Prove there is a compact self adjoint $S:H\to H$ such that $S^3=T$.

Question

Let $T:H\to H$ be compact and self adjoint. Prove there is a compact self adjoint $S:H\to H$ such that $S^3=T$.

Is the $S^3$ means power of 3 or applying the operator 3 times? What is there to prove here? Why wouldn't there be souch $S$?

Where do I start here? I have no idea.

Answer 1

If $T=0$, then $S=0$ works. If $T$ has a finite-dimensional range and is selfadjoint, then there is an orthonormal set $\{ e_n \}_{n=1}^{N}$ and non-zero real numbers $\{ \lambda_n \}_{n=1}^{N}$ such that $$ Tf = \sum_{n=1}^{N}\lambda_n (f,e_n)e_n. $$ Then $$ Sf = \sum_{n=1}^{N}\frac{\lambda_n}{|\lambda_n|}|\lambda_n|^{1/3}(f,e_n)e_n $$ is compact, selfadjoint, and satisfies $S^3=T$. A non-finite rank $T$ is compact and selfadjoint iff there exists an orthonormal sequence $\{ e_n \}_{n=1}^{\infty}$ and non-zero real $\{ \lambda_n \}_{n=1}^{\infty}$ such that $\lim_n \lambda_n = 0$ and $$ Tf = \sum_{n=1}^{\infty}\lambda_n(f,e_n)e_n . $$ The same $S$ given above works for $N=\infty$, and is compact because $|\lambda_n|^{1/3}\rightarrow 0$ as $n\rightarrow\infty$.