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

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?

• Looks good. Alternatively, $A(A^{m-1})=I=(A^{m-1})A$ so $A^{-1}=A^{m-1}$. – Chrystomath Mar 18 '19 at 11:32

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 $$-1$$.
I think it is easier to see it this way: what is $$AA^{m-1} = A^{m-1}A$$?
It's invertible since it has a inverse, $$A^{m-1}$$ that is. By the way, $$\det(A)$$ could also be -1.