Linear transformation and its matrix I have two bases:
$A = \{v_1, v_2, v_3\}$ and $B = \{2v_1, v_2+v_3, -v_1+2v_2-v_3\}$
There is also a linear transformation: $T: \mathbb R^3 \rightarrow \mathbb R^3$
Matrix in base $A$:
$M_{T}^{A} = \begin{bmatrix}1 & 2 &3\\4 & 5 & 6\\1 & 1 & 0\end{bmatrix}$
Now I am to find matrix of the linear transformation $T$ in base B.
I have found two transition matrixes (from base $A$ to $B$ and from $B$ to $A$):
$P_{A}^{B} = \begin{bmatrix}2 & 0 & -1\\0 & 1 & 2\\0 & 1 & -1\end{bmatrix}$
$(P_{A}^{B})^{-1} = P_{B}^{A} = \begin{bmatrix}\frac{1}{2} & \frac{1}{6} & \frac{-1}{6}\\0 & \frac{1}{3} & \frac{2}{3}\\0 & \frac{1}{3} & \frac{-1}{3}\end{bmatrix}$
How can I find $M_{T}^{B}$?
Is it equal to:
$(P_{A}^{B})^{-1}M_{T}^{A}P_{A}^{B}$?
If yes why?
 A: Some notation: for a vector $v \in \Bbb R^3$, let $[v]_A$ denote the coordinate vector of $v$ with respect to the basis $A$, and let $[v]_B$ denote the coordinate vector of $v$ with respect to the basis $B$.  both of these are column vectors. To put this another way,
$$
[v]_A = \pmatrix{a_1\\a_2\\a_3} \iff v = a_1v_1 + a_2 v_2 + a_3v_3
$$
We can think of $M_T^A$ as a "machine" with the property that, with the usual matrix multiplication, $M_T^A [v]_A = [T(v)]_A$.  Similarly, $P^B_A$ satisfies $P^B_A [v]_B = [v]_A$, whereas $P^A_B$ satisfies $P^A_B[v]_A = [v]_B$.  What we want is to "build" is a machine $M_T^B$ for which $M_T^B[v]_B = [T(v)]_B$.
We can break the process of going from $[v]_B$ to $[T(v)]_B$ into three steps, each of which uses machinery that we already have.  First, go from $[v]_B$ to $[v]_A$ with $P^B_A[v]_B = [v]_A$.  Then, go from $[v]_A$ to $[T(v)]_A$ using $M_T^A [v]_A = [T(v)]_A$.  Then, go from $[T(v)]_A$ to $[T(v)]_B$ using $P^A_B[T(v)]_A = [T(v)]_B$.
Putting it all together, we have
$$
[T(v)]_B = P^A_B[T(v)]_A = P^A_B(M_T^A [v]_A) = 
P^A_BM_T^A (P^B_A[v]_B) = (P^A_BM_T^A P^B_A) [v]_B
$$
What we have found, then, is that the matrix which takes us from $[v]_B$ to $[T(v)]_B$ is the product $P^A_BM_T^A P^B_A = (P^B_A)^{-1} M_T^A P^B_A$.  So, this is our matrix $M_T^B$.
A: \begin{align}
V &- M^A \to V\\
\uparrow\,\,   & \,\qquad \,\qquad \uparrow\\
P_A^B\,\,   & \,\,\qquad \, \,\qquad P_A^B\\
|\,\,   & \,\,\qquad \,\,\qquad |\\
V &- M^B\to V
\end{align}
First of all, you need to see that this diagram commutes. Once you see that, it's obvious.
Try proving that the diagram commutes from basic definitions of what a linear map is.
I am assuming that $P^B_A$ takes from basis $B$ to $A$.
As stated by @Omnomnomnom in his comment, this is a good answer for those that know how to chase/follow arrows. While this is in general harder than linear algebra, I think it's a nice second view of the question, it can be used more generally. Furthermore, just going through the diagram and writing down what it means for it to commute more concretely, you would find something similar to Omnomnomnom's answer. 
