Commit 80e0323c authored by Jim Hefferon's avatar Jim Hefferon

more map4 edits

parent 22ca3de2
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......@@ -15279,7 +15279,7 @@
\end{ans}
\subsection{Subsection Three.IV.4: Inverses}
\begin{ans}{Three.IV.4.13}
\begin{ans}{Three.IV.4.12}
Here is one way to proceed.
\begin{multline*}
\grstep{\rho_1\leftrightarrow\rho_2}\;
......@@ -15317,7 +15317,7 @@
\end{multline*}
\end{ans}
\begin{ans}{Three.IV.4.14}
\begin{ans}{Three.IV.4.13}
\begin{exparts*}
\partsitem Yes, it has an inverse: $ad-bc=2\cdot 1-1\cdot(-1)\neq 0$.
\partsitem Yes.
......@@ -15325,7 +15325,7 @@
\end{exparts*}
\end{ans}
\begin{ans}{Three.IV.4.15}
\begin{ans}{Three.IV.4.14}
\begin{exparts}
\partsitem
$\displaystyle \frac{1}{2\cdot 1-1\cdot (-1)}
......@@ -15355,7 +15355,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Three.IV.4.16}
\begin{ans}{Three.IV.4.15}
\begin{exparts}
\partsitem The reduction is routine.
\begin{equation*}
......@@ -15548,7 +15548,7 @@
\end{exparts}
\end{ans}
\begin{ans}{Three.IV.4.17}
\begin{ans}{Three.IV.4.16}
We can use \nearbycorollary{cor:TwoByTwoInv}.
\begin{equation*}
\frac{1}{1\cdot 5-2\cdot 3}\cdot
......@@ -15563,7 +15563,7 @@
\end{equation*}
\end{ans}
\begin{ans}{Three.IV.4.18}
\begin{ans}{Three.IV.4.17}
\begin{exparts}
\partsitem The proof that the inverse is
\( r^{-1}H^{-1}=(1/r)\cdot H^{-1} \)
......@@ -15632,17 +15632,17 @@
\end{exparts}
\end{ans}
\begin{ans}{Three.IV.4.19}
\begin{ans}{Three.IV.4.18}
Yes:
\( T^k(T^{-1})^k=(TT\cdots T)\cdot (T^{-1}T^{-1}\cdots T^{-1})
=T^{k-1}(TT^{-1})(T^{-1})^{k-1}=\dots=I \).
\end{ans}
\begin{ans}{Three.IV.4.20}
\begin{ans}{Three.IV.4.19}
Yes, the inverse of \( H^{-1} \) is \( H \).
\end{ans}
\begin{ans}{Three.IV.4.21}
\begin{ans}{Three.IV.4.20}
One way to check that the first is true is with
the angle sum formulas from trigonometry.
\begin{align*}
......@@ -15672,7 +15672,7 @@
vectors about the origin through an angle of \( \theta \)~radians.
\end{ans}
\begin{ans}{Three.IV.4.22}
\begin{ans}{Three.IV.4.21}
There are two cases.
For the first case we assume that \( a \) is nonzero.
Then
......@@ -15734,7 +15734,7 @@
\( ad-bc=-bc \)).
\end{ans}
\begin{ans}{Three.IV.4.23}
\begin{ans}{Three.IV.4.22}
With $H$ a $\nbym{2}{3}$ matrix,
in looking for a matrix $G$ such that the combination $HG$
acts as the $\nbyn{2}$ identity we
......@@ -15810,10 +15810,10 @@
and so there is no left inverse.
\end{ans}
\begin{ans}{Three.IV.4.24}
\begin{ans}{Three.IV.4.23}
With respect to the standard bases we have
\begin{equation*}
\rep{\eta}{\stdbasis_2,\stdbasis_3}
\rep{\iota}{\stdbasis_2,\stdbasis_3}
=\begin{mat}[r]
1 &0 \\
0 &1 \\
......@@ -15871,17 +15871,40 @@
\;\mapsunder{f_{2,3}}\;
\colvec{x+2z \\ y+3z}
\end{equation*}
The check that $\composed{f_{2,3}}{\eta}$ is the identity map on
The check that $\composed{f_{2,3}}{\iota}$ is the identity map on
$\Re^2$ is easy.
\end{ans}
\begin{ans}{Three.IV.4.25}
\begin{ans}{Three.IV.4.24}
By \nearbylemma{le:LeftAndRightInvEqual} it cannot have infinitely many
left inverses, because
a matrix with both left and right inverses has only one of each (and
that one of each is one of both\Dash the left and right inverse matrices
are equal).
\end{ans}
\begin{ans}{Three.IV.4.25}
\begin{exparts}
\partsitem True,
It must be linear,
as the proof from Theorem~II.\ref{th:OOHomoEquivalence} shows.
\partsitem False.
It may be linear, but it need not be.
Consider the projection map $\map{\pi}{\Re^3}{\Re^2}$ described
at the start of this subsection.
Define $\map{\eta}{\Re^2}{\Re^3}$ in this way.
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{x \\ y \\ 1}
\end{equation*}
It is a right inverse of $\pi$ because $\composed{\pi}{\eta}$
does this.
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{x \\ y \\ 1}\mapsto\colvec{x \\ y}
\end{equation*}
It is not linear because it does not map the zero vector to the zero
vector.
\end{exparts}
\end{ans}
\begin{ans}{Three.IV.4.26}
The associativity of matrix multiplication gives
This diff is collapsed.
......@@ -2383,7 +2383,10 @@ In a product of two matrices $G$ and $H$,
the columns of $GH$ are formed by taking $G$ times the columns of $H$
\begin{equation*}
G\cdot \begin{pmat}{c|@{\hspace{1em}}c@{\hspace{1em}}|c}
\vdots & &\vdots \\
% \MTFlushSpaceAbove
% \vdotswithin{\vec{h}_1} & &\vdotswithin{\vec{h}_n}
% \MTFlushSpaceBelow
\vdots & &\vdots \\
\vec{h}_1 &\cdots &\vec{h}_n \\
\vdots & &\vdots
\end{pmat}
......@@ -3751,50 +3754,52 @@ clearer idea then we will go with it.
\subsection{Inverses}
We now consider how to represent the
We finish this section by considering how to represent the
inverse of a linear map.
We start by recalling some facts about function
inverses.\appendrefs{function inverses} %\spacefactor=1000\
Some functions have no inverse, or have an inverse on the left side
or right side only.
\begin{example} \label{ex:ProjLeftInvOfEmbed}
We start by remembering some things about function
inverses.\appendrefs{function inverses} % \spacefactor=1000\
% Some functions have no inverse, or have an inverse on the left side
%or right side only.
%
% \begin{example} \label{ex:ProjLeftInvOfEmbed}
Where
\( \map{\pi}{\Re^3}{\Re^2} \) is the projection map
and \( \map{\iota}{\Re^2}{\Re^3} \) is the embedding
\begin{equation*}
\colvec{x \\ y \\ z}
\mapsto
\mapsunder{\pi}
\colvec{x \\ y}
\end{equation*}
and \( \map{\eta}{\Re^2}{\Re^3} \) is the embedding
\begin{equation*}
\qquad
\colvec{x \\ y}
\mapsto
\mapsunder{\iota}
\colvec{x \\ y \\ 0}
\end{equation*}
the composition $\composed{\pi}{\eta}$ is the identity map on $\Re^2$.
then the composition $\composed{\pi}{\iota}$ is the identity map on $\Re^2$.
\begin{equation*}
%\composed{\pi}{\eta}:\qquad
\colvec{x \\ y}
\mapsunder{\eta}
\mapsunder{\iota}
\colvec{x \\ y \\ 0}
\mapsunder{\pi}
\colvec{x \\ y}
\end{equation*}
We say $\pi$ is a \definend{left inverse map}\index{function!left inverse}
of $\eta$ or, what is the same thing,
that $\eta$ is a \definend{right inverse map}\index{function!right inverse}
of $\pi$.
However, composition in the other order $\composed{\eta}{\pi}$
We say that
$\iota$ is a \definend{right inverse map}\index{function!right inverse}
of $\pi$
or, what is the same thing,
that $\pi$ is a \definend{left inverse}\index{function!left inverse}
of $\iota$.
However, composition in the other order $\composed{\iota}{\pi}$
doesn't give the identity map\Dash here is a vector that is not
sent to itself under $\composed{\eta}{\pi}$.
sent to itself under $\composed{\iota}{\pi}$.
\begin{equation*}
%\composed{\eta}{\pi}:\qquad
\colvec[r]{0 \\ 0 \\ 1}
\mapsunder{\pi}
\colvec[r]{0 \\ 0}
\mapsunder{\eta}
\mapsunder{\iota}
\colvec[r]{0 \\ 0 \\ 0}
\end{equation*}
In fact, the projection
......@@ -3810,30 +3815,33 @@ then we would have
\colvec{x \\ y \\ z}
\end{equation*}
for all of the infinitely many $z$'s.
But no function $f$ can send a single argument to more than one value.
\end{example}
\par\noindent (An example of a function with no inverse on either side
But no function can send a single argument to more than one value.
% \end{example}
(An example of a function with no inverse on either side
is the zero transformation on $\Re^2$.)
Some functions have a
\definend{two-sided inverse map},\index{function!two-sided inverse}
another function
that is the inverse of the first, both from the left and from the right.
that is the inverse of the first both from the left and from the right.
For instance, the map given by
$\vec{v}\mapsto 2\cdot \vec{v}$ has the two-sided inverse
$\vec{v}\mapsto (1/2)\cdot\vec{v}$.
In this subsection we will focus on two-sided inverses.
% In this subsection we will focus on two-sided inverses.
%
The appendix shows that a function
has a two-sided inverse if and only if it is both one-to-one and onto.
The appendix also shows that if a function $f$ has a two-sided inverse then
it is unique, and so it is called
it is unique, and so we call it
`the'~inverse,\index{inverse function}\index{function!inverse}
and is denoted $f^{-1}$.
So our purpose in this subsection is, where a linear map $h$ has an inverse,
and denote it $f^{-1}$.
Also, recall that we have shown in Theorem~II.\ref{th:OOHomoEquivalence}
that if a linear map has a two-sided inverse
then that inverse is also linear.
Thus, our goal in this subsection is, where a linear $h$ has an inverse,
to find the relationship between $\rep{h}{B,D}$ and $\rep{h^{-1}}{D,B}$
(recall that we have shown, in \nearbytheorem{th:OOHomoEquivalence}
of Section~II of this chapter, that if a linear map has an inverse
then the inverse is a linear map also).
\begin{definition}
A matrix \( G \) is a \definend{left inverse matrix}\index{inverse!left}
......@@ -3841,7 +3849,7 @@ of the matrix \( H \) if \( GH \) is the identity matrix.
It is a \definend{right inverse matrix}\index{inverse!right}
if \( HG \) is the identity.
A matrix $H$ with a two-sided inverse is an \definend{invertible matrix}.
That two-sided inverse is called
That two-sided inverse is
\definend{the inverse matrix}\index{inverse} and
is denoted \( H^{-1} \).
\end{definition}
......@@ -4847,12 +4855,12 @@ elementary one can be interesting and useful.
and so there is no left inverse.
\end{answer}
\item
In \nearbyexample{ex:ProjLeftInvOfEmbed},
how many left inverses has \( \eta \)?
In the review of inverses example, % \nearbyexample{ex:ProjLeftInvOfEmbed},
how many left inverses has \( \iota \)?
\begin{answer}
With respect to the standard bases we have
\begin{equation*}
\rep{\eta}{\stdbasis_2,\stdbasis_3}
\rep{\iota}{\stdbasis_2,\stdbasis_3}
=\begin{mat}[r]
1 &0 \\
0 &1 \\
......@@ -4910,7 +4918,7 @@ elementary one can be interesting and useful.
\;\mapsunder{f_{2,3}}\;
\colvec{x+2z \\ y+3z}
\end{equation*}
The check that $\composed{f_{2,3}}{\eta}$ is the identity map on
The check that $\composed{f_{2,3}}{\iota}$ is the identity map on
$\Re^2$ is easy.
\end{answer}
\item
......@@ -4924,6 +4932,37 @@ elementary one can be interesting and useful.
that one of each is one of both\Dash the left and right inverse matrices
are equal).
\end{answer}
\item Assume that $\map{g}{V}{W}$ is linear.
One of these is true, the other is false.
Which is which?
\begin{exparts}
\partsitem If $\map{f}{W}{V}$ is a left inverse of $g$ then $f$
must be linear.
\partsitem If $\map{f}{W}{V}$ is a right inverse of $g$ then $f$
must be linear.
\end{exparts}
\begin{answer}
\begin{exparts}
\partsitem True,
It must be linear,
as the proof from Theorem~II.\ref{th:OOHomoEquivalence} shows.
\partsitem False.
It may be linear, but it need not be.
Consider the projection map $\map{\pi}{\Re^3}{\Re^2}$ described
at the start of this subsection.
Define $\map{\eta}{\Re^2}{\Re^3}$ in this way.
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{x \\ y \\ 1}
\end{equation*}
It is a right inverse of $\pi$ because $\composed{\pi}{\eta}$
does this.
\begin{equation*}
\colvec{x \\ y}\mapsto\colvec{x \\ y \\ 1}\mapsto\colvec{x \\ y}
\end{equation*}
It is not linear because it does not map the zero vector to the zero
vector.
\end{exparts}
\end{answer}
\recommended \item
Assume that \( H \) is invertible and that \( HG \) is the zero matrix.
Show that \( G \) is a zero matrix.
......
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