Commit 9c703288 authored by Jim Hefferon's avatar Jim Hefferon

misc corrections to entire book

parent b8e47a3b
......@@ -9,7 +9,7 @@
R.\ H.\ Ackerson,
\emph{A Note on Vector Spaces},
American Mathematical Monthly,
volume 62 number 10 (Dec.\ 1955),
vol.\ 62 no.\ 10 (Dec.\ 1955),
p.~721.
\bibitem[Agnew]{Agnew}
Jeanne Agnew,
......@@ -20,61 +20,61 @@
C.\ A.\ Rupp (proposer), H.\ T.\ R.\ Aude (solver),
problem 3468,
American Mathematical Monthly,
volume 37 number 6 (June-July 1931),
vol.\ 37 no.\ 6 (June-July 1931),
p.~355.
\bibitem[Am.\ Math.\ Mon., Feb.\ 1933]{Monthly33p118}
V.\ F.\ Ivanoff (proposer), T.\ C.\ Esty (solver),
problem 3529,
American Mathematical Monthly,
volume 39 number 2 (Feb.\ 1933),
vol.\ 39 no.\ 2 (Feb.\ 1933),
p.~118.
\bibitem[Am.\ Math.\ Mon., Jan.\ 1935]{Monthly35p47}
W.\ R.\ Ransom (proposer), Hansraj Gupta (solver),
Elementary problem 105,
American Mathematical Monthly,
volume 42 number 1 (Jan. 1935),
vol.\ 42 no.\ 1 (Jan. 1935),
p.~47.
\bibitem[Am.\ Math.\ Mon., Jan.\ 1949]{Monthly49p33}
C.\ W.\ Trigg (proposer), R.\ J.\ Walker (solver),
Elementary problem 813,
American Mathematical Monthly,
volume 56 number 1 (Jan.\ 1949),
vol.\ 56 no.\ 1 (Jan.\ 1949),
p.~33.
\bibitem[Am.\ Math.\ Mon., Jun.\ 1949]{Monthly49p409}
Don Walter (proposer), Alex Tytun (solver),
Elementary problem 834,
American Mathematical Monthly,
volume 56 number 6 (June-July 1949),
vol.\ 56 no.\ 6 (June-July 1949),
p.~409.
\bibitem[Am.\ Math.\ Mon., Nov.\ 1951]{Monthly51p614}
Albert Wilansky,
\emph{The Row-Sums of the Inverse Matrix},
American Mathematical Monthly,
volume 58 number 9 (Nov.\ 1951),
vol.\ 58 no.\ 9 (Nov.\ 1951),
p.~614.
\bibitem[Am.\ Math.\ Mon., Feb.\ 1953]{Monthly53p115}
Norman Anning (proposer), C.\ W.\ Trigg (solver),
Elementary problem 1016,
American Mathematical Monthly,
volume 60 number 2 (Feb.\ 1953),
vol.\ 60 no.\ 2 (Feb.\ 1953),
p.~115.
\bibitem[Am.\ Math.\ Mon., Apr.\ 1955]{Monthly55p257}
Vern Haggett (proposer), F.\ W.\ Saunders (solver),
Elementary problem 1135,
American Mathematical Monthly,
volume 62 number 4 (Apr.\ 1955),
vol.\ 62 no.\ 4 (Apr.\ 1955),
p.~257.
\bibitem[Am.\ Math.\ Mon., Jan.\ 1963]{Monthly63p93}
Underwood Dudley, Arnold Lebow (proposers), David Rothman (solver),
Elemantary problem 1151,
Elementary problem 1151,
American Mathematical Monthly,
volume 70 number 1 (Jan.\ 1963),
vol.\ 70 no. 1 (Jan.\ 1963),
p.~93.
\bibitem[Am.\ Math.\ Mon., Dec.\ 1966]{Monthly66p1114}
Hans Liebeck,
\emph{A Proof of the Equality of Column Rank and Row Rank of a Matrix}
American Mathematical Monthly,
volume 73 number 10 (Dec.\ 1966),
vol.\ 73 no.\ 10 (Dec.\ 1966),
p.~1114.
\bibitem[Anton]{Anton}
Howard Anton,
......@@ -104,7 +104,7 @@
% \emph{Wall Street Journal},
% 15 March, 1993.
\bibitem[Beck]{Beck}
Matthias Beck \& Ross Geoghegan,
Matthias Beck, Ross Geoghegan,
\emph{The Art of Proof},
\url{http://math.sfsu.edu/beck/papers/aop.noprint.pdf},
2011-Aug-08.
......@@ -112,7 +112,7 @@
A.F.\ Beardon,
\emph{The Dimension of the Space of Magic Squares},
The Mathematical Gazette,
vol~87, no~508 (Mar~2003), p~112-114.
vol.\ 87, no.\ 508 (Mar.\ 2003), p.~112-114.
\bibitem[Birkhoff \&~MacLane]{BirkhoffMaclane}
Garrett Birkhoff, Saunders MacLane,
\emph{Survey of Modern Algebra},
......@@ -552,7 +552,7 @@
\textit{How Euler Did It},
\url{http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2041%20factoring%20F5.pdf} (version: 2012-Jan-27).
\bibitem[Schmidt]{SchmidtSO}
Jack Schmidt (\url{http://math.stackexchange.com/users/583/jack-schmidt}),
Jack Schmidt % (\url{http://math.stackexchange.com/users/583/jack-schmidt}),
\url{http://math.stackexchange.com/a/98558/12012} (version: 2012-01-12).
\bibitem[Shepelev]{Shepelev}
Anton Shepelev,
......@@ -637,6 +637,10 @@
\emph{The Square-cube law},
\url{http://en.wikipedia.org/wiki/Square-cube_law},
2011-Jan-17.
\bibitem[Wikipedia Google Page Rank]{WikipediaPageRank}
\emph{Page Rank},
\url{http://en.wikipedia.org/wiki/PageRank},
2012-Feb-27.
\bibitem[Wills]{Wills}
Rebecca S.~Wills,
\emph{Google's Page Rank},
......
This diff is collapsed.
......@@ -22,7 +22,7 @@ gr1,gr2,gr3,cas,leontif,ppivot,network,%
vs1,vs2,vs3,fields,crystal,voting,dimen,%
map1,map2,map3,map4,map5,map6,lstsqs,homogeom,magicsqs,markov,erlang,%
det1,det2,det3,cramer,detspeed,projplane,%
jc1,jc2,jc3,jc4,powers,pops,recur,search,%eigengeom,prinaxis,%
jc1,jc2,jc3,jc4,powers,pops,search,recur,%eigengeom,prinaxis,%
appen,%
bib%
%test%
......@@ -66,7 +66,7 @@ bib%
\include{jc3}
\include{jc4}
\pagestyle{booktopic}
\include{powers} \include{pops} \include{recur} \include{search} \pagestyle{bookbody}
\include{powers} \include{pops} \include{search} \include{recur} \pagestyle{bookbody}
%\include{eigengeom}
\include{appen}
%\typeout{ BIBLIOGRAPHY COMMENTED OUT}
......
......@@ -218,11 +218,11 @@
\setlength{\leftmargin}{\parindent}
\setlength{\rightmargin}{0em}
\setlength{\parsep}{\parskip}
\setlength{\itemsep}{.4ex plus .1ex minus .15ex}
\setlength{\itemsep}{.4ex plus .15ex minus .15ex}
\setlength{\itemindent}{0em}
\setlength{\listparindent}{\parindent}
\setlength{\topsep}{0ex plus 5pt}
\setlength{\partopsep}{0ex plus 5pt}
\setlength{\topsep}{0ex plus 5pt minus 3pt}
\setlength{\partopsep}{0ex plus 5pt minus 3pt}
\setlength{\labelsep}{.5em}
\setlength{\labelwidth}{0em}
\setlength{\abovedisplayskip}{3pt plus2pt minus3pt}%
......@@ -251,7 +251,7 @@
% 1999-Oct-23 jh
\newenvironment{exercises}{%
\bigbreak
\noindent\textbf{Exercises}\nopagebreak
\noindent\shortstack{\textbf{Exercises}\\ \rule{0pt}{.25ex}}\nopagebreak
\announcesectioninginanswerfile % note the section change in the answer file
\begingroup\@beginparpenalty 10000
\small
......
......@@ -29470,6 +29470,16 @@ ans = 0.017398
\end{ans}
\subsection{Topic: Stable Populations}
\subsection{Topic: Page Ranking}
\begin{ans}{1}
The sum of the entries in column~$j$ is
$\sum_i \alpha h_{i,j}+(1-\alpha)s_{i,j}
=\sum_i \alpha h_{i,j}+\sum_i (1-\alpha)s_{i,j}
=\alpha\sum_i \alpha h_{i,j} +(1-\alpha)\sum_i s_{i,j}
=\alpha\cdot 1+(1-\alpha)\cdot 1$,
which is one.
\end{ans}
\subsection{Topic: Linear Recurrences}
\begin{ans}{1}
\begin{exparts}
......@@ -29536,13 +29546,3 @@ ans = 0.017398
\end{exparts}
\end{ans}
\subsection{Topic: Page Ranking}
\begin{ans}{1}
The sum of the entries in column~$j$ is
$\sum_i \alpha h_{i,j}+(1-\alpha)s_{i,j}
=\sum_i \alpha h_{i,j}+\sum_i (1-\alpha)s_{i,j}
=\alpha\sum_i \alpha h_{i,j} +(1-\alpha)\sum_i s_{i,j}
=\alpha\cdot 1+(1-\alpha)\cdot 1$,
which is one.
\end{ans}
......@@ -477,7 +477,7 @@
\vskip 0\p@ \fi %was 20\p@
\interlinepenalty\@M
\Huge \bfseries \textcolor{black}{#1}\par\nobreak
\vskip 30\p@
\vskip 40\p@ % was 30\p@
}}
......
......@@ -2322,15 +2322,15 @@ keeping track of the sign changes.
\end{example}
That example captures the new calculation scheme.
We apply multilinearity to a determinant to get
Multilinearity gives us
many separate determinants, each with one entry per row from the
original matrix.
Most of these matrices have
Most of these have
one row that is a multiple of another
so we can omit these determinants from the calculation.
We are left with the determinants that have one entry per row and
per column from the original matrix.
Then by factoring out the scalars we can further reduce the
so we can omit them.
We are left with those determinants that have one entry per row and
column from the original matrix.
By factoring out the scalars we can further reduce the
determinants that we must compute to those
one-entry-per-row-and-column matrices where all the entries are $1$'s.
......@@ -2368,8 +2368,8 @@ $\iota$'s.
\begin{definition}
An \definend{\( n \)-permutation}\index{permutation}
is a sequence consisting of an arrangement of the numbers
$1$, $2$, \ldots, $n$.
is an arrangement of the numbers
$1$, \ldots, $n$.
\end{definition}
\begin{example} \label{ex:AllTwoThreePerms}
......
......@@ -661,7 +661,7 @@ then the system has many solutions.
The next subsection deals with the third case.
We will see that such a system must have infinitely many solutions
and we will see how to describe the
and we will describe the
solution set.
......
This diff is collapsed.
......@@ -37,7 +37,7 @@ and scalar multiplication
\( h(r\cdot\vec{v})=r\cdot h(\vec{v}) \)
\end{center}
is a \definend{homomorphism}\index{homomorphism}%
\index{function!structure preserving!see{homomorphism}}%
\index{function!structure preserving!\see{homomorphism}}%
\index{vector space!homomorphism}\index{vector space!map}
or \definend{linear map}\index{linear map!see{homomorphism}}.
\end{definition}
......
......@@ -1032,7 +1032,7 @@ arrow diagram.\index{arrow diagram}
\begin{equation*}
\begin{CD}
V_\wrt{B} @>h>H> W_\wrt{D} \\
@V{\scriptsize\identity} VV @V\scriptsize\identity VV \\
@V{\text{\scriptsize$\identity$}} VV @V{\text{\scriptsize$\identity$}} VV \\
V_\wrt{\hat{B}} @>h>\hat{H}> W_\wrt{\hat{D}}
\end{CD}
\end{equation*}
......@@ -1090,7 +1090,7 @@ by using the arrow diagram and formula~($*$) above.
\begin{equation*}
\begin{CD}
\Re^2_\wrt{\stdbasis_2} @>t>T> \Re^2_\wrt{\stdbasis_2} \\
@V\scriptsize\identity VV @V\scriptsize\identity VV \\
@V\text{\scriptsize$\identity$} VV @V\text{\scriptsize$\identity$} VV \\
\Re^2_\wrt{\hat{B}} @>t>\hat{T}> \Re^2_\wrt{\hat{D}}
\end{CD}
\qquad
......
......@@ -6,11 +6,11 @@
In 1202 Leonardo of Pisa, known as Fibonacci, posed this problem.
\begin{quotation}
\noindent A certain man put a pair of rabbits in a place
\noindent A certain man put a pair of rabbits in a place
surrounded on all sides by a wall.
How many pairs of rabbits can be produced from that pair in a year if it
is supposed that every month each pair begets a new pair which from the
second month on becomes productive?
is supposed that every month each pair begets a new pair which from the
second month on becomes productive?
\end{quotation}
This moves past an elementary exponential growth model for
populations
......@@ -125,7 +125,8 @@ The calculation is ugly but not hard.
\left(\frac{1+\sqrt{5}}{2}\right)^n &0 \\
0 &\left(\frac{1-\sqrt{5}}{2}\right)^n
\end{mat}
\colvec{\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}}} \\ &=\begin{mat}
\colvec{\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}}} \\
&=\begin{mat}
\frac{1+\sqrt{5}}{2} &\frac{1-\sqrt{5}}{2} \\
1 &1
\end{mat}
......@@ -144,7 +145,7 @@ We want the second component of that equation.
\end{equation*}
This formula finds the value of any member of the sequence
without having to first find the intermediate values.
Notice that $(1-\sqrt{5})/2\approx 0.618$
Notice that $(1-\sqrt{5})/2\approx -0.618$
has absolute value less than one and so its powers go to zero.
Thus the formula giving $f(n)$ is dominated by its first term.
......
......@@ -8,20 +8,19 @@ Imagine that you are trying to find the best book
on Linear Algebra.
You probably would try a web search engine such as Google.
These lists pages ranked by importance.
The ranking is, as Google's founders said,
The ranking is defined, as Google's founders have said in \cite{BrinPage},
that a page is important if other important
pages link to it:
``a page can have a high PageRank if there are many pages that point
to it, or if there are some pages that point to it and have a high PageRank''
\cite{BrinPage}.
to it, or if there are some pages that point to it and have a high PageRank.''
But isn't that circular\Dash how can they tell whether a page is
important without first deciding on the important pages?
With eigenvalues and eigenvectors.
We will present a simplified version of the Page Rank algorithm.
We model the World Wide Web as a collection of pages connected by
For that we will model the World Wide Web as a collection of pages connected by
links.
This diagram from \cite{Wills}
This diagram, from \cite{Wills},
shows the pages as circles, and the links as arrows;
for instance, page~$p_1$ has a link to page~$p_2$.
\begin{center} % add a little vertical spacing; looked tight to me.
......@@ -29,13 +28,13 @@ for instance, page~$p_1$ has a link to page~$p_2$.
\end{center}
The key idea is that pages that should be highly ranked if they are
often cited by other pages.
cited often by other pages.
% (In practice people have tried to game the ranking system by
% setting up link farms of many pages that all point to each other.
% But our model will be simple.)
That is, we raise the importance of a page~$p_i$
if it is linked-to from page~$p_j$.
The increment is the importance of the linking page~$p_j$
The increment depends on the importance of the linking page~$p_j$
divided by how many out-links $a_j$ are on that page.
\begin{equation*}
\mathcal{I}(p_i)=\sum_{\text{in-linking pages $p_j$}} \frac{\mathcal{I}(p_j)}{a_j}
......@@ -50,7 +49,7 @@ This matrix stores the information.
\end{mat}
\end{equation*}
The algorithm's inventors describe a way to think about the matrix~$H$.
The algorithm's inventors describe a way to think about that matrix.
\begin{quotation}
PageRank can be thought of as a model of user behavior.
We assume there is a `random surfer' who is
......@@ -64,19 +63,13 @@ The probability that the random surfer visits a page is its PageRank.
In the diagram, a surfer on page~$p_3$ has a probability $1/3$ of going
next to each of the other pages.
We will find vector $\vec{\mathcal{I}}$ whose components are the
importance rankings of each page $\mathcal{I}(p_i)$.
With this notation,
our requirements for the page rank are: $H\vec{\mathcal{I}}=\vec{\mathcal{I}}$.
That is, we want an eigenvector of the matrix associated with the
eigenvalue~$\lambda=1$.
To that basic strategy we will add two refinements.
The first is the problem of page~$p_4$.
That leads us to the problem of page~$p_4$.
Many targets of links are
\definend{dangling} or \definend{sink links},
without any outbound links.
The simplest thing is to imagine that when the surfer gets to a page like this
(For instance, a page may link to an image.)
The simplest way to model what could happen next is to
imagine that when the surfer gets to a page like this
then they go to a next page entirely at random.
\begin{equation*}
H=\begin{mat}
......@@ -87,7 +80,15 @@ then they go to a next page entirely at random.
\end{mat}
\end{equation*}
This is \textit{Sage}'s calculation of the eigenvectors
We will find vector $\vec{\mathcal{I}}$ whose components are the
importance rankings of each page $\mathcal{I}(p_i)$.
With this notation,
our requirements for the page rank are that
$H\vec{\mathcal{I}}=\vec{\mathcal{I}}$.
That is, we want an eigenvector of the matrix associated with the
eigenvalue~$\lambda=1$.
Here is \textit{Sage}'s calculation of the eigenvectors
(slightly edited to fit on the page).
\begin{lstlisting}
sage: H=matrix([[0,0,1/3,1/4], [1,0,1/3,1/4], [0,1,0,1/4], [0,0,1/3,1/4]])
......@@ -97,11 +98,11 @@ sage: H.eigenvectors_right()
], 1), (0, [
(0, 1, 3, -4)
], 1), (-0.3750000000000000? - 0.4389855730355308?*I,
[(1, -0.1250000000000000? + 1.316956719106593?*I,
-1.875000000000000? - 1.316956719106593?*I, 1)], 1),
(-0.3750000000000000? + 0.4389855730355308?*I,
[(1, -0.1250000000000000? - 1.316956719106593?*I,
-1.875000000000000? + 1.316956719106593?*I, 1)], 1)]
[(1, -0.1250000000000000? + 1.316956719106593?*I,
-1.875000000000000? - 1.316956719106593?*I, 1)], 1),
(-0.3750000000000000? + 0.4389855730355308?*I,
[(1, -0.1250000000000000? - 1.316956719106593?*I,
-1.875000000000000? + 1.316956719106593?*I, 1)], 1)]
\end{lstlisting}
The eigenvector that \textit{Sage} gives
associated with the eigenvalue~$\lambda=1$ is this.
......@@ -142,10 +143,10 @@ Let this happen with probability~$\alpha$.
\end{mat}
\end{equation*}
This is the \definend{Google matrix}\index{Google matrix}\index{matrix!Google}.
In practice $\alpha$ is often set in between $0.85$ and~$0.99$.
Here are the ranks for the four pages with a couple of different values
of $\alpha$.
In practice $\alpha$ is often set between $0.85$ and~$0.99$.
Here are the ranks for the four pages with a spread of
$\alpha$'s.
\begin{center}
\begin{tabular}{r|cccc}
\multicolumn{1}{c}{$\alpha$}
......@@ -180,9 +181,10 @@ of $\alpha$.
\medskip
The details of the algorithms used by commercial search engines are
secret, and no doubt have many refinements, and also change frequently.
secret, no doubt have many refinements, and also change frequently.
But the inventors of Google were gracious enough to outline the basis for
their work in \cite{BrinPage}.
A more current source is \cite{WikipediaPageRank}.
Two additional excellent expositions are
\cite{Wills} and
\cite{Austin}.
......
......@@ -2228,7 +2228,7 @@ and Gauss' method in particular, is the following vector space.
\begin{definition}
The \definend{row space\/}\index{matrix!row space}\index{row space}
of a matrix is the span of the set of its rows.
The \definend{row rank\/}\index{row!rank}\index{matrix!row rank}
The \definend{row rank\/}\index{row!rank}\index{matrix!row rank}\index{row rank}
is the dimension of the row space, the number of linearly independent rows.
\end{definition}
......@@ -5334,4 +5334,4 @@ needed to do the Jordan Form construction in the fifth chapter.
\end{exparts}
\end{answer}
\end{exercises}
\index{direct sum!)}
\index{direct sum|)}
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