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:
If yes why?

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

  • $\begingroup$ 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 $B$? $\endgroup$ – Hendrra Mar 28 '17 at 9:22
  • $\begingroup$ I've told you what it does. That should be more than enough. $\endgroup$ – Omnomnomnom Mar 28 '17 at 10:53
  • $\begingroup$ 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). $\endgroup$ – Hendrra Mar 28 '17 at 12:19
  • $\begingroup$ You're right. Fixed it. $\endgroup$ – Omnomnomnom Mar 28 '17 at 12:46

\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.

  • $\begingroup$ Also if anyone knows a better way to create commutative diagrams, then I would instantly love you. $\endgroup$ – mdave16 Mar 27 '17 at 23:42
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Mar 27 '17 at 23:54
  • $\begingroup$ Also, this is a fine answer for anyone who knows how to chase arrows, which is not usually a prerequisite for linear algebra. $\endgroup$ – Omnomnomnom Mar 27 '17 at 23:55
  • $\begingroup$ @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 $\endgroup$ – mdave16 Mar 28 '17 at 0:19
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    $\begingroup$ 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. $\endgroup$ – Omnomnomnom Mar 28 '17 at 0:31

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