How to find the transition matrix for ordered basis of 2x2 diagonal matrices 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.
 A: 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
$$
\begin{pmatrix}
x&y\\z&t
\end{pmatrix}
\begin{pmatrix}
-1\\1
\end{pmatrix}=
\begin{pmatrix}
-1\\5
\end{pmatrix}
$$
A: Thank you for your direction.  I was able to use your ideas to find the correct solution to the problem.  First I expressed B and C in terms of the basis 
$\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]$,$\left[ \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right]$
The basis B is thus equivalent to $\left[ \begin{matrix} -5\\ 4 \end{matrix} \right]$,$\left[ \begin{matrix} -1 \\ 1\end{matrix} \right]$
And the basis C to   $\left[ \begin{matrix} -5\\ -4 \end{matrix} \right]$,$\left[ \begin{matrix} -1 \\ 5\end{matrix} \right]$
Then I computed the transition matrix from C to B by multiplying the inverse of B times C: 
$\left[ \begin{matrix} -5 & -1 \\ 4 & 1 \end{matrix} \right]^{-1}$$\left[ \begin{matrix} -5 & -1 \\ -4 & 5 \end{matrix} \right]$=$\left[ \begin{matrix} 9 & -4 \\ -40 & 21 \end{matrix} \right]$
