Are the corresponding eigenspaces of $A$ and $A^n$ equal? Let $A$ be a square matrix and $n$ be a positive integer. If $\mu$ is an eigenvalue of $A^n$ such that there is a unique eigenvalue $\lambda$ of $A$ with $\lambda^n=\mu$, can we say that the eigenspaces of $A^n$ and $A$ corresponding to $\mu$ and $\lambda$ respectively, are same?
Edit: If I further assume that all the eigenvalues of $A$ are non-zero, can I claim the result? 

$$A=\left(\begin{matrix}0&1\\0&0\end{matrix}\right)$$
$$n=2$$
$$A^2=\left(\begin{matrix}0&0\\0&0\end{matrix}\right)$$
$$\lambda=\mu=0$$
$Ax=0$ for $x=(x_1,0)^T$, while $A^2x=0$ for all $x$.

If $\lambda\neq0$ then we can argue as follows:
It is enough to look at one Jordan block $A=\lambda I+N$ where $N$ has zeros everywhere except the diagonal above the main diagonal where it has $1$'s.
The eigenspace of $A$ is of dimension $1$ since it is the kernel of $N$.
On the other hand, $A^n=(\lambda I+N)^n=\mu I+n\lambda^{n-1}N+...$, where $...$ are matrices that are zero starting from the diagonal above the main diagonal. The eigenspace of $A^n$ is the kernel of $n\lambda^{n-1}N+...$.
Since $\lambda\neq0$ this kernel is also one-dimensional.
