Find Transition matrix given two basis Consider the ordered bases $B=(\begin{bmatrix}2 & -1\\0 & −1\end{bmatrix},\begin{bmatrix}3 & -3\\0& −1\end{bmatrix}, \begin{bmatrix}-2 & -3\\0 & 2\end{bmatrix})$ and $C=(\begin{bmatrix}-2 & 2\\0& 2\end{bmatrix},\begin{bmatrix}2 & -3\\0& −3\end{bmatrix},\begin{bmatrix}0 & -2\\0& 0\end{bmatrix})$ for the vector space $V$ of upper triangular $2×2$ matrices.
Find the transition matrix from $C$ to $B$. 
I have tried solving for linear combinations of $C$ that would create each matrix in $B$. However, the result was not right. Please help with how to proceed.
 A: I believe the following approach may be what you are looking for.  Consider adding the missing basis matrix to each set, unrolling into $4 \times 4$ matrices, then taking the appropriate products.
The procedure I am suggesting is adding $\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}$ to each set:
$$
B=
\begin{bmatrix}
2 & -1 \\
0 & -1 \\
\end{bmatrix},
\begin{bmatrix}
3 & -3 \\
0 & -1 \\
\end{bmatrix},
\begin{bmatrix}
-2 & -3 \\
0 & 2 \\
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}
$$
$$
C=
\begin{bmatrix}
-2 & 2 \\
0 & 2 \\
\end{bmatrix},
\begin{bmatrix}
2 & -3 \\
0 & -3 \\
\end{bmatrix},
\begin{bmatrix}
0 & -2 \\
0 & 0 \\
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
1 & 0 \\
\end{bmatrix}
$$.
Now, unpacking each basis to form a larger matrix by extracting each entry in a clockwise order from top left to form each column:
$$
B_{4}=
\begin{bmatrix}
2 & 3 & -2 &0 \\
-1 & -3 & -3 & 0\\
-1 & -1& 2 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$
$$
C_{4}=
\begin{bmatrix}
-2 & 2 & 0 &0 \\
2 & -3 & -2 & 0\\
2 & -3& 0 & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$
We want to change the basis from $B \rightarrow C$, which requires $C_{4}^{-1}$:
$$
C_{4}^{-1}=
\begin{bmatrix}
-\frac{3}{2} & 0 & -1 &0 \\
-1 & 0 & -1 & 0\\
0 & -\frac{1}{2}& \frac{1}{2} & 0\\
0 &0 & 0 & 1\\
\end{bmatrix}
$$
Lastly, form the product $M=
C_{4}^{-1}B_{4}$:
$$
M=
C_{4}^{-1}B_{4}=
\begin{bmatrix}
-2 & -\frac{7}{2} & 1 &0 \\
-1 & -2 & 0 & 0\\
0 & 1& \frac{5}{2} & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
$$.
I believe this final matrix $M$ is the transition matrix from $B\rightarrow C$.  
To test this, consider what the first basis in $B$ would be in $C$.  In otherwords, we want to compute:
$$
M\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}=
\begin{bmatrix}
-2 & -\frac{7}{2} & 1 &0 \\
-1 & -2 & 0 & 0\\
0 & 1& \frac{5}{2} & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
-1 \\
0 \\
0 \\
\end{bmatrix}
$$.
This means, we take $-2$ of the first basis in $C$ and add it to $-1$ of the second basis in $C$:
$$
\begin{bmatrix}
2 & -1 \\
0 & -1 \\
\end{bmatrix}
=
-2
\begin{bmatrix}
-2 & 2 \\
0 & 2 \\
\end{bmatrix}
-
\begin{bmatrix}
2 & -3 \\
0 & -3 \\
\end{bmatrix}
?
$$
$$
\begin{bmatrix}
2 & -1 \\
0 & -1 \\
\end{bmatrix}
=
\begin{bmatrix}
4 & -4 \\
0 & -4 \\
\end{bmatrix}
+
\begin{bmatrix}
-2 & 3 \\
0 & 3 \\
\end{bmatrix}
$$
This seems to check out.  
I hope this helps.
