TITLE: Is every eigenvector of AA an eigenvector of A?
QUESTION [5 upvotes]: Let $V$ be a (finite-dimensional) vector space and $A \colon V \to V$ a linear map. 
Is it true that, if $v$ is an eigenvector of $A\circ A$, then $v$ is an eigenvector of $A$?
I know the converse statement is true.

REPLY [3 votes]: A rotation by 90° as represented by a matrix e.g.
$$A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$
has no real eigenvalue ($\chi_A(\lambda)=\lambda^2+1$ is the characteristic polynomial) and so no eigenvectors. The rotation by 180°
$$A^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$
has the whole space as a space of eigenvectors and zero vector
.