Compact and self-adjoint operator It is true that if $T:H \to H$ is a compact operator ($H$ Hilbert space) then $T^\ast T$ is algo compact and indeed self-adjoint.
Conversely, is it true that every compact and self-adjoint operator $S$ can be decomposed like $S=A^\ast A$ with $A$ compact?
Thanks!

You also need $S$ to be positive semidefinite, i.e. $\langle x, S x \rangle \ge 0$ for all $x \in H$.  Then you can take $A = \sqrt{S}$ using the continuous functional calculus.  Note that any continuous function $f$ on $\sigma(S)$ with $f(0)=0$ is the uniform limit on $\sigma(S)$ of a sequence of polynomials $p_n$ with $p_n(0)=0$, and so $f(S)$ is the norm limit of $p_n(S)$.  If $S$ is compact, then so are $p_n(S)$ and therefore $f(S)$. 
