Begin with two vector spaces, $S_2$, the subspace of $M_{22}$ containing all $2\times 2$ symmetric matrices, and $P_3$ (Example VSP), the vector space of all polynomials of degree 3 or less. Then define the linear transformation $\ltdefn{Q}{S_2}{P_3}$ by \begin{equation*} \lt{Q}{\begin{bmatrix}a&b\\b&c\end{bmatrix}} = (5a-2b+6c)+(3a-b+2c)x+(a+3b-c)x^2+(-4a+2b+c)x^3 \end{equation*}
Here are two bases for each vector space, one nice, one nasty. First for $S_2$, \begin{align*} B&= \set{ \begin{bmatrix}5&-3\\-3&-2\end{bmatrix},\, \begin{bmatrix}2&-3\\-3&0\end{bmatrix},\, \begin{bmatrix}1&2\\2&4\end{bmatrix} } & C&= \set{ \begin{bmatrix}1&0\\0&0\end{bmatrix},\, \begin{bmatrix}0&1\\1&0\end{bmatrix},\, \begin{bmatrix}0&0\\0&1\end{bmatrix} } \end{align*} and then for $P_3$, \begin{align*} D&=\set{ 2+x-2x^2+3x^3,\, -1-2x^2+3x^3,\, -3-x+x^3,\, -x^2+x^3 }\\ E&=\set{1,\,x,\,x^2,\,x^3} \end{align*}
We will begin with a matrix representation of $Q$ relative to $C$ and $E$. We first find vector representations of the elements of $C$ relative to $E$, \begin{align*} \vectrep{E}{\lt{Q}{\begin{bmatrix}1&0\\0&0\end{bmatrix}}} &=\vectrep{E}{5+3x+x^2-4x^3}=\colvector{5\\3\\1\\-4}\\ \vectrep{E}{\lt{Q}{\begin{bmatrix}0&1\\1&0\end{bmatrix}}} &=\vectrep{E}{-2-x+3x^2+2x^3}=\colvector{-2\\-1\\3\\2}\\ \vectrep{E}{\lt{Q}{\begin{bmatrix}0&0\\0&1\end{bmatrix}}} &=\vectrep{E}{6+2x-x^2+x^3}=\colvector{6\\2\\-1\\1}\\ \end{align*}
So \begin{align*} \matrixrep{Q}{C}{E} = \begin{bmatrix} 5 & -2 & 6\\ 3 & -1 & 2\\ 1 & 3 & -1\\ -4 & 2 & 1 \end{bmatrix} \end{align*}
Now we construct two change-of-basis matrices. First, $\cbm{B}{C}$ requires vector representations of the elements of $B$, relative to $C$. Since $C$ is a nice basis, this is straightforward, \begin{align*} \vectrep{C}{\begin{bmatrix}5&-3\\-3&-2\end{bmatrix}} &=\vectrep{C}{ (5)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (-3)\begin{bmatrix}0&1\\1&0\end{bmatrix}+ (-2)\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{5\\-3\\-2}\\ \vectrep{C}{\begin{bmatrix}2&-3\\-3&0\end{bmatrix}} &=\vectrep{C}{ (2)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (-3)\begin{bmatrix}0&1\\1&0\end{bmatrix}+ (0)\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{2\\-3\\0}\\ \vectrep{C}{\begin{bmatrix}1&2\\2&4\end{bmatrix}} &=\vectrep{C}{ (1)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (2)\begin{bmatrix}0&1\\1&0\end{bmatrix}+ (4)\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{1\\2\\4} \end{align*}
So \begin{align*} \cbm{B}{C}&= \begin{bmatrix} 5 & 2 & 1\\ -3 & -3 & 2\\ -2 & 0 & 4 \end{bmatrix} \end{align*}
The other change-of-basis matrix we will compute is $\cbm{E}{D}$. However, since $E$ is a nice basis (and $D$ is not) we will turn it around and instead compute $\cbm{D}{E}$ and apply Theorem ICBM to use an inverse to compute $\cbm{E}{D}$. \begin{align*} \vectrep{E}{2+x-2x^2+3x^3} &=\vectrep{E}{(2)1+(1)x+(-2)x^2+(3)x^3} =\colvector{2\\1\\-2\\3}\\ \vectrep{E}{-1-2x^2+3x^3} &=\vectrep{E}{(-1)1+(0)x+(-2)x^2+(3)x^3} =\colvector{-1\\0\\-2\\3}\\ \vectrep{E}{-3-x+x^3} &=\vectrep{E}{(-3)1+(-1)x+(0)x^2+(1)x^3} =\colvector{-3\\-1\\0\\1}\\ \vectrep{E}{-x^2+x^3} &=\vectrep{E}{(0)1+(0)x+(-1)x^2+(1)x^3} =\colvector{0\\0\\-1\\1} \end{align*}
So, we can package these column vectors up as a matrix to obtain $\cbm{D}{E}$ and then, \begin{align*} \cbm{E}{D} &=\inverse{\left(\cbm{D}{E}\right)} &&\knowl{./knowls/theorem.ICBM.knowl}{\text{Theorem ICBM}}\\ &=\inverse{ \begin{bmatrix} 2 & -1 & -3 & 0 \\ 1 & 0 & -1 & 0 \\ -2 & -2 & 0 & -1 \\ 3 & 3 & 1 & 1 \end{bmatrix} }\\ &= \begin{bmatrix} 1 & -2 & 1 & 1 \\ -2 & 5 & -1 & -1 \\ 1 & -3 & 1 & 1 \\ 2 & -6 & -1 & 0 \end{bmatrix} \end{align*}
We are now in a position to apply Theorem MRCB. The matrix representation of $Q$ relative to $B$ and $D$ can be obtained as follows, \begin{align*} \matrixrep{Q}{B}{D} &=\cbm{E}{D}\matrixrep{Q}{C}{E}\cbm{B}{C} &&\knowl{./knowls/theorem.MRCB.knowl}{\text{Theorem MRCB}}\\ &= \begin{bmatrix} 1 & -2 & 1 & 1 \\ -2 & 5 & -1 & -1 \\ 1 & -3 & 1 & 1 \\ 2 & -6 & -1 & 0 \end{bmatrix} \begin{bmatrix} 5 & -2 & 6\\ 3 & -1 & 2\\ 1 & 3 & -1\\ -4 & 2 & 1 \end{bmatrix} \begin{bmatrix} 5 & 2 & 1\\ -3 & -3 & 2\\ -2 & 0 & 4 \end{bmatrix}\\ &= \begin{bmatrix} 1 & -2 & 1 & 1 \\ -2 & 5 & -1 & -1 \\ 1 & -3 & 1 & 1 \\ 2 & -6 & -1 & 0 \end{bmatrix} \begin{bmatrix} 19 & 16 & 25 \\ 14 & 9 & 9 \\ -2 & -7 & 3 \\ -28 & -14 & 4 \end{bmatrix}\\ &= \begin{bmatrix} -39 & -23 & 14 \\ 62 & 34 & -12 \\ -53 & -32 & 5 \\ -44 & -15 & -7 \end{bmatrix} \end{align*}
Now check our work by computing $\matrixrep{Q}{B}{D}$ directly (Exercise CB.C21).