3
$\begingroup$

Suppose $H$ is a Hilbert space and $A \in L(H)$ is such that $A^2$ is compact and self-adjoint. Prove that $A$ has an eigenvalue.

(Here $L(H)$ is the set of bounded linear operators on a Hilbert space $H$.)

$\endgroup$
4
$\begingroup$

(This proves a bit more than asked: there is no need to $A^2$ to be compact nor selfadjoint, only that it has an eigenvalue)

Let $\alpha$ be an eigenvalue of $A^2$. Choose $\lambda\in\mathbb C$ with $\lambda^2=\alpha$.

As $\alpha$ is an eigenvalue for $A^2$, we have that there exists nonzero $v$ with $(A^2-\alpha I)v=0$. Then $$ 0=(A^2-\alpha I)v=(A-\lambda I)(A+\lambda I)v=0. $$ If $(A+\lambda I)v=0$, then $-\lambda$ is an eigenvalue of $A$. If $(A+\lambda I)v\ne0$, then $\lambda$ is an eigenvalue of $A$, with eigenvector $(A+\lambda I)v$.

$\endgroup$
3
$\begingroup$

For this to hold it is crucial that $H$ is a complex Hilbert space since on $\mathbb{R}^2$ a rotation $T$ by $\pi/2$ has no eigenvalues, yet $T^2 = -I$ is compact and self-adjoint.

The spectral theorem for compact self-adjoint operators yields an eigenvector $w$ of $A^2$ with real eigenvalue $\lambda$. Without loss of generality we can rescale $A$ (possibly by a complex factor) such that $\lambda = 1$. Now either $w$ is an eigenvector of $A$ and we are done, or it is not. In the latter case, we must have $Aw = v$ and $Av = w$ for some $v \neq w$. But then $$A(v-w) = Av - Aw = w - v = -(v - w)$$ so that $v-w$ is an eigenvector of $A$. (Note $v-w \neq 0$ since $v \neq w$).

$\endgroup$
  • 1
    $\begingroup$ At the start of the second paragraph, you mean "yields an eigenvector $w$ of $A^2$", right? $\endgroup$ – user21467 Dec 17 '13 at 22:46
  • $\begingroup$ @StevenTaschuk Of course. Edited. $\endgroup$ – user38355 Dec 17 '13 at 22:49
1
$\begingroup$

This is the first thought that I had. I do not see how it leads to what you want but maybe you have an idea with it: Since $A^2$ is a self-adjoint compact operator either $||A||^2$ or $-||A||^2$ is an eigenvalue for $A^2$.

$\endgroup$
  • $\begingroup$ I think too this is the direction, tried to develop it but without success.. By the way, it $||A^2||$ or $-||A^2||$, isn't it ? $\endgroup$ – daPollak Dec 17 '13 at 20:39
  • $\begingroup$ I think eigen space for $A^2$ corresponding to non-zero eigenvalue is finite dimensional and A invariant, hence from the finite dimensional case ..., that's good unless $A^2 = 0$ in which case there is easily a 0 eigenvalue. $\endgroup$ – mike Dec 17 '13 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.