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?

• Note that $P^B_A$ takes you from basis $A$ to basis $B$, so it can't be $(P^B_A)^{-1}M^A_TP^B_A$. – B. Pasternak Mar 27 '17 at 22:27
• I think the notation is $P_A^B$ takes you from basis $B$ to basis $A$. Although your point stands, it's crappy notation. – mdave16 Mar 27 '17 at 23:36

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 $M_T^B$.

• Thank you! That is a very clear explanation. However I think I'm not sure what $P_A^B$ means. Is it the matrix from basis $A$ to basis $B$? – Hendrra Mar 28 '17 at 9:22
• I've told you what it does. That should be more than enough. – Omnomnomnom Mar 28 '17 at 10:53
• Of course it is but you used two different notations. I think it's just a typo but it's better to be sure (in line 5 and 8). – Hendrra Mar 28 '17 at 12:19
• You're right. Fixed it. – Omnomnomnom Mar 28 '17 at 12:46

I am assuming that $P^B_A$ takes from basis $B$ to $A$.
• Note, for instance, that $(1,0,0)$ wrt the basis $B$ should be mapped to $(2,0,0)$ with respect to the basis $A$. Now, looking at the first columns, it's clear which is which. – Omnomnomnom Mar 27 '17 at 23:54
• @Omnomnomnom, Why is $P_A^B$ the matrix from basis $B$ to $A$? I've literally never seen either notation before for such a concept. Wait nvm, I made a typo several times, it's fine. Thanks for the correction – mdave16 Mar 28 '17 at 0:19
• see my own answer for a detailed explanation. I think this is the choice of notation in Lay's text, if I remember correctly. You sometimes encounter $[I]^B_A$ for the same ($I$ denoting the identity transformation) or $[I]_{B \to A}$, or $[I]_{A \leftarrow B}$ (I prefer this last notation). The convention $P^B_A$ follows the set theory convention of $V^U = \{f \mid f:U \to V\}$. In set/category theory, arrows go down by default. – Omnomnomnom Mar 28 '17 at 0:31