Is every eigenvector of AA an eigenvector of A?

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.

Hint: Consider $$A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ Or $$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}.$$
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 .
• I love this answer. It's very intuitive, since $A\cdot v$ is orthogonal to $v$, yet $A^2 \cdot v$ is anti parallel. – infinitezero Sep 15 at 20:56
No, it is false: consider in a real vector space $$A=\begin{bmatrix} 0&1\\1&0\end{bmatrix}\implies A\circ A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$ Then the vector $$\begin{bmatrix} 1\\0\end{bmatrix}$$ is an eigenvector of $$A^2$$ and not of $$A$$.