"$1$ is an eigenvalue of $A^n$" implies an eigenvalue of $A$ is a root of unity?

Question

Let $A$ be a square matrix. If $1$ is an eigenvalue of $A^n$, then is it true that there is an eigenvalue of $A$ which is a root of unity?

Answer 1

Let the eigenvalues of $A$ are $\lambda_1, \dots, \lambda_k$. That implise, that $A^n$ has the eigenvalues $\lambda_1^n, \dots, \lambda_k^n$.

If some of $\lambda_i^n =1$, that implies $\lambda_i$ is the $n$-th root of unity.

Answer 2

Yes it is true. To see it you should only consider eigenvalues in a Jordan form of $A$.

In two other answers authors do not proof that each eigenvalue of $A^n$ is a n-degree of an eigenvalue of $A$. But you can see it consider matrix $A$ in a Jordan form and take its n-th degree and for this triangle matrix calculate characteristic polynomial which contains all information about eigenvalues repetition factors.

Answer 3

We prove easily that if $\lambda$ is an eigenvalue of $A$ then $P(\lambda$) is an eigenvalue of $P(A)$ where $P$ is a polynomial.

Now take $P(x)=x^n$ and by the hypothesis we have for some $\lambda$, $\lambda^n=1$. Conclude.