Negative powers of matrix

Question

For any matrix $A$ is it true that, $A^{-n} = (A^{-1})^n = (A^n)^{-1}$?

Does this also then apply to powers of diagonalizable matrices? That is, if $A = PDP^{-1}$ then $A^{-n} = PD^{-n}P^{-1}$.

Answer 1

Yes. If $A$ is an invertble square matrix, then we have $A^{-n} := (A^{-1})^n = (A^n)^{-1}$ for all natural $n$, which can easily be proved by induction.

If $A$ is diagoalizable and $A = PDP^{-1}$, then again by induction, we have $A^{-n} = PD^{-n}P^{-1}$ for all $n \in \mathbb N.$

Answer 2

In view of the first question, for each commutative ring $R$, if $a\in R$ is invertible, then define the negative powers as $a^{-n} = (a^{-1})^n$, $n\geq 0$.

The equation $(a^{-1})^n = (a^n)^{-1}$ needs a proof (by induction on $n$).