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The formula of change of basis $[T]_{B'} = P_{B'\leftarrow B}[T]_{B}P_{B\leftarrow B'}$.

I don't understand why you need $P_{B\leftarrow B'}$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_{B}$ you just need to translate to $B'$ by using $P_{B'\leftarrow B}$ to get $[T]_{B'}$ rendering $P_{B\leftarrow B'}$ as useless. Can someone explain what I am missing here?

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    $\begingroup$ @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it. $\endgroup$
    – WindBreeze
    Apr 20, 2019 at 21:01

3 Answers 3

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Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.

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Write $B=\{e_1,...,e_n\}, B' =\{e_1',...,e_n'\}$

If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.

So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_{B'\to B}(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_{B\to B'}$ comes from on the left. This gives the formula

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Remember that $T_{B'}$ is a function from $B'$ to $B'$.

Consider the following diagram:

commutative diagram

The arrow $T_{B'}$ can be found by the following steps:

  1. Follow the arrow $P_{B' \to B}$.
  2. Follow the arrow $T_B$.
  3. Follow the arrow $P_{B \to B'}$.

Then $T_{B'}$ is simply the composition of these 3 arrows.

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