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?

• @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. Apr 20, 2019 at 21:01

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.

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

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:

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.