Prove that if $B=P^{-1}AP$, then $q(B)=P^{-1}q(A)P$

Question

Is it possible to prove this using a similarity invariant? For example showing that $$\det(q(B))=\det(q(A))$$

Answer 1

You can prove part (a) simply by subbing $B = P^{-1}AP$ into the definition for $q()$ and then noting the following:

$B^n = (P^{-1}AP)^n = (P^{-1}AP)^{n-1}(P^{-1}AP) = (P^{-1}AP)^{n-2}(P^{-1}AP)(P^{-1}AP) = (P^{-1}AP)^{n-2}(P^{-1}A^2P) $

$ = ... = P^{-1}A^nP$, since $P^{-1}P = I$.

Then for part (b), if $A$ is diagonalizable, $A = PBP^{-1}$, by definition, for diagonal $B$ containing eigenvalues of $A$ and eigenvector matrix $P$. Thus, $q(A) = P^{-1}q(B)P$ by part (a). Also, since $B$ is diagonal, $q(B)$ will also be diagonal (should be easy to check since $B^n$ will be diagonal for all $n$). Therefore, $q(A)$ is diagonalizable by definition.

Answer 2

For the first part, one must simply note that $(P^{-1}AP)^k = P^{-1}A^kP$ for any $k$. The second part follows easily, as diagonal matrices are closed under addition and multiplication with other diagonal matrices.