TITLE: Finding a matrix with respect to a basis
QUESTION [5 upvotes]: Let $T: R^2 \to R^2$ be represented by $\begin{bmatrix}5 & -3\\2 & -2\end{bmatrix}$ with respect to the standard basis. Find the matrix T with respect to the basis B = { $\begin{bmatrix}3 \\1\end{bmatrix}$ , $\begin{bmatrix}1\\2 \end{bmatrix}$ }.
I found T$\begin{bmatrix}3 \\1\end{bmatrix}$ and T$\begin{bmatrix}1\\2 \end{bmatrix}$ by:
$\begin{bmatrix}5 & -3\\2 & -2\end{bmatrix}$$\begin{bmatrix}3 \\1\end{bmatrix}$ = $\begin{bmatrix}12 \\4\end{bmatrix}$ and
$\begin{bmatrix}5 & -3\\2 & -2\end{bmatrix}$$\begin{bmatrix}1 \\2\end{bmatrix}$ = $\begin{bmatrix}-1 \\-2\end{bmatrix}$
so that $[T]_B$ = $\begin{bmatrix}12 & -1\\4 & -2\end{bmatrix}$
but I'm not sure if this is correct. Am I doing the right thing or are my steps wrong?

REPLY [6 votes]: Note: I have assumed that the basis was changed in both the domain and the range in this answer.
If $y$ is a point expressed in the new basis, then it corresponds to a point  $x=By$ in the old basis.
So, to compute the new form of the operator $T$, first map from the new coordinates to the old (multiply by $B$), then perform the mapping to get the mapped point in the old coordinates (multiply by $T$) and then map into the new coordinates (multiply by $B^{-1}$).
Putting these together gives $\tilde{T} = B^{-1} T B$.
Note that in this particular example, $T$ behaves as multiplication on the rows of $B$ (that is, $B$ is a matrix of eigenvectors), this should help considerably with the computations. In fact, if you think carefully, little computation will be needed (other than multiplying the columns of $B$ by $T$)...
Note that $B^{-1} T B = \operatorname{diag} (4, -1)$.