# 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.

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can the dimension be infinite ? – Yann Hamdaoui May 6 '14 at 16:30
Yes, nothing about dimension in the problem. – Mateusz May 6 '14 at 16:33
By "$\lambda \vee -\lambda$", do you mean "$\lambda$ or $-\lambda$"? $\vee$ is a boolean operator; it doesn't do all the things the English conjunction does. – user2357112 May 6 '14 at 23:50
If you want the infinite dimensional case handled, why do you accept an answer that does not? Please unaccept that anwser, at least temporarily to give the answerer the option to delete the question. – Marc van Leeuwen May 7 '14 at 8:35

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!!!

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+1 Nice and simple, and no mention of dimension. – Prahlad Vaidyanathan May 6 '14 at 16:44
@Prahlad Vaidyanathan: Indeed, no mention of dimension! This little rhyme expresses what I too like about this quick and easy argument. Thanks for the +1! Cheers! – Robert Lewis May 6 '14 at 16:48
The left hand side of $(1)$ is missing $v$. – Marc van Leeuwen May 7 '14 at 8:56
@Marc van Leeuwen: right you are! Thanks for noticing! Will repair right now! – Robert Lewis May 7 '14 at 8:57

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. $-\lambda$)

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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.

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It's basically the same, but you can even simply write the matrix in upper triangular form in some basis (being on $\mathbb{C}$), seeing that the diagonal element of $F^2$ in this basis, which are its eigenvalues, are indeed the diagonal elements of $F$ squared – Yann Hamdaoui May 6 '14 at 16:36
That would do, becuase I didn't have Jordan Decomposition Theorem presented on my lectures (yet, I presume), so I would have to prove it in the class if I wanted to use it. Thanks for help! – Mateusz May 6 '14 at 16:39
+1, that's the way. – Andreas Caranti May 6 '14 at 16:41

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.

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