Proving that $(A^n)^{-1} = (A^{-1})^n$ for invertible matrix $A$.

Question

I have seen a proof of the fact that for an invertible matrix $A$, $A^n$ is also invertible and $$ (A^n)^{-1} = (A^{-1})^n. $$ The proof was by induction and it was mentioned that one has to use induction because one has that it is true for all $n$.

I am wondering why one has to use induction. Why can't one just say that $$ (A^n)(A^{-1})^n = (AA^{-1})^n = I^n =I $$ where one has used that $A$ and $A^{-1}$ by definition commute. Isn't this enough to show that $A^n$ is also invertible and $(A^n)^{-1} = (A^{-1})^n$?

Answer 1

your proof is fine (to me, at least). I think if you wanted to be super rigorous, you'd need to use induction anyways to fully show that $(A^n)(A^{-1})^n=(AA^{-1})^n$ because even though $A$ and $A^{-1}$ commute, you need to be sure it holds for arbitrarily large products of the two.

Honestly though, like I said, your proof is fine how it is.

Answer 2

We can use induction to prove this.

• For $n=1$ this is trivial.

• Suppose that this is true for $n-1$, that is $(A^{n-1})^{-1}= (A^{-1})^{n-1}$

  Now we can write $A^n$ as $A\cdot A^{n-1}$. Hence \begin{align}(A^n)^{-1}&= (A\cdot A^{n-1})^{-1}=(A^{n-1})^{-1}\cdot A^{-1}= &\left[\text{as }\ (AB)^{-1}=B^{-1}A^{-1}\right]\\&= (A^{-1})^{n-1}\cdot A^{-1}=(A^{-1})^{n}\end{align}

Hence proved