Do eigenvectors of a squared matrix A tell us anything about the eigenvectors of A? 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!
 A: Hint


*

*Consider a nonzero matrix $A$ whose square is the zero matrix.

*Alternatively, if you are considering a real matrix $A$ and only considering real eigenvalues, consider a matrix $A$ such that $A^2$ has a negative eigenvalue.



Additional hints Consider, respectively, $$\textrm{(1)} \qquad \pmatrix{0&1\\0&0} \qquad \textrm{and} \qquad \textrm{(2)} \qquad \pmatrix{0&-1\\1&0} .$$

That said, at least for matrices $A$ over an algebraically closed field (like $\Bbb C$), the answer to the titular question is still yes, that knowing the eigenvectors of $A^2$ gives us at least some information about the eigenvectors of $A$:
Considering the Jordan block decomposition of $A$ shows that in this setting, for each eigenvalue $\lambda$ of $A^2$ there is at least one eigenvector $\bf v$ of $A^2$ that is an eigenvector of $A$. In particular, if all of the eigenvalues of $A^2$ are distinct, then all of the eigenvectors of $A^2$ are eigenvectors of $A$.
A: 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 $A$.
