I'm struggling a little bit with this question and would really appreciate a nod in the right direction!
I need to show that if $A=PBP^{-1}$, then the centralizers of $A$ and $B$ have the same dimension.
I've tried representing one by the other (but it doesn't really seem doable) and I realise that they have the same eigenvalues which might be useful.
I understand that I probably need to prove a bijection from one to the other but I'm not really sure where to start. Any help would be much appreciated, thanks.
Let us denote $$ \forall A \in \Bbb M_n ,Z(A) = \{M \in \Bbb M_n \mid MA=AM\}. $$ If $A=PBP^{-1}$ and $M \in Z(A)$, we have \begin{align} MA=AM &\iff MPBP^{-1}=PBP^{-1}M\\ &\iff P^{-1}MPB=BP^{-1}MP\\ &\iff M'B=BM' \end{align} with $M'=P^{-1}MP$. This gives us a hint: let $$ \phi_P : Z(A) \to Z(B) $$ such that $\phi_P(M)=M'=P^{-1}MP$. We see that $\phi$ is linear and that $$ \phi_{P^{-1}} \circ \phi_{P} =\phi_{P} \circ \phi_{P^{-1}}=\text{id}. $$ So $\phi$ is an isomorphism. This shows that $Z(A) \cong Z(B)$.