Diagonalizable similar matrices proof

Question

Prove that if A is diagonalizable and B is similar to A, then B is also diagonalizable.

$A=PDP^{-1}$

$B=PAP^{-1}$

$B=PPDP^{-1}P^{-1}$

$B=P^2DP^{-2}$

Would it be incorrect if I left the proof like that? Should I go further and substitute $P^2$ for a variable $Q$?

Answer 1

It is incorrect to use "$P$" in both places. There's no reason they would be the same matrix. You know $A=PDP^{-1}$ for some invertible $P$ and diagonal $D$, and $B=QAQ^{-1}$ for some invertible $Q$, so $B=QPDP^{-1}Q^{-1}=(QP)D(QP)^{-1}$, which shows that $B$ is similar to $D$.

Answer 2

Let $A=PDP^{-1}$ and $B=QAQ^{-1}$, where $P$ and $Q$ are non-singular matrices. Now plug $A$ in $B=QAQ^{-1}$ to get $B=(QP)D(QP)^{-1}$ which shows that $B$ is also diagonalizable with same $D$ consisiting of eigenvalues of $A$.