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Jim Hefferon
linearalgebra
Commits
148032c0
Commit
148032c0
authored
Dec 08, 2016
by
Jim Hefferon
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four_i.tex
slides/four_i.tex
+65
65
four_ii.tex
slides/four_ii.tex
+7
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slides/four_i.tex
View file @
148032c0
...
...
@@ 26,6 +26,10 @@
\addheadbox
{
filler
}{
\
}
% create extra space at top of slide
\hypersetup
{
colorlinks=true,linkcolor=blue
}
\usepackage
{
tikz
}
\usetikzlibrary
{
backgrounds
}
\usetikzlibrary
{
matrix
}
\tikzset
{
node style ge/.style=
{
circle
}}
\title
[Determinants]
% (optional, use only with long paper titles)
{
Four.I Definition of Determinant
}
...
...
@@ 55,7 +59,8 @@
\section
{
Properties of Determinants
}
%..........
\begin{frame}
{
Nonsingular matrices
}
\noindent
An
\(
\nbyn
{
n
}
\)
matrix
\(
T
\)
is nonsingular if and only if
\noindent
For any matrix, whether or not it is nonsingular is a key question.
Recall that an
\(
\nbyn
{
n
}
\)
matrix
\(
T
\)
is nonsingular if and only if
each of these holds:
%
\ExecuteMetaData
[../det1.tex]
{
EquivalentOfNonsingular
}
This chapter develops a formula to determine whether a
...
...
@@ 72,30 +77,28 @@ matrix is nonsingular.
\begin{frame}
{
Remark: h
ow we will proceed
}
\begin{frame}
{
H
ow we will proceed
}
The prior slide gives a formula for the
$
\nbyn
{
1
}$
,
$
\nbyn
{
2
}$
, and~
$
\nbyn
{
3
}$
determinants.
We can give a formula for the general
$
\nbyn
{
n
}$
~case,
the permutation expansion.
However, this formula has a number of disadvantages, including
its complexity and that
it is too slow for practical computations.
Instead, we will define a
determinant function as one that satisfies some conditions.
These conditions let us compute the determinant via Gauss's Method,
But while those three formulas are a help, they don't make clear what
should be the general formula, for the
$
\nbyn
{
n
}$
~case.
So we will proceed by stating some conditions that a
determinant function must satisfy.
These will let us compute the determinant via Gauss's Method,
which we know to be fast and easy.
(The conditions extrapolate from the
$
\nbyn
{
1
}$
,
$
\nbyn
{
2
}$
,
and~
$
\nbyn
{
3
}$
cases; see the discussion in the book.)
and~
$
\nbyn
{
3
}$
cases; see the book's discussion.)
We will give an algorithm that, following the conditions, goes from
any square matrix input to a real number output.
Defining a function by giving a list of conditions it must satisfy is
common in more advanced courses.
But it
has the downside that we must
\pause
However, defining a function by giving a list of conditions
has the downside that we must
show that there is one and only one function satisfying those conditions.
We will first give an algorithm that, following the conditions, goes from
any square matrix input to a real number output.
Later we will develop the formula for the determinant, which will
show that the determinant is welldefined (that is,
After giving the algorithm, we will develop a formula,
the permutation expansion,
which will show that the determinant is welldefined (that is,
for any input there is exactly one associated output).
\end{frame}
...
...
@@ 295,32 +298,35 @@ that has to do with multplying diagonals.
d
&
e
&
f
\\
g
&
h
&
i
\end{vmat}
=
aei+bfg+cdh
gechfaidb
=
{
\color
{
green
}
aei+bfg+cdh
}
{
\color
{
red
}
gechfaidb
}
\qquad
\begin{pmat}
{
ccccc
}
\begin{tikzpicture}
[baseline=(A.center), scale=1]
\path
[draw, line width=4pt, color=green, opacity=0.5]
(1.35,.6)(.3,.6);
\path
[draw, line width=4pt, color=green, opacity=0.5]
(.75,.6)(.8,.6);
\path
[draw, line width=4pt, color=green, opacity=0.5]
(.25,.6)(1.3,.6);
\path
[draw, line width=4pt, color=red, opacity=0.5]
(1.3,.6)(.3,.6);
\path
[draw, line width=4pt, color=red, opacity=0.5]
(.75,.6)(.8,.6);
\path
[draw, line width=4pt, color=red, opacity=0.5]
(.25,.6)(1.3,.6);
\node
(A) at (0,0)
{$
\begin
{
pmat
}{
ccccc
}
a
&
b
&
c
&
a
&
b
\\
d
&
e
&
f
&
d
&
e
\\
g
&
h
&
i
&
g
&
h
\end{pmat}
\end
{
pmat
}$}
;
\end{tikzpicture}
\end{equation*}
Don't try to extend to
$
\nbyn
{
4
}$
or larger sizes.
For those cases instead use Gauss's Method.
Don't try to extend to
$
\nbyn
{
4
}$
or larger sizes; there is no general
pattern here.
Instead, for larger matrices use Gauss's Method.
\end{frame}
\begin{frame}
{
The determinant is unique
}
\begin{frame}
{
Uniqueness
}
Recall our definition, that a function is a determinant if
it satisfies four conditions.
This approach does not make evident that
such function is unique.
(An analogy: imagine defining a function
$
\map
{
f
}{
\N
}{
\N
}$
to be an `evenmaker' under the condition that its
output is an even constant.
There is such a function, but also there is more than one.)
We now handle that issue; later we will handle the issue of showing that such
a function exists at all.
We now handle that question.
\pause
\lm
[lm:DetFcnIsUnique]
...
...
@@ 330,45 +336,38 @@ a function exists at all.
\pf
\ExecuteMetaData
[../det1.tex]
{
pf:DetFcnIsUnique
}
\qed
% \medskip
% So if there is a function mapping $\matspace_{\nbyn{n}}$ to $\Re$ that
% satisfies the four conditions of the definition then there is only one such
% function.
\end{frame}
\begin{frame}
{
More process discussion
}
We are left with the possibility that such a function does not exist.
How could there fail to be such a function, when we have been using
Gauss's Method to compute its outputs?
\ExecuteMetaData
[../det1.tex]
{
DifferentGaussMethodReductions
}
\begin{frame}
{
Existence
}
We still must show that determinants exist, that there is a
welldefined function satisfying all the conditions.
That we get consistent results in this one case
does not ensure that all determinant computations using Gauss's Method
give welldefined values.
How could there fail to be such a function, when we have been
computing its outputs?
\ExecuteMetaData
[../det1.tex]
{
DifferentGaussMethodReductions
}
But just because we get consistent results in this case
does not mean that all determinant computations
give the same value.
\end{frame}
\begin{frame}
In particular, recall that the definition's condition~(2),
that row swaps change the sign of the determinant, is
redundant.
Imagine that we did not notice the need for a sign change and had mistakenly
proposed a definition where row swaps leave the determinant unchanged.
Then the prior slide's pair of
$
\nbyn
{
2
}$
calculations would yield the
conflicting output values of
$

2
$
and~
$
2
$
.
We must prove that our definition does not lead to
such a thing,
%
In particular, recall that the definition's condition~(2),
%
that row swaps change the sign of the determinant, is
%
redundant.
%
Imagine that we did not notice the need for a sign change and had mistakenly
%
proposed a definition where row swaps leave the determinant unchanged.
%
Then the prior slide's pair of $\nbyn{2}$ calculations would yield the
%
conflicting output values of $2$ and~$2$.
%
We must prove that our definition does not lead to
%
such a thing,
\pause
%
\pause
The rest of this section gives an alternative way to compute
the determinant, a formula.
This formula does not involve Gauss's Method and
makes plain that the determinant is a function,
that it returns welldefined outputs.
As mentioned earlier, computing a determinant with this formula
is less practical than using the algorithm of Gauss's Method since it
Computing a determinant with this formula
is less practical than using Gauss's Method since this formula
is slow.
But it nonetheless is invaluable for the theory.
\end{frame}
...
...
@@ 862,9 +861,10 @@ the determinant (not involving Gauss's Method) is to give a formula for
the determinant of such matrices.
We do that in the next subsection.
\pause
That subsection is optional
so we state its results here.
\pause\bigskip
The next subsection is optional
so for those not going on, we state its results here.
We will follow up on the second result.
\th
[th:DetsExist]
\ExecuteMetaData
[../det1.tex]
{
th:DetsExist
}
...
...
slides/four_ii.tex
View file @
148032c0
...
...
@@ 75,8 +75,7 @@ This parallelogram is defined by the two vectors.
&
\quad
x
_
2y
_
2/2x
_
2y
_
2/2x
_
1y
_
1/2x
_
2y
_
1
\\
&
=x
_
1y
_
2x
_
2y
_
1
\end{align*}
We will argue that
the determinant of a square matrix
The determinant of this matrix
gives the size of the box formed by the matrix's columns.
\begin{equation*}
\begin{vmat}
...
...
@@ 90,14 +89,16 @@ gives the size of the box formed by the matrix's columns.
We now switch from considering the determinant as a function of the
rows to considering it as a function of the columns.
For that observe again that, b
ecause the determinant of the transpose equals
B
ecause the determinant of the transpose equals
the determinant of the matrix,
the row operation conditions in the definition
translate over to column operation conditions.
That is, (1)~a
determinant is unchanged by a column combination:~where
$
A
$
is square and
$
A
\!\!\raisebox
{

.
5
ex
}{
\smash
{
\grstep
{
k
\text
{
col
}_
i
+
\text
{
col
}_
j
}}}
\!\!
\hat
{
A
}$
\begin{equation*}
A
\!\!\raisebox
{
.5ex
}{
\smash
{
\grstep
{
k
\text
{
col
}_
i+
\text
{
col
}_
j
}}}
\!\!
\hat
{
A
}
\end{equation*}
(with~
$
i
\neq
j
$
)
then
$
\det
(
A
)=
\det
{
\hat
{
A
}}$
,
(2)~a column swap
...
...
@@ 105,7 +106,7 @@ changes the determinant's sign,
and (3)~multiplying a column by a
scalar multiplies the entire determinant by that scalar.
Condition~(4), that the determinant of the identity matrix is~
$
1
$
,
isn't about row o
perations so there is no
translation.
isn't about row o
r column operations so it still applies, without
translation.
\end{frame}
...
...
@@ 151,7 +152,7 @@ others so we leave it aside for a moment.
The final condition is that the determinant of the identity matrix is~
$
1
$
.
This also fits
with our program of arguing that the conditions in the
defin
nition of determinant are good ones for defining
the function
defin
ition of determinant are good ones for
the function
giving the size of the box formed by the columns of the matrix.
\centergraphic
{
../ch4.36
}
\end{frame}
...
...
@@ 163,8 +164,6 @@ giving the size of the box formed by the columns of the matrix.
\re
[re:PropertyTwoGivesSign]
The second condition in the definition is that
swapping changes the sign of the determinant.
Although this condition is redundant,
it is nonetheless notable.
Consider these pictures, using the same pair of vectors.
\begin{center}
\small
\begin{tabular}
{
c@
{
\hspace*
{
8em
}}
c
}
...
...
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