Let $A$ and $B$ be $n\times n$ matrices. Suppose $A$ is similar to $B$. Prove trace($A$) = trace($B$).

Let $A$ and $B$ be $n\times n$ matrices. Suppose $A$ is similar to $B$. Prove $\operatorname{trace}(A) = \operatorname{trace}(B)$.

I'm not sure where to go on this. So far I have this:

If $A$ is similar to $B$, then

$B=P^{-1}AP$ and $A=PBP^{-1}$

This implies that:

$\operatorname{trace}(B) = \operatorname{trace}(P^{-1}AP)$ and $\operatorname{trace}(A) = \operatorname{trace}(PBP^{-1})$

Not sure where to go from here

Because $A$ and $B$ are similar, we know that $B=P^{-1}AP$. Thus

$$tr(B)=tr(P^{-1}AP)=tr(P^{-1}(AP))=tr((AP)P^{-1})=tr(A(PP^{-1}))=tr(AI)=tr(A)$$

Use the fact that $\operatorname{trace}(AB) = \operatorname{trace}(BA)$.