Commit b23a639f authored by Jim Hefferon's avatar Jim Hefferon

more final touches

parent 630868a1
This diff is collapsed.
......@@ -75,5 +75,5 @@ bib%
%\typeout{ BIBLIOGRAPHY COMMENTED OUT}
{\pagestyle{empty}\include{bib}}
%\typeout{ INDEX COMMENTED OUT}
\printindex
{\printindex}
\end{document}
\chapter{Chapter One: Linear Systems}
\section{Solving Linear Systems}
\subsection{Gauss' Method}
\subsection{One.I.1: Linear Systems}
\begin{ans}{One.I.1.17}
We can perform Gauss' method can in different ways, so these simply
We can perform Gauss' method in different ways so these
exhibit one possible way to get the answer.
\begin{exparts}
\partsitem Gauss' method
......@@ -708,6 +710,7 @@
then \( x=-1 \), \( y=2 \).
\end{ans}
\subsection{Describing the Solution Set}
\subsection{One.I.2: Linear Systems}
\begin{ans}{One.I.2.15}
\begin{exparts*}
......@@ -1356,6 +1359,7 @@
Ag and \( 501.2\mbox{ cm}^3 \) of Al.
\end{ans}
\subsection{General = Particular + Homogeneous}
\subsection{One.I.3: Linear Systems}
\begin{ans}{One.I.3.15}
For the arithmetic to these, see the answers from the prior
......@@ -1770,6 +1774,8 @@
all-rational solution.
\end{ans}
\section{Linear Geometry}
\subsection{Vectors in Space}
\subsection{One.II.1: Linear Systems}
\begin{ans}{One.II.1.1}
\begin{exparts*}
......@@ -2047,6 +2053,7 @@
that definition.
\end{ans}
\subsection{Length and Angle Measures}
\subsection{One.II.2: Linear Systems}
\begin{ans}{One.II.2.10}
\begin{exparts*}
......@@ -2744,6 +2751,8 @@
to derive the left-hand side.
\end{ans}
\section{Reduced Echelon Form}
\subsection{Gauss-Jordan Reduction}
\subsection{One.III.1: Linear Systems}
\begin{ans}{One.III.1.8}
These answers show only the Gauss-Jordan reduction.
......@@ -3287,6 +3296,7 @@
\end{exparts}
\end{ans}
\subsection{The Linear Combination Lemma}
\subsection{One.III.2: Linear Systems}
\begin{ans}{One.III.2.9}
Bring each to reduced echelon form and compare.
......@@ -4214,6 +4224,8 @@
\end{ans}
\chapter{Chapter Two: Vector Spaces}
\section{Definition of Vector Space}
\subsection{Definition and Examples}
\subsection{Two.I.1: Vector Spaces}
\begin{ans}{Two.I.1.17}
\begin{exparts}
......@@ -4721,6 +4733,7 @@
\end{exparts}
\end{ans}
\subsection{Subspaces and Spanning Sets}
\subsection{Two.I.2: Vector Spaces}
\begin{ans}{Two.I.2.20}
By \nearbylemma{th:SubspIffClosed}, to see if each
......@@ -5575,6 +5588,8 @@
\end{equation*}
\end{ans}
\section{Linear Independence}
\subsection{Definition and Examples}
\subsection{Two.II.1: Vector Spaces}
\begin{ans}{Two.II.1.19}
For each of these, when the subset is independent you must prove it, and
......@@ -6490,6 +6505,8 @@
scalar multiplication operations from \( V \).
\end{ans}
\section{Basis and Dimension}
\subsection{Basis}
\subsection{Two.III.1: Vector Spaces}
\begin{ans}{Two.III.1.16}
By \nearbytheorem{th:BasisIffUniqueRepWRT}, each is a basis if and only
......@@ -7193,6 +7210,7 @@
Checking the span is similar.
\end{ans}
\subsection{Dimension}
\subsection{Two.III.2: Vector Spaces}
\begin{ans}{Two.III.2.15}
One basis is \( \sequence{1,x,x^2} \), and so
......@@ -7663,6 +7681,7 @@
so $V=\Re^n$ and $\dim V=n$.
\end{ans}
\subsection{Vector Spaces and Linear Systems}
\subsection{Two.III.3: Vector Spaces}
\begin{ans}{Two.III.3.16}
\begin{exparts*}
......@@ -8445,6 +8464,7 @@
happening.)
\end{ans}
\subsection{Combining Subspaces}
\subsection{Two.III.4: Vector Spaces}
\begin{ans}{Two.III.4.20}
With each of these we can apply \nearbylemma{le:DirectSumTwoSp}.
......@@ -9772,6 +9792,8 @@
\end{ans}
\chapter{Chapter Three: Maps Between Spaces}
\section{Isomorphisms}
\subsection{Def{}inition and Examples}
\subsection{Three.I.1: Maps Between Spaces}
\begin{ans}{Three.I.1.11}
\begin{exparts}
......@@ -10791,6 +10813,7 @@
\end{exparts}
\end{ans}
\subsection{Dimension Characterizes Isomorphism}
\subsection{Three.I.2: Maps Between Spaces}
\begin{ans}{Three.I.2.9}
Each pair of spaces is isomorphic if and only if the two have the
......@@ -11009,6 +11032,8 @@
representations.
\end{ans}
\section{Homomorphisms}
\subsection{Definition}
\subsection{Three.II.1: Maps Between Spaces}
\begin{ans}{Three.II.1.17}
\begin{exparts}
......@@ -11703,6 +11728,7 @@
\end{exparts}
\end{ans}
\subsection{Rangespace and Nullspace}
\subsection{Three.II.2: Maps Between Spaces}
\begin{ans}{Three.II.2.21}
First, to answer whether a polynomial is in the nullspace,
......@@ -12277,6 +12303,8 @@
\( \rangespace{t^{j+2}}=\rangespace{t^{j+1}} \).
\end{ans}
\section{Computing Linear Maps}
\subsection{Representing Linear Maps with Matrices}
\subsection{Three.III.1: Maps Between Spaces}
\begin{ans}{Three.III.1.12}
\begin{exparts*}
......@@ -13277,6 +13305,7 @@
\end{exparts}
\end{ans}
\subsection{Any Matrix Represents a Linear Map}
\subsection{Three.III.2: Maps Between Spaces}
\begin{ans}{Three.III.2.11}
\begin{exparts}
......@@ -13752,6 +13781,8 @@
See the following section.
\end{ans}
\section{Matrix Operations}
\subsection{Sums and Scalar Products}
\subsection{Three.IV.1: Maps Between Spaces}
\begin{ans}{Three.IV.1.8}
\begin{exparts*}
......@@ -14013,6 +14044,7 @@
\end{exparts}
\end{ans}
\subsection{Matrix Multiplication}
\subsection{Three.IV.2: Maps Between Spaces}
\begin{ans}{Three.IV.2.14}
\begin{exparts*}
......@@ -14711,6 +14743,7 @@
\end{exparts}
\end{ans}
\subsection{Mechanics of Matrix Multiplication}
\subsection{Three.IV.3: Maps Between Spaces}
\begin{ans}{Three.IV.3.24}
\begin{exparts}
......@@ -15283,6 +15316,7 @@
\dim(\rangespace{A}\intersection\nullspace{A}) \).
\end{ans}
\subsection{Inverses}
\subsection{Three.IV.4: Maps Between Spaces}
\begin{ans}{Three.IV.4.12}
Here is one way to proceed.
......@@ -16050,6 +16084,8 @@
(\( A \) is singular if \( k=0 \)).
\end{ans}
\section{Change of Basis}
\subsection{Changing Representations of Vectors}
\subsection{Three.V.1: Maps Between Spaces}
\begin{ans}{Three.V.1.6}
For the matrix to change bases from $D$ to $\stdbasis_2$ we need that
......@@ -16576,6 +16612,7 @@
\end{exparts}
\end{ans}
\subsection{Changing Map Representations}
\subsection{Three.V.2: Maps Between Spaces}
\begin{ans}{Three.V.2.10}
\begin{exparts}
......@@ -17115,6 +17152,8 @@
\end{exparts}
\end{ans}
\section{Projection}
\subsection{Orthogonal Projection Into a Line}
\subsection{Three.VI.1: Maps Between Spaces}
\begin{ans}{Three.VI.1.7}
Each is a straightforward application of the formula from
......@@ -17500,6 +17539,7 @@
\end{center}
\end{ans}
\subsection{Gram-Schmidt Orthogonalization}
\subsection{Three.VI.2: Maps Between Spaces}
\begin{ans}{Three.VI.2.10}
\begin{exparts}
......@@ -18271,6 +18311,7 @@
the vectors $\vec{\kappa}_j$ and $\vec{\kappa}_m$ are orthogonal).
\end{ans}
\subsection{Projection Into a Subspace}
\subsection{Three.VI.3: Maps Between Spaces}
\begin{ans}{Three.VI.3.10}
\begin{exparts}
......@@ -21438,6 +21479,8 @@ octave:6> gplot z
\end{ans}
\chapter{Chapter Four: Determinants}
\section{Def{}inition}
\subsection{Exploration}
\subsection{Four.I.1: Determinants}
\begin{ans}{Four.I.1.1}
\begin{exparts*}
......@@ -21820,6 +21863,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\subsection{Properties of Determinants}
\subsection{Four.I.2: Determinants}
\begin{ans}{Four.I.2.8}
\begin{exparts}
......@@ -22236,6 +22280,7 @@ octave:6> gplot z
degree.
\end{ans}
\subsection{The Permutation Expansion}
\subsection{Four.I.3: Determinants}
\begin{ans}{Four.I.3.15}
\begin{exparts}
......@@ -22785,6 +22830,7 @@ octave:6> gplot z
\end{equation*}
\end{ans}
\subsection{Determinants Exist}
\subsection{Four.I.4: Determinants}
\begin{ans}{Four.I.4.9}
This is the permutation expansion of the determinant of a
......@@ -22986,6 +23032,8 @@ octave:6> gplot z
\end{exparts}
\end{ans}
\section{Geometry of Determinants}
\subsection{Determinants as Size Functions}
\subsection{Four.II.1: Determinants}
\begin{ans}{Four.II.1.8}
For each, find the determinant and take the absolute value.
......@@ -23466,6 +23514,8 @@ octave:6> gplot z
\end{exparts}
\end{ans}
\section{Laplace's Expansion}
\subsection{Laplace's Expansion Formula}
\subsection{Four.III.1: Determinants}
\begin{ans}{Four.III.1.13}
\begin{exparts*}
......@@ -24543,6 +24593,11 @@ ans = 0.017398
\end{ans}
\chapter{Chapter Five: Similarity}
\section{Complex Vector Spaces}
\subsection{Factoring and Complex Numbers; A Review}
\subsection{Complex Representations}
\section{Similarity}
\subsection{Definition and Examples}
\subsection{Five.II.1: Similarity}
\begin{ans}{Five.II.1.4}
One way to proceed is left to right.
......@@ -25125,6 +25180,7 @@ ans = 0.017398
the same \( P \).)
\end{ans}
\subsection{Diagonalizability}
\subsection{Five.II.2: Similarity}
\begin{ans}{Five.II.2.6}
Because we chose the basis vectors arbitrarily, many different answers
......@@ -25859,6 +25915,7 @@ ans = 0.017398
\end{exparts}
\end{ans}
\subsection{Eigenvalues and Eigenvectors}
\subsection{Five.II.3: Similarity}
\begin{ans}{Five.II.3.22}
\begin{exparts}
......@@ -26629,6 +26686,8 @@ ans = 0.017398
Thus \( c \) is a characteristic root of \( A \).
\end{ans}
\section{Nilpotence}
\subsection{Self-Composition}
\subsection{Five.III.1: Similarity}
\begin{ans}{Five.III.1.9}
For the zero transformation,
......@@ -26790,6 +26849,7 @@ ans = 0.017398
or any nonsingular map.
\end{ans}
\subsection{Strings}
\subsection{Five.III.2: Similarity}
\begin{ans}{Five.III.2.18}
Three.
......@@ -27471,6 +27531,8 @@ ans = 0.017398
is invertible.
\end{ans}
\section{Jordan Form}
\subsection{Polynomials of Maps and Matrices}
\subsection{Five.IV.1: Similarity}
\begin{ans}{Five.IV.1.13}
For each,
......@@ -28379,6 +28441,7 @@ ans = 0.017398
\end{equation*}
\end{ans}
\subsection{Jordan Canonical Form}
\subsection{Five.IV.2: Similarity}
\begin{ans}{Five.IV.2.18}
We must check that
......@@ -380,7 +380,7 @@
\usepackage[titles]{tocloft}
\tocloftpagestyle{booktoc}
\cftpagenumbersoff{chapter}
\setlength{\cftchapnumwidth}{8em}
\setlength{\cftchapnumwidth}{7em}
% \renewcommand*\l@chapter[2]{%
% \ifnum \c@tocdepth >\m@ne
% \addpenalty{-\@highpenalty}%
......@@ -646,6 +646,9 @@
% The declarations to set the style of the heading (e.g. \itshape for
% an italic heading.) The last command in <style> may be a
% command which takes a single argument but all the others must be declarations.
% \def\SetSectioningName#1{\gdef\sectioningname{#1}%
% \typeout{SUBSECTION\space START:\space\sectioningname}}
\renewcommand{\section}{\@startsection%
{section}{1}{0em}{-14ex plus1ex minus2ex}{1em}%
{\usefont{T1}{fvs}{b}{n}\Large}}
......@@ -679,6 +682,42 @@
% \subsecheader{#2}}
% \newcommand{\subseccmdb}[1]{\subsecheader{#1}}
\def\@sect#1#2#3#4#5#6[#7]#8{%
\ifnum #2>\c@secnumdepth
\let\@svsec\@empty
\else
\refstepcounter{#1}%
\protected@edef\@svsec{\@seccntformat{#1}\relax}%
\fi
\@tempskipa #5\relax
\ifdim \@tempskipa>\z@
\begingroup
#6{%
\@hangfrom{\hskip #3\relax\@svsec}%
\interlinepenalty \@M #8\@@par}%
\endgroup
\csname #1mark\endcsname{#7}%
% JH 2012-Mar-10
\Writetofile{\ans@fn}{\expandafter\protect\csname #1\endcsname\protect{#8\protect}}
\addcontentsline{toc}{#1}{%
\ifnum #2>\c@secnumdepth \else
\protect\numberline{\csname the#1\endcsname}%
\fi
#7}%
\else
\def\@svsechd{%
#6{\hskip #3\relax
\@svsec #8}%
\csname #1mark\endcsname{#7}%
\addcontentsline{toc}{#1}{%
\ifnum #2>\c@secnumdepth \else
\protect\numberline{\csname the#1\endcsname}%
\fi
#7}}%
\fi
\@xsect{#5}}
\usepackage{trimspaces}
% \newcommand{\subsecheader}[1]{\pagebreak[3]%
......
......@@ -2299,19 +2299,19 @@ beginfig(46) % f_1(x)=e^x
u:=.2in; w:=u; v:=.175in;
save codomain_shift; pair codomain_shift; codomain_shift=(3w,0v);
z0=(0w,0v); z1=(0w,7.5v); z2=(0w,-5.5v);
z0=(0w,0v); z1=(0w,7.5v); z2=(0w,-2.5v);
z10=z0+codomain_shift; z11=z1+codomain_shift; z12=z2+codomain_shift;
% the axes
pickup pensquare scaled line_width_light;
draw z2--z1; % input axis
sidetoside_ticks(13,(0w,-5v),(0w,1v));
label.lft(btex {\tiny -5} etex,(0w,-5v));
sidetoside_ticks(10,(0w,-2v),(0w,1v));
% label.lft(btex {\tiny -5} etex,(0w,-5v));
label.lft(btex {\tiny 0} etex,(0w,0v));
label.lft(btex {\tiny 5} etex,(0w,5v));
draw z12--z11; % output axis
sidetoside_ticks(13,codomain_shift+(ticklength,-5v),(0w,1v));
label.rt(btex {\tiny -5} etex,(0w,-5v)+codomain_shift);
sidetoside_ticks(10,codomain_shift+(ticklength,-2v),(0w,1v));
% label.rt(btex {\tiny -5} etex,(0w,-5v)+codomain_shift);
label.rt(btex {\tiny 0} etex,(0w,0v)+codomain_shift);
label.rt(btex {\tiny 5} etex,(0w,5v)+codomain_shift);
......@@ -2333,19 +2333,19 @@ beginfig(47) % f_2(x)=x^2
u:=.2in; w:=u; v:=.175in;
save codomain_shift; pair codomain_shift; codomain_shift=(3w,0v);
z0=(0w,0v); z1=(0w,7.5v); z2=(0w,-5.5v);
z0=(0w,0v); z1=(0w,7.5v); z2=(0w,-2.5v);
z10=z0+codomain_shift; z11=z1+codomain_shift; z12=z2+codomain_shift;
% the axes
pickup pensquare scaled line_width_light;
draw z2--z1; % input axis
sidetoside_ticks(13,(0w,-5v),(0w,1v));
label.lft(btex {\tiny -5} etex,(0w,-5v));
sidetoside_ticks(10,(0w,-2v),(0w,1v));
% label.lft(btex {\tiny -5} etex,(0w,-5v));
label.lft(btex {\tiny 0} etex,(0w,0v));
label.lft(btex {\tiny 5} etex,(0w,5v));
draw z12--z11; % output axis
sidetoside_ticks(13,codomain_shift+(ticklength,-5v),(0w,1v));
label.rt(btex {\tiny -5} etex,(0w,-5v)+codomain_shift);
sidetoside_ticks(10,codomain_shift+(ticklength,-2v),(0w,1v));
% label.rt(btex {\tiny -5} etex,(0w,-5v)+codomain_shift);
label.rt(btex {\tiny 0} etex,(0w,0v)+codomain_shift);
label.rt(btex {\tiny 5} etex,(0w,5v)+codomain_shift);
......@@ -2893,7 +2893,7 @@ beginfig(60) % composition; chain rule
drawinterval((0w,.8v)--(0w,1.2v));
label.lft(btex {\tiny $x$} etex,(0w,1v));
drawinterval(((0w,1.6v)--(0w,2.4v)) shifted codomain_shift);
label.rt(btex {\tiny $f(x)$} etex,((0w,2v)+codomain_shift));
label.rt(btex {\tiny \ $f(x)$} etex,((0w,2v)+codomain_shift));
drawinterval(((0w,3.4v)--(0w,4.6v)) shifted 2codomain_shift);
label.rt(btex {\tiny $g(f(x))$} etex,((0w,4v)+2codomain_shift));
......
......@@ -232,7 +232,7 @@ beginfig(6) % Tower of Hanoi (three disks at start)
save needle_shift; pair needle_shift; needle_shift = (1.2disk_length,0v);
save needle_length, needle_width, needle_hgt;
numeric needle_length, needle_width, needle_hgt; % how thick is body of bee?
needle_length = .4u; needle_width = .4needle_length; needle_hgt = 6u;
needle_length = .4u; needle_width = .4needle_length; needle_hgt = 4u;
% left needle
drawdisk(disk_length,disk_width,disk_hgt);
drawdisk(.9disk_length,.9disk_width,disk_hgt) shifted (0,1.5disk_hgt);
......@@ -263,7 +263,7 @@ beginfig(7) % Tower of Hanoi (three disks at start)
save needle_shift; pair needle_shift; needle_shift = (1.2disk_length,0v);
save needle_length, needle_width, needle_hgt;
numeric needle_length, needle_width, needle_hgt; % how thick is body of bee?
needle_length = .4u; needle_width = .4needle_length; needle_hgt = 6u;
needle_length = .4u; needle_width = .4needle_length; needle_hgt = 4u;
% left needle
drawdisk(disk_length,disk_width,disk_hgt);
drawdisk(needle_length,needle_width,needle_hgt-disk_hgt)
......
......@@ -127,8 +127,9 @@ Cramer's Rule allows us to solve
simple two equations/two unknowns systems by eye
(they must be simple in that we can mentally compute with the numbers
in the system).
With practice a person can also do simple equations/three unknowns systems.
But computing large determinants takes a long time, so solving
With practice a person can also do simple three equations/three unknowns
systems.
But computing large determinants takes a long time so solving
large systems by Cramer's Rule is not practical.
\begin{exercises}
......
......@@ -2454,15 +2454,14 @@ for determinants is
t_{1,\phi_1(1)}t_{2,\phi_1(2)}\cdots
t_{n,\phi_1(n)}\deter{P_{\phi_1}} \\[.5ex]
\>\hbox{}+t_{1,\phi_2(1)}t_{2,\phi_2(2)}\cdots
t_{n,\phi_2(n)}\deter{P_{\phi_2}} \\[.5ex]
\>\alignedvdots \\
t_{n,\phi_2(n)}\deter{P_{\phi_2}} \\
\>\vdotswithin{+} \\
\>\hbox{}+t_{1,\phi_k(1)}t_{2,\phi_k(2)}\cdots
t_{n,\phi_k(n)}\deter{P_{\phi_k}}
\end{array}
\end{equation*}
where \( \phi_1,\ldots,\phi_k \) are all of the \( n \)-permutations.
\end{definition}
This formula is often written in
\definend{summation
notation}\index{summation notation!for permutation expansion}
......@@ -2472,7 +2471,7 @@ notation}\index{summation notation!for permutation expansion}
t_{1,\phi(1)}t_{2,\phi(2)}\cdots t_{n,\phi(n)}
\deter{P_{\phi}}
\end{equation*}
read aloud as
read aloud as,
``the sum, over all permutations \( \phi \), of terms having the form
\( t_{1,\phi(1)}t_{2,\phi(2)}\cdots t_{n,\phi(n)} \deter{P_{\phi}} \).''
......
......@@ -712,9 +712,9 @@ $\det(AB)$ equals $\det(A)\cdot \det(B)$.
using \nearbytheorem{th:MatChVolByDetMat}.
(The observation that increasing the linear size of a three-dimensional
object by a factor of $c$ will increase its volume by a factor of
$c^3$ (while only increasing its surface area by an amount proportional
$c^3$ while only increasing its surface area by an amount proportional
to a factor of
$c^2$) is the \definend{Square-cube law}~\cite{Wikipedia}.)
$c^2$ is the \definend{Square-cube law}~\cite{Wikipedia}.)
\begin{answer}
Where the sides of the box are \( c \) times longer, the box
has \( c^3 \) times as many cubic units of volume.
......
......@@ -79,14 +79,15 @@ array \lstinline[style=inline]!A!.
The program's outer loop runs through
each \lstinline[style=inline]!ROW! between \lstinline[style=inline]!1! and
\lstinline[style=inline]!N-1!
and does the entry-by-entry combination with the lower rows:
and does the entry-by-entry combination
% \begin{equation*}
\( -\text{\lstinline[style=inline]!PIVINV!}\cdot \rho_{\,\text{\lstinline!ROW!}}
+\rho_{\,\text{\lstinline!I!}} \)
% \end{equation*}
(there are no rows below
the \lstinline[style=inline]!N!-th
so it is not included in the loop).
with the lower rows.
% (there are no rows below
% the \lstinline[style=inline]!N!-th
% so it is not included in the loop).
\begin{lstlisting}
DO 10 ROW=1, N-1
PIVINV=1.0/A(ROW,ROW)
......
......@@ -689,7 +689,7 @@ a no response by showing that no solution exists.}
\end{linsys}$
\end{exparts*}
\begin{answer}
We can perform Gauss' method can in different ways, so these simply
We can perform Gauss' method in different ways so these
exhibit one possible way to get the answer.
\begin{exparts}
\partsitem Gauss' method
......
......@@ -1064,7 +1064,7 @@ The matrix
has an index of nilpotency of two, as this calculation shows.
\begin{center}
\begin{tabular}{c|cc}
\multicolumn{1}{c}{\( p \)} &\( M^p \) &\( \nullspace{M^p} \) \\
\multicolumn{1}{c}{\textit{power} \( p \)} &\( M^p \) &\( \nullspace{M^p} \) \\
\hline
\( 1 \)
&\(
......@@ -1178,7 +1178,7 @@ is nilpotent.
% }}}
\begin{center}
\begin{tabular}{c|cc}
\multicolumn{1}{c}{\( p \)} &\( N^p \) &\( \nullspace{N^p} \) \\
\multicolumn{1}{c}{\textit{power} \( p \)} &\( N^p \) &\( \nullspace{N^p} \) \\
\hline
\( 1 \)
&\matrixvenlarge{
......
......@@ -2417,7 +2417,7 @@ with respect to the standard basis.)
\begin{center}
\renewcommand{\arraystretch}{1.25}
\begin{tabular}{r|ccc}
\multicolumn{1}{c}{\textit{power} \( p \)}
\multicolumn{1}{c}{\( p \)}
&\( (T-2I)^p \) &\( \nullspace{(t-2)^p} \)
&\textit{nullity} \\
\hline
......@@ -2520,28 +2520,25 @@ While the characteristic polynomial is the same,
\begin{center}
% \renewcommand{\arraystretch}{1.25}
\begin{tabular}{r|ccc}
\multicolumn{1}{c}{\textit{power} \( p \)}
&\( (T-2I)^p \) &\( \nullspace{(t-2)^p} \)
&\textit{nullity} \\
\hline
\( 1 \)
&\matrixvenlarge{\begin{mat}[r]
0 &2 &1 \\
0 &4 &2 \\
0 &0 &0
\end{mat}}
&\( \set{\matrixvenlarge{\colvec{x \\ -z/2 \\ z}}
\suchthat x,z\in\C} \)
&$2$ \\
\( 2 \)
&\matrixvenlarge{\begin{mat}[r]
0 &8 &4 \\
0 &16 &8 \\
0 &0 &0
\end{mat}}
&\textit{--same--}
&\textit{--same--}
\end{tabular}
\multicolumn{1}{c}{\( p \)} &\( (T-6I)^p \) &\( \nullspace{(t-6)^p} \)
&\textit{nullity} \\ \hline
\( 1 \)
&\matrixvenlarge{\begin{mat}[r]
-4 &3 &1 \\
0 &0 &2 \\
0 &0 &-4
\end{mat}}
&\( \set{\matrixvenlarge{\colvec{x \\ (4/3)x \\ 0}}\suchthat x\in\C} \)
&$1$ \\
\( 2 \)
&\matrixvenlarge{\begin{mat}[r]
16 &-12&-2 \\
0 &0 &-8 \\
0 &0 &16
\end{mat}}
&\textit{--same--}
&\textit{---}
\end{tabular}
\end{center}
here the action of $t-2$ is stable after only one application\Dash the
restriction
......@@ -2577,27 +2574,7 @@ and the associated Jordan block $J_6$ is the $\nbyn{1}$ matrix with entry $6$.
%\begin{center}
% % \renewcommand{\arraystretch}{1.25}
% \begin{tabular}{c|cc}
% \multicolumn{1}{c}{\textit{power} \( p \)} &\( (T-6I)^p \) &\( \nullspace{(t-6)^p} \)
% &\textit{nullity} \\ \hline
% \( 1 \)
% &\matrixvenlarge{\begin{mat}[r]
% -4 &3 &1 \\
% 0 &0 &2 \\
% 0 &0 &-4
% \end{mat}}
%%\( \set{\matrixvenlarge{\colvec{x \\ (4/3)x \\ 0}}\suchthat x\in\C} \)
% &$1$ \\
% \( 2 \)
% &\matrixvenlarge{\begin{mat}[r]
% 16 &-12&-2 \\
% 0 &0 &-8 \\
% 0 &0 &16
% \end{mat}}
% &\textit{--same--}
% &\textit{---}
% \end{tabular}
%\end{center}
Therefore, \( T \) is diagonalizable.
% \multicolumn{1}{c}{Therefore, \( T \) is diagonalizable.
\begin{equation*}
\rep{t}{B,B}=
\begin{mat}[r]
......@@ -2611,8 +2588,8 @@ Therefore, \( T \) is diagonalizable.
\colvec[r]{0 \\ 1 \\ -2},
\colvec[r]{2 \\ 4 \\ 0}}
\end{equation*}
(Checking that the third vector in $B$ is in the nullspace of $t-6$ is
routine.)
Checking that the third vector in $B$ is in the nullspace of $t-6$ is
routine.
\end{example}
\begin{example} \label{ThirdJordanForm}
......
No preview for this file type
......@@ -13,13 +13,11 @@ the founder of Chinese civilization,
interpreted this to mean that
the river was still annoyed.
Fortunately, a child noticed
that on its shell the turtle had what is today
called the pattern of Lo Shu (``river scroll'').
\begin{center}
\includegraphics[height=1in]{LoShu.png}
\end{center}
The dots make a matrix.
that on its shell the turtle had the pattern on the right below, which is today
called Lo Shu (``river scroll'').
\begin{center}
\vcenteredhbox{\includegraphics[height=.8in]{LoShu.png}} % http://en.wikipedia.org/wiki/File:Lo_Shu_3x3_magic_square.svg
\hspace{1in}
\begin{tabular}{|c|c|c|}
\hline
$4$ &$9$ &$2$ \\ \hline
......@@ -27,12 +25,11 @@ The dots make a matrix.
$8$ &$1$ &$6$ \\ \hline
\end{tabular}
\end{center}
Each row, column,
and diagonal adds to $15$
(the number of days in each of the twenty four cycles of the
Chinese solar year).
The dots make the matrix on the right where
each row, column,
and diagonal adds to $15$.
Now that the people knew how much to sacrifice,
at last the river's anger cooled.
the river's anger cooled.
% (http://en.wikipedia.org/wiki/Lo_Shu_Square)
A square matrix is
......@@ -48,15 +45,15 @@ Another example of a magic square appears in the engraving
\end{center}
One interpretation is that it depicts melancholy, a depressed state.
The figure, genius,
has a wealth of fascinating things to explore.
These include
has a wealth of fascinating things to explore including
the compass, the geometrical solid, the scale, and the hourglass.
But the figure is unmoved; all of these things lie unused.
One of the potential delights is the $\nbyn{4}$ matrix in the upper right.
The rows, columns, and diagonals add to $34$.
One of the potential delights, in the upper right,
is a $\nbyn{4}$ matrix whose
rows, columns, and diagonals add to~$34$.
\begin{center}
\vcenteredhbox{\includegraphics[height=1.25in]{Melencoliadetail.jpg}} % wikipedia http://upload.wikimedia.org/wikipedia/commons/7/7e/Albrecht_D%C3%BCrer_-_Melencolia_I_%28detail%29.jpg
\qquad
\hspace{1in}
\begin{tabular}{|c|c|c|c|}
\hline
$16$ &$3$ &$2$ &$13$ \\ \hline
......@@ -80,16 +77,16 @@ see \cite{OnlineEncyclopedia}.
If we don't require that the squares be normal then we can say much more.
Every $\nbyn{1}$ square is magic, trivially.
If the rows, columns, and diagonals of a $\nbyn{2}$ matrix add to
the magic number~$s$
If the rows, columns, and diagonals of a $\nbyn{2}$ matrix
\begin{equation*}
\begin{mat}
a &b \\
c &d
\end{mat}
\end{equation*}
add to~$s$
then $a+b=s$, $c+d=s$, $a+c=s$, $b+d=s$, $a+d=s$, and $b+c=s$.
\nearbyexercise{exer:TwoByTwoMagicSqUnique} shows that
\nearbyexercise{exer:TwoByTwoMagicSqUnique} shows that this
system has the unique solution $a=b=c=d=s/2$.
So the set of $\nbyn{2}$ magic squares
is a one-dimensional subspace of $\matspace_{\nbyn{2}}$.
......@@ -102,10 +99,10 @@ This Topic shows that for $n\geq 3$ the
dimension of
$\magicsquares_n$ is $n^2-n$.
The set $\magicsquares_{n,0}$ of $\nbyn{n}$~magic squares with magic number~$0$
is another subspace, and we will find the formula for its dimension also,
is another subspace, and we will find the formula for its dimension also:
$n^2-2n-1$ when $n\geq 3$.
We will first show that $\dim\magicsquares_n=\dim\magicsquares_{n,0}+1$.
We will first prove that $\dim\magicsquares_n=\dim\magicsquares_{n,0}+1$.
Define the
\definend{trace}\index{trace}\index{matrix!trace} of a matrix to be
the sum down its upper-left to lower-right diagonal
......@@ -153,8 +150,8 @@ add to zero then we have an $\nbym{(2n+2)}{n^2}$ linear system.
\end{linsys}
\end{equation*}
The matrix of coefficients for the particular cases of $n=3$ and~$n=4$
are below (with the rows and columns numbered to help in reading
the proof).
are below, with the rows and columns numbered to help in reading
the proof.
With respect to the standard basis, each represents a linear map
$\map{h}{\Re^{n^2}}{\Re^{2n+2}}$.
The domain has dimension~$n^2$ so if we show that the
......@@ -221,7 +218,7 @@ Now it gets messy.
In the final two rows, in the first~$n$ columns, is a subrow that