# Product of compact, bounded and self adjoint operator.

$T \in B(H)$, and $T = S^2$ for some self adjoint operator $S \in B(H)$. I need to prove that T is compact if and only if S is compact.

If S is compact, it is easy to show that T is compact since S is also bounded and the product of compact operator and bounded linear operator is compact. But how can I prove the converse of it?

Also, is self-adjoint necessary here, if we drop the self-ajoint property, does the statement still hold?

• See the following link!!! this helps for the converse part (math.stackexchange.com/questions/148162/…) – Urban PENDU Apr 30 '15 at 1:24
• A counterexample in the non self-adjoint case: define a map on $\ell^2$ by $$e_k \mapsto \begin{cases} e_{k+1} & k \text{ is even}\\ \frac 1{2^k}e_k & k \text{ is odd} \end{cases}$$ – Omnomnomnom Apr 30 '15 at 1:31
• @UrbanPENDU Thank you for the help. But why can we represent a compact operator T as $T = \sum_{n \ge 1} \lambda_n \langle \cdot, x_n\rangle x_n$ for some $\lambda_n \to 0$, is there any theorem stating this fact? – TH000 Apr 30 '15 at 1:41

I think there is rather a simple way to solve this problem. Let us assume that $T$ is compact and let $T=S^2$ where $S$ is a self-adjoint operator. We need to show that $S$ is compact. Observe that (Take this as an exercise) an operator $A$ is compact if and only if $AA^*$ and $A^*A$ both are compact.
Now since $T$ is compact it follows that $S^2 = SS^* = S^*S$ (because $S$ is self-adjoint) are both compact. So, by the above observation/exercise it follows that $S$ is compact.
$\textbf{Note}:$ As a hint for the above exercise I suggest you to use the sequential criterion for compactness operator of an operator. Though the exercise is not very hard but if you are not able to solve the exercise then just leave a comment.
If $S^2$ is compact, then $|S|=(S^2)^{1/2}$ is compact. By looking at the polar decomposition $S=U|S|$, we deduce that $S$ is compact.
For non-normal, the statement is not true. Consider the shift-like operator $S$ that, for a fixed orthonormal basis $\{e_n\}$, sends $e_{2n}\longmapsto e_{2n+1}$, $e_{2n+1}\longmapsto0$. The $S$ is not compact, but $S^2=0$.