# 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.

Let $$\{x_n \}$$ is a bounded sequence with bound $$M$$. If $$T^{\star}T$$ is compact, then there exists $$\{ x_{n_{k}}\}$$ such that $$\{ T^{\star}Tx_{n_{k}}\}$$ converges. Then \begin{align} \|Tx_{n_k}-Tx_{n_j}\|^2 & =(T^{\star}Tx_{n_{k}}-T^{\star}Tx_{n_j},x_{n_k}-x_{n_j}) \\ & \le 2M\|T^{\star}Tx_{n_k}-T^{\star}Tx_{n_j}\|. \end{align} This forces $$\{Tx_{n_{k}}\}$$ to be a Cauchy sequence and, hence, to converge.
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.
If $S = T^{\ast}T$ is compact, then so is $p(S)$ for any polynomial p such that $p (0) = 0$. By taking a norm limit of such expressions you can conclude that $\sqrt {S} = |T|$ is compact. Now write $T = V|T|$ by the polar decomposition. Hence $T$ is compact.