If $\lambda^n $ is an eigenvalue of $A^n$ then $\lambda $ is an eigenvalue of $A$? I'm trying to figure out whether the following statement is correct:

If $\lambda^n $ is an eigenvalue of $A^n$ associated to a certain
  eigenvector $v$ then $\lambda $ is an eigenvalue of $A$ associated to
  the same eigenvector $v$.

I know that the opposite statement is correct, that is, if $\lambda $ is an eigenvalue of $A$ associated to a certain eigenvector $v$ then $\lambda^n $ is an eigenvalue of $A^n$ associated to the same eigenvector $v$.
I tried to guess some matrices to find out a counter example but I couldn't find anything. It seems that if we take $A$ to be diagonal matrix then the statement is always correct. In particular if we take $A$ such that $A^n$=$\lambda^n I $ then it must be that $A=\lambda I$.
 A: This is one of those problems that can be solved with a little matrix algebra. Let's work with complex matrices for simplicity (because then we can take $n$-th roots without worrying whether or not they exist). Suppose $\lambda$ is an eigenvalue of $A^2$. Then
$$\det(A^2-\lambda I)=0$$
Because $A$ and $\sqrt \lambda I$ commute, we have the identity:
$$A^2-\lambda I = (A-\sqrt\lambda I)(A+\sqrt\lambda I)$$
Therefore
$$\det(A^2-\lambda I)=\det(A-\sqrt\lambda I)\det(A+\sqrt\lambda I)=0$$
So at least one of $\sqrt\lambda$ and $-\sqrt\lambda$ is an eigenvalue of $A$. You can easily find an example to show that both need not be eigenvalues by letting $A$ be a diagonal matrix with eigenvalues $\sqrt\lambda$ and, say, $6$.
A generalization of this argument shows that if $\lambda$ is an eigenvalue of $A^n$, then at least one of the $n$-th roots of $\lambda$ is an eigenvalue of $A$, but not necessarily all of them.
A: 
In particular, if we take $A$ such that $A^n = \lambda^n I$, then it must be that $A = \lambda I$.

This is not the case.  Two examples:
$$
A = \pmatrix{1&0\\0&-1}
$$
Note that $v = (1,1)$ is an eigenvector of $A^2 = I$, but not of $A$.
$$
A = \pmatrix{0&1\\0&0}
$$
Note that $v = (0,1)$ is an eigenvector of $A^2 = 0$, but not of $A$.
