TITLE: Do eigenvectors of a squared matrix A tell us anything about the eigenvectors of A?
QUESTION [5 upvotes]: I have a following task:

Let $X$ be an eigenvector of $A$.
a) Prove that $X$ is also an eigenvector of $A^2$
b) Is the convere theorem true?

I had no problem with the first part. Ive tried disproving the second part this way:
Let $e$ be the eigenvalue of $A^2$ corresponding to the eigenvector $X$. By definition we know that:
$A^2X=eX \to (A^2-eI)X=0 \to (A-eI)(A+eI)X=0$
If im not mistaken that proves that $X$ is an eigenvector either for $A$ or $-A$ (since $X$ is a nonzero vector). How can I evaluate which of those two matrices share the common eigenvector with $A^2$?
Thanks in advance!

REPLY [3 votes]: Consider $$A = \pmatrix{0 & 1 \\ 1 & 0}$$
We have $A^2 = I$ so $$A^2\pmatrix{1 \\ 0} = \pmatrix{1 & 0 \\ 0& 1}\pmatrix{1 \\ 0} = \pmatrix{1 \\ 0}$$
and therefore $\pmatrix{1 \\ 0}$ is an eigenvector for $A^2$. However,
$$A\pmatrix{1 \\ 0} = \pmatrix{0 & 1 \\ 1 & 0}\pmatrix{1 \\ 0} = \pmatrix{0 \\ 1}$$
is not a scalar multiple of $\pmatrix{1 \\ 0}$ so $\pmatrix{1 \\ 0}$ is not an eigenvector for $A$.