T compact if and only if $T^*T$ is compact. I have an operator $T \in B(\mathcal{H})$.
I need to prove that T is comapct if and only if $T^*T$ is compact.
One way is ok, because if $A$ or $B$ is compact then $AB$ is compact, so I get at once that if $T$ is compact then $T^*T$ is compact.
But how do I go the other way? If I assume that $T^*T$ is compact I am not quite sure how to see that $T$ is compact. If I assume for contradiction that $T$ is not compact I must also have that $T^*$ is not compact. If I knew that either $T$ or $T^*$ was invertible it would be ok, because then I could find a bounded subsequence that did not converge. But when I do not have invertibility I am not quite sure how to proceed.

Assume that $S := T^\ast T$ is compact. By the spectral theorem for compact self-serving operators, this implies that $|T|=\sqrt{S}$ is also compact.
But the polar decomposition theorem yields $T=V \cdot |T|$ for some partial isometry $V$. Since the compact operators form an ideal in the space of bounded operators, we are done.
