If A is similar to B then $A^2$ is similar to $B^2$

Question

im not sure how to begin to prove this, all i know is that for two matrices to be similar, the following equation must be true

$A = PBP^{-1}$

any help will be appreciated

Answer 1

The definition of $A \sim B$ is that there exists an invertible matrix $P$ of the same dimensions as $A$ and $B$ that satisfies $A = PBP^{-1}$.

You just need to show that:

$$A^{2} = AA = (PBP^{-1})(PBP^{-1}) = PB(P^{-1}P)BP^{-1} = PBIBP^{-1} = PBBP^{-1} = PB^{2}P^{-1}$$

And then conclude that, in fact, there exists an invertible matrix $P$ such that

$$A^{2} = PB^{2}P^{-1}\Rightarrow A^{2} \sim B^{2}$$

And you're done.

Answer 2

$A^2 = AA = (PBP^{-1})(PBP^{-1}) = PB^2P^{-1}$