Ch.8 Homomorphisms

The left is a homomorphism and the right is not:
$\begin{array}{cc}\begin{pmatrix}x\\y\end{pmatrix}\xmapsto{h}2x-3y&\begin{pmatrix}x\\y\end{pmatrix}\xmapsto{g}2x-3y+1\end{array}$
$h$ respects addition and scalar multiplication:
$h(\begin{pmatrix}rx_1+sx_2\\ry_1+sy_2\end{pmatrix})=2(rx_1+sx_2)-3(ry_1+sy_2)\\=r(2x_1-3y_1)+s(2x_2-3y_2)=r\cdot h(\begin{pmatrix}x_1\\y_1\end{pmatrix})+s\cdot h(\begin{pmatrix}x_2\\y_2\end{pmatrix})$
The right does not respect addition
$g(\begin{pmatrix}x_1+x_2\\y_1+y_2\end{pmatrix})=2(x_1+x_2)-3(y_1+y_2)+1\\\ne g(\begin{pmatrix}x_1\\y_1\end{pmatrix})+g(\begin{pmatrix}x_2\\y_2\end{pmatrix})=2(x_1+x_2)-3(y_1+y_2)+2$

$\iota(\begin{pmatrix}rx_1+sx_2\\ry_1+sy_2\end{pmatrix})=\begin{pmatrix}rx_1+sx_2\\ry_1+sy_2\\0\end{pmatrix}=r\begin{pmatrix}x_1\\y_1\\0\end{pmatrix}+s\begin{pmatrix}x_2\\y_2\\0\end{pmatrix}=r\cdot\iota(\begin{pmatrix}x_1\\y_1\end{pmatrix})+s\cdot\iota(\begin{pmatrix}x_2\\y_2\end{pmatrix})$

$\text{Tr}(\begin{pmatrix}ra_1+sa_2&rb_1+sb_2\\rc_1+sc_2&rd_1+sd_2\end{pmatrix})=ra_1+sa_2+rd_1+sd_2=r(a_1+d_1)+s(a_2+d_2)\\=r\cdot\text{Tr}(\begin{pmatrix}a_1&b_1\\c_1&d_1\end{pmatrix})+s\cdot\text{Tr}(\begin{pmatrix}a_2&b_2\\c_2&d_2\end{pmatrix})$

Suppose we want a rotation map $t_\theta:\mathbb{R}^2\to\mathbb{R}^2$ rotating all vectors in the plane by $\theta$. We can take a basis of the domain, say $\mathcal{E}_2$ and find where they map to:
$\begin{array}{cc}\begin{pmatrix}1\\0\end{pmatrix}\mapsto\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}&\begin{pmatrix}0\\1\end{pmatrix}\mapsto\begin{pmatrix}-\sin\theta\\\cos\theta\end{pmatrix}\end{array}$
This can therefore be linearly extended to:
$t_\theta(\begin{pmatrix}x\\y\end{pmatrix})=x\cdot\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}+y\cdot\begin{pmatrix}-\sin\theta\\\cos\theta\end{pmatrix}=\begin{pmatrix}x\cos\theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{pmatrix}$

The set is nonempty because it contains the zero homomorphism $Z:V\to W$, $Z(\vec{v})=\vec{0}_W$. To show it is a subspace it suffices to show it is closed under addition and scalar multiplication.
Let $f,g:V\to W$ be linear functions. Then
$\begin{array}{lcl}(f+g)(c_1\vec{v}_1+c_2\vec{v}_2)&=&f(c_1\vec{v}_1+c_2\vec{v}_2)+g(c_1\vec{v}_1+c_2\vec{v}_2)\\&=&c_1f(\vec{v}_1)+c_2f(\vec{v}_2)+c_1g(\vec{v}_1)+c_2g(\vec{v}_2)\\&=&c_1(f+g)(\vec{v}_1)+c_2(f+g)(\vec{v}_2)\end{array}$
so addition is preserved and since
$\begin{array}{lcl}(r\cdot f)(c_1\vec{v}_1+c_2\vec{v}_2)&=&r(c_1f(\vec{v}_1)+c_2f(\vec{v}_2))\\&=&c_1(r\cdot f)(\vec{v}_1)+c_2(r\cdot f)(\vec{v}_2)\end{array}$
scalar multiplication is preserved. Hence, this space is a subspace.

Consider $\mathcal{L}(\mathbb{R},\mathbb{R}^2)$. A linear map in this space $t\in\mathcal{L}(\mathbb{R},\mathbb{R}^2)$ is determined by its action on the basis
$\begin{array}{cc}B_\mathbb{R}=1&B_{\mathbb{R}^2}=\langle\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\rangle\end{array}$
So every function is determined by $c_1,c_2$ here:
$1\xmapsto{t}c_1\begin{pmatrix}1\\0\end{pmatrix}+c_2\begin{pmatrix}0\\1\end{pmatrix}$
Thus $\mathcal{L}(\mathbb{R},\mathbb{R}^2)$ is a 2-dimensional vector space.

Let $h(S)$ be the image of subspace $S$. It is nonempty because $S$ is nonempty. It suffices to show $h(S)$ is closed under addition and scalar multiplication. For $c_1,c_2\in\mathbb{R}$ and $h(\vec{s}_1),h(\vec{s}_2)\in h(S)$
$c_1h(\vec{s}_1)+c_2h(\vec{s}_2)=h(c_1\vec{s}_1+c_2\vec{s}_2)\in h(S)$
because it is the image of $c_1\vec{s}+c_2\vec{s}_2\in S$.

Consider the linear map $h:\mathcal{M}_{2\times2}\to\mathbb{R}^2$
$\begin{pmatrix}1&b\\c&d\end{pmatrix}\xmapsto{h}\begin{pmatrix}a+b\\2a+2b\end{pmatrix}$
The range space is the line through the origin
$\{t\begin{pmatrix}1\\2\end{pmatrix}\space|\space t\in\mathbb{R}\}$
Every member of that set is the image of a $2\times2$ matrix
$\begin{pmatrix}t\\2t\end{pmatrix}=h(\begin{pmatrix}t&0\\0&0\end{pmatrix})$
The space is one dimensional, so the rank is 1.

The projection map $\pi:\mathbb{R}^3\to\mathbb{R}^2$
$\begin{pmatrix}x\\y\\z\end{pmatrix}\xmapsto{\pi}\begin{pmatrix}x\\y\end{pmatrix}$
has the range space
$\{t\begin{pmatrix}1\\0\end{pmatrix}+s\begin{pmatrix}0\\1\end{pmatrix}\space|\space t,s\in\mathbb{R}\}$

Consider the linear projection map $\pi:\mathbb{R}^2\to\mathbb{R}$
$\pi(\begin{pmatrix}x\\y\end{pmatrix})=x$
The inverse image $\pi^{-1}(c)$ is a set of vectors $(\begin{pmatrix}c\\y\end{pmatrix}\space|\space y\in\mathbb{R}^2)$
The endpoints make up the line $x=c$

Let $h:\mathcal{P}_2\to\mathbb{R}^2$ be
$ax^2+bx+c\mapsto\begin{pmatrix}b\\b\end{pmatrix}$
Consider three output vectors such that $\vec{w}_1+\vec{w}_2=\vec{w}_3$, for example
$\begin{array}{ccc}\vec{w}_1=\begin{pmatrix}1\\1\end{pmatrix}&\vec{w}_2=\begin{pmatrix}-1\\-1\end{pmatrix}&\vec{w}_3=\begin{pmatrix}0\\0\end{pmatrix}\end{array}$
The inverse images are respectively
$h^{-1}(\vec{w}_1)=\{a_1x^2+1x+c_1\space|\space a_1,c_1\in\mathbb{R}\}\\ h^{-1}(\vec{w}_2)=\{a_2x^2-1x+c_2\space|\space a_2,c_2\in\mathbb{R}\}\\ h^{-1}(\vec{w}_3)=\{a_3x^2+0x+c_3\space|\space a_3,c_3\in\mathbb{R}\}$
Since this is a homomorphism, we can take any $\vec{v}_1\in h^{-1}(\vec{w}_1)$ and add it to $\vec{v}_2\in h^{-1}(\vec{w}_2)$ to get an element $\vec{v}_3\in h^{-1}(\vec{w}_3)$

The trivial subspace of $W$ is the set$\{\vec{0}_W\}$, not just $\vec{v}_W$, so we may write the null space as $h^{-1}(\{\vec{0}_W\})$, however we have defined this to be equivalent to $h^{-1}(\vec{0}_W)$ and that's easier to write so we just use that.

Consider the derivative map $d/dx:\mathcal{P}_2\to\mathcal{P}_1$. The nullspace is a subset of the domain
$\mathscr{N}(d/dx)=\{ax^2+bx+c\space|\space 2a+b=0\}$
$2a+b=0x+0$ iff $a=b=0$, which means the nullspace is equal to
$\mathscr{N}(d/dx)=\{ax^2+bx+c\space|\space a=b=0,c\in\mathbb{R}\}=\{c\space|\space c\in\mathbb{R}\}$
The nullity is 1.

Consider the homomorphism $f:\mathcal{M}_{2\times2}\to\mathbb{R}^2$
$f(\begin{pmatrix}a&b\\c&d\end{pmatrix})=\begin{pmatrix}a+b\\c+d\end{pmatrix}$
The nullspace is
$\begin{array}{rcl}\mathscr{N}(f)&=&\left\{\begin{pmatrix}a&b\\c&d\end{pmatrix}\space\middle|\space a+b=0\text{ and }c+d=0\right\}\\\\&=&\left\{\begin{pmatrix}a&-a\\b&-b\end{pmatrix}\space\middle|\space a,b\in\mathbb{R}\right\}\end{array}$
The nullity is 2.

Let $h:V\to W$ be a linear map and let $B_N=\langle\textcolor{#EE4266}{\vec{\beta}_1},...,\textcolor{#EE4266}{\vec{\beta}_k}\rangle$ be the basis for the null space. The size of this is the nullity. Since the nullspace is a subspace, the basis $B_N$ can be expanded to a basis $B_V=\langle\textcolor{#EE4266}{\vec{\beta}_1},...,\textcolor{#EE4266}{\vec{\beta}_k},\textcolor{#F79256}{\vec{\beta}_{k+1}},...,\textcolor{#F79256}{\vec{\beta}_n}\rangle$ for the domain $V$. The size of $B_V$ is the dimension of the domain. We will show that $B_R=\langle h(\textcolor{#F79256}{\vec{\beta}_{k+1}}),...,h(\textcolor{#F79256}{\vec{\beta}_n})\rangle$ is the basis of the range space.

First we check that $B_R$ is linearly independent.
$\begin{array}{rcl}\vec{0}_W&=&c_{k+1}h(\vec{\beta}_{k+1})+\cdots+c_nh(\vec{\beta}_n)\\&=&h(c_{k+1}\vec{\beta}_{k+1}+\cdots+c_nh(\vec{\beta}_n))\end{array}$
Thus $c_{k+1}\vec{\beta}_{k+1}+\cdots+c_n\vec{\beta}_n$ is in the nullspace of $h$. Therefore, we can express it as a linear combination of $B_N$
$c_1\vec{\beta}_1+\cdots+c_k\vec{\beta}_k=c_{k+1}\vec{\beta}_{k+1}+\cdots+c_n\vec{\beta}_n$
Letting $c'_i=-c_i$ for $i=k+1,...,n$, this can be rearranged to
$c_1\vec{\beta}_1+\cdots+c_k\vec{\beta}_k+c'_{k+1}\vec{\beta}_{k+1}+\cdots+c'_n\vec{\beta}_n=\vec{0}$
This is a linear combination of $B_V$, which means the only solution is $c_i=0$ for $i=1,...,n$. Therefore, $B_R$ is linearly independent.

Next, we show that $B_R$ spans the range space.
Consider a member of the range space $h(\vec{v})$.
Express $\vec{v}$ as a linear combination of members of $B_V$ and apply the map:
$\vec{v}=c_1\vec{\beta}_1+\cdots+c_k\vec{\beta}_k+c_{k+1}\vec{\beta}_{k+1}+\cdots+c_n\vec{\beta}_n\\ h(\vec{v})=c_1h(\vec{\beta}_1)+\cdots+c_kh(\vec{\beta}_k)+c_{k+1}h(\vec{\beta}_{k+1})+\cdots+c_nh(\vec{\beta}_n)$
Since $\vec{\beta}_1,...,\vec{\beta}_k$ are in the nullspace, $h(\vec{\beta}_1),...,h(\vec{\beta}_k)=0$, so
$h(\vec{v})=\vec{0}+c_{k+1}\vec{\beta}_{k+1}+\cdots+c_n\vec{\beta}_n$
Thus, we have $h(\vec{v})$ as a linear combination of members of $B_R$, so $B_R$ spans the range space.

The projection $\pi:\mathbb{R}^3\to\mathbb{R}^2$
$\begin{pmatrix}a\\b\\c\end{pmatrix}\mapsto\begin{pmatrix}a\\b\end{pmatrix}$
takes a 3 dimensional domain to a 2 dimensional range, so the nullspace is 1 dimensional, i.e. the z-axis

Suppose $c_1\vec{v}_1+\cdots+c_n\vec{v}_n=\vec{0}_V$ with some $c_i\ne0$.
Apply $h$ to both sides: $h(c_1\vec{v}_1+\cdots+c_n\vec{v}_n)=c_1h(\vec{v}_1)+\cdots+c_nh(\vec{v}_n)$ and $h(\vec{0}_V)=\vec{0}_W$.
So we have $c_1h(\vec{v}_1)+\cdots+c_nh(\vec{v}_n)=\vec{0}_W$ with some $c_i\ne0$

First, we prove $(1)\implies(2)$
Suppose $h$ is one-to-one. It must have an inverse $h^{-1}:\mathscr{R}(h)\to V$. Taking a linear combination of two elements of $\mathscr{R}(h)$, we have $c_1h(\vec{v}_1)+c_2h(\vec{v}_2)$. Taking the inverse:
$\begin{array}{rcl}h^{-1}(c_1h(\vec{v}_1)+c_2h(\vec{v}_2))&=&h^{-1}(h(c_1\vec{v}_1+c_2\vec{v}_2))\\&=&c_1\vec{v}_1+c_2\vec{v}_2\\&=&c_1h^{-1}(h(\vec{v}_1))+c_2h^{-1}(h(\vec{v}_2))\end{array}$
Thus $h^{-1}$ is a linear map.
$(2)\implies(1)$ because $h$ is a bijection from $V$ to $\mathscr{R}(h)$.

It remains to show that $(1)\implies(3)\implies(4)\implies(5)\implies(2)$
$(1)\implies(3)$ because any homomorphism maps $\vec{0}_V$ to $\vec{0}_W$, but if $h$ is one-to-one, only$\vec{0}_V$ maps to $\vec{0}_W$.
For $(3)\implies(4)$, since rank plus nullity equals dimension and nulltiy is 0, rank must equal dimension, i.e. $\text{rank}(h)=n$
For $(4)\implies(5)$, we must show that $\langle h(\vec{\beta}_1),...,h(\vec{\beta}_n)\rangle$ spans the range space, since by assumption the dimension is $n$, implying these are linearly independent.
Consider $h(\vec{v})\in\mathscr{R}(h)$. Expressing $\vec{v}$ in terms of the basis of $V$ gives $h(\vec{v})=h(c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n)=c_1h(\vec{\beta}_1)+\cdots+c_nh(\vec{\beta}_n)$ as desired.
Finally for $(5)\implies(2)$, for every $\vec{w}\in\mathscr{R}(h)$, there is a unique representation $\vec{w}=c_1h(\vec{\beta}_1)+\cdots+c_nh(\vec{\beta}_n)$. Define a map from $\mathscr{R}(h)$ to $V$ by
$\vec{w}\mapsto c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n$
This is linear and is the inverse of $h$ as desired.

A line through the origin is a set $\{r\cdot\vec{v}\space|\space r\in\mathbb{R}\}$
Under a linear transformation $t:\mathbb{R}^n\to\mathbb{R}^n$
$r\cdot\vec{v}\xmapsto{t} t(r\cdot\vec{v})=r\cdot t(\vec{v})$
So the range space is $\{s\cdot t(\vec{v})\space|\space s\in\mathbb{R}\}$, a line through the origin