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!!!
Eigenvalues are always in the field over which we are working. But since real numbers are also complex numbers, you can see your linear transformation as a linear transformation of vector spaces over $\mathbb{C}$, hence it makes sense to talk about complex Eigenvalues.
Best Answer
In addition to other good points made in comments and answer: yes, there is some ambiguity in the idea of "eigenvalue" for linear maps/operators on vector spaces over fields (such as $\mathbb R$) that are not algebraically closed. And we should not get distracted by the idea that $\mathbb C^n$ can be a vector space over $\mathbb R$... that is mostly a distraction, without genuine mathematical content.
For example, consider operator $T=\pmatrix{0&-1\cr 1&0}$ on $\mathbb R^2$. If we look at the characteristic equation $\lambda^2+1=0$, we see that there are no real eigenvalues. But $\pm i$ are eigenvalues... if we extend the vector space to its complexification...
(Choosing a basis also doesn't really have much to do with this phenomenon...)