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.
 A: 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.
A: Well, here is a last one, that is not using dimension and also gives you an eigenvector for $\lambda$ or $-\lambda$ :
let $v$ be an eigenvector for eigenvalue $\lambda^2$, let $u=F(v)$, you know that $F(u)=F^2(v)=\lambda^2v$, then try to find an eigenvector for $F$ looking like $w=\alpha u+\beta v$. Let's try $w = u + \lambda v$ then $$F(w)= F(u) + \lambda F(v)= \lambda^2 v + \lambda u = \lambda w$$, and the same holds for $w'=u - \lambda v$ that $F(w')=-\lambda w'$. 
Now assume that both $w$ and $w'$ are null, then $-\lambda v = \lambda v$, $v$ being an eigenvector, $v \neq 0$ and it implies that $\lambda=0$. But then since we assumed $w=0$ and $\lambda=0$ we have $u=0$ and we showed that $u=F(v)=0$ and $v$ is an eigenvector for eigenvalue $0$ and $F$.
Otherwise, either $w$ (or $w'$) is not null, and is an eigenvector for eigenvalue $\lambda$ (resp. $-\lambda$)
A: 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.
A: If
$F^2v = \lambda^2 v \tag{1}$
with $v \ne 0$, then
$(F^2 - \lambda^2)v = 0, \tag{2}$
so that
$(F + \lambda)(F - \lambda)v = 0; \tag{3}$
if now
$(F -\lambda)v = 0, \tag{4}$
then
$Fv = \lambda v, \tag{5}$
and we are done.  If
$(F - \lambda)v \ne 0, \tag{6}$
then (3) shows that
$F(F - \lambda)v = -\lambda(F - \lambda)v, \tag{7}$
showing, with the aid of (6), that $(F - \lambda)v$ is an eigenvector of $F$ with eigenvalue $-\lambda$, and now we are done. QED.
Note Added in Edit, Wednesday 7 May 2014 11:00 PM PST:  I'm always drawn to properties of infinite dimensional vector spaces $V$, and operators on them, which make no reference either to any topology on $V$ or continuity or boundedness of the operators.  In the present case we can illustrate by means of an example:  let $V = C^\infty(\Bbb R, \Bbb C)$ be the space of infinitely differentiable complex valued functions on the real line $\Bbb R$, and let $F = (d/dx):C^\infty(\Bbb R, \Bbb C) \to C^\infty(\Bbb R, \Bbb C)$ be the derivative operator.  Then for any $0 \ne k \in \Bbb R$, we can consider the eigenvalue $-k^2$ of $F^2 = (d^2/dx^2)$; we have $F^2(A\sin kt + B\cos kt) = -k^2(A\sin kt + B\cos kt)$ for $A, B \in \Bbb R$, and we see that $F - ik = (d/dx) - ik$ does not annihilate $(A\sin kt + B\cos kt)$; it follows that $(d/dx - ik)(A\sin kt + B\cos kt)$ is an eigenfunction of $d/dx$ with eigenvalue $-ik$; of course, this is a close-to-trivial example, but I think it points in an interesting direction.  And neither a topology on $C^\infty(\Bbb R, \Bbb C)$ nor continuity of $F$ enter in to this "little" result.  End of Note.
Final Note, Added Saturday 10 May 2014 1:28 PM PST:  This is the answer which put me over 10k!  Yessss!!!  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
