Why is the change of basis formula counter-intuitive? [See details] 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?
 A: 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.
A: 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
A: Remember that $T_{B'}$ is a function from $B'$ to $B'$.
Consider the following diagram:

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


*

*Follow the arrow $P_{B' \to B}$.

*Follow the arrow $T_B$.

*Follow the arrow $P_{B \to B'}$.


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