Shared eigenvectors between $A$ and $A^k$

Solution 1:

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.

Solution 2:

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 $A^2=I$.

Solution 3:

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 $-1$).