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Jim Hefferon
linearalgebra
Commits
9c703288
Commit
9c703288
authored
Feb 27, 2012
by
Jim Hefferon
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misc corrections to entire book
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bib.tex
bib.tex
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book.pdf
book.pdf
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book.tex
book.tex
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bookans.sty
bookans.sty
+4
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bookans.tex
bookans.tex
+10
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bookjh.sty
bookjh.sty
+1
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det1.tex
det1.tex
+8
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gr1.tex
gr1.tex
+1
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jhanswer.pdf
jhanswer.pdf
+24417
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map2.tex
map2.tex
+1
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map5.tex
map5.tex
+2
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recur.tex
recur.tex
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search.tex
search.tex
+30
28
vs3.tex
vs3.tex
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No files found.
bib.tex
View file @
9c703288
...
...
@@ 9,7 +9,7 @@
R.
\
H.
\
Ackerson,
\emph
{
A Note on Vector Spaces
}
,
American Mathematical Monthly,
vol
ume 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,
vol
ume 37 number
6 (JuneJuly 1931),
vol
.
\
37 no.
\
6 (JuneJuly 1931),
p.~355.
\bibitem
[Am.\ Math.\ Mon., Feb.\ 1933]
{
Monthly33p118
}
V.
\
F.
\
Ivanoff (proposer), T.
\
C.
\
Esty (solver),
problem 3529,
American Mathematical Monthly,
vol
ume 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,
vol
ume 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,
vol
ume 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,
vol
ume 56 number
6 (JuneJuly 1949),
vol
.
\
56 no.
\
6 (JuneJuly 1949),
p.~409.
\bibitem
[Am.\ Math.\ Mon., Nov.\ 1951]
{
Monthly51p614
}
Albert Wilansky,
\emph
{
The RowSums of the Inverse Matrix
}
,
American Mathematical Monthly,
vol
ume 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,
vol
ume 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,
vol
ume 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),
Elem
a
ntary problem 1151,
Elem
e
ntary problem 1151,
American Mathematical Monthly,
vol
ume 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,
vol
ume 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
}
,
2011Aug08.
...
...
@@ 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
~112114.
vol
.
\
87, no.
\
508 (Mar.
\
2003), p.
~112114.
\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: 2012Jan27).
\bibitem
[Schmidt]
{
SchmidtSO
}
Jack Schmidt (
\url
{
http://math.stackexchange.com/users/583/jackschmidt
}
),
Jack Schmidt
%
(\url{http://math.stackexchange.com/users/583/jackschmidt}),
\url
{
http://math.stackexchange.com/a/98558/12012
}
(version: 20120112).
\bibitem
[Shepelev]
{
Shepelev
}
Anton Shepelev,
...
...
@@ 637,6 +637,10 @@
\emph
{
The Squarecube law
}
,
\url
{
http://en.wikipedia.org/wiki/Squarecube
_
law
}
,
2011Jan17.
\bibitem
[Wikipedia Google Page Rank]
{
WikipediaPageRank
}
\emph
{
Page Rank
}
,
\url
{
http://en.wikipedia.org/wiki/PageRank
}
,
2012Feb27.
\bibitem
[Wills]
{
Wills
}
Rebecca S.~Wills,
\emph
{
Google's Page Rank
}
,
...
...
book.pdf
View file @
9c703288
This diff is collapsed.
Click to expand it.
book.tex
View file @
9c703288
...
...
@@ 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}
...
...
bookans.sty
View file @
9c703288
...
...
@@ 218,11 +218,11 @@
\setlength
{
\leftmargin
}{
\parindent
}
\setlength
{
\rightmargin
}{
0em
}
\setlength
{
\parsep
}{
\parskip
}
\setlength
{
\itemsep
}{
.4ex plus .1ex minus .15ex
}
\setlength
{
\itemsep
}{
.4ex plus .1
5
ex 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 @@
% 1999Oct23 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
...
...
bookans.tex
View file @
9c703288
...
...
@@ 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}
bookjh.sty
View file @
9c703288
...
...
@@ 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@
}}
...
...
det1.tex
View file @
9c703288
...
...
@@ 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 the
se determinants from the calculation
.
We are left with the determinants that have one entry per row and
per
column from the original matrix.
Then b
y factoring out the scalars we can further reduce the
so we can omit the
m
.
We are left with th
os
e determinants that have one entry per row and
column from the original matrix.
B
y factoring out the scalars we can further reduce the
determinants that we must compute to those
oneentryperrowandcolumn 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 a
n arrangement of the numbers
$
1
$
,
$
2
$
,
\ldots
,
$
n
$
.
is an arrangement of the numbers
$
1
$
,
\ldots
,
$
n
$
.
\end{definition}
\begin{example}
\label
{
ex:AllTwoThreePerms
}
...
...
gr1.tex
View file @
9c703288
...
...
@@ 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.
...
...
jhanswer.pdf
View file @
9c703288
This diff is collapsed.
Click to expand it.
map2.tex
View file @
9c703288
...
...
@@ 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}
...
...
map5.tex
View file @
9c703288
...
...
@@ 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
...
...
recur.tex
View file @
9c703288
...
...
@@ 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.
...
...
search.tex
View file @
9c703288
...
...
@@ 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 linkedto 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 outlinks
$
a
_
j
$
are on that page.
\begin{equation*}
\mathcal
{
I
}
(p
_
i)=
\sum
_{
\text
{
inlinking 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 th
e matrix~
$
H
$
.
The algorithm's inventors describe a way to think about th
at 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}
{
rcccc
}
\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
}
.
...
...
vs3.tex
View file @
9c703288
...
...
@@ 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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