TITLE: Shared eigenvectors between $A$ and $A^k$
QUESTION [11 upvotes]: $\newcommand\la{\lambda}$
Thanks to the spectral mapping theorem,
we know that if $\la_1,\ldots,\la_n$ are the eigenvalues of a $n\times n$ complex matrix $A$, then $\la_1^k,\ldots,\la_n^k$ are the eigenvalues of $A^k$.
Furthermore,
if $v_i$ is an eigenvector of $A$ associated to $\lambda_i$,
then $v_i$ is also an eigenvector of $A^k$ associated to $\la_i^k$.
Indeed,
$A^kv=A^{k-1}\la_i v_i=\cdots=\la_i^kv_i$.
So my question is the following: If $v_1,\ldots,v_n$ are the eigenvectors of $A$, are $v_1,\ldots,v_n$ also all the eigenvectors of $A^k$? In other words, is it possible that there is some vector $v$ that is an eigenvector of $A^k$ but not an eigenvector of $A$? It seems intuitive that this should not happen, but I am unable to find an argument to show it.

REPLY [2 votes]: Here's yet another flavor of example (over $\mathbb{R}$, anyway):
The matrix
$$J := \begin{pmatrix}0 & -1\\1 & 0\end{pmatrix}$$
has no real (nonzero) eigenvectors at all but $$J^2 = \begin{pmatrix}-1 & 0\\0 & -1\end{pmatrix}$$ is a scalar matrix, and in particular every vector in $\mathbb{R}^2$ is an eigenvector of $J^2$.
Remark If we regard $\Bbb C$ as a vector space over $\Bbb R$, then $J$ is the matrix (with respect to the standard basis) for multiplication by $i$, or equivalently, anticlockwise rotation by $\frac{\pi}{2}$; this latter interpretation immediately gives a geometric explanation for why $J$ has no real eigenvalues (and why $J^2$ is the scalar matrix with scalar $-1$).