TITLE: Eigenvalue of $F^2$ and$ F$
QUESTION [6 upvotes]: Let $F$ be a linear operator on vector space $V$ over $\mathbb{C}$ and let $\lambda \in \mathbb{C}$
Show that if $\lambda^2$ is eigenvalue of linear operator $F^2$ then at least one of $\lambda,-\lambda$ is eigenvalue of $F$.
What I tried to do was to write simply:
$\exists v \ F^2(v)=\lambda^2v \Rightarrow \exists v \ F(F(v))=\lambda^2v \Rightarrow \exists v \ F(F(v))=\lambda\cdot\lambda\cdot v \ \vee F(F(v))=(-\lambda)\cdot(-\lambda)\cdot v$
But I am stuck here. Thought of applying $F^{-1}$ from both sides, but we don't know if such an operator exists and besides, it would lead me to nothing, I presume. 
I will appreciate any help.

REPLY [2 votes]: You don't need finite dimension, as you can restrict to a subspace of finite dimension.
It is given that $F^2$ has an eigenvector for $\lambda^2$, say $v\neq0$ is one. Now the subspace $V=\langle v,F(v)\rangle$ is $F$-stable, and we can restrict $F$ to $V$, which is clearly of dimension $1$ or $2$; call the restriction$~\tilde F$. One easily checks $\tilde F^2-\lambda^2 I=0$. Since we are over the complex numbers, $\tilde F$ has at least one eigenvalue, which in view of the given relation must be a root of $(X^2-\lambda^2)=(X+\lambda)(X-\lambda)$.
Actually you don't need an algebraically closed field either, as long as $\lambda$ is in your field, since $(\tilde F+\lambda I)\circ(\tilde F-\lambda I)=0$, so at least one of those two factors fails to be injective.