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$H$ is a Hilbert space and $A$ is a bounded operator on $H$. If $A^*A$ is compact, is it necessarily that $A$ is compact?

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Yes. If $A^\ast A$ is compact, then its positive square root $|A|=(A^\ast A)^{1/2}$ is also compact. Then one can consider the polar decomposition of $A$, \begin{equation} A = U|A|, \end{equation} where $U$ is a partial isometry. Since the compact operators form an ideal in $B(H)$, it follows that $A$ is compact as well.

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