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Jim Hefferon
linearalgebra
Commits
a739640a
Commit
a739640a
authored
Nov 23, 2016
by
Jim Hefferon
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correct reported errors for third edition
parent
ad9bc4b3
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635
Acknowledgements
Acknowledgements
+9
1
bookans.tex
bookans.tex
+249
238
det1.tex
det1.tex
+7
7
det3.tex
det3.tex
+1
1
gr1.tex
gr1.tex
+13
17
gr3.tex
gr3.tex
+36
14
problems_7.tex
homeworks/problems_7.tex
+1
1
homogeom.tex
homogeom.tex
+265
255
map2.tex
map2.tex
+1
1
map4.tex
map4.tex
+3
3
three_iv.tex
slides/three_iv.tex
+13
13
voting.tex
voting.tex
+64
82
vs1.tex
vs1.tex
+1
1
vs3.tex
vs3.tex
+1
1
No files found.
Acknowledgements
View file @
a739640a
...
...
@@ 166,4 +166,12 @@ General:
William Karstens for much help with the Topic on Coupled Pendula.
2016Aug20
Christelle Vincent Add slides that use handout option
\ No newline at end of file
Christelle Vincent Add slides that use handout option.
2016Nov23
Darij Grinberg Typo.
David Guichard Many typos and suggestions.
Mitch Rothstein Exercise error, and proof suggestion.
James Moore Typo.
Richard Satterfield Thoughtprovoking report about free variables.
Christopher Thron Incorrect exercise in voting Topic.
\ No newline at end of file
bookans.tex
View file @
a739640a
This diff is collapsed.
Click to expand it.
det1.tex
View file @
a739640a
...
...
@@ 1074,13 +1074,13 @@ write it as a function of the rows
We want a formula
to determine whether an
$
\nbyn
{
n
}$
matrix is nonsingular.
We will not begin by stating such a formula.
Instead we will begin by considering
the function that such a formula
calculates.
We will define this function by
its properties,
then prove that the
function with these properties exists and is unique,
Instead we will begin by considering
, for eacn~
$
n
$
,
the function that such a formula
calculates.
We will define this function by
a list of properties.
We will then prove that a
function with these properties exists and is unique,
and also describe how to compute it.
(Because we will eventually
show that the function exists and is unique,
from the start
we will just say `
\(
\det
(
T
)
\)
' instead of
(Because we will eventually
prove this, from the start
we will just say `
\(
\det
(
T
)
\)
' instead of
`if there is a unique determinant function then
\(
\det
(
T
)
\)
'.)
\begin{definition}
\label
{
def:Det
}
...
...
@@ 1325,7 +1325,7 @@ quickly with the Gauss's Method procedure.
The prior example illustrates an important point.
Although we have not yet found a
$
\nbyn
{
4
}$
determinant formula,
if one exists then we know what value it gives to the matrix
\Dash
if there is a function with properties (1)

(4) then on the above
if there is a function with properties (1)
\,

\,
(4) then on the above
matrix the function must return
$
5
$
.
\begin{lemma}
\label
{
lm:DetFcnIsUnique
}
...
...
det3.tex
View file @
a739640a
...
...
@@ 330,7 +330,7 @@ their associated cofactors.
\tag*
{
(
$
*
$
)
}
\end{equation*}
Recall that a matrix with two identical rows has a zero determinant.
Thus,
weighing the cofactors by entries from
weigh
t
ing the cofactors by entries from
row~
$
k
$
with
$
k
\neq
i
$
gives zero
\begin{equation*}
t
_{
i,1
}
\cdot
T
_{
k,1
}
+t
_{
i,2
}
\cdot
T
_{
k,2
}
+
\dots
+t
_{
i,n
}
\cdot
T
_{
k,n
}
=0
...
...
gr1.tex
View file @
a739640a
...
...
@@ 2125,7 +2125,7 @@ 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
}
.
%
\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
}
%
...
...
@@ 2420,11 +2420,10 @@ is a particular solution of the system.
\end{example}
Before the exercises, we will consider what we have accomplished
and what we
have yet to do
.
and what we
will do in the remainder of the chapter
.
So far we have done the mechanics of Gauss's Method.
We
have not stopped to consider any of the interesting questions
We have not stopped to consider any of the questions
that arise,
except for proving
\nearbytheorem
{
th:GaussMethod
}
\Dash
which
justifies the method by showing that it gives
...
...
@@ 2436,10 +2435,13 @@ a particular solution vector added to an unrestricted linear combination of
some other vectors?
We've noted that the solution sets we described in this way
have infinitely many solutions
so an
answer to
this question
so an
swering
this question
would tell us about the size of solution sets.
The following subsection shows that the answer is ``yes.''
This chapter's second section then
uses that answer to describe the geometry of solution sets.
Many questions arise from our
observation that we can do Gauss's Method in
Other questions arise from the
observation that we can do Gauss's Method in
more than one way (for instance, when swapping rows we may have a choice of
more than one row).
\nearbytheorem
{
th:GaussMethod
}
says that we must get the same solution set
...
...
@@ 2450,18 +2452,12 @@ Must those be the same variables, that is, is it impossible to
solve a problem
one way to get
$
y
$
and~
$
w
$
free and solve it another way to get
$
y
$
and~
$
z
$
free?
The third section of this chapter answers ``yes,'' that
from any starting linear system,
all derived echelon form versions
have the same free variables.
In the rest of this chapter we will answer these questions.
The answer to each is `yes'.
In the next subsection
we do the first one: we will prove that we can always describe solution sets
in that way.
Then, in this chapter's second section,
we will use that understanding to describe the geometry of solution sets.
In this chapter's final section,
we will settle the questions about the parameters.
When we are done, we will not only have a
Thus, by the end of the chapter we will not only have a
solid grounding in the practice of Gauss's Method but
we will also have a solid grounding in the theory.
We will know exactly what can and cannot happen in a reduction.
...
...
gr3.tex
View file @
a739640a
...
...
@@ 36,8 +36,8 @@ from $2\rho_1+\rho_2$ followed by $2\rho_2+\rho_1$
echelon form so the second one is extra work
but it is nonetheless a legal row operation).
The fact that echelon form
is not unique
raises questions.
In this chapter's first section we noted that
this
raises questions.
Will any two echelon form versions of a linear system have the same number of
free variables?
If yes,
...
...
@@ 45,8 +45,8 @@ will the two have exactly the same free variables?
In this section we will
give a way to decide if one linear system
can be derived from another by row operations.
The answers to both questions, both `
yes',
will follow
from this
.
The answers to both questions, both `
`yes,''
will follow.
...
...
@@ 1794,22 +1794,44 @@ So \( B=C \).
%</pf:ReducedEchelonFormIsUnique4>
\end{proof}
That result answers the two questions from this section's introduction:~do
We have asked whether
any two echelon form versions of a linear system
have the same number of free variables, and if so are they
exactly the same variables?
With the prior result we can answer both questions ``yes.''
There is no linear system such
that, say, we could apply Gauss's Method
one way and get
$
y
$
and
$
z
$
free but apply it another way
and get
$
y
$
and
$
w
$
free.
Before the proof,
recall the distinction between free variables and parameters.
This system
\begin{equation*}
\begin{linsys}
{
3
}
x
&
+
&
y
&
&
&
=
&
1
\\
&
&
y
&
+
&
z
&
=
&
2
\end{linsys}
\end{equation*}
has one free variable,~
$
z
$
, because it is the only variable not leading a row.
We have the habit of parametrizing using the free variable
$
y
=
2

z
$
,
$
x
=
1
+
z
$
,
but we could also parametrize using another variable, such as
$
z
=
2

y
$
,
$
x
=
1

y
$
.
So the set of parameters is not unique, it is the set of free variables
that is unique.
\begin{corollary}
If from a starting linear systems we derive by Gauss's Method two
different echelon form systems, then the two have the same free variables.
\end{corollary}
\begin{proof}
The prior result says that the reduced echelon form is unique.
We get from any echelon form version to the reduced echelon form by
eliminating up,
so any echelon form version of a system has the same free variables as the
reduced echelon form,
and therefore uniqueness of reduced echelon form gives
that the same variables
are free in all echelon form version of a system.
Thus both questions are answered ``yes.''
There is no linear system and no combination of row operations such
that, say, we could solve the system
one way and get
$
y
$
and
$
z
$
free but solve it
another way and get
$
y
$
and
$
w
$
free.
reduced echelon form version.
\end{proof}
We close with a recap.
In Gauss's Method we start with a matrix and then
...
...
homeworks/problems_7.tex
View file @
a739640a
...
...
@@ 64,7 +64,7 @@ quote any result from the book).}
$
\map
{
\composed
{
g
}{
h
}}{
\Re
^
3
}{
\matspace
_{
\nbyn
{
2
}}}$
directly from the above definition.
\item
Represent
$
h
$
and~
$
g
$
with respect to the appropriate bases.
\item
Represent
$
\composed
{
g
}{
h
}$
with res
ep
ct to the appropriate bases.
\item
Represent
$
\composed
{
g
}{
h
}$
with res
pe
ct to the appropriate bases.
\item
Check that the two matrices from the second part multiply to the
matrix from the third part.
\end{enumerate}
...
...
homogeom.tex
View file @
a739640a
This diff is collapsed.
Click to expand it.
map2.tex
View file @
a739640a
...
...
@@ 964,7 +964,7 @@ is more fruitful and more central to progress.
is a linear transformation of that space.
\begin{equation*}
f
\mapsto
\frac
{
d
^
k
}{
dx
^
k
}
f+c
_{
k1
}
\frac
{
d
^{
k1
}}{
dx
^{
k1
}}
f
c
_{
k
}
\frac
{
d
^
k
}{
dx
^
k
}
f+c
_{
k1
}
\frac
{
d
^{
k1
}}{
dx
^{
k1
}}
f
+
\dots
+
c
_
1
\frac
{
d
}{
dx
}
f+c
_
0f
\end{equation*}
...
...
map4.tex
View file @
a739640a
...
...
@@ 997,8 +997,8 @@ function composition is possible.
\text
{
dimension
\(
m
\)
space
}
\tag
{$
*
$}
\end{equation*}
Thus, matrix product combines
an
$
\nbym
{
m
}{
r
}$
matrix~
$
G
$
with
an
$
\nbym
{
r
}{
n
}$
matrix~
$
F
$
to yield the
Thus, matrix product combines
the
$
\nbym
{
m
}{
r
}$
matrix~
$
G
$
with
the
$
\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
}$
.
...
...
@@ 3077,7 +3077,7 @@ 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
on
e Gaussian operation to an identity matrix.
result from applying a
singl
e Gaussian operation to an identity matrix.
\begin{enumerate}
\item
\(
I
\grstep
{
k
\rho
_
i
}
M
_
i
(
k
)
\)
for
\(
k
\neq
0
\)
\item
\(
I
\grstep
{
\rho
_
i
\leftrightarrow\rho
_
j
}
P
_{
i,j
}
\)
for
...
...
slides/three_iv.tex
View file @
a739640a
...
...
@@ 506,7 +506,7 @@ Square matrices of the same size have a defined product.
\end{mat}
\end{equation*}
This reflects the fact that we can compose two
functions from a space to itself
$
\map
{
f,g
}{
V
}{
V
}$
.
functions from a space to itself
$
\map
{
g,h
}{
V
}{
V
}$
.
\end{frame}
...
...
@@ 571,10 +571,10 @@ Fix these bases.
0
&
4
\end{mat}
}
\end{gather*}
Suppose that
$
\map
{
f
}{
\Re
^
2
}{
\polyspace
_
2
}$
and
Suppose that
$
\map
{
h
}{
\Re
^
2
}{
\polyspace
_
2
}$
and
$
\map
{
g
}{
\polyspace
_
2
}{
\matspace
_{
\nbyn
{
2
}}}$
have these actions.
\begin{equation*}
\colvec
{
a
\\
b
}
\mapsunder
{
f
}
ax
^
2+(a+b)
\colvec
{
a
\\
b
}
\mapsunder
{
h
}
ax
^
2+(a+b)
\qquad
px
^
2+qx+r
\mapsunder
{
g
}
\begin{mat}
...
...
@@ 584,7 +584,7 @@ $\map{g}{\polyspace_2}{\matspace_{\nbyn{2}}}$ have these actions.
\end{equation*}
Then the composition does this.
\begin{equation*}
\colvec
{
a
\\
b
}
\mapsunder
{
f
}
ax
^
2+(a+b)
\mapsunder
{
g
}
\colvec
{
a
\\
b
}
\mapsunder
{
h
}
ax
^
2+(a+b)
\mapsunder
{
g
}
\begin{mat}
a
&
a
\\
0
&
a+b
...
...
@@ 592,7 +592,7 @@ Then the composition does this.
\end{equation*}
Here is the same statement in the other notation.
\begin{equation*}
\composed
{
g
}{
f
}
\,
(
\colvec
{
a
\\
b
}
)=
\composed
{
g
}{
h
}
\,
(
\colvec
{
a
\\
b
}
)=
\begin{mat}
a
&
a
\\
0
&
a+b
...
...
@@ 601,19 +601,19 @@ Here is the same statement in the other notation.
\end{frame}
\begin{frame}
So far in this example we have given the maps above the arrows,
$
f
$
,
$
g
$
,
and
$
\composed
{
g
}{
f
}$
.
and
$
\composed
{
g
}{
h
}$
.
\centergraphic
{
../ch3.20
}
We next compute the matrices representing those maps, and we will
finish by checking that the product of
$
H
$
and~
$
G
$
is the matrix
finish by checking that the product of
$
G
$
and~
$
H
$
is the matrix
representing
$
\composed
{
g
}{
h
}$
.
\pause
First, find
$
H
=
\rep
{
f
}{
B,D
}$
: compute the action of
$
h
$
First, find
$
H
=
\rep
{
h
}{
B,D
}$
: compute the action of
$
h
$
on the domain basis vectors,
\begin{equation*}
\colvec
{
1
\\
1
}
\mapsunder
{
f
}
x
^
2+2
\colvec
{
1
\\
1
}
\mapsunder
{
h
}
x
^
2+2
\quad
\colvec
{
1
\\
1
}
\mapsunder
{
f
}
x
^
2
\colvec
{
1
\\
1
}
\mapsunder
{
h
}
x
^
2
\end{equation*}
represent the results with respect to
$
D
$
, and make the matrix.
\begin{equation*}
...
...
@@ 662,7 +662,7 @@ and represent those with respect to its codomain basis.
\end{equation*}
\pause
Next,
$
\composed
{
g
}{
f
}$
has this action.
Next,
$
\composed
{
g
}{
h
}$
has this action.
\begin{equation*}
\colvec
{
1
\\
1
}
\mapsto
\begin{mat}
...
...
@@ 1267,7 +1267,7 @@ can be expressed as this product; note that the order is righttoleft.
\begin{frame}
\ex
To can bring this augmented matrix to echelon form with matrix multiplication
You can bring this augmented matrix to echelon form with matrix multiplication.
\begin{equation*}
\begin{amat}
{
3
}
1
&
1
&
2
&
4
\\
...
...
@@ 1277,7 +1277,7 @@ To can bring this augmented matrix to echelon form with matrix multiplication
\end{equation*}
\end{frame}
\begin{frame}
f
irst perform
$

2
\rho
_
1
+
\rho
_
2
$
via left multiplication by
$
C
_{
1
,
2
}
(
2
)
$
.
F
irst perform
$

2
\rho
_
1
+
\rho
_
2
$
via left multiplication by
$
C
_{
1
,
2
}
(
2
)
$
.
\begin{equation*}
\begin{mat}
1
&
0
&
0
\\
...
...
voting.tex
View file @
a739640a
%\documentclass{article}
%\usepackage{verbatim}
%
%\begin{document}
%\newcommand{\votepref}[1]{\overrightarrow{#1}}
%\newcommand{\votepreflist}[3]{#1\,\votepref{DT}
% +#2\,\votepref{TR}+#3\,\votepref{RS}}
% \newcommand{\votepreflist}[3]{\colvec{#1 \\ #2 \\ #3}}
% \newcommand{\votecycle}[4]{\psset{xunit=#4pt,yunit=#4pt,runit=#4pt}
% \pspicture[.4](7.25,6)(7.25,5) %\psgrid(1,1)(1,1)
% \SpecialCoor
% \rput(3;90){\scriptsize $D$}
% \psarc{>}(0,0){3}{10}{70} %rd>
% \rput[lb](3.3;20){\scriptsize $#1$}
% \rput[B](3;210){\scriptsize $T$}
% \psarc{>}(0,0){3}{220}{320} %tr>
% \rput[t](3.1;270){\scriptsize $#2$}
% \rput[B](3;330){\scriptsize $R$}
% \psarc{>}(0,0){3}{110}{175} %dt>
% \rput[rb](3.3;160){\scriptsize $#3$}
% \endpspicture}
% %save it to the answer file.
% \begin{Filesave}{bookans}
% \newcommand{\votepreflist}[3]{\colvec{#1 \\ #2 \\ #3}}
% \newcommand{\votecycle}[4]{\psset{xunit=#4pt,yunit=#4pt,runit=#4pt}
% \pspicture[.4](7.25,6)(7.25,5) %\psgrid(1,1)(1,1)
% \SpecialCoor
% \rput(3;90){\scriptsize $D$}
% \psarc{>}(0,0){3}{10}{70} %rd>
% \rput[lb](3.3;20){\scriptsize $#1$}
% \rput[B](3;210){\scriptsize $T$}
% \psarc{>}(0,0){3}{220}{320} %tr>
% \rput[t](3.1;270){\scriptsize $#2$}
% \rput[B](3;330){\scriptsize $R$}
% \psarc{>}(0,0){3}{110}{175} %dt>
% \rput[rb](3.3;160){\scriptsize $#3$}
% \endpspicture}
% \end{Filesave}
%\newcommand{\votinggraphic}[1]{\hspace*{.8em}\mathord{\raisebox{.2in}[.3in][.2in]{\includegraphics{voting.#1}}}\hspace*{.8em}}
\topic
{
Voting Paradoxes
}
\index
{
voting paradox(
}
%\emph{(Optional material from this chapter is discussed here,
...
...
@@ 336,10 +295,22 @@ voting paradox can happen only when the
tendencies toward cyclic preference reinforce each other.
For the proof, assume that we have canceled
opposite preference orders and we are left with one set
of preference lists from each of the three rows.
Consider the sum of these three
(here, the numbers
$
a
$
,
$
b
$
, and
$
c
$
could be positive, negative, or zero).
opposite preference orders and are left with one set
of preference lists for each of the three rows.
Consider the first row's remaining preference lists.
They could be from the first row's left or right
(or neither, since the lists could have canceled exactly).
We shall write
\begin{equation*}
\votinggraphic
{
27
}
\end{equation*}
where
$
a
$
is an integer that is positive if the remaining lists are on the
left, where
$
a
$
is negative if the lists are on the right, and zero
if the cancellation was perfect.
Similiarly we have integers
$
b
$
and
$
c
$
for the second and third rows,
which can each be positive, negative, or zero.
Then the election is determined by the sum of these three.
\begin{equation*}
\votinggraphic
{
27
}
%\votecycle{a}{a}{a}{4}
+
...
...
@@ 348,23 +319,27 @@ Consider the sum of these three
\votinggraphic
{
29
}
%\votecycle{c}{c}{c}{4}
=
\hspace
{
.30in
}
\votinggraphic
{
30
}
%\votecycle{ab+c}{a+bc}{a+b+c}{4}
\hspace
{
.30in
}
\hbox
{}
\hspace
{
.30in
}
\hbox
{}
\end{equation*}
A voting paradox occurs when the three numbers on the right,
$
a

b
+
c
$
and
$
a
+
b

c
$
and
$

a
+
b
+
c
$
, are all nonnegative or all nonpositive.
On the left, at least two of the three numbers
$
a
$
and
$
b
$
and
$
c
$
are both nonnegative or both nonpositive.
We can assume that they are
$
a
$
and
$
b
$
.
That makes four cases:~the cycle is nonnegative and
$
a
$
and
$
b
$
are nonnegative, the cycle is
nonpositive and
$
a
$
and
$
b
$
are nonpositive, etc.
We will do only the first case, since the second is similar
and the other two are also easy.
A voting paradox occurs when the three numbers in the total cycle on the right,
% $a+b+c$ and $ab+c$ and $a+bc$,
are all nonnegative or all nonpositive.
We will prove this occurs only when either all
three of
$
a
$
,
$
b
$
, and~
$
c
$
are nonnegative or all three are nonpositive.
So assume that the cycle is nonnegative and
that
$
a
$
and
$
b
$
are nonnegative.
The conditions
$
0
\leq
a

b
+
c
$
and
$
0
\leq

a
+
b
+
c
$
add to give
that
$
0
\leq
2
c
$
, which implies that
$
c
$
is also nonnegative, as desired.
Suppose that the total cycle is nonnegative; the other case is similar.
\begin{equation*}
\begin{linsys}
{
3
}
a
&
+
&
b
&
+
&
c
&
\geq
&
0
\\
a
&

&
b
&
+
&
c
&
\geq
&
0
\\
a
&
+
&
b
&

&
c
&
\geq
&
0
\end{linsys}
\end{equation*}
Adding the first two rows shows that
$
c
\geq
0
$
.
Adding the first and third rows gives
$
b
\geq
0
$
.
And the sum of the second and third shows
$
a
\geq
0
$
.
Thus if the total cycle is nonnegative then in each row the remaining
preference lists are from the table's left.
That ends the proof.
This result says only that having all three spin in the same direction is a
...
...
@@ 379,12 +354,14 @@ this work, showing that no voting system is entirely fair
(for a reasonable definition of ``fair'').
Some good introductory articles are
\cite
{
Gardner70
}
,
\cite
{
Gardner74
}
,
\cite
{
Gardner80
}
, and
\cite
{
NeimiRiker
}
.
\cite
{
Taylor
}
is a readable
recent book
.
\cite
{
Taylor
}
is a readable
text
.
The long list of cases from recent American political history in
\cite
{
GamingVote
}
shows these paradoxes are
routinely manipulated in practice.
This Topic is largely drawn from
\cite
{
Zwicker
}
.
(On the other hand, quite recent research shows that
computing how to manipulate elections can in general be unfeasible,
but this is beyond our scope.)
This Topic is drawn from
\cite
{
Zwicker
}
.
\emph
{
(Author's Note:~I would like to thank Professor~Zwicker
for his kind and illuminating discussions.)
}
...
...
@@ 491,7 +468,7 @@ for his kind and illuminating discussions.)}
gives
$
c
_
3
=
2
/
3
$
,
$
c
_
2
=
4
/
3
$
, and
$
c
_
1
=
1
/
3
$
.
\end{answer}
\item
\label
{
exer:CancelPolSci
}
Do
the cancellations of opposite preference orders
Perform
the cancellations of opposite preference orders
for the Political Science class's mock election.
Are all the remaining preferences from the left three rows of the
table or from the right?
...
...
@@ 622,15 +599,16 @@ for his kind and illuminating discussions.)}
to get the same outcome.
\end{answer}
\item
\label
{
exer:VoterCondNotSuff
}
The necessary condition that
is proved above
\Dash
a
The necessary condition that a
voting paradox can happen only if all three preference lists remaining
after cancellation have the same spin

is not also sufficient.
after cancellation have the same spin
is not also sufficient.
\begin{exparts}
\partsitem
Continuing the positive cycle case considered in the proof,
use the two inequalities
$
0
\leq
a

b
+
c
$
and
$
0
\leq

a
+
b
+
c
$
to show that
$
\absval
{
a

b
}
\leq
c
$
.
\partsitem
Also show that
$
c
\leq
a
+
b
$
, and hence that
$
\absval
{
a

b
}
\leq
c
\leq
a
+
b
$
.
% \partsitem Continuing the nonnegative total cycle case
% considered in the proof,
% use the two inequalities $0\leq ab+c$ and $0\leq a+b+c$
% to show that $\absval{ab}\leq c$.
% \partsitem Also show that $c\leq a+b$, and hence that
% $\absval{ab}\leq c\leq a+b$.
\partsitem
Give an example of a vote where there is a majority cycle,
and addition of one more voter with the same spin causes the cycle to
go away.
...
...
@@ 641,17 +619,18 @@ for his kind and illuminating discussions.)}
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem
We can rewrite the two as
$

c
\leq
a

b
$
and
$

c
\leq
b

a
$
.
Either
$
a

b
$
or
$
b

a
$
is nonpositive and so
$

c
\leq

\absval
{
a

b
}$
,
as required.
\partsitem
This is immediate from the supposition that
$
0
\leq
a
+
b

c
$
.
\partsitem
A trivial example starts with the zerovoter election
and
adds any one voter.
%
\partsitem We can rewrite the two as $c\leq ab$ and $c\leq ba$.
%
Either $ab$ or $ba$ is nonpositive and so $c\leq\absval{ab}$,
%
as required.
%
\partsitem This is immediate from the supposition that $0\leq a+bc$.
\partsitem
A trivial example starts with the zerovoter election
,
which has a trivial majority cycle, and
adds any one voter.
A more interesting example takes the Political Science mock
election and add two
$
T>D>R
$
voters
(they can be added one at a time, to satisfy the
``addition of one more voter'' criteria in the question).
The additional voters have positive spin, which is the
election and adds two
$
T>D>R
$
voters
(to satisfy the
``addition of one more voter'' criteria in the question
we can add them one at a time).
The new voters have positive spin, which is the
spin of the votes remaining after cancellation in the original mock
election.
This is the resulting table of voters and next to it is
...
...
@@ 775,11 +754,14 @@ for his kind and illuminating discussions.)}
% \endpspicture
\tag*
{}
\end{equation*}
The majority cycle has indeed disappeared.
\partsitem
Reverse the prior exercise.
That is, add a voter to the result of the prior exercise that
cancels the one who made the cycle disappear.
\partsitem
One such condition is that, after cancellation,
all three be nonnegative or all three be nonpositive,
and:~
$
\absval
{
c
}
<
\absval
{
a
+
b
}$
and
$
\absval
{
b
}
<
\absval
{
a
+
c
}$
and
$
\absval
{
a
}
<
\absval
{
b
+
c
}$
.
That follows from this diagram.
That follows from this diagram.
\begin{equation*}
\votinggraphic
{
27
}
%\votecycle{a}{a}{a}{4}
+
...
...
vs1.tex
View file @
a739640a
...
...
@@ 2175,7 +2175,7 @@ the inherited ones.
=
\colvec
{
rx
\\
0
}
\end{equation*}
As in the prior example, to verify directly from the definition
that this is a subspace we simply that it is a
that this is a subspace we simply
note
that it is a
subset and then check that it satisfies
the conditions in definition of a vector space.
For instance the two closure conditions are
...
...
vs3.tex
View file @
a739640a
...
...
@@ 2712,7 +2712,7 @@ set.
Lemma~One.III.
\ref
{
le:EchFormNoLinCombo
}
says that no nonzero row of an echelon form matrix is
a linear combination of the other rows.
This result
just restates that in
this chapter's terminology.
This result
restates that using
this chapter's terminology.
%</pf:RowsEchMatLI>
\end{proof}
...
...
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