# Suppose that $A\in M_{n}(\mathbb{C})$ and $A$ is nonsingular

Suppose that $$A\in M_{n}(\mathbb{C})$$ and $$A$$ is nonsingular and $$A$$ is similar to $$A^k$$ for each $$k=1,2,3,\ldots$$. Can we conclude that $$1$$ is the only eigenvalue of $$A$$?

I have showed that any eigenvalue of $$A$$ is of modulus $$1$$. From this how to proceed?

Let $$E(A)$$ be the set of eigenvalues of $$A$$. Then $$E(A^k) = \{\lambda^k: \lambda \in E(A)\}$$. But if $$A$$ and $$A^k$$ are similar, $$E(A) = E(A^k)$$. Suppose $$\lambda \in E(A)$$ and $$\lambda \ne 1$$. The $$\lambda^k \in E(A)$$ for each $$k$$. But $$E(A)$$ is finite, so the $$\lambda^k$$ are not distinct: there must be $$m < n$$ such that $$\lambda^m = \lambda^n$$. Since $$A$$ is nonsingular, $$\lambda \ne 0$$, so $$\lambda^{n-m} = 1$$, i.e. $$\lambda$$ is a root of unity.
If $$M$$ is the least common multiple of the orders of these roots of unity, we have $$\lambda^M = 1$$ for all $$\lambda \in E(A)$$. But that means $$E(A^M) = \{1\}$$; since by assumption $$E(A^M) = E(A)$$, that says $$E(A) = \{1\}$$.
Hint: Probably you have already use that: If $$A$$ is similar to $$B$$ then the eigenvalues of $$A$$ and $$B$$ are the same.
Now: The only complex numbers that are solution to $$x^k=x$$ simultaneously for every $$k$$ are one and zero. If there is another one, $$z$$ notice that there should exist a number $$r$$ such that $$z^r=1$$ is a primitive root of unity. So for $$\ell < r$$ $$z^{\ell}\neq 1.$$