$A^2$ self-adjoint and Compact, prove $A$ has an eigenvalue

Question

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 $H$.)

Answer 1

(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 $(A+\lambda I)v$.

Answer 2

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 $A^2$.

Answer 3

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 $v \neq w$).