} \).
Show that the product of this matrix with itself is defined;
what the map does it represent?
what map does it represent?
\begin{answer}
The action of $d/dx$ on $B$ is
$1\mapsto0$, $x\mapsto 1$, $x^2\mapsto 2x$, \ldots and so
......@@ -3037,14 +3037,14 @@ clearer idea then we will go with it.
\begin{exparts*}
\partsitem \( \begin{mat}[r]
3 &0 \\
0 &0
0 &1
\end{mat}
\begin{mat}[r]
1 &2 \\
3 &4
\end{mat} \)
\partsitem \( \begin{mat}[r]
4 &0 \\
1 &0 \\
0 &2
\end{mat}
\begin{mat}[r]
......@@ -3079,20 +3079,17 @@ clearer idea then we will go with it.
\begin{answer}
\begin{exparts}
\partsitem The second matrix has its first row multiplied
by \( 3 \) and
its second row multiplied by \( 0 \).
by \( 3 \).
\begin{equation*}
\begin{mat}[r]
3 &6 \\
0 &0
3 &4
\end{mat}
\end{equation*}
\partsitem The second matrix has its first row multiplied
by \( 4 \) and
its second row multiplied by \( 2 \).
\partsitem The second matrix has its second row multiplied by \( 2 \).
\begin{equation*}
\begin{mat}[r]
4 &8 \\
1 &2 \\
6 &8
\end{mat}
\end{equation*}
......@@ -3124,6 +3121,49 @@ clearer idea then we will go with it.
\end{equation*}
\end{exparts}
\end{answer}
\item
Predict the result of each multiplication by a
diagonal matrix, and then check by multiplying it out.
\begin{exparts*}
\partsitem \( \begin{mat}[r]
3 &0 \\
0 &0
\end{mat}
\begin{mat}[r]
1 &2 \\
3 &4
\end{mat} \)
\partsitem \( \begin{mat}[r]
4 &0 \\
0 &2
\end{mat}
\begin{mat}[r]
1 &2 \\
3 &4
\end{mat} \)
\end{exparts*}
\begin{answer}
\begin{exparts}
\partsitem The second matrix has its first row multiplied
by \( 3 \) and
its second row multiplied by \( 0 \).
\begin{equation*}
\begin{mat}[r]
3 &6 \\
0 &0
\end{mat}
\end{equation*}
\partsitem The second matrix has its first row multiplied
by \( 4 \) and
its second row multiplied by \( 2 \).
\begin{equation*}
\begin{mat}[r]
4 &8 \\
6 &8
\end{mat}
\end{equation*}
\end{exparts}
\end{answer}
\recommended \item
This table gives the number of hours of each
type done by each worker, and the associated pay rates.
......@@ -3175,8 +3215,8 @@ clearer idea then we will go with it.
\parbox{2in}{\includegraphics{ch3.99}}
\end{center}
\begin{exparts}
\partsitem The \definend{incidence matrix}\index{incidence matrix}%
\index{matrix!incidence}
\partsitem The \definend{adjacency matrix}\index{adjacency matrix}%
\index{matrix!adjacency}
of a map is the square matrix
whose \( i,j \) entry is the number of roads from city \( i \)
to city \( j \).
......@@ -3185,7 +3225,7 @@ clearer idea then we will go with it.
\partsitem A matrix is
\definend{symmetric}\index{matrix!symmetric}
if it equals its transpose.
Show that an incidence matrix is symmetric.
Show that an adjacency matrix is symmetric.
(These are all two-way streets.
Vermont doesn't have many one-way streets.)
\partsitem What is the significance of the square of the
......@@ -3194,7 +3234,7 @@ clearer idea then we will go with it.
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem The incidence matrix is this
\partsitem The adjacency matrix is this
(e.g, the first row shows that there is only one connection
including Burlington, the road to Winooski).
\begin{equation*}
......@@ -3210,7 +3250,7 @@ clearer idea then we will go with it.
any road connecting city~\( i \)
to city~\( j \) gives a connection between city~\( j \) and
city~\( i \).
\partsitem The square of the incidence matrix tells how cities
\partsitem The square of the adjacency matrix tells how cities
are connected
by trips involving two roads.
\end{exparts}
......@@ -3587,8 +3627,8 @@ clearer idea then we will go with it.
\end{align*}
\end{answer}
\recommended \item
Give an example of two matrices of the same rank with squares of
differing rank.
Give an example of two matrices of the same rank and size
with squares of differing rank.
\begin{answer}
Matrices representing (say, with respect to
\( \stdbasis_2,\stdbasis_2\subset\Re^2 \)) the maps that send
......
......@@ -795,7 +795,7 @@ the same space, and where the map is the identity map.
are all bases.)
\end{answer}
\recommended \item
Let \( H \) be a \( \nbyn{n} \) nonsingular matrix.
Let \( H \) be an \( \nbyn{n} \) nonsingular matrix.
What basis of \( \Re^n \) does \( H \) change to the standard basis?
\begin{answer}
Taking $H$ as a change of basis matrix
......
......@@ -157,6 +157,16 @@ I periodically issue revisions.
My contact information is on the web page.
\medskip
\noindent{\bf Acknowledgements.}
I am grateful to all of the people who have helped with this work,
from class-testers and adopters to people who sent bug reports.
I am also very grateful to Saint Michael's College,
for supporting my work on this text over many years.
\newcommand{\classday}[1]{\textsc{#1}}
\newcommand{\colwidth}{1.25in}
......@@ -168,34 +178,12 @@ My contact information is on the web page.
This book's emphasis on motivation and development,
and its availability, make it widely used for self-study.
If you are an independent student then good for you; I admire your industry.
You may find some advice useful.
However, you may find some advice useful.
While an experienced instructor knows what subjects and
pace suit their class, a suggested semester's timetable
may help you estimate how long sections typically take to cover.
(This is adapted from one graciously shared by George Ashline.)
% \begin{center} \small
% \begin{tabular}{r|*{2}{p{\colwidth}}l}
% \textit{week}
% &\textit{Monday}
% &\textit{Wednesday}
% &\textit{Friday} \\ \hline
% 1 &One.I.1 &One.I.1, 2 &One.I.2, 3 \\
% 2 &One.I.3 &One.II.1 &One.II.2 \\
% 3 &One.III.1, 2 &One.III.2 &Two.I.1 \\
% 4 &Two.I.2 &Two.II &Two.III.1 \\
% 5 &Two.III.1, 2 &Two.III.2 &\classday{exam} \\
% 6 &Two.III.2, 3 &Two.III.3 &Three.I.1 \\
% 7 &Three.I.2 &Three.II.1 &Three.II.2 \\
% 8 &Three.II.2 &Three.II.2 &Three.III.1 \\
% 9 &Three.III.1 &Three.III.2 &Three.IV.1, 2 \\
% 10 &Three.IV.2, 3, 4 &Three.IV.4 &\classday{exam} \\
% 11 &Three.IV.4, Three.V.1 &Three.V.1, 2 &Four.I.1, 2 \\
% 12 &Four.I.3 &Four.II &Four.II \\
% 13 &Four.III.1 &Five.I &Five.II.1 \\
% 14 &Five.II.2 &Five.II.3 &\classday{review}
% \end{tabular}
% \end{center}
This one one was graciously shared by George Ashline.
\begin{center} % George Ashline's
\begin{tabular}{r|*{2}{p{\colwidth}}l}
\textit{week}
......@@ -227,29 +215,6 @@ take-home problem sets that include proofs.
That is,
the computations are important but so are the proofs.
% The second is more ambitious.
% \begin{center} \small
% \begin{tabular}{r|*{2}{p{\colwidth}}l}
% \textit{week}
% &\textit{Monday}
% &\textit{Wednesday}
% &\textit{Friday} \\ \hline
% 1 &One.I.1 &One.I.2 &One.I.3 \\
% 2 &One.I.3 &One.III.1, 2 &One.III.2 \\
% 3 &Two.I.1 &Two.I.2 &Two.II \\
% 4 &Two.III.1 &Two.III.2 &Two.III.3 \\
% 5 &Two.III.4 &Three.I.1 &\classday{exam} \\
% 6 &Three.I.2 &Three.II.1 &Three.II.2 \\
% 7 &Three.III.1 &Three.III.2 &Three.IV.1, 2 \\
% 8 &Three.IV.2 &Three.IV.3 &Three.IV.4 \\
% 9 &Three.V.1 &Three.V.2 &Three.VI.1 \\
% 10 &Three.VI.2 &Four.I.1 &\classday{exam} \\
% 11 &Four.I.2 &Four.I.3 &Four.I.4 \\
% 12 &Four.II &Four.II, Four.III.1 &Four.III.2, 3 \\
% 13 &Five.II.1, 2 &Five.II.3 &Five.III.1 \\
% 14 &Five.III.2 &Five.IV.1, 2 &Five.IV.2
% \end{tabular}
% \end{center}
In the table of contents
I have marked subsections as optional if
some instructors will pass over them in favor of spending more time elsewhere.
......@@ -273,7 +238,7 @@ Try to find a knowledgeable person to work with you on these.
\bigskip
Finally, a caution for all students, independent or not:~I
cannot overemphasize that the
statement, ``I understand the material but it's only
statement, ``I understand the material but it is only
that I have trouble with the problems''\spacefactor=1000\ %
shows a misconception.
Being able to do things with the ideas is their entire point.
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......@@ -1212,7 +1212,7 @@ tells us that a linearly independent set is maximal when it spans the space.
\partsitem \( \set{0,x,x^2} \)
\end{exparts*}
\begin{answer}
In each case, if the set is independent then you must prove that,
In each case, if the set is independent then you must prove that
and if it is
dependent then you must exhibit a dependence.
\begin{exparts}
......@@ -1224,7 +1224,8 @@ tells us that a linearly independent set is maximal when it spans the space.
Consider the relationship
$c_1\cdot 1+c_2\cdot\sin(x)+c_3\cdot\sin(2x)=0$
(that `$0$' is the zero function).
Taking $x=0$, $x=\pi/2$ and $x=\pi/4$ gives this system.
Taking three suitable points such as $x=\pi$, $x=\pi/2$, $x=\pi/4$
gives a system
\begin{equation*}
\begin{linsys}{3}
c_1 & & & & &= &0 \\
......
......@@ -406,7 +406,7 @@ We will see that in the next subsection.
\end{exparts*}
\begin{answer}
By \nearbytheorem{th:BasisIffUniqueRepWRT}, each is a basis if and only
if we can express each vector in the space can in a unique way as a linear
if we can express each vector in the space in a unique way as a linear
combination of the given vectors.
\begin{exparts}
\partsitem Yes this is a basis.
......@@ -548,7 +548,7 @@ We will see that in the next subsection.
\begin{equation*}
\rep{x^2+x^3}{D}=\colvec[r]{0 \\ -1 \\ 0 \\ 1}_D
\end{equation*}
\partsitem \( \rep[r]{\colvec{0 \\ -1 \\ 0 \\ 1}}{\stdbasis_4}
\partsitem \( \rep{\colvec[r]{0 \\ -1 \\ 0 \\ 1}}{\stdbasis_4}
=\colvec[r]{0 \\ -1 \\ 0 \\ 1}_{\stdbasis_4} \)
\end{exparts}
\end{answer}
......