The problem:
For the vector space of lower triangular matrices with zero trace,
given ordered basis:
$B=${$$
\begin{bmatrix}
-5 & 0 \\
4 & 5 \\
\end{bmatrix},
$$
\begin{bmatrix}
-1 & 0 \\
1 & 1 \\
\end{bmatrix}}
and $C=${$$
\begin{bmatrix}
-5 & 0 \\
-4 & 5 \\
\end{bmatrix},
$$
\begin{bmatrix}
-1 & 0 \\
5 & 1 \\
\end{bmatrix}}
find the transition matrix $C$ to $B$.
I know how to find a transition matrix when the basis consists of $n \times 1 $ vectors, but my textbook doesn't address this scenario where the basis consists of a set of $2 \times 2$ matrices and haven't found applicable guidance online.
Best Answer
Hint:
If you know how to solve the problem for $n\times 1$ vectors than consider that your matrices can be considered as vectors of a vector space with standard basis the $2\times 2$ matrices that have only one element $=1$ and the other elements $=0$. In this basis, the matrix: $$ \begin{bmatrix} -5 & 0 \\ 4 & 5 \\ \end{bmatrix} $$ is the vector $$ \begin{bmatrix} -5 \\ 0 \\ 4\\ 5 \\ \end{bmatrix} $$
You can do the same for the other matrices and solve the problem as for usual vectors, but note that yours sets $B$ and $C$ are not basis for $M_{2\times 2}(\mathbb{R})$.
If we work in the space of lower triangular and null trace matrix in $M_{2\times 2}(\mathbb{R})$, than this subspace has dimension $2$ and any matrix in it can be expressed as $$ \begin{bmatrix} a&0\\ b&-a \end{bmatrix} =a \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix} +b \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix} =a\hat i +b \hat j $$ so $a$ $b$ can be seen as the componets of a vector $(a,b)^T$ in the basis $\{\hat i, \hat j\}$.
In this notation your basis are: $$ B=\{(-5,4)^T,(-1,1)^T\} \qquad B=\{(-5,-4)^T,(-1,5)^T\} $$
Now you can find the $2\times2$ matrix that represents the transformation (in the basis$\{\hat i, \hat j\}$) solving:
$$ \begin{pmatrix} x&y\\z&t \end{pmatrix} \begin{pmatrix} -5\\4 \end{pmatrix}= \begin{pmatrix} -5\\-4 \end{pmatrix} $$ and
$$ \begin{pmatrix} x&y\\z&t \end{pmatrix} \begin{pmatrix} -1\\1 \end{pmatrix}= \begin{pmatrix} -1\\5 \end{pmatrix} $$