Similar matrices and conjugation outside of the center of GL(n,R) (a basic linear algebra question)

Question

I'm teaching an introductory linear algebra course and have confused myself about something that is probably very straightforward; I would love someone to help me see where I'm going wrong.

We have just started talking about eigenanalysis and have recently shown that two matrices are similar if and only if they represent the same linear transformation. That is, we have $A = P B P^{-1}$ if and only if $A$ and $B$ are both matrix representations of $T$ in the bases $\mathcal{B}_A$ and $\mathcal{B}_B$. This is all fine; the proof is straightforward and totally unobjectionable.

My confusion comes from the fact that conjugation ($\phi_g: a \mapsto gag^{-1}$) is a nontrivial action for most subgroups of linear transformations. For an example, let's look at $r$ conjugated by $s$ in the dihedral group $D_n$. I have $r^{-1} = s r s$, and $r$ and $r^{-1}$ are not the same linear transformation for any $n > 2$. But $s = s^{-1}$, so we certainly have $r^{-1} = P r P^{-1}$ (matrix similarity) for a matrix representation of $D_n$ in whatever basis we might choose. It seems like these would then be a pair of similar matrices that represent different linear transformations, which is obviously not possible.

I must be operating under some kind of silly misconception or misunderstanding some basic definition; can somebody tell me what it is?

Answer 1

Relative to a given basis, the matrices for $r$ and $r^{-1}$ will, as expected, be inverses. In particular, since we are in $D_n,\,n\gt2$, they will represent different linear transformations.

The relation $srs=r^{-1}$, or, as a relator just $(rs)^2$, ensures that their representations as matrices will be similar.

Remember, that just means that under a certain change of basis the matrices represent the same linear transformation.

In sum, there's no contradiction.

Answer 2

The confusion comes from taking an implicit can to mean must, which is not right. If a same linear operator has matrices $A$ respectively $B$ with respect to two different basis, then $A$ and $B$ are similar. Moreover all cases of similarity come about in this way: is $A$ and $B$ are similar square matrices then there exists a linear operator and two basis for the space it acts on such that $A$ and $B$ are its matrices with respect to these respective bases. So (distinct) similar matrices can be matrices of a same operator with respect to different bases. This does not mean however that when $A$ and $B$ arise as matrices of linear operators with respect to some pair of bases, then those linear operators must be the same. Indeed, if the two bases happen to be the same, then certainly different matrices will correspond to different linear operators. And in the way it was stated, the two linear operators might even operate on different spaces (in which case the bases are bases of the respective, distinct, spaces), and then the two linear operators cannot even be compared for equality.