Ch.9 Linear Maps as Matrices

Let $V=\mathcal{P}_2$ and $W=\mathbb{R}^2$ with the bases
$\begin{array}{cc}B_V=\{1,1+x,1+x+x^2\}&B_W=\{\begin{pmatrix}2\\0\end{pmatrix},\begin{pmatrix}-1\\1\end{pmatrix}\}\end{array}$
Let $h:\mathcal{P}_2\to\mathbb{R}^2$ have this action on $B_V$
$\begin{array}{ccc}h(1)=\begin{pmatrix}0\\1\end{pmatrix}&h(1+x)=\begin{pmatrix}3\\2\end{pmatrix}&h(1+x+x^2)=\begin{pmatrix}-2\\-1\end{pmatrix}\end{array}$
Note that these coordinates are under $\mathcal{E}_2$, not $B_W$
So, next we find the representation of these in terms of $B_W$
$\begin{array}{rrr}h(1)=&\begin{pmatrix}0\\1\end{pmatrix}=&1/2\begin{pmatrix}2\\0\end{pmatrix}+1\begin{pmatrix}-1\\1\end{pmatrix}\\ h(1+x)=&\begin{pmatrix}3\\2\end{pmatrix}=&5/2\begin{pmatrix}2\\0\end{pmatrix}+2\begin{pmatrix}-1\\1\end{pmatrix}\\ h(1+x+x^2)=&\begin{pmatrix}-2\\-1\end{pmatrix}=&-3/2\begin{pmatrix}2\\0\end{pmatrix}-1\begin{pmatrix}-1\\1\end{pmatrix} \end{array}$
So we have
$\begin{array}{rcl}\text{Rep}_{B_W}(h(1))&=&\begin{pmatrix}1/2\\1\end{pmatrix}\\ \text{Rep}_{B_W}(h(1+x))&=&\begin{pmatrix}5/2\\2\end{pmatrix}\\ \text{Rep}_{B_W}(h(1+x+x^2))&=&\begin{pmatrix}-3/2\\-1\end{pmatrix}\end{array}$
Concatenate these into one matrix
$\begin{pmatrix}1/2&5/2&-3/2\\1&2&-1\end{pmatrix}$
This is the matrix representation of $h$ with respect to $B_V$, $B_W$

If we have a vector in the domain $\vec{v}=c_1\cdot\vec{\beta}_1+c_2\cdot\vec{\beta}_2+c_3\cdot\vec{\beta}_3$ and apply the transformation,
$\begin{array}{rcl}h(c_1\cdot\vec{\beta}_1+c_2\cdot\vec{\beta}_2+c_3\cdot\vec{\beta}_3)&=&c_1\cdot h(\vec{\beta}_1)+c_2\cdot h(\vec{\beta}_2)+c_3\cdot h(\vec{\beta}_3)\\ &=&c_1\cdot(1/2\cdot\begin{pmatrix}2\\0\end{pmatrix}+1\cdot\begin{pmatrix}-1\\1\end{pmatrix})\\ &&+c_2\cdot(5/2\cdot\begin{pmatrix}2\\0\end{pmatrix}+2\cdot\begin{pmatrix}-1\\1\end{pmatrix})\\ &&+c_3\cdot(-3/2\begin{pmatrix}2\\0\end{pmatrix}-1\begin{pmatrix}-1\\1\end{pmatrix})\end{array}$
This can be regrouped to
$=(1/2c_1+5/2c_2-3/2c_3)\cdot\begin{pmatrix}2\\0\end{pmatrix}+(1c_1+2c_2-1c_3)\cdot\begin{pmatrix}-1\\1\end{pmatrix}$
So the coordinate representation of $h(\vec{v})$ is
$\text{Rep}_{B_W}(h(\vec{v}))=\begin{pmatrix}1/2c_1+5/2c_2-3/2c_3\\1c_1+2c_2-1c_3\end{pmatrix}=\begin{pmatrix}1/2&5/2&-3/2\\1&2&-1\end{pmatrix}_{B_V,B_W}\begin{pmatrix}c_1\\c_2\\c_3\end{pmatrix}_{B_V}$
The computation is performed by taking the dot product of the rows of the matrix with the column representing the input

Find the product of
$\begin{pmatrix}3&2&4\\2&5&3\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix}$

$\begin{pmatrix}3&2&4\\2&5&3\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}3(1)+2(2)+4(3)\\2(1)+5(2)+3(3)\end{pmatrix}=\begin{pmatrix}19\\21\end{pmatrix}$

Suppose matrix $H$ is $m\times n$. Fix domain and codomain spaces $V$ and $W$ of dimension $n$ and $m$ with basis $B=\langle\vec{\beta}_1,...,\vec{\beta}_n\rangle$ and $D$. Then $H$ represents some linear map $h$ whose range space is the span of $\{h(\vec{\beta}_1),...,h(\vec{\beta}_n)\}$. The rank of map $h$ is the dimension of this range space.
The rank of the matrix is the rank of the column space, the span of $\{\text{Rep}_D(h(\vec{\beta}_1)),...,\text{Rep}_D(h(\vec{\beta}_n))\}$
Next check if the dimensions are equal.
Under the isomorphism $\text{Rep}_D:W\mathbb{R}^m$, $\vec{0}=c_1\cdot h(\vec{\beta}_1)+\cdots+c_n\cdot h(\vec{\beta}_n)$ iff $\vec{0}=c_1\cdot\text{Rep}_D(h(\vec{\beta}_1))+\cdots+c_n\cdot\text{Rep}_D(h(\vec{\beta}_n))$.
Therefore, the size of the largest linearly independent subset of the range space equals the size of the largest linearly independent subset of the column space, so the two spaces have the same dimension.

For onto half:
$\text{rank}(h)=\text{rank}(H)$ by previous theorem
dimension of $W$ = number of rows in $H$
So
$\text{rank}(h)$ = number of rows $\implies\text{dim}(\mathscr{R}(h))=\text{dim}(W)$
But $\mathscr{R}(h)\subseteq\text{dim}(W)$, so if their dimensions are equal, they msut be equal sets.
The range space covers the entire codomain, so $h$ is onto.

For one-to-one half:
A linear map is one-to-one iff $\text{dim}(V)=\text{dim}(\mathscr{R}(h))$
The dimension of the domain is the number of columns of $H$ and by the theorem the rank of $H$ equals the dimension of $\mathscr{R}(h)$

Assume $h:V\to W$ is nonsingular. Then for a matrix $H$, because $h$ is one-to-one and onto, the number of rows equals the number of columns equals the rank of $H$, so it is square.
Next assume $H$ is a square $n\times n$ matrix.
Matrix $H$ is nonsingular $\iff$ row rank is $n$$\iff$$H$'s rank is $n$$\iff$$h$'s rank is $n$, $\iff$$h$ is an isomorphism (dimension of domain and codomain are both $n$).

The matrix
$\begin{pmatrix}0&3\\-1&2\end{pmatrix}$
is nonsingular (we can check by Gauss's Method or seeing that the rows are linearly indepenedent), so any map represented by this matrix is an isomorphism.

The matrix
$\begin{pmatrix}2&1&-2\\3&2&1\\-1&0&5\end{pmatrix}$
is singular ($\rho_2-2\rho_1=\rho_3$), so a map represented by this matrix is a homomorphism that isn't an isomorphism.

Consider the map $h:\mathbb{R}^3\to\mathbb{R}^3$
$\begin{pmatrix}x\\y\\z\end{pmatrix}\xmapsto{h}\begin{pmatrix}x\\x+z\\x-z\end{pmatrix}$
The basis vectors get mapped to
$\begin{array}{ccc}\begin{pmatrix}1\\0\\0\end{pmatrix}\xmapsto{h}\begin{pmatrix}1\\1\\1\end{pmatrix}&\begin{pmatrix}0\\1\\0\end{pmatrix}\xmapsto{h}\begin{pmatrix}0\\0\\0\end{pmatrix}&\begin{pmatrix}0\\0\\1\end{pmatrix}\xmapsto{h}\begin{pmatrix}0\\1\\-1\end{pmatrix}\end{array}$
The matrix representing this map is thus
$\begin{pmatrix}1&0&0\\1&0&1\\1&0&-1\end{pmatrix}$
The mapping can be represented by this equation for input vector $\begin{pmatrix}x\\y\\z\end{pmatrix}$ and output vector $\begin{pmatrix}a\\b\\c\end{pmatrix}$
$\begin{pmatrix}1&0&0\\1&0&1\\1&0&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$
First check if there is a solution for any $a,b,c$
$\left(\begin{array}{ccc|c}1&0&0&a\\1&0&1&b\\1&0&-1&c\end{array}\right)\sim\left(\begin{array}{ccc|c}1&0&0&a\\0&0&1&-a+b\\0&0&0&-2a+b+c\end{array}\right)$
This only has a solution if $b+c=2a$, so this is not onto.
In particular, the range space $\mathscr{R}(h)=\left\{t\begin{pmatrix}1/2\\1\\0\end{pmatrix}+s\begin{pmatrix}1/2\\0\\1\end{pmatrix}\space|\space t,s\in\mathbb{R}\right\}$
The associated homogeneous system is when $a=b=c=0$, so the matrix is
$\left(\begin{array}{ccc|c}1&0&0&0\\0&0&1&0\\0&0&0&0\end{array}\right)$
The solution set is non trivial, so it is not one-to-one.
In particular, the null space $\mathscr{N}(h)=\left\{t\begin{pmatrix}0\\1\\0\end{pmatrix}\space\middle|\space t\in\mathbb{R}\right\}$

Consider the map $h$ represented by
$H=\begin{pmatrix}1&2&1\\2&3&0\end{pmatrix}$
This takes a 3-dimensional input and a 2-dimensional output, so for any $a,b$,
$\left(\begin{array}{ccc|c}1&2&1&a\\2&3&0&b\end{array}\right)\sim\left(\begin{array}{ccc|c}1&0&-3&-3a+2b\\0&1&2&2a-b\end{array}\right)$
Since there clearly exists an input that maps to $a,b$, so this is onto.
To see if it is one-to-one, plug in $a=b=0$ and we get the solution set
$\left\{\begin{pmatrix}3\\-2\\1\end{pmatrix}z\space\middle|\space z\in\mathbb{R}\right\}$
There is more than one solution, so it is not one-to-one
This is the nullspace; its dimension is 1. This could be determined by the fact that the output is 2-dimensional, so $\text{rank}(h)=2$, meaning the nullity is $3-2=1$.