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% Chapter 3, Section 2 _Linear Algebra_ Jim Hefferon
%  http://joshua.smcvt.edu/linalg.html
%  2001-Jun-11
\section{Homomorphisms}
The definition of isomorphism has two conditions.
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In this section we will consider the second one.
We will study maps that 
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are required only to preserve structure,
maps that are not also required to be correspondences. 
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Experience shows that these maps are 
tremendously useful.
For one thing we shall see in the second subsection below
that while isomorphisms describe how spaces are the same,
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we can think of these maps as describe how spaces are alike.
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\subsection{Definition}

\begin{definition}  \label{def:Homo}
A function between vector spaces \( \map{h}{V}{W} \) that 
preserves\index{preserves structure}\index{structure! preservation} 
the operations of addition
\begin{center}
  if \( \vec{v}_1,\vec{v}_2\in V \) then
      \( h(\vec{v}_1+\vec{v}_2)=h(\vec{v}_1)+h(\vec{v}_2) \)
\end{center}
and scalar multiplication
\begin{center}
      if \( \vec{v}\in V \) and \( r\in\Re \) then
      \( h(r\cdot\vec{v})=r\cdot h(\vec{v}) \)
\end{center}
is a \definend{homomorphism}\index{homomorphism}%
\index{function!structure preserving!see{homomorphism}}%
\index{vector space!homomorphism}\index{vector space!map}
or \definend{linear map}\index{linear map!see{homomorphism}}.
\end{definition}

\begin{example}    \label{ex:RThreeHomoRTwoFirst}
The projection\index{projection}
map \( \map{\pi}{\Re^3}{\Re^2} \)
\begin{equation*}
   \colvec{x \\ y \\ z}
    \mapsunder{\pi}
   \colvec{x \\ y}
\end{equation*}
is a homomorphism.
It preserves addition
\begin{equation*}
  \pi(\colvec{x_1 \\ y_1 \\ z_1}\!+\!\colvec{x_2 \\ y_2 \\ z_2})
  =
  \pi(\colvec{x_1+x_2 \\ y_1+y_2 \\ z_1+z_2})
  =
  \colvec{x_1+x_2 \\ y_1+y_2}
  =
  \pi(\colvec{x_1 \\ y_1 \\ z_1})
  +
  \pi(\colvec{x_2 \\ y_2 \\ z_2})
\end{equation*}
and scalar multiplication.
\begin{equation*}
  \pi(r\cdot\colvec{x_1 \\ y_1 \\ z_1})
  =
  \pi(\colvec{rx_1 \\ ry_1 \\ rz_1})
  =
  \colvec{rx_1 \\ ry_1}
  =
  r\cdot\pi(\colvec{x_1 \\ y_1 \\ z_1})
\end{equation*}
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This is not an isomorphism since it is not one-to-one. 
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For instance, both $\zero$ and $\vec{e}_3$ in $\Re^3$ map to
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the zero vector in $\Re^2$.
\end{example}

\begin{example} \label{exam:TwoMapsHomoNotIso}
Of course, the domain and codomain 
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can be other than spaces of column vectors.
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Both of these are homomorphisms;
the verifications are straightforward.
\begin{enumerate}
  \item \( \map{f_1}{\polyspace_2}{\polyspace_3} \) given by
    \begin{equation*}
      a_0+a_1x+a_2x^2 \;\mapsto\; a_0x+(a_1/2)x^2+(a_2/3)x^3 
    \end{equation*}
  \item \( \map{f_2}{M_{\nbyn{2}}}{\Re} \) given by
    \begin{equation*}
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      \begin{mat}
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        a  &b  \\
        c  &d
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      \end{mat}
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        \mapsto
      a+d
    \end{equation*}
\end{enumerate}
\end{example}

\begin{example}
Between any two spaces there is a \definend{zero homomorphism},%
\index{zero homomorphism}\index{homomorphism!zero}\index{function!zero}
mapping every vector in the domain to the zero vector in the codomain.
\end{example}

\begin{example}
These two suggest why we use the term `linear map'.
\begin{enumerate}
  \item The map \( \map{g}{\Re^3}{\Re} \) given by
    \begin{equation*}
      \colvec{x \\ y \\ z}
        \mapsunder{g}
      3x+2y-4.5z
    \end{equation*}
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    is linear, that is, is a homomorphism.
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    In contrast, the map \( \map{\hat{g}}{\Re^3}{\Re} \) given by
    \begin{equation*}
      \colvec{x \\ y \\ z}
        \mapsunder{\hat{g}}
      3x+2y-4.5z+1
    \end{equation*}
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    is not.
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    \begin{equation*}
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      \hat{g}(\colvec[r]{0 \\ 0 \\ 0}+\colvec[r]{1 \\ 0 \\ 0})=4
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      \qquad
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      \hat{g}(\colvec[r]{0 \\ 0 \\ 0})
      +\hat{g}(\colvec[r]{1 \\ 0 \\ 0})=5
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    \end{equation*}
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    To show that a map is not linear we need only produce a single
    linear combination that the map does not preserve.
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  \item The first of these two maps 
    \( \map{t_1,t_2}{\Re^3}{\Re^2} \)
    is linear while the second is not.
    \begin{equation*}
      \colvec{x \\ y \\ z}
        \mapsunder{t_1}
      \colvec{5x-2y \\ x+y}
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      \qquad
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      \colvec{x \\ y \\ z}
        \mapsunder{t_2}
      \colvec{5x-2y \\ xy}
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    \end{equation*}
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    Finding a linear combination that the second map does not 
    preserve is easy. 
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\end{enumerate}
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The homomorphisms have 
coordinate functions that are linear combinations of the arguments.
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% See also \nearbyexercise{exer:GrpahNotALine}.
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\end{example}

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Any isomorphism is a
homomorphism, since an isomorphism is a 
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homomorphism that is also a correspondence.
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So one way to think of `homomorphism' 
is as a generalization of `isomorphism'
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motivated by the observation that many of the properties of
isomorphisms have only to do with the map's structure preservation property
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and not to do with being a correspondence.
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The next two results are examples of that thinking.
The proof for each given in the prior section
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does not use one-to-one-ness or onto-ness  
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and therefore applies here.
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\begin{lemma}       \label{le:HomoSendsZeroToZero}
A homomorphism sends a zero vector to a zero vector.
\end{lemma}

\begin{lemma}  \label{le:HomoPreserveLinCombo}
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For any map 
\( \map{f}{V}{W} \) 
between vector spaces, the following are equivalent.
\begin{tfae}
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  \item 
      $f$ is a homomorphism 
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  \item 
      $f(c_1\cdot\vec{v}_1+c_2\cdot\vec{v}_2)
      =c_1\cdot f(\vec{v}_1)+c_2\cdot f(\vec{v}_2)$
      for any \( c_1,c_2\in\Re \) and \( \vec{v}_1,\vec{v}_2\in V \)
  \item
    $f(c_1\cdot\vec{v}_1+\dots+c_n\cdot\vec{v}_n)
    =c_1\cdot f(\vec{v}_1)+\dots+c_n\cdot f(\vec{v}_n)$ 
    for any \( c_1,\dots,c_n\in\Re \) and
    \( \vec{v}_1,\ldots,\vec{v}_n\in V \)
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\end{tfae}
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\end{lemma}

\begin{example}
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The function \( \map{f}{\Re^2}{\Re^4} \) given by
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\begin{equation*}
  \colvec{x \\ y}
    \mapsunder{f}
  \colvec{x/2 \\ 0 \\ x+y \\ 3y}
\end{equation*}
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is linear since it satisfies item~(2).
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\begin{equation*}
  \colvec{r_1(x_1/2)+r_2(x_2/2) \\ 0 \\ 
                 r_1(x_1+y_1)+r_2(x_2+y_2) \\ r_1(3y_1)+r_2(3y_2)}
   =
  r_1\colvec{x_1/2 \\ 0 \\ x_1+y_1 \\ 3y_1}
   +
  r_2\colvec{x_2/2 \\ 0 \\ x_2+y_2 \\ 3y_2}
\end{equation*}
\end{example}

However,
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some of the things that we have seen for isomorphisms fail to hold for
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homomorphisms in general.
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One example is the proof of Lemma~I.\ref{lem:IsoImpliesSameDim}, 
which shows that an isomorphism between spaces gives 
a correspondence between their bases.
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Homomorphisms do not give any such correspondence;
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\nearbyexample{ex:RThreeHomoRTwoFirst} shows this and another example is
the zero map between two nontrivial spaces. 
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Instead, for homomorphisms a weaker but still very useful result holds.

\begin{theorem}
\label{th:HomoDetActOnBasis}
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A homomorphism is determined by its action on a basis:~if
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\( \sequence{\vec{\beta}_1,\dots,\vec{\beta}_n} \)
is a basis of a vector space \( V \) and
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\( \vec{w}_1,\dots,\vec{w}_n \) are elements of
a vector space \( W \)  (perhaps not distinct elements) then
there exists a homomorphism from \( V \) to \( W \) sending each
\( \vec{\beta}_i \) to \( \vec{w}_i \), and that homomorphism is unique.
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\end{theorem}

\begin{proof}
We will define the map by 
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associating $\vec{\beta}_i$ with $\vec{w}_i$
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and then extending linearly to all of the domain.
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That is, given the input $\vec{v}$, we find its coordinates with
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respect to the basis
\( \vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n \) and define
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the associated output by using the same $c_i$ coordinates
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$h(\vec{v})=c_1\vec{w}_1+\dots+c_n\vec{w}_n$.
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This is a well-defined function because, with respect to the basis,
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the representation of each domain vector \( \vec{v} \) is unique.

This map is a homomorphism since it preserves linear combinations;
where \( \vec{v_1}=c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n \) and
\( \vec{v_2}=d_1\vec{\beta}_1+\cdots+d_n\vec{\beta}_n \),
we have this.
\begin{align*}
  h(r_1\vec{v}_1+r_2\vec{v}_2)
  &=h((r_1c_1+r_2d_1)\vec{\beta}_1+\dots+(r_1c_n+r_2d_n)\vec{\beta}_n)  \\
  &=(r_1c_1+r_2d_1)\vec{w}_1+\dots+(r_1c_n+r_2d_n)\vec{w}_n   \\
  &=r_1h(\vec{v}_1)+r_2h(\vec{v}_2)
\end{align*}

And, this map is unique since if \( \map{\hat{h}}{V}{W} \) 
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is another homomorphism satisfying that \( \hat{h}(\vec{\beta}_i)=\vec{w}_i \) 
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for each \( i \)
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then \( h \) and \( \hat{h} \) agree on all of the vectors in the domain. 
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\begin{multline*}
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  \hat{h}(\vec{v})
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  =\hat{h}(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n)  
  =c_1\hat{h}(\vec{\beta}_1)+\dots+c_n\hat{h}(\vec{\beta}_n)  \\  
  =c_1\vec{w}_1+\dots+c_n\vec{w}_n 
  =h(\vec{v})
\end{multline*}
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Thus, $h$ and $\hat{h}$ are the same map.
\end{proof}

\begin{example}
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If we specify a map \( \map{h}{\Re^2}{\Re^2} \)
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that acts on the standard basis $\stdbasis_2$ in this way 
\begin{equation*}
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  h(\colvec[r]{1 \\ 0})=\colvec[r]{-1 \\ 1}
  \qquad
  h(\colvec[r]{0 \\ 1})=\colvec[r]{-4 \\ 4}
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\end{equation*}
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then we have also specified the action of $h$ on any other member of the domain.
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For instance,
the value of $h$ on this argument
\begin{equation*}
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  h(\colvec[r]{3 \\ -2})=h(3\cdot \colvec[r]{1 \\ 0}-2\cdot \colvec[r]{0 \\ 1})
                      =3\cdot h(\colvec[r]{1 \\ 0})-2\cdot h(\colvec[r]{0 \\ 1})
                      =\colvec[r]{5 \\ -5}
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\end{equation*}
is a direct consequence of the value of $h$ on the basis vectors.
\end{example}

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So we can construct a homomorphism 
by selecting a basis for the 
domain and specifying where the map sends those basis vectors.
The prior lemma shows that we can always extend the action on the map 
linearly to the entire domain.
Later in this chapter we shall develop a convenient scheme for computations 
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like this one, using matrices.

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Just as the isomorphisms of a space with itself are useful and interesting, 
so too are the homomorphisms of a space with itself.

\begin{definition}
A linear map from a space into itself \( \map{t}{V}{V} \) is a
{\em linear transformation}\index{linear transformation!see{transformation}}.
\end{definition}

\begin{remark}
In this book we use `linear transformation' only in the case where 
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the codomain equals the domain but it is often used instead as a
synonym for `homomorphism'.
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\end{remark}

\begin{example}
The map on $\Re^2$ that projects all vectors down to the $x$-axis
\begin{equation*}
  \colvec{x \\ y}\mapsto\colvec{x \\ 0}
\end{equation*}
is a linear transformation.
\end{example}

\begin{example}
The derivative map \( \map{d/dx}{\polyspace_n}{\polyspace_n} \)
\begin{equation*}
  a_0+a_1x+\cdots+a_nx^n
    \mapsunder{d/dx}
  a_1+2a_2x+3a_3x^2+\cdots+na_nx^{n-1}
\end{equation*}
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is a linear transformation as this result from calculus shows:
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\( d(c_1f+c_2g)/dx=c_1\,(df/dx)+c_2\,(dg/dx) \).
\end{example}

\begin{example} \label{ex:MatTransMapLinear}
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The matrix transpose operation
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\begin{equation*}
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  \begin{mat}
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    a  &b  \\
    c  &d
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  \end{mat}
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  \;\mapsto\;
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  \begin{mat}
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    a  &c  \\
    b  &d
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  \end{mat}
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\end{equation*}
is a linear transformation of \( \matspace_{\nbyn{2}} \).
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(Transpose is one-to-one and onto and so in fact it is
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an automorphism.)
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\end{example}

We finish this subsection about maps by recalling that
we can linearly combine maps.
For instance, for these maps from \( \Re^2 \) to itself
\begin{equation*}
  \colvec{x \\ y}
   \mapsunder{f}
  \colvec{2x \\ 3x-2y}
  \quad\text{and}\quad
  \colvec{x \\ y}
   \mapsunder{g}
  \colvec{0 \\ 5x}
\end{equation*}
the linear combination \( 5f-2g \) is also a map from 
$R^2$ to itself.
\begin{equation*}
  \colvec{x \\ y}
   \mapsunder{5f-2g}
  \colvec{10x \\ 5x-10y}
\end{equation*}

\begin{lemma} \label{le:SpLinFcns}
For vector spaces \( V \) and \( W \),
the set of linear functions from \( V \) to
\( W \) is itself a vector space, a subspace of the space of all functions
from \( V \) to \( W \).
\end{lemma}

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\noindent We denote the space of linear maps with
\( \linmaps{V}{W} \)\index{linear maps!space of}.

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\begin{proof}
This set is non-empty because it contains the zero homomorphism.
So to show that it is a subspace we need only check that it is 
closed under linear combinations.
Let \( \map{f,g}{V}{W} \) be linear. 
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Then the sum of the two is linear
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\begin{align*}
   (f+g)(c_1\vec{v}_1+c_2\vec{v}_2)
   &=f(c_1\vec{v}_1+c_2\vec{v}_2) + 
   g(c_1\vec{v}_1+c_2\vec{v}_2)       \\
   &=c_1f(\vec{v}_1)+c_2f(\vec{v}_2)
   +c_1g(\vec{v}_1)+c_2g(\vec{v}_2)   \\
   &=c_1\bigl(f+g\bigr)(\vec{v}_1)+c_2\bigl(f+g\bigr)(\vec{v}_2)
\end{align*}
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and any scalar multiple of a map is also linear.
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\begin{align*}
   (r\cdot f)(c_1\vec{v}_1+c_2\vec{v}_2)
   &=r(c_1f(\vec{v}_1)+c_2f(\vec{v}_2))  \\
   &=c_1(r\cdot f)(\vec{v}_1)+c_2(r\cdot f)(\vec{v}_2)
\end{align*}
Hence \( \linmaps{V}{W} \) is a subspace.
\end{proof}

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We started this section by 
defining homomorphisms as a generalization of isomorphisms,
isolating the structure preservation property.
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Some of the properties of isomorphisms carried over unchanged
while we adapted others.
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However, if we thereby get an impression that 
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the idea of 
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`homomorphism' is in some way secondary to
that of `isomorphism' then that is mistaken.
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In the rest of this chapter we shall work mostly with homomorphisms.
This is
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partly because any statement made about homomorphisms is automatically true 
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about isomorphisms but more because, 
while the isomorphism concept is more natural, 
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experience shows that the homomorphism concept 
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is more fruitful and more central to further progress.
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\begin{exercises}
  \recommended \item
    Decide if each \( \map{h}{\Re^3}{\Re^2} \) is linear.
    \begin{exparts*}
      \partsitem \( h(\colvec{x \\ y \\ z})=\colvec{x \\ x+y+z}  \)
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      \partsitem \( h(\colvec{x \\ y \\ z})=\colvec[r]{0 \\ 0}  \)
      \partsitem \( h(\colvec{x \\ y \\ z})=\colvec[r]{1 \\ 1}  \)
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      \partsitem \( h(\colvec{x \\ y \\ z})=\colvec{2x+y \\ 3y-4z}  \)
    \end{exparts*}
    \begin{answer}
      \begin{exparts}
        \partsitem Yes.
          The verification is straightforward.
          \begin{align*}
            h( c_1\cdot\colvec{x_1 \\ y_1 \\ z_1}
               +c_2\cdot\colvec{x_2 \\ y_2 \\ z_2} )
            &=h( \colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2} ) \\
            &=\colvec{c_1x_1+c_2x_2 \\ 
                     c_1x_1+c_2x_2+c_1y_1+c_2y_2+c_1z_1+c_2z_2}          \\
            &=c_1\cdot\colvec{x_1 \\ x_1+y_1+z_1}
               +c_2\cdot\colvec{x_2 \\ c_2+y_2+z_2}                       \\  
            &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) 
               +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2})
          \end{align*}
        \partsitem Yes.
          The verification is easy.
          \begin{align*}
            h(c_1\cdot\colvec{x_1 \\ y_1 \\ z_1}
              +c_2\cdot\colvec{x_2 \\ y_2 \\ z_2})
            &=h( \colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2} ) \\
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            &=\colvec[r]{0 \\ 0}                                      \\
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            &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) 
               +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2})            
          \end{align*}
        \partsitem No.
          An example of an addition that is not respected is this.
          \begin{equation*}
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            h(\colvec[r]{0 \\ 0 \\ 0}+\colvec[r]{0 \\ 0 \\ 0})
            =\colvec[r]{1 \\ 1}
            \neq h(\colvec[r]{0 \\ 0 \\ 0})+h(\colvec[r]{0 \\ 0 \\ 0})
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          \end{equation*}
        \partsitem Yes.
           The verification is straightforward.
          \begin{align*}
            h( c_1\cdot\colvec{x_1 \\ y_1 \\ z_1}
               +c_2\cdot\colvec{x_2 \\ y_2 \\ z_2} )
            &=h( \colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2} ) \\
            &=\colvec{2(c_1x_1+c_2x_2)+(c_1y_1+c_2y_2) \\ 
                      3(c_1y_1+c_2y_2)-4(c_1z_1+c_2z_2)}          \\
            &=c_1\cdot\colvec{2x_1+y_1 \\ 3y_1-4z_1}
               +c_2\cdot\colvec{2x_2+y_2 \\ 3y_2-4z_2}            \\  
            &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) 
               +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2})
          \end{align*}
      \end{exparts}  
    \end{answer}
  \recommended \item
    Decide if each map \( \map{h}{\matspace_{\nbyn{2}}}{\Re} \) is linear.
    \begin{exparts}
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      \partsitem \( h(\begin{mat} a &b  \\ c &d \end{mat})=a+d  \)
      \partsitem \( h(\begin{mat} a &b  \\ c &d \end{mat})=ad-bc  \)
      \partsitem \( h(\begin{mat} a &b \\ c &d \end{mat})=2a+3b+c-d  \)
      \partsitem \( h(\begin{mat} a &b  \\ c &d \end{mat})=a^2+b^2  \)
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    \end{exparts}
    \begin{answer} 
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      For each, we must either check that the map preserves linear combinations
      or give an example of a linear combination that is 
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      not.
      \begin{exparts*}
        \partsitem Yes.
           The check that it preserves combinations is routine.
           \begin{align*}
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             h(r_1\cdot\begin{mat}
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                 a_1  &b_1  \\
                 c_1  &d_1
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               \end{mat}
              +r_2\cdot\begin{mat}
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                 a_2  &b_2  \\
                 c_2  &d_2
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               \end{mat})
              &=h(\begin{mat}
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                 r_1a_1+r_2a_2  &r_1b_1+r_2b_2  \\
                 r_1c_1+r_2c_2  &r_1d_1+r_2d_2
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               \end{mat})                    \\
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              &=(r_1a_1+r_2a_2)+(r_1d_1+r_2d_2)    \\
              &=r_1(a_1+d_1)+r_2(a_2+d_2)          \\
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              &=r_1\cdot h(\begin{mat}
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                             a_1  &b_1  \\
                             c_1  &d_1
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                            \end{mat})
                +r_2\cdot h(\begin{mat}
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                             a_2  &b_2  \\
                             c_2  &d_2
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                            \end{mat})
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           \end{align*}
        \partsitem No.
          For instance, not preserved is multiplication by the scalar $2$.
          \begin{equation*}
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            h(2\cdot\begin{mat}[r]
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                1  &0  \\
                0  &1  
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              \end{mat})
            =h(\begin{mat}[r]
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                2  &0  \\
                0  &2  
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              \end{mat})
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            =4
            \quad\text{while}\quad
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            2\cdot h(\begin{mat}[r]
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                1  &0  \\
                0  &1  
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              \end{mat})
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            =2\cdot 1=2
          \end{equation*}
        \partsitem Yes.
           This is the check that it preserves combinations of two members of
           the domain.
           \begin{align*}
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             h(r_1\cdot\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat}
               +r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat})
             &=h(\begin{mat} 
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                   r_1a_1+r_2a_2 &r_1b_1+r_2b_2   \\ 
                   r_1c_1+r_2c_2 &r_1d_1+r_2d_2 
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                 \end{mat})                                 \\
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             &=2(r_1a_1+r_2a_2)+3(r_1b_1+r_2b_2)
                    +(r_1c_1+r_2c_2)-(r_1d_1+r_2d_2)              \\
             &=r_1(2a_1+3b_1+c_1-d_1)
               +r_2(2a_2+3b_2+c_2-d_2)              \\
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             &=r_1\cdot h(\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat}
               +r_2\cdot h(\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat})
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           \end{align*}
        \partsitem No.
          An example of a combination that is not preserved is this.
          \begin{equation*}
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            h(\begin{mat}[r]
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              1  &0  \\
              0  &0
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            \end{mat}
            +\begin{mat}[r]
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              1  &0  \\
              0  &0
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            \end{mat})
           =h(\begin{mat}[r]
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              2  &0  \\
              0  &0
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            \end{mat})
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           =4
           \quad\text{while}\quad
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            h(\begin{mat}[r]
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              1  &0  \\
              0  &0
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            \end{mat})
            +h(\begin{mat}[r]
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              1  &0  \\
              0  &0
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            \end{mat})
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            =1+1
            =2
          \end{equation*}
      \end{exparts*}   
    \end{answer}
  \recommended \item
    Show that these two maps are homomorphisms.
    \begin{exparts}
      \partsitem \( \map{d/dx}{\polyspace_3}{\polyspace_2} \)
        given by \( a_0+a_1x+a_2x^2+a_3x^3 \) maps to
        \( a_1+2a_2x+3a_3x^2 \)
      \partsitem \( \map{\int}{\polyspace_2}{\polyspace_3} \) given by
        \( b_0+b_1x+b_2x^2 \) maps to \( b_0x+(b_1/2)x^2+(b_2/3)x^3 \)
    \end{exparts}
    Are these maps inverse to each other?
    \begin{answer}
      The check that each is a homomorphisms is routine.
      Here is the check for the differentiation map.
      \begin{multline*}
        \frac{d}{dx}(r\cdot (a_0+a_1x+a_2x^2+a_3x^3)
                    +s\cdot (b_0+b_1x+b_2x^2+b_3x^3))                 \\
        \begin{aligned}
           &=\frac{d}{dx}((ra_0+sb_0)+(ra_1+sb_1)x+(ra_2+sb_2)x^2
                                                    +(ra_3+sb_3)x^3)  \\
           &=(ra_1+sb_1)+2(ra_2+sb_2)x+3(ra_3+sb_3)x^2         \\
           &=r\cdot (a_1+2a_2x+3a_3x^2)+s\cdot (b_1+2b_2x+3b_3x^2)         \\
           &=r\cdot \frac{d}{dx}(a_0+a_1x+a_2x^2+a_3x^3)
                    +s\cdot \frac{d}{dx} (b_0+b_1x+b_2x^2+b_3x^3)
        \end{aligned}
      \end{multline*}
      (An alternate proof is to simply note that this is a 
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      property of differentiation that is familiar from calculus.)
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      These two maps are not inverses as this composition 
      does not act as the identity map on 
      this element of the domain.
      \begin{equation*}
        1\in\polyspace_3\;\mapsunder{d/dx}\;
        0\in\polyspace_2\;\mapsunder{\int}\;
        0\in\polyspace_3
      \end{equation*}   
     \end{answer}
  \item  
    Is (perpendicular) projection from \( \Re^3 \) to the \( xz \)-plane
    a homomorphism?
    Projection to the \( yz \)-plane?
    To the \( x \)-axis?
    The \( y \)-axis?
    The \( z \)-axis?
    Projection to the origin?
    \begin{answer}
      Each of these projections is a homomorphism.
      Projection to the $xz$-plane and to the $yz$-plane are these maps.
      \begin{equation*}
         \colvec{x \\ y \\ z}\mapsto\colvec{x \\ 0 \\ z}
           \qquad
         \colvec{x \\ y \\ z}\mapsto\colvec{0 \\ y \\ z}
      \end{equation*}
      Projection to the $x$-axis, to the $y$-axis, and to the $z$-axis are
      these maps. 
      \begin{equation*}
         \colvec{x \\ y \\ z}\mapsto\colvec{x \\ 0 \\ 0}
           \qquad
         \colvec{x \\ y \\ z}\mapsto\colvec{0 \\ y \\ 0}
           \qquad
         \colvec{x \\ y \\ z}\mapsto\colvec{0 \\ 0 \\ z}
      \end{equation*}
      And projection to the origin is this map.
      \begin{equation*}
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         \colvec{x \\ y \\ z}\mapsto\colvec[r]{0 \\ 0 \\ 0}
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      \end{equation*}
      Verification that each is a homomorphism is straightforward.
      (The last one, of course, is the zero transformation on $\Re^3$.)  
     \end{answer}
  \item 
    Show that, while the maps from \nearbyexample{exam:TwoMapsHomoNotIso}
    preserve linear operations, they are not isomorphisms.
    \begin{answer}
      The first is not onto; for instance, there is no polynomial that is
      sent the constant polynomial $p(x)=1$.
      The second is not  one-to-one; both of these members of the domain
      \begin{equation*}
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        \begin{mat}[r]
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          1  &0  \\
          0  &0  
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        \end{mat}
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        \quad\text{and}\quad
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        \begin{mat}[r]
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          0  &0  \\
          0  &1  
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        \end{mat}
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      \end{equation*}
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      map to the same member of the codomain, $1\in\Re$.
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     \end{answer}
  \item 
    Is an identity map a linear transformation?
    \begin{answer}
      Yes;~in any space \( \text{id}(c\cdot \vec{v}+d\cdot \vec{w})
      = c\cdot \vec{v}+d\cdot \vec{w}
      = c\cdot\text{id}(\vec{v})+d\cdot\text{id}(\vec{w}) \).
    \end{answer}
  \recommended \item \label{exer:GrpahNotALine}
    Stating that a function is `linear' is different than 
    stating that its graph is a line.
    \begin{exparts}
      \partsitem The function \( \map{f_1}{\Re}{\Re} \) given by 
        \( f_1(x)=2x-1 \) has a graph that is a line.
        Show that it is not a linear function.
      \partsitem The function \( \map{f_2}{\Re^2}{\Re} \) given by
        \begin{equation*}
          \colvec{x \\ y} \mapsto x+2y
        \end{equation*}
        does not have a graph that is a line.
        Show that it is a linear function.
    \end{exparts}
    \begin{answer}
      \begin{exparts}
         \partsitem This map does not preserve structure since
           \( f(1+1)=3 \), while \( f(1)+f(1)=2 \).
         \partsitem The check is routine.
           \begin{align*}
             f(r_1\cdot\colvec{x_1 \\ y_1}+r_2\cdot\colvec{x_2 \\ y_2})
             &=f(\colvec{r_1x_1+r_2x_2 \\ r_1y_1+r_2y_2})               \\
             &=(r_1x_1+r_2x_2)+2(r_1y_1+r_2y_2)                          \\
             &=r_1\cdot (x_1+2y_1)+r_2\cdot (x_2+2y_2)                   \\
             &=r_1\cdot f(\colvec{x_1 \\ y_1})+r_2\cdot f(\colvec{x_2 \\ y_2})
           \end{align*}
      \end{exparts}   
     \end{answer}
  \recommended \item 
    Part of the definition of a linear function is that it respects
    addition. 
    Does a linear function respect subtraction?
    \begin{answer}
      Yes.
      Where \( \map{h}{V}{W} \) is linear,
      \( h(\vec{u}-\vec{v})
         =h(\vec{u}+(-1)\cdot\vec{v})
         =h(\vec{u})+(-1)\cdot h(\vec{v})
         =h(\vec{u})-h(\vec{v}) \).
    \end{answer}
  \item 
    Assume that \( h \) is a linear transformation of \( V \) and that
    \( \sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n} \) is a basis of \( V \).
    Prove each statement.
    \begin{exparts}
      \partsitem If \( h(\vec{\beta}_i)=\zero \) for each basis vector
        then \( h \) is the zero map.
      \partsitem If \( h(\vec{\beta}_i)=\vec{\beta}_i \) for each basis
        vector then \( h \) is the identity map.
      \partsitem If there is a scalar \( r \) such that
        \( h(\vec{\beta}_i)=r\cdot\vec{\beta}_i \) for each
        basis vector then \( h(\vec{v})=r\cdot\vec{v} \) for all vectors
        in $V$.
    \end{exparts}
    \begin{answer}
      \begin{exparts}
       \partsitem Let \( \vec{v}\in V \) be represented with respect to the 
         basis as \( \vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n \).
         Then \( h(\vec{v})=h(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n)
                  =c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n)
                  =c_1\cdot\zero+\dots+c_n\cdot\zero
                  =\zero \).
       \partsitem This argument is similar to the prior one.
         Let \( \vec{v}\in V \) be represented with respect to the 
         basis as \( \vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n \).
         Then \( h(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n)
            =c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n)
            =c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n
            =\vec{v} \).
       \partsitem As above, only 
         \( c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n)
            =c_1r\vec{\beta}_1+\dots+c_nr\vec{\beta}_n
            =r(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n)
            =r\vec{v} \).
      \end{exparts}  
     \end{answer}
  \recommended \item
    Consider the vector space \( \Re^+ \) where vector addition
    and scalar multiplication are not the ones inherited from $\Re$
    but rather are these:
    \( a+b \) is the product of \( a \) and \( b \), and \( r\cdot a \) is the
    \( r \)-th power of \( a \).
    (This was shown to be a vector space in an earlier exercise.)
    Verify that the natural logarithm map \( \map{\ln}{\Re^+}{\Re} \) 
    is a homomorphism between these two spaces.
    Is it an isomorphism?
    \begin{answer}
      That it is a homomorphism follows from the familiar rules that
      the logarithm of a product is the sum of the logarithms
      $\ln(ab)=\ln(a)+\ln(b)$ 
      and that the logarithm of a power is the multiple of the logarithm
      $\ln(a^r)=r\ln(a)$.
      This map is an isomorphism because it has an inverse, namely, 
      the exponential map, so it is a correspondence,
      and therefore it is an isomorphism.  
     \end{answer}
  \recommended \item 
     Consider this transformation of \( \Re^2 \).
     \begin{equation*}
       \colvec{x \\ y} \mapsto \colvec{x/2 \\ y/3}
     \end{equation*}
     Find the image under this map of this ellipse.
     \begin{equation*}
       \set{\colvec{x \\ y} \suchthat (x^2/4)+(y^2/9)=1}
     \end{equation*}
     \begin{answer}
       Where \( \hat{x}=x/2 \) and \( \hat{y}=y/3 \),
       the image set is
       \begin{equation*}
         \set{\colvec{\hat{x} \\ \hat{y}} \suchthat
           \frac{\displaystyle (2\hat{x})^2}{\displaystyle 4}
           +\frac{\displaystyle (3\hat{y})^2}{\displaystyle 9}=1}
         =\set{\colvec{\hat{x} \\ \hat{y}} \suchthat
            \hat{x}^2+\hat{y}^2=1}
       \end{equation*}
       the unit circle in the \( \hat{x}\hat{y} \)-plane.
     \end{answer}
  \recommended \item
    Imagine a rope wound around the earth's equator so that it fits snugly
    (suppose that the earth is a sphere).
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    How much extra rope must we add to raise the circle to a constant
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    six feet off the ground?
    \begin{answer}
      The circumference function $r\mapsto 2\pi r$ is linear.  
      Thus we have
      $2\pi\cdot (r_{\text{earth}}+6)-
         2\pi\cdot (r_{\text{earth}})=12\pi$.
      Observe that
      it takes the same amount of extra rope to raise the circle from tightly 
      wound around a basketball to six feet above that basketball as it does
      to raise it from tightly wound around the earth to six feet above the
      earth.
     \end{answer}
  \recommended \item 
    Verify that this map \( \map{h}{\Re^3}{\Re} \)
    \begin{equation*}
      \colvec{x \\ y \\ z}\;\mapsto\;
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      \colvec{x \\ y \\ z}\dotprod\colvec[r]{3 \\ -1 \\ -1}=3x-y-z
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    \end{equation*}
    is linear.
    Generalize.
    \begin{answer}
      Verifying that it is linear is routine.
      \begin{align*}
        h(c_1\cdot \colvec{x_1 \\ y_1 \\ z_1}
          +c_2\cdot \colvec{x_2 \\ y_2 \\ z_2})
        &=h(\colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2})   \\
        &=3(c_1x_1+c_2x_2)-(c_1y_1+c_2y_2)-(c_1z_1+c_2z_2)             \\
        &=c_1\cdot (3x_1-y_1-z_1)+c_2\cdot (3x_2-y_2-z_2)             \\
        &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1})
          +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2})
      \end{align*}
      The natural guess at a generalization 
      is that for any fixed \( \vec{k}\in\Re^3 \)
      the map \( \vec{v}\mapsto\vec{v}\dotprod\vec{k} \) is linear.
      This statement is true.
      It follows from properties of the dot product we have seen earlier:
      \( (\vec{v}+\vec{u})\dotprod\vec{k}=\vec{v}\dotprod\vec{k}+
         \vec{u}\dotprod\vec{k} \)
      and \( (r\vec{v})\dotprod\vec{k}=r(\vec{v}\dotprod\vec{k}) \).
      (The natural guess at a generalization of this generalization, 
      that the map from \( \Re^n \)
      to \( \Re \) whose action consists of taking the dot product of its
      argument with a  fixed vector \( \vec{k}\in\Re^n \) is linear, 
      is also true.) 
    \end{answer}
  \item \label{exer:HomoRONeMultByScalar}
    Show that every homomorphism from \( \Re^1 \) to \( \Re^1 \) 
    acts via multiplication by a scalar.
    Conclude that every nontrivial linear transformation of \( \Re^1 \) is an
    isomorphism.
    Is that true for transformations of \( \Re^2 \)?
    \( \Re^n \)?
    \begin{answer}
      Let \( \map{h}{\Re^1}{\Re^1} \) be linear.
      A linear map is determined by its action on a basis, so fix the basis
      \( \sequence{1} \)  for \( \Re^1 \).
      For any \( r\in\Re^1 \) we have that \( h(r)=h(r\cdot 1)=r\cdot h(1) \)
      and so \( h \) acts on any argument $r$ 
      by multiplying it by the constant \( h(1) \).
      If \( h(1) \) is not zero then the map is a correspondence\Dash its 
      inverse is division by \( h(1) \)\Dash so any nontrivial transformation 
      of $\Re^1$ is an isomorphism.

      This projection map is an example that shows that not every
      transformation of \( \Re^n \) acts via multiplication by a constant
      when \( n>1 \), including when $n=2$.
      \begin{equation*}
        \colvec{x_1 \\ x_2 \\ \vdots \\ x_n}
          \mapsto\colvec{x_1 \\ 0 \\ \vdots \\ 0}
      \end{equation*}  
    \end{answer}
  \item 
    %(This will be used in \nearbyexercise{exer:Cosets} below.)
    \begin{exparts}
      \partsitem Show that for any scalars \( a_{1,1},\dots, a_{m,n}  \) 
        this map
        \( \map{h}{\Re^n}{\Re^m} \) is a homomorphism.
        \begin{equation*}
          \colvec{x_1 \\ \vdots \\ x_n}
            \mapsto
          \colvec{a_{1,1}x_1+\dots+a_{1,n}x_n \\ 
                       \vdots \\
                       a_{m,1}x_1+\cdots+a_{m,n}x_n}
        \end{equation*}
      \partsitem   Show that for each $i$, the \( i \)-th derivative operator 
        $d^i/dx^i$ is a linear transformation of \( \polyspace_n \).
        Conclude that for any scalars \( c_k,\ldots, c_0 \) this map
        is a linear transformation of that space.
        \begin{equation*}
          f\mapsto
          \frac{d^k}{dx^k}f+c_{k-1}\frac{d^{k-1}}{dx^{k-1}}f
             +\dots+
          c_1\frac{d}{dx}f+c_0f
        \end{equation*}
    \end{exparts}
    \begin{answer}
      \begin{exparts}
        \partsitem Where \( c \) and \( d \) are scalars,
          we have this.
          \begin{align*}
            h(c\cdot \colvec{x_1 \\ \vdots \\ x_n} 
              +d\cdot \colvec{y_1 \\ \vdots \\ y_n})
            &=h(\colvec{cx_1+dy_1 \\ \vdots \\ cx_n+dy_n})        \\
            &=\colvec{a_{1,1}(cx_1+dy_1)+\dots+a_{1,n}(cx_n+dy_n) \\
                         \vdots                                   \\
                         a_{m,1}(cx_1+dy_1)+\dots+a_{m,n}(cx_n+dy_n)} \\
            &=c\cdot\colvec{a_{1,1}x_1+\dots+a_{1,n}x_n \\ 
                         \vdots                     \\ 
                         a_{m,1}x_1+\dots+a_{m,n}x_n}
            +d\cdot\colvec{a_{1,1}y_1+\dots+a_{1,n}y_n \\ 
                         \vdots \\ 
                         a_{m,1}y_1+\dots+a_{m,n}y_n} \\
            &=c\cdot h(\colvec{x_1 \\ \vdots \\ x_n})
              +d\cdot h(\colvec{y_1 \\ \vdots \\ y_n})
          \end{align*}
        \partsitem Each power $i$ of the derivative operator is linear 
          because of these rules familiar from calculus.
          \begin{equation*}
            \frac{d^i}{dx^i}(\,f(x)+g(x)\,)=\frac{d^i}{dx^i}f(x)
                                         +\frac{d^i}{dx^i}g(x)
            \quad\text{and}\quad
            \frac{d^i}{dx^i}\,r\cdot f(x)=r\cdot\frac{d^i}{dx^i}f(x)
          \end{equation*}
          Thus the given map is a linear transformation of \( \polyspace_n \)
          because any linear combination of linear maps is also a linear map.
      \end{exparts}  
     \end{answer}
  \item 
    \nearbylemma{le:SpLinFcns} shows that a sum of linear functions is
    linear and that a scalar multiple of a linear function is linear.
    Show also that a composition of linear functions is linear.
    \begin{answer}
      (This argument has already appeared, as part of the proof that
      isomorphism is an equivalence.)
      Let $\map{f}{U}{V}$ and $\map{g}{V}{W}$ be linear.
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      The composition preserves linear combinations
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      \begin{multline*}
        \composed{g}{f}(c_1\vec{u}_1+c_2\vec{u}_2)
        =g(\,f(c_1\vec{u}_1+c_2\vec{u}_2)\,)          
        =g(\,c_1f(\vec{u}_1)+c_2f(\vec{u}_2)\,)              \\          
        =c_1\cdot g(f(\vec{u}_1))+c_2\cdot g(f(\vec{u}_2))          
        =c_1\cdot \composed{g}{f}(\vec{u}_1)
           +c_2\cdot \composed{g}{f}(\vec{u}_2)          
      \end{multline*}
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      where $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$ 
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    \end{answer}
  \recommended \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 \( W \).
    \begin{exparts}
      \partsitem If the set of \( \vec{w}\, \)'s is independent, must
        the set of \( \vec{v}\, \)'s also be independent?
      \partsitem If the set of \( \vec{v}\, \)'s is 
        independent, must the
        set of \( \vec{w}\, \)'s also be independent?
      \partsitem If the set of \( \vec{w}\, \)'s spans \( W \), must the set of
        \( \vec{v}\, \)'s span \( V \)?
      \partsitem If the set of \( \vec{v}\, \)'s spans \( V \), must the set of
        \( \vec{w}\, \)'s span \( W \)?
    \end{exparts}
    \begin{answer}
      \begin{exparts}
        \partsitem Yes.
          The set of $\vec{w}\,$'s cannot be linearly independent if the
          set of $\vec{v}\,$'s is linearly dependent because
          any nontrivial relationship in the domain
          \( \zero_V=c_1\vec{v}_1+\dots+c_n\vec{v}_n \)
          would give a nontrivial relationship in the range
          \( f(\zero_V)=\zero_W=f(c_1\vec{v}_1+\dots+c_n\vec{v}_n)
             =c_1f(\vec{v}_1)+\dots+c_nf(\vec{v}_n)
             =c_1\vec{w}+\dots+c_n\vec{w}_n \).
        \partsitem Not necessarily.
          For instance, the transformation of \( \Re^2 \) given by
          \begin{equation*}
            \colvec{x \\ y} \mapsto \colvec{x+y \\ x+y}
          \end{equation*}
          sends this linearly independent set in the domain 
          to a linearly dependent image.
          \begin{equation*}
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            \set{\vec{v}_1,\vec{v}_2}=\set{\colvec[r]{1 \\ 0},\colvec[r]{1 \\ 1}}
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            \;\mapsto\;
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            \set{\colvec[r]{1 \\ 1},\colvec[r]{2 \\ 2}}=\set{\vec{w}_1,\vec{w}_2}
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          \end{equation*}
        \partsitem Not necessarily.
          An example is the projection map 
          \( \map{\pi}{\Re^3}{\Re^2} \)
          \begin{equation*}
            \colvec{x \\ y \\ z}\mapsto\colvec{x \\ y}
          \end{equation*}
          and this set that does not span the domain but 
          maps to a set that does span the codomain.
          \begin{equation*}
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            \set{\colvec[r]{1 \\ 0 \\ 0},\colvec[r]{0 \\ 1 \\ 0}}
            \mapsunder{\pi}\set{\colvec[r]{1 \\ 0},\colvec[r]{0 \\ 1}}
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          \end{equation*}
        \partsitem Not necessarily.
          For instance, the injection map $\map{\iota}{\Re^2}{\Re^3}$ sends
          the standard basis $\stdbasis_2$ for the domain to a set that 
          does not span
          the codomain. 
          (\textit{Remark.}
          However, the set of $\vec{w}$'s does span the range.
          A proof is easy.)
      \end{exparts}  
     \end{answer}
  \item  
    Generalize \nearbyexample{ex:MatTransMapLinear}
    by proving that the matrix transpose map is linear.
    What is the domain and codomain?
    \begin{answer}
      Recall that the entry in row~$i$ and column~$j$ of the transpose of $M$
      is the entry $m_{j,i}$ from row~$j$ and column~$i$ of $M$.
      Now, the check is routine. 
      \begin{align*}
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        \trans{[r\cdot\begin{mat}
                \    &\vdots          \\
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             \cdots &a_{i,j} &\cdots \\
                    &\vdots
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           \end{mat}
         +s\cdot\begin{mat}
                \    &\vdots          \\
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             \cdots &b_{i,j} &\cdots \\
                    &\vdots
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           \end{mat}]}
        &=\trans{\begin{mat}
                 \       &\vdots                    \\
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                  \cdots &ra_{i,j}+sb_{i,j} &\cdots \\
                         &\vdots
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                \end{mat}}                               \\
        &=\begin{mat}
              \     &\vdots                    \\
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            \cdots &ra_{j,i}+sb_{j,i} &\cdots \\
                   &\vdots
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          \end{mat}                               \\
        &=r\cdot\begin{mat}
             \     &\vdots          \\
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            \cdots &a_{j,i} &\cdots \\
                   &\vdots
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          \end{mat}
         +s\cdot\begin{mat}
             \      &\vdots          \\
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            \cdots &b_{j,i} &\cdots \\
                   &\vdots
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          \end{mat}                                \\
        &=r\cdot\trans{\begin{mat}
             \      &\vdots          \\
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            \cdots &a_{j,i} &\cdots \\
                   &\vdots
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          \end{mat} }
         +s\cdot\trans{\begin{mat}
              \     &\vdots          \\
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            \cdots &b_{j,i} &\cdots \\
                   &\vdots
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          \end{mat} }
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      \end{align*}
      The domain is \( \matspace_{\nbym{m}{n}} \) 
      while the codomain is \( \matspace_{\nbym{n}{m}} \).  
     \end{answer}
  \item 
    \begin{exparts}
     \partsitem Where \( \vec{u},\vec{v}\in \Re^n \), 
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        by definition the line segment connecting them is the set
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        \( \ell=\set{t\cdot\vec{u}+(1-t)\cdot\vec{v}\suchthat t\in [0..1]} \).
        Show that the image, under a homomorphism $h$, of the segment between
        $\vec{u}$ and $\vec{v}$ is the segment between $h(\vec{u})$ and
        $h(\vec{v})$.
     \partsitem A subset of \( \Re^n \) is \definend{convex}\index{convex set} 
       if, for any two points in
       that set, the line segment joining them lies entirely in that set.
       (The inside of a sphere is convex
       while the skin of a sphere is not.)
       Prove that linear maps from \( \Re^n \) to \( \Re^m \) preserve 
       the property of set convexity.
     \end{exparts}
    \begin{answer}
      \begin{exparts}
        \partsitem  For any homomorphism \( \map{h}{\Re^n}{\Re^m} \) we have
          \begin{equation*}
            h(\ell)
            =\set{h(t\cdot\vec{u}+(1-t)\cdot\vec{v})\suchthat t\in [0..1]}
            =\set{t\cdot h(\vec{u})+(1-t)\cdot h(\vec{v})\suchthat t\in [0..1]}
          \end{equation*}
          which is the line segment from $h(\vec{u})$ to $h(\vec{v})$.
        \partsitem We must show that if a subset of the domain is convex then
          its image, as a subset of the range, is also convex. 
          Suppose that \( C\subseteq \Re^n \) is convex
          and consider its image $h(C)$.
          To show $h(C)$ is convex we must show that for any two of its 
          members, $\vec{d}_1$ and $\vec{d}_2$, 
          the line segment connecting them
          \begin{equation*}
            \ell=\set{t\cdot\vec{d}_1+(1-t)\cdot\vec{d}_2\suchthat t\in [0..1]}
          \end{equation*}
          is a subset of $h(C)$.

          Fix any member $\hat{t}\cdot\vec{d}_1+(1-\hat{t})\cdot\vec{d}_2$
          of that line segment.
          Because the endpoints of $\ell$ are in the image of $C$, there are
          members of $C$ that map to them, say $h(\vec{c}_1)=\vec{d}_1$
          and $h(\vec{c}_2)=\vec{d}_2$.
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          Now, where $\hat{t}$ is the scalar that we fixed in the first
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          sentence of this paragraph, observe that
          $h(\hat{t}\cdot\vec{c}_1+(1-\hat{t})\cdot\vec{c}_2)
          =\hat{t}\cdot h(\vec{c}_1)+(1-\hat{t})\cdot h(\vec{c}_2)
          =\hat{t}\cdot\vec{d}_1+(1-\hat{t})\cdot\vec{d}_2$
          Thus, any member of $\ell$ is a member of $h(C)$, and so $h(C)$ is
          convex.
      \end{exparts}
    \end{answer}
  \recommended \item \label{exer:HomosPresLinStruc}
    Let \( \map{h}{\Re^n}{\Re^m} \) be a homomorphism.
    \begin{exparts}
      \partsitem Show that the image under \( h \) of a line in 
        \( \Re^n \) is a (possibly degenerate) line in \( \Re^m \).
      \partsitem What happens to a \( k \)-dimensional linear surface?
    \end{exparts}
    \begin{answer}
      \begin{exparts}
        \partsitem For \( \vec{v}_0,\vec{v}_1\in\Re^n \), the line through 
          \( \vec{v}_0 \) with direction \( \vec{v}_1 \) is the set 
          $\set{\vec{v}_0+t\cdot \vec{v}_1\suchthat t\in\Re}$.
          The image under $h$ of that line
          $\set{h(\vec{v}_0+t\cdot \vec{v}_1)\suchthat t\in\Re}
             =\set{h(\vec{v}_0)+t\cdot h(\vec{v}_1)\suchthat t\in\Re}$
          is the line through $h(\vec{v}_0)$ with direction $h(\vec{v}_1)$.
          If \( h(\vec{v}_1) \) is the zero vector then this line is 
          degenerate.
        \partsitem A \( k \)-dimensional linear surface in \( \Re^n \) maps to
          a (possibly degenerate) \( k \)-dimensional linear surface in
          \( \Re^m \).
          The proof is just like that the one for the line.
      \end{exparts}  
     \end{answer}
  \item 
    Prove that the restriction of a homomorphism to a subspace of its
    domain is another homomorphism.
    \begin{answer}
      Suppose that \( \map{h}{V}{W} \) is a homomorphism and suppose
      that \( S \) is a subspace of \( V \).
      Consider the map \( \map{\hat{h}}{S}{W} \) defined by
      \( \hat{h}(\vec{s})=h(\vec{s}) \).
      (The only difference between $\hat{h}$ and $h$ is the difference in  
      domain.)
      Then this new map is linear: 
      \( \hat{h}(c_1\cdot\vec{s}_1+c_2\cdot\vec{s}_2)=
              h(c_1\vec{s}_1+c_2\vec{s}_2)=c_1h(\vec{s}_1)+c_2h(\vec{s}_2)=
              c_1\cdot\hat{h}(\vec{s}_1)+c_2\cdot\hat{h}(\vec{s}_2) \).  
    \end{answer}
  \item 
    Assume that \( \map{h}{V}{W} \) is linear.
    \begin{exparts}
      \partsitem Show that the \definend{rangespace} of this map 
        \( \set{h(\vec{v})\suchthat \vec{v}\in V} \) is a subspace of 
        the codomain \( W \).
      \partsitem Show that the \definend{nullspace} of this map
        \( \set{\vec{v}\in V\suchthat h(\vec{v})=\zero_W} \) 
        is a subspace of the domain \( V \).
      \partsitem Show that if \( U \) is a subspace of the domain \( V \) then
        its image \( \set{h(\vec{u})\suchthat \vec{u}\in U} \) is a subspace 
        of the codomain \( W \).
        This generalizes the first item.
      \partsitem Generalize the second item.
    \end{exparts}
    \begin{answer} This will appear as a lemma in the next subsection.
      \begin{exparts}
        \partsitem The range is nonempty because \( V \) is nonempty.
          To finish we need to show that it is closed under combinations.
          A combination of range vectors has the form,
          where \( \vec{v}_1,\dots,\vec{v}_n\in V \),
          \begin{equation*}
            c_1\cdot h(\vec{v}_1)+\dots+c_n\cdot h(\vec{v}_n)
            =
            h(c_1\vec{v}_1)+\dots+h(c_n\vec{v}_n)
            =
            h(c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n),
          \end{equation*}
          which is itself in the range as
          \( c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n \) is a member of
          domain \( V \).
          Therefore the range is a subspace.
        \partsitem The nullspace is nonempty since it contains $\zero_V$, as
          \( \zero_V \) maps to \( \zero_W \).
          It is closed under linear combinations because, where
          \( \vec{v}_1,\dots,\vec{v}_n\in V \) are elements 
          of the inverse image set
          \( \set{\vec{v}\in V\suchthat h(\vec{v})=\zero_W} \),
          for \( c_1,\ldots,c_n\in\Re \)
          \begin{equation*}
            \zero_W=c_1\cdot h(\vec{v}_1)+\dots+c_n\cdot h(\vec{v}_n)
            =h(c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n)
          \end{equation*}
          and so \( c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n \) is also in
          the inverse image of \( \zero_W \).
        \partsitem This image of \( U \) nonempty because \( U \) is nonempty.
          For closure under combinations, 
          where \( \vec{u}_1,\ldots,\vec{u}_n\in U \),
          \begin{equation*}
            c_1\cdot h(\vec{u}_1)+\dots+c_n\cdot h(\vec{u}_n)
            =
            h(c_1\cdot \vec{u}_1)+\dots+h(c_n\cdot \vec{u}_n)
            =
            h(c_1\cdot \vec{u}_1+\dots+c_n\cdot \vec{u}_n)
          \end{equation*}
          which is itself in \( h(U) \) as
          \( c_1\cdot \vec{u}_1+\dots+c_n\cdot \vec{u}_n \) is in \( U \).
          Thus this set is a subspace.
        \partsitem The natural generalization is that the inverse image of a
          subspace of is a subspace.

          Suppose that \( X \) is a subspace of \( W \).
          Note that \( \zero_W\in X \) so the set
          \( \set{\vec{v}\in V \suchthat h(\vec{v})\in X} \) is not empty.
          To show that this set is closed under combinations, let
          \( \vec{v}_1,\dots,\vec{v}_n \) be elements of \( V \)
          such that \( h(\vec{v}_1)=\vec{x}_1 \), \ldots,
          \( h(\vec{v}_n)=\vec{x}_n \) and note that 
          \begin{equation*}
            h(c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n)
            =c_1\cdot h(\vec{v}_1)+\dots+c_n\cdot h(\vec{v}_n)
            =c_1\cdot \vec{x}_1+\dots+c_n\cdot \vec{x}_n
          \end{equation*}
          so a linear combination of elements of \( h^{-1}(X) \) is also in
          \( h^{-1}(X) \).
      \end{exparts}  
     \end{answer}
  \item 
    Consider the set of isomorphisms from a vector space to itself.
    Is this a subspace of the space \( \linmaps{V}{V} \)
    of homomorphisms from the space to itself?
    \begin{answer}
      No; the set of isomorphisms does not contain the zero map
      (unless the space is trivial).
    \end{answer}
  \item 
   Does \nearbytheorem{th:HomoDetActOnBasis} need that
   $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$
   is a basis? 
   That is, can we still get a well-defined and unique homomorphism if we
   drop either the condition that the set of $\vec{\beta}$'s 
   be linearly independent, 
   or the condition that it span the domain?
   \begin{answer}
     If $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$ doesn't span the space
     then the map needn't be unique.
     For instance, if we try to define a map from $\Re^2$ to itself by 
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     specifying only that $\vec{e}_1$ maps to itself, then 
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     there is more than
     one homomorphism possible; both the identity map and the projection map 
     onto the first component fit this condition.

     If we drop the condition that 
     $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$
     is linearly independent then we risk an inconsistent specification
     (i.e, there could be no such map).
     An example is if we consider 
     $\sequence{\vec{e}_2,\vec{e}_1,2\vec{e}_1}$, and try 
     to define a map from $\Re^2$ to itself that 
     sends $\vec{e}_2$ to itself, and sends both
     $\vec{e}_1$ and $2\vec{e}_1$ to $\vec{e}_1$.
     No homomorphism can satisfy these three conditions. 
   \end{answer}
  \item 
    Let \( V \) be a vector space and assume that 
    the maps \( \map{f_1,f_2}{V}{\Re^1} \) are linear.
    \begin{exparts}
      \partsitem Define a map \( \map{F}{V}{\Re^2} \) whose component
        functions are the given linear ones.
        \begin{equation*}
           \vec{v}\mapsto\colvec{f_1(\vec{v}) \\ f_2(\vec{v})}
        \end{equation*}
        Show that \( F \) is linear.
      \partsitem Does the converse hold\Dash is any linear map from \( V \) to
        \( \Re^2 \) made up of two linear component maps to \( \Re^1 \)?
      \partsitem Generalize.
    \end{exparts}
    \begin{answer}
      \begin{exparts}
        \partsitem Briefly, the check of linearity is this.
          \begin{equation*}
            F(r_1\cdot \vec{v}_1+r_2\cdot \vec{v}_2)
            =\colvec{f_1(r_1\vec{v}_1+r_2\vec{v}_2) \\ 
                        f_2(r_1\vec{v}_1+r_2\vec{v}_2)}
            =r_1\colvec{f_1(\vec{v}_1) \\ f_2(\vec{v}_1)}
            +r_2\colvec{f_1(\vec{v}_2) \\ f_2(\vec{v}_2)}
            =r_1\cdot F(\vec{v}_1)+r_2\cdot F(\vec{v}_2)
          \end{equation*}
        \partsitem Yes.
          Let \( \map{\pi_1}{\Re^2}{\Re^1} \) and
          \( \map{\pi_2}{\Re^2}{\Re^1} \) be the projections
          \begin{equation*}
            \colvec{x \\ y}\mapsunder{\pi_1} x
              \quad\text{and}\quad
            \colvec{x \\ y}\mapsunder{\pi_2} y
          \end{equation*}
          onto the two axes.
          Now, where \( f_1(\vec{v})=\pi_1(F(\vec{v})) \) and
          \( f_2(\vec{v})=\pi_2(F(\vec{v})) \) 
          we have the desired component functions.
          \begin{equation*}
            F(\vec{v})=
            \colvec{f_1(\vec{v}) \\ f_2(\vec{v})}
          \end{equation*}
          They are linear because they are the composition of linear functions,
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          and the fact that the composition of linear functions is linear
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          was part of the proof that isomorphism is an equivalence
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          relation (alternatively, the check that they are linear is
          straightforward). 
        \partsitem In general, a map from a vector space \( V \) to an 
          \( \Re^n \) is linear if and only if each of the component 
          functions is linear.
          The verification is as in the prior item.
      \end{exparts}  
     \end{answer}
\end{exercises}















\subsection{Rangespace and Nullspace}
Isomorphisms and homomorphisms both preserve structure.
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The difference is that homomorphisms are subject to fewer restrictions 
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because they needn't be onto and
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needn't be one-to-one.
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We will examine what can happen with homomorphisms
that cannot happen to isomorphisms. 
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We first consider the effect of 
not requiring that a homomorphism be
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onto its codomain.
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Of course, 
each homomorphism is onto some set, namely its range.
For example, the injection map \( \map{\iota}{\Re^2}{\Re^3} \)
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\begin{equation*}
  \colvec{x \\ y} \mapsto \colvec{x \\ y \\ 0}
\end{equation*}
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is a homomorphism 
that is not onto.
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But, $\iota$ is onto the $xy$-plane subset of \( \Re^3\).
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\begin{lemma}   \label{le:RangeIsSubSp}
Under a homomorphism, the image of
any subspace of the domain is a subspace of the codomain.
In particular, the image of the entire space,
the range of the homomorphism, is a subspace of the codomain.
\end{lemma}

\begin{proof}
Let $\map{h}{V}{W}$ be linear and let $S$ be a subspace of the domain 
$V$.
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The image $h(S)$ is a subset of the codomain $W$, which is
nonempty because $S$ is nonempty.
Thus, to show that $h(S)$ is a subspace of  $W$
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we need only show that it is closed under linear combinations
of two vectors.
If $h(\vec{s}_1)$ and $h(\vec{s}_2)$ are members of $h(S)$ then
$c_1\cdot h(\vec{s}_1)+c_2\cdot h(\vec{s}_2)
  =
  h(c_1\cdot \vec{s}_1)+h(c_2\cdot \vec{s}_2)
  =
  h(c_1\cdot \vec{s}_1+c_2\cdot \vec{s}_2)$
is also a member of $h(S)$ because it is the image of
\( c_1\cdot \vec{s}_1+c_2\cdot \vec{s}_2 \) from \( S \).
\end{proof}

\begin{definition}
The \definend{rangespace}\index{rangespace}\index{homomorphism!rangespace}
of a homomorphism \( \map{h}{V}{W} \) is
\begin{equation*}
  \rangespace{h}=\set{h(\vec{v})\suchthat \vec{v}\in V}
\end{equation*}
sometimes denoted \( h(V) \).
The dimension of the rangespace is the map's
\definend{rank}.\index{rank!of a homomorphism}
\end{definition}
\noindent 
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We shall soon see the connection between the rank of a map and the rank of a 
matrix.
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\begin{example}  \label{ex:DerivMapRnge}
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For the derivative map 
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\( \map{d/dx}{\polyspace_3}{\polyspace_3} \)
given by \( a_0+a_1x+a_2x^2+a_3x^3 \mapsto a_1+2a_2x+3a_3x^2 \)
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the rangespace 
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\( \rangespace{d/dx} \) is the set of quadratic polynomials
\( \set{r+sx+tx^2\suchthat r,s,t\in\Re } \).
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Thus, this map's rank is~\( 3 \).
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\end{example}

\begin{example}  \label{ex:MatToPolyRnge}
With this homomorphism \( \map{h}{M_{\nbyn{2}}}{\polyspace_3} \)
\begin{equation*}
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  \begin{mat}
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    a  &b  \\
    c  &d
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  \end{mat}
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  \mapsto
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  (a+b+2d)+cx^2+cx^3
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\end{equation*}
an image vector in the range
can have any constant term, must have an $x$ coefficient of zero,
and must have the same coefficient of $x^2$ as of $x^3$.
That is, the rangespace is
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\( \rangespace{h}=\set{r+sx^2+sx^3\suchthat r,s\in\Re} \)
and so the rank is~\( 2 \).
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\end{example}

The prior result shows that, 
in passing from the definition of isomorphism to the more
general definition of homomorphism,  
omitting the `onto' requirement doesn't make an essential difference.
Any homomorphism is onto its rangespace.

However, omitting the `one-to-one' condition does make a difference. 
A homomorphism may have many elements
of the domain that map to one element of the codomain.
Below is a ``bean'' sketch of a many-to-one 
map between sets.\appendrefs{many-to-one maps}\spacefactor=1000 %
It shows three elements of the codomain that are each the image of
many members of the domain.
\begin{center}  
  \includegraphics{ch3.5}  % bean to bean; many to one
\end{center}
Recall that for any function $\map{h}{V}{W}$, 
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the set of elements of $V$ that map to \( \vec{w}\in W \)
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is the  \definend{inverse image\/}\index{inverse image}%
\index{function! inverse image}   
$h^{-1}(\vec{w})=\set{\vec{v}\in V\suchthat h(\vec{v})=\vec{w}}$.
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Above, the left bean shows three inverse image sets.
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\begin{example}
Consider the projection\index{projection}
\( \map{\pi}{\Re^3}{\Re^2} \)
\begin{equation*}
   \colvec{x \\ y \\ z}
    \mapsunder{\pi}
   \colvec{x \\ y}
\end{equation*}
which is a homomorphism that is many-to-one.
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An inverse image set is a vertical line of vectors
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in the domain.
\begin{center}
  \includegraphics{ch3.11}
\end{center}
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One example is this.
\begin{equation*}
  \pi^{-1}(\colvec[r]{1 \\ 3})=\set{\colvec[r]{1 \\ 3 \\ z}\suchthat z\in\Re}
\end{equation*}
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\end{example}

\begin{example} \label{ex:RTwoHomoREasyOneMap}
This homomorphism $\map{h}{\Re^2}{\Re^1}$
\begin{equation*}
  \colvec{x \\ y}
   \mapsunder{h}
  x+y
\end{equation*}
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is also many-to-one.
For a fixed $w\in\Re^1$, 
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the inverse image $h^{-1}(w)$
\begin{center}
  \includegraphics{ch3.12}
\end{center}
is the set of plane vectors whose components add to $w$.
\end{example}

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% The above examples have only to do with the
% fact that we are considering functions,
% specifically, many-to-one functions.
% They show the inverse images 
% as sets of vectors that are 
% related to the image vector $\vec{w}$.
% But these are more than just arbitrary functions, they are
% homomorphisms; what do the two preservation
% conditions say about the relationships?
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In generalizing from isomorphisms to homomorphisms by 
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dropping the one-to-one condition,
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we lose the property that we've stated intuitively as that the 
domain is ``the same'' as the range.
We lose that the domain
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corresponds perfectly to the range.
What we retain, as the examples below illustrate,
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is that a homomorphism describes how
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the domain is ``like'' or ``analogous to'' the range.
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\begin{example}    \label{ex:RThreeHomoRTwo} %\label{exPicProj}
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We think of $\Re^3$ as like $\Re^2$ except that vectors have an extra
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component. 
That is, we think of the vector with components $x$, $y$, and~$z$ 
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as somehow like the vector with components $x$ and~$y$.
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In defining the projection map $\pi$, we make precise which
members of the domain we are thinking of as related to which members
of the codomain.

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To understanding how the
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preservation conditions in the definition of homomorphism
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show that the domain elements are like the codomain elements, 
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we start by picturing $\Re^2$ as the $xy$-plane inside of $\Re^3$.
(Of course, $\Re^2$ is not the 
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$xy$~plane inside of $\Re^3$ since the $xy$~plane
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is a set of three-tall vectors with a 
third component of
zero, but there is a natural correspondence.)
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Then
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the preservation of addition property says that
vectors in \( \Re^3 \) act like their shadows in the plane.
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\begin{center}  \small
  \begin{tabular}{@{}c@{}c@{}c@{}c@{}c@{}}
    \includegraphics{ch3.1}
    &&\includegraphics{ch3.2}
    &&\includegraphics{ch3.3} \\[1.5ex]
    {\small  $\colvec{x_1 \\ y_1 \\ z_1}$ above $\colvec{x_1 \\ y_1}$}
    &{\small \ plus\ }
    &{\small $\colvec{x_2 \\ y_2 \\ z_2}$ above $\colvec{x_2 \\ y_2}$}
    &{\small \ equals\ }
    &{\small $\colvec{x_1+y_1 \\ y_1+y_2 \\ z_1+z_2}$ 
              above $\colvec{x_1+x_2 \\ y_1+y_2}$}
  \end{tabular}
\end{center}
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\noindent 
Thinking of $\pi(\vec{v})$ as the ``shadow'' of $\vec{v}$ in the plane 
gives this restatement:
the sum of the shadows $\pi(\vec{v}_1)+\pi(\vec{v}_2)$ equals
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the shadow of the sum 
$\pi(\vec{v}_1+\vec{v}_2)$.
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Preservation of scalar multiplication is similar.
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Redrawing by showing the codomain $\Re^2$ on the right
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gives a picture that is uglier but is more faithful to the
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``bean'' sketch.
\begin{center}  \small
  \includegraphics{ch3.4}
\end{center}
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Again, the domain vectors that map to $\vec{w}_1$ lie
in a vertical line;
the picture shows one in gray.
Call any member of this inverse image $\pi^{-1}(\vec{w}_1)$
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a ``$\vec{w}_1$~vector.''
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Similarly, there is a vertical line of ``$\vec{w}_2$~vectors'' and
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a vertical line of ``$\vec{w}_1+\vec{w}_2$~vectors.''
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Now, saying that $\pi$ is a homomorphism is recognizing that 
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if $\pi(\vec{v}_1)=\vec{w}_1$ and $\pi(\vec{v}_2)=\vec{w}_2$ 
then $\pi(\vec{v}_1+\vec{v}_2)=\pi(\vec{v}_1)+\pi(\vec{v}_2)
  =\vec{w}_1+\vec{w}_2$.
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That is, the classes add:~any 
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\( \vec{w}_1 \)~vector plus any
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\( \vec{w}_2 \)~vector 
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equals a \( \vec{w}_1+\vec{w}_2 \)~vector.
Scalar multiplication is similar.
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So although $\Re^3$ and $\Re^2$ are not isomorphic
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$\pi$ describes a way in which they are alike:~vectors 
in $\Re^3$ add as do the associated vectors in $\Re^2$\Dash vectors 
add as their shadows add. 
\end{example}

\begin{example}   \label{ex:RTwoHomoRHardOne}
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A homomorphism can express
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an analogy between spaces that is more subtle than the prior one.
For the map 
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from \nearbyexample{ex:RTwoHomoREasyOneMap}
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\begin{equation*}
  \colvec{x \\ y}
   \mapsunder{h}
  x+y
\end{equation*}
fix two numbers $w_1, w_2$ in the range \( \Re \).
A $\vec{v}_1$ that maps to $w_1$ has components that 
add to $w_1$,
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so the inverse image $h^{-1}(w_1)$ is the set of vectors 
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with endpoint on the diagonal line $x+y=w_1$.
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Think of these as ``$w_1$ vectors.''
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Similarly we have ``$w_2$ vectors'' and ``$w_1+w_2$ vectors.''  
The addition preservation property says this.
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\begin{center}  \small
  \begin{tabular}{@{}ccccc@{}}
    \includegraphics{ch3.6}
    &&\includegraphics{ch3.7}
    &&\includegraphics{ch3.8}  \\[1.5ex]
    {\small a ``$w_1$ vector''}
    &{\small plus}
    &{\small a ``$w_2$ vector''}
    &{\small equals}
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    &{\small a ``$w_1+w_2$ vector''}
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  \end{tabular}
\end{center}
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Restated, if we add a
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$w_1$~vector to a $w_2$~vector then 
$h$ maps the result to a $w_1+w_2$ vector.
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Briefly, the sum of the images is the image of the sum.
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Even more briefly, \( h(\vec{v}_1)+h(\vec{v}_2)=h(\vec{v}_1+\vec{v}_2) \).
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% (The preservation of scalar multiplication condition has a
% similar restatement.)
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\end{example}

\begin{example}   \label{ex:PicRThreeToRTwo}
The inverse images can be structures other than lines.
For the linear map \( \map{h}{\Re^3}{\Re^2} \)
\begin{equation*}
  \colvec{x \\ y \\ z}
    \mapsto
  \colvec{x \\ x}
\end{equation*}
the inverse image sets are planes $x=0$, $x=1$, etc.,
perpendicular to the \( x \)-axis.
\begin{center}  \small
  \includegraphics{ch3.9}
\end{center}
\end{example}

We won't describe how every homomorphism that we will use
is an analogy because the formal
sense that we make of ``alike in that~\ldots'' is 
`a homomorphism exists such that~\ldots'.
Nonetheless, the idea that a homomorphism between two spaces expresses how
the domain's vectors fall into classes that act like
the range's vectors is a good way to view homomorphisms.

Another reason that we won't treat all of the homomorphisms that
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we see as above is that many vector spaces are hard to draw,
e.g., a space of polynomials.
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But there is nothing wrong with leveraging those spaces
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that we can draw.
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We derive two insights from the three examples
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\ref{ex:RThreeHomoRTwo}, \ref{ex:RTwoHomoRHardOne}, 
and~\ref{ex:PicRThreeToRTwo}.

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The first insight is that in all three examples 
the inverse image of the range's zero vector is a line or plane
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through the origin, a subspace of the domain.
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\begin{lemma}  \label{le:NullspIsSubSp}
For any homomorphism, the inverse image of a subspace of the range 
is a subspace of the domain.
In particular, the inverse image of the trivial subspace of the range
is a subspace of the domain.
\end{lemma}

\begin{proof}
Let $\map{h}{V}{W}$ be a homomorphism
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and let $S$ be a subspace of the rangespace of $h$.
Consider the inverse image
$h^{-1}(S)=\set{\vec{v}\in V\suchthat h(\vec{v})\in S}$.
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It is nonempty because it contains $\zero_V$, since
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\( h(\zero_V)=\zero_W \) and \( \zero_W \) is an element $S$, 
as $S$ is a subspace.
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To finish we show that it is closed under linear combinations.
Let \( \vec{v}_1 \) and \( \vec{v}_2 \) be two elements of $h^{-1}(S)$.
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Then
$h(\vec{v}_1)$ and $h(\vec{v}_2)$ are elements of $S$.
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That implies that
$c_1\vec{v}_1+c_2\vec{v}_2$
is an element of the inverse image $h^{-1}(S)$ 
because
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$h(c_1\vec{v}_1+c_2\vec{v}_2)
  =c_1h(\vec{v}_1)+c_2h(\vec{v}_2)$
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is a member of $S$. 
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\end{proof}

\begin{definition}
The \definend{nullspace}\index{homomorphism!nullspace}\index{nullspace}
or \definend{kernel}\index{kernel} of a linear map
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\( \map{h}{V}{W} \) is the inverse image of $0_W$.
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\begin{equation*}
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  \nullspace{h}=h^{-1}(\zero_W)=\set{\vec{v}\in V\suchthat h(\vec{v})=\zero_W}
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\end{equation*}
The dimension of the nullspace is the map's
{\em nullity}\index{nullity}\index{homomorphism!nullity}.
\end{definition}

\begin{center}
  \includegraphics{ch3.10}
\end{center}

\begin{example}
The map from \nearbyexample{ex:DerivMapRnge} has this nullspace
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\( \nullspace{d/dx}=\set{a_0+0x+0x^2+0x^3\suchthat a_0\in\Re} \)
so its nullity is $1$.
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\end{example}

\begin{example}
The map from \nearbyexample{ex:MatToPolyRnge}
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has this nullspace and nullity $2$.
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\begin{equation*}
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   \nullspace{h}=\set{\begin{mat}
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                         a  &b          \\
                         0  &-(a+b)/2
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                      \end{mat}\suchthat a,b\in\Re}
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\end{equation*}
\end{example}

Now for the second insight from the above pictures.
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In \nearbyexample{ex:RThreeHomoRTwo} each of the vertical lines  
squashes down 
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