We return to the linear transformation $\ltdefn{T}{M_{22}}{M_{22}}$ of Example ELTBM defined by \begin{equation*} \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}} = \begin{bmatrix} -17a+11b+8c-11d & -57a+35b+24c-33d \\ -14a+10b+6c-10d & -41a+25b+16c-23d \end{bmatrix} \end{equation*}
In Example ELTBM we showcased four eigenvectors of $T$. We will now put these four vectors in a set, \begin{equation*} B=\set{\vect{x}_1,\,\vect{x}_2,\,\vect{x}_3,\,\vect{x}_4} =\set{ \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} ,\, \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} ,\, \begin{bmatrix} 1 & 3 \\ 2 & 3 \end{bmatrix} ,\, \begin{bmatrix} 2 & 6 \\ 1 & 4 \end{bmatrix} } \end{equation*}
Check that $B$ is a basis of $M_{22}$ by first establishing the linear independence of $B$ and then employing Theorem G to get the spanning property easily. Here is a second set of $2\times 2$ matrices, which also forms a basis of $M_{22}$ (Example BM), \begin{equation*} C=\set{\vect{y}_1,\,\vect{y}_2,\,\vect{y}_3,\,\vect{y}_4} =\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{equation*}
We can build two matrix representations of $T$, one relative to $B$ and one relative to $C$. Each is easy, but for wildly different reasons. In our computation of the matrix representation relative to $B$ we borrow some of our work in Example ELTBM. Here are the representations, then the explanation. \begin{align*} \vectrep{B}{\lt{T}{\vect{x}_1}} &= \vectrep{B}{2\vect{x}_1} =\vectrep{B}{2\vect{x}_1+0\vect{x}_2+0\vect{x}_3+0\vect{x}_4} =\colvector{2\\0\\0\\0}\\ \vectrep{B}{\lt{T}{\vect{x}_2}} &= \vectrep{B}{2\vect{x}_2} =\vectrep{B}{0\vect{x}_1+2\vect{x}_2+0\vect{x}_3+0\vect{x}_4} =\colvector{0\\2\\0\\0}\\ \vectrep{B}{\lt{T}{\vect{x}_3}} &= \vectrep{B}{(-1)\vect{x}_3} =\vectrep{B}{0\vect{x}_1+0\vect{x}_2+(-1)\vect{x}_3+0\vect{x}_4} =\colvector{0\\0\\-1\\0}\\ \vectrep{B}{\lt{T}{\vect{x}_4}} &= \vectrep{B}{(-2)\vect{x}_4} =\vectrep{B}{0\vect{x}_1+0\vect{x}_2+0\vect{x}_3+(-2)\vect{x}_4} =\colvector{0\\0\\0\\-2} \end{align*}
So the resulting representation is \begin{align*} \matrixrep{T}{B}{B} = \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -2\\ \end{bmatrix} \end{align*}
Very pretty.
Now for the matrix representation relative to $C$ first compute, \begin{align*} &\vectrep{C}{\lt{T}{\vect{y}_1}} =\vectrep{C}{\begin{bmatrix}-17&-57\\-14&-41\end{bmatrix}}\\ &=\vectrep{C}{ (-17)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (-57)\begin{bmatrix}0&1\\0&0\end{bmatrix}+ (-14)\begin{bmatrix}0&0\\1&0\end{bmatrix}+ (-41)\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{-17\\-57\\-14\\-41}\\ &\vectrep{C}{\lt{T}{\vect{y}_2}} =\vectrep{C}{\begin{bmatrix}11&35\\10&25\end{bmatrix}}\\ &=\vectrep{C}{ 11\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 35\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 10\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 25\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{11\\35\\10\\25}\\ &\vectrep{C}{\lt{T}{\vect{y}_3}} =\vectrep{C}{\begin{bmatrix}8&24\\6&16\end{bmatrix}}\\ &=\vectrep{C}{ 8\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 24\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 6\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 16\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{8\\24\\6\\16}\\ &\vectrep{C}{\lt{T}{\vect{y}_4}} =\vectrep{C}{\begin{bmatrix}-11&-33\\-10&-23\end{bmatrix}}\\ &=\vectrep{C}{ (-11)\begin{bmatrix}1&0\\0&0\end{bmatrix}+ (-33)\begin{bmatrix}0&1\\0&0\end{bmatrix}+ (-10)\begin{bmatrix}0&0\\1&0\end{bmatrix}+ (-23)\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{-11\\-33\\-10\\-23}\\ \end{align*}
So the resulting representation is \begin{align*} \matrixrep{T}{C}{C} = \begin{bmatrix} -17 & 11 & 8 & -11 \\ -57 & 35 & 24 & -33 \\ -14 & 10 & 6 & -10 \\ -41 & 25 & 16 & -23 \end{bmatrix} \end{align*}
Not quite as pretty.
The purpose of this example is to illustrate Theorem SCB. This theorem says that the two matrix representations, $\matrixrep{T}{B}{B}$ and $\matrixrep{T}{C}{C}$, of the one linear transformation, $T$, are related by a similarity transformation using the change-of-basis matrix $\cbm{B}{C}$. Let us compute this change-of-basis matrix. Notice that since $C$ is such a nice basis, this is fairly straightforward, \begin{align*} \vectrep{C}{\vect{x}_1} &=\vectrep{C}{\begin{bmatrix}0 & 1 \\ 0 & 1\end{bmatrix}} =\vectrep{C}{ 0\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 1\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 0\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 1\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{0\\1\\0\\1}\\ \vectrep{C}{\vect{x}_2} &=\vectrep{C}{\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}} =\vectrep{C}{ 1\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 1\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 1\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 0\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{1\\1\\1\\0}\\ \vectrep{C}{\vect{x}_3} &=\vectrep{C}{\begin{bmatrix}1 & 3 \\ 2 & 3\end{bmatrix}} =\vectrep{C}{ 1\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 3\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 2\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 3\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{1\\3\\2\\3}\\ \vectrep{C}{\vect{x}_4} &=\vectrep{C}{\begin{bmatrix}2 & 6 \\ 1 & 4\end{bmatrix}} =\vectrep{C}{ 2\begin{bmatrix}1&0\\0&0\end{bmatrix}+ 6\begin{bmatrix}0&1\\0&0\end{bmatrix}+ 1\begin{bmatrix}0&0\\1&0\end{bmatrix}+ 4\begin{bmatrix}0&0\\0&1\end{bmatrix} } =\colvector{2\\6\\1\\4} \end{align*}
So we have, \begin{equation*} \cbm{B}{C} = \begin{bmatrix} 0 & 1 & 1 & 2 \\ 1 & 1 & 3 & 6 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 3 & 4 \end{bmatrix} \end{equation*}
Now, according to Theorem SCB we can write, \begin{align*} \matrixrep{T}{B}{B}&=\inverse{\cbm{B}{C}}\matrixrep{T}{C}{C}\cbm{B}{C}\\ \begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -2\\ \end{bmatrix} &= \inverse{ \begin{bmatrix} 0 & 1 & 1 & 2 \\ 1 & 1 & 3 & 6 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 3 & 4 \end{bmatrix} } \begin{bmatrix} -17 & 11 & 8 & -11 \\ -57 & 35 & 24 & -33 \\ -14 & 10 & 6 & -10 \\ -41 & 25 & 16 & -23 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 & 2 \\ 1 & 1 & 3 & 6 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 3 & 4 \end{bmatrix} \end{align*}
This should look and feel exactly like the process for diagonalizing a matrix, as was described in Section SD. And it is.