Prove or disprove: If $A$ is $n\times n$ and $\exists\;m\in \Bbb{N}:\;A^m=I_n$, then $A$ is invertible.

Question

Is this statement true? If $A$ is an $n\times n$ matrix and $A^m=I_n$ for some $m\in \Bbb{N}$, then $A$ is invertible.

My trial

Let $n\in \Bbb{N}$ be fixed. Then, $$[\det(A)]^m=\det(A^m)=I_n=1.$$ Hence, $$\det(A)=1\neq 0.$$ Thus, $A$ is invertible since $\det(A)\neq 0.$. I'm I right or is there a counter-example?

Answer 1

The right conclusion is $\det(A) \ne 0$, hence it is invertible.

Notice that we can't conclude that $\det(A)=1$. After all, it can take value $-1$.

Answer 2

I think it is easier to see it this way: what is $AA^{m-1} = A^{m-1}A$?

Answer 3

It's invertible since it has a inverse, $A^{m-1}$ that is. By the way, $\det(A)$ could also be -1.