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?

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 $M_T^B$.
