Eigenvalue of $F^2$ and$ F$ 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.

Maybe this is overkill, but assuming the space is finite dimensional, you can use the Jordan Decomposition Theorem. Choose a basis and write the corresponding matrix of $F$ (also denoted by $F$) as
$$
F = D+N
$$
where $D$ is diagonal, $N$ is nilpotent and $DN = ND$. Then
$$
F^2 = D^2 + (2DN+N^2)
$$
and $D^2$ is diagonalizable, and $(2DN+N^2)$ is nilpotent (Note that $DN$ is nilpotent since $DN=ND$).
Hence, the eigen values of $F^2$ are precisely the diagonal entries of $D^2$, which proves what you want.
