Shared eigenvectors between $A$ and $A^k$ $\newcommand\la{\lambda}$
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.
 A: No. Consider the matrix
$$N := \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}:$$
Its only eigenvalue is $0$, and the corresponding eigenspace $L$ is spanned by the first basis vector,
$$\begin{pmatrix}1 \\ 0\end{pmatrix}.$$
On the other hand, $N^2$ is the zero matrix, and so every vector in $\mathbb{R}^2$ is an eigenvector of $N^2$ (of eigenvalue zero), and in particular every vector in $\mathbb{R}^2 - L$ is an eigenvector of $N^2$ but not $N$.
On the other hand, all of the eigenvectors of $N$ are generalized eigenvectors of $N^2$. Edit This property does not generally hold for counterexamples, however, as avid19's instructive example shows.
A: Let $$ A=\left( \begin{array}{ccc}
1 & 0\\
0 & -1 \end{array} \right)$$
Then $$ A^2=\left( \begin{array}{ccc}
1 & 0\\
0 & 1 \end{array} \right)$$
Then $$ A[1,1]^t=\left( \begin{array}{ccc}
1 & 0\\
0 & -1 \end{array} \right)[1,1]^t=[1,-1]$$
Which shows that $[1,1]^t$ is not an eigenvector for $A$. But it certainly is an eigenvector for $A^2=I$.
A: 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 $-1$).
