Commit 3a171de3 authored by Jim Hefferon's avatar Jim Hefferon

map1 and map2 exercise adjustments

parent d9f0a83a
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......@@ -715,7 +715,7 @@ partitions the collection of vector spaces.
\end{split}
\end{multline*}
\end{answer}
\item Verify that this map is an isomorphism:
\recommended \item Verify that this map is an isomorphism:
$\map{h}{\Re^4}{\matspace_{\nbyn{2}}}$ given by
\begin{equation*}
\colvec{a \\ b \\ c \\ d}
......@@ -803,9 +803,9 @@ partitions the collection of vector spaces.
\end{align*}
\end{answer}
\recommended \item
Decide whether each map is an isomorphism
(if it is an isomorphism then prove it and if it isn't then
state a condition that it fails to satisfy).
Decide whether each map is an isomorphism.
If it is an isomorphism then prove it and if it isn't then
state a condition that it fails to satisfy.
\begin{exparts}
\partsitem \( \map{f}{\matspace_{\nbyn{2}}}{\Re} \) given by
\begin{equation*}
......@@ -1210,7 +1210,7 @@ partitions the collection of vector spaces.
(Incidentally, any such map $a\mapsto ka$ is an isomorphism,
as is easy to check.)
\end{answer}
\recommended \item
\item
These prove that isomorphism is an equivalence relation.
\begin{exparts}
\partsitem Show that the identity map $\map{\mbox{id}}{V}{V}$ is an
......@@ -2218,7 +2218,7 @@ See \nearbyexercise{exer:FcnWellDef}.
\begin{corollary} \label{co:FiniteDimensionalIsoToReN}
%<*co:FiniteDimensionalIsoToReN>
A finite-dimensional vector space is isomorphic to one and only one
Each finite-dimensional vector space is isomorphic to one and only one
of the $\Re^n$.
%</co:FiniteDimensionalIsoToReN>
\end{corollary}
......@@ -2425,6 +2425,26 @@ Any finite-dimensional vector space is actually
\partsitem Yes, both have dimension \( 2k \).
\end{exparts}
\end{answer}
\item Which of these spaces are isomorphic to each other?
\begin{exparts*}
\partsitem $\Re^3$
\partsitem $\matspace_{\nbyn{2}}$
\partsitem $\polyspace_3$
\partsitem $\Re^4$
\partsitem $\polyspace_2$
\end{exparts*}
\begin{answer}
Just find the dimension of each space, for instance by finding a basis,
and then spaces with the same dimension are isomorphic.
This lists the dimension of each space.
\begin{exparts*}
\partsitem $3$
\partsitem $4$
\partsitem $4$
\partsitem $4$
\partsitem $3$
\end{exparts*}
\end{answer}
\recommended \item
Consider the isomorphism
\( \map{\rep{\cdot}{B}}{\polyspace_1}{\Re^2} \)
......@@ -2442,6 +2462,58 @@ Any finite-dimensional vector space is actually
\partsitem \( \colvec[r]{-1 \\ 1} \)
\end{exparts*}
\end{answer}
\item For which $n$ is the space isomorphic to~$\Re^n$?
\begin{exparts}
\partsitem $\polyspace_4$
\partsitem $\polyspace_1$
\partsitem $\matspace_{\nbym{2}{3}}$
\partsitem the plane $2x-y+z=0$ subset of~$\Re^3$
\partsitem the vector space of linear combinations of
three letters $\set{ax+by+cz\suchthat a,b,c\in\Re}$
\end{exparts}
\begin{answer}
For each, the simplest thing is to find the dimension of the space
by finding a basis.
For each basis given below, We will omit the verification that it
is a basis.
\begin{exparts}
\partsitem It is isomorphic to $\Re^5$.
One basis for $\polyspace_4$
is $\set{x^4,x^3,x^2,x,1}$ so the space has dimension~$5$.
\partsitem It is isomorphic to $\Re^2$ since one basis for the
space $\polyspace_1=\set{a+bx\suchthat a,b\in\Re}$ is
$\set{1,x}$.
\partsitem It is isomorphic to $\Re^6$.
One basis has these six matrices.
\begin{equation*}
\begin{mat}
1 &0 &0 \\
0 &0 &0
\end{mat},\;
\begin{mat}
0 &1 &0 \\
0 &0 &0
\end{mat},\;
\cdots\;
\begin{mat}
0 &0 &0 \\
0 &0 &1
\end{mat}
\end{equation*}
\partsitem It is a plane so it is isomorphic to~$\Re^2$.
For a more extensive answer, parametrizing the plane gives
this vector description
\begin{equation*}
\set{\colvec{x \\ y \\ z}=\colvec{1/2 \\ 1 \\ 0}y
+\colvec{-1/2 \\ 0 \\ 1}z
\suchthat y,z\in\Re}
\end{equation*}
and so it has a basis consisting of those two vectors.
\partsitem It is isomorphic to $\Re^3$.
One basis is the set of linear combinations
$\set{x,y,z}$, that is, $\set{x+0y+0z, 0x+y+0z,0x+0y+z}$.
\end{exparts}
\end{answer}
\recommended \item
Show that if \( m\neq n \) then \( \Re^m\not\isomorphicto\Re^n \).
\begin{answer}
......
......@@ -829,7 +829,7 @@ is more fruitful and more central to progress.
=r\vec{v} \).
\end{exparts}
\end{answer}
\recommended \item
\item
Consider the vector space \( \Re^+ \) where vector addition
and scalar multiplication are not the ones inherited from $\Re$
but rather are these:
......@@ -849,8 +849,8 @@ is more fruitful and more central to progress.
the exponential map, so it is a correspondence,
and therefore it is an isomorphism.
\end{answer}
\recommended \item
Consider this transformation of \( \Re^2 \).
\item
Consider this transformation of the plane \( \Re^2 \).
\begin{equation*}
\colvec{x \\ y} \mapsto \colvec{x/2 \\ y/3}
\end{equation*}
......@@ -1022,7 +1022,7 @@ is more fruitful and more central to progress.
\end{multline*}
where $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$
\end{answer}
\recommended \item
\item
Where \( \map{f}{V}{W} \) is linear, suppose that
\( f(\vec{v}_1)=\vec{w}_1 \), \ldots, \( f(\vec{v}_n)=\vec{w}_n \)
for some vectors \( \vec{w}_1 \), \ldots, \( \vec{w}_n \) from \( W \).
......@@ -2313,6 +2313,51 @@ point and the map is an isomorphism between the domain and the range.
\end{equation*}
for \( k\leq n \).
\end{answer}
\item For the map $\map{h}{\Re^3}{\Re^2}$ given by
\begin{equation*}
\colvec{x \\ y \\ z}
\mapsto
\colvec{x+y \\ x+z}
\end{equation*}
find the range space, rank, null space, and nullity.
\begin{answer}
To see that the range space is all of~$\Re^2$ note that for any
$u,v\in\Re$ this system has a solution.
\begin{equation*}
\begin{linsys}{3}
x &+ &y & & &= &u \\
x & & &+ &z &= &v
\end{linsys}
\grstep{-\rho_1+\rho_2}
\begin{linsys}{3}
x &+ &y & & &= &u \\
& &-y &+ &z &= &-u+v
\end{linsys}
\end{equation*}
(In fact, because there is a free variable, $z$, it has infinitely
many solutions.)
Thus the rank, the dimension of the range space, is~$2$.
Since the rank plus the nullity equals the dimension of the domain,
we know that the nullity is~$1$.
Finding the null space verifies that:
\begin{equation*}
\begin{linsys}{3}
x &+ &y & & &= &0 \\
x & & &+ &z &= &0
\end{linsys}
\grstep{-\rho_1+\rho_2}
\begin{linsys}{3}
x &+ &y & & &= &0 \\
& &-y &+ &z &= &0
\end{linsys}
\end{equation*}
the nullspace is this set
\begin{equation*}
\set{\colvec{-1 \\ 1 \\ 1}\cdot z\suchthat z \in \Re}
\end{equation*}
which is one-dimensional.
\end{answer}
\item \label{exer:CondTwoProjMap}
\nearbyexample{ex:RThreeHomoRTwo} restates the first condition in the
definition of homomorphism as `the shadow of a sum is the sum of the
......@@ -2604,7 +2649,7 @@ point and the map is an isomorphism between the domain and the range.
&=\spanof{h(S)}
\end{align*}
\end{answer}
\recommended \item \label{exer:Cosets}
\item \label{exer:Cosets}
\begin{exparts}
\partsitem Prove that for any linear map \( \map{h}{V}{W} \) and any
\( \vec{w}\in W \), the set \( h^{-1}(\vec{w}) \) has the form
......
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