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.
Best Answer
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!!!