Consider the linear transformation, $\ltdefn{S}{P_3}{M_{22}}$, given by \begin{align*} \lt{S}{a+bx+cx^2+dx^3}= \begin{bmatrix} 3a+7b-2c-5d & 8a+14b-2c-11d\\ -4a-8b+2c+6d & 12a+22b-4c-17d \end{bmatrix} \end{align*}
First, we build a representation relative to the bases, \begin{align*} B&=\set{1+2x+x^2-x^3,\,1+3x+x^2+x^3,\,-1-2x+2x^3,\,2+3x+2x^2-5x^3}\\ C&=\set{ \begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix},\, \begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix},\, \begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix},\, \begin{bmatrix}-1 & -4 \\ -2 & -4\end{bmatrix} } \end{align*}
We evaluate $S$ with each element of the basis for the domain, $B$, and coordinatize the result relative to the vectors in the basis for the codomain, $C$. Notice here how we take elements of vector spaces and decompose them into linear combinations of basis elements as the key step in constructing coordinatizations of vectors. There is a system of equations involved almost every time, but we will omit these details since this should be a routine exercise at this stage. \begin{align*} &\vectrep{C}{\lt{S}{1+2x+x^2-x^3}} =\vectrep{C}{\begin{bmatrix}20 & 45 \\ -24 & 69\end{bmatrix}}\\ &\quad=\vectrep{C}{ (-90)\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+ 37\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+ (-40)\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+ 4\begin{bmatrix}-1 & -4 \\ -2 & -4\end{bmatrix} } =\colvector{-90\\37\\-40\\4}\\ &\vectrep{C}{\lt{S}{1+3x+x^2+x^3}} =\vectrep{C}{\begin{bmatrix}17 & 37 \\ -20 & 57\end{bmatrix}}\\ &\quad=\vectrep{C}{ (-72)\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+ 29\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+ (-34)\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+ 3\begin{bmatrix}-1 & -4 \\ -2 & -4\end{bmatrix} } =\colvector{-72\\29\\-34\\3}\\ &\vectrep{C}{\lt{S}{-1-2x+2x^3}} =\vectrep{C}{\begin{bmatrix}-27 & -58 \\ 32 & -90\end{bmatrix}}\\ &\quad=\vectrep{C}{ 114\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+ (-46)\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+ 54\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+ (-5)\begin{bmatrix}-1 & -4 \\ -2 & -4\end{bmatrix} } =\colvector{114\\-46\\54\\-5}\\ &\vectrep{C}{\lt{S}{2+3x+2x^2-5x^3}} =\vectrep{C}{\begin{bmatrix}48 & 109 \\ -58 & 167\end{bmatrix}}\\ &\quad=\vectrep{C}{ -220\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}+ 91\begin{bmatrix}2 & 3 \\ 2 & 5\end{bmatrix}+ (-96)\begin{bmatrix}-1 & -1 \\ 0 & -2\end{bmatrix}+ 10\begin{bmatrix}-1 & -4 \\ -2 & -4\end{bmatrix} } =\colvector{-220\\91\\-96\\10}\\ \end{align*}
Thus, employing Definition MR \begin{equation*} \matrixrep{S}{B}{C}= \begin{bmatrix} -90 & -72 & 114 & -220 \\ 37 & 29 & -46 & 91 \\ -40 & -34 & 54 & -96 \\ 4 & 3 & -5 & 10 \end{bmatrix} \end{equation*}
Often we use “nice” bases to build matrix representations and the work involved is much easier. Suppose we take bases \begin{align*} D&=\set{1,\,x,\,x^2,\,x^3} & E&=\set{ \begin{bmatrix}1&0\\0&0\end{bmatrix},\, \begin{bmatrix}0&1\\0&0\end{bmatrix},\, \begin{bmatrix}0&0\\1&0\end{bmatrix},\, \begin{bmatrix}0&0\\0&1\end{bmatrix}} \end{align*}
The evaluation of $S$ at the elements of $D$ is easy and coordinatization relative to $E$ can be done on sight, \begin{align*} \vectrep{E}{\lt{S}{1}} &=\vectrep{E}{\begin{bmatrix}3 & 8 \\ -4 & 12\end{bmatrix}}\\ &=\vectrep{E}{ 3\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 8\begin{bmatrix}0&1\\0&0\end{bmatrix}+ (-4)\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 12\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{3 \\ 8 \\ -4 \\ 12}\\ \vectrep{E}{\lt{S}{x}} &=\vectrep{E}{\begin{bmatrix}7 & 14 \\ -8 & 22\end{bmatrix}}\\ &=\vectrep{E}{ 7\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 14\begin{bmatrix}0&1\\0&0\end{bmatrix}+ (-8)\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 22\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{7 \\ 14 \\ -8 \\ 22}\\ \vectrep{E}{\lt{S}{x^2}} &=\vectrep{E}{\begin{bmatrix}-2 & -2 \\ 2 & -4\end{bmatrix}}\\ &=\vectrep{E}{ (-2)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (-2)\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 2\begin{bmatrix}0&0\\1&0\end{bmatrix}+ (-4)\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{-2 \\ -2 \\ 2 \\ -4}\\ \vectrep{E}{\lt{S}{x^3}} &=\vectrep{E}{\begin{bmatrix}-5 & -11 \\ 6 & -17\end{bmatrix}}\\ &=\vectrep{E}{ (-5)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (-11)\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 6\begin{bmatrix}0&0\\1&0\end{bmatrix}+ (-17)\begin{bmatrix}0&0\\0&1\end{bmatrix} }\\ &=\colvector{-5 \\ -11 \\ 6 \\ -17} \end{align*}
So the matrix representation of $S$ relative to $D$ and $E$ is \begin{equation*} \matrixrep{S}{D}{E}= \begin{bmatrix} 3 & 7 & -2 & -5 \\ 8 & 14 & -2 & -11 \\ -4 & -8 & 2 & 6 \\ 12 & 22 & -4 & -17 \end{bmatrix} \end{equation*}
One more time, but now let us use bases \begin{align*} F&=\set{1+x-x^2+2x^3,\,-1+2x+2x^3,\,2+x-2x^2+3x^3,\,1+x+2x^3}\\ G&=\set{ \begin{bmatrix}1&1\\-1&2\end{bmatrix},\, \begin{bmatrix}-1&2\\0&2\end{bmatrix},\, \begin{bmatrix}2&1\\-2&3\end{bmatrix},\, \begin{bmatrix}1&1\\0&2\end{bmatrix} } \end{align*} and evaluate $S$ with the elements of $F$, then coordinatize the results relative to $G$, \begin{align*} \vectrep{G}{\lt{S}{1+x-x^2+2x^3}} &=\vectrep{G}{\begin{bmatrix}2 & 2 \\ -2 & 4\end{bmatrix}} =\vectrep{G}{ 2\begin{bmatrix}1&1\\-1&2\end{bmatrix} } =\colvector{2\\0\\0\\0}\\ \vectrep{G}{\lt{S}{-1+2x+2x^3}} &=\vectrep{G}{\begin{bmatrix}1 & -2 \\ 0 & -2\end{bmatrix}} =\vectrep{G}{ (-1)\begin{bmatrix}-1&2\\0&2\end{bmatrix} } =\colvector{0\\-1\\0\\0}\\ \vectrep{G}{\lt{S}{2+x-2x^2+3x^3}} &=\vectrep{G}{\begin{bmatrix}2 & 1 \\ -2 & 3\end{bmatrix}} =\vectrep{G}{ \begin{bmatrix}2&1\\-2&3\end{bmatrix} } =\colvector{0\\0\\1\\0}\\ \vectrep{G}{\lt{S}{1+x+2x^3}} &=\vectrep{G}{\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}} =\vectrep{G}{ 0\begin{bmatrix}1&1\\0&2\end{bmatrix} } =\colvector{0\\0\\0\\0}\\ \end{align*}
So we arrive at an especially economical matrix representation, \begin{equation*} \matrixrep{S}{F}{G}= \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}