Im looking at the solution to one of my questions. Basically, we started off with a matrix $A$ (in the elementary basis) which we want to convert into a diagonal matrix $B$ of another basis.
Question 1:
The teacher uses $B=PAP^{-1}$ and $B=P^{-1}AP$ interchangeably, can I really do that if $P$ is the basis of $B$? I would think they would give different values for $B$ wouldn't they?
Question 2:
Since Matrix $A$ is in the elementary basis, isn't $BP=A$? Why can't we find $B$ by $AP^{-1}$ and why do we have $BP=P^{-1}A$?
Well, in some ways they are equivalent; it just depends on your point of view. If $B = P^{-1} A P$ then $A = P B P^{-1}$, so in terms of speaking about similar matrices, it makes no difference.
However: You are usually interested in taking a matrix $A$, which represents a linear transformation with respect to the standard basis, and writing a new (diagonal, sometimes) matrix $B$ which represents the transformation with respect to some other basis $\beta$. You think of this new basis making the columns of your matrix $P$. Then you want $B = P^{-1} A P$, because $P$ will take a vector $[v]_{\beta}$ (so written with respect to the basis $\beta$) and then $P [v]_{\beta}$ will be that vector in standard basis; then $A$ performs the transformation wrt the standard basis giving $Av$, and $P^{-1}$ then changes back to the basis $\beta$ giving $[Av]_{\beta}$.
The flip side of course is when you want to think of the matrix $B$ being diagonal for convenience, but acting on standard vectors in which case you want $P B P^{-1}$ so you can think about things happening in the opposite order ($P^{-1}$ converting to $\beta$, then transform, then change back to standard basis).