Let $A$ be a trace class operator on a separable Hilbert space $H$, I would like to show that $|A| = \sqrt{A^* A}$ is also a trace class operator. This seems to be a simple result, but I could not find a proof for it so I am trying to prove it myself.
Since $A$ is trace class, that means $$tr|A| < \infty \implies \sum_{i=1}^\infty \langle \varphi_i, |A| \varphi_i\rangle < \infty$$ where $\{\varphi_i\}$ is any orthonormal basis for $H$. Thus for $|A|$ to be trace class we would like to show that $$tr \big||A|\big| < \infty \implies \sum_{i=1}^\infty \langle \varphi_i, \big||A|\big| \varphi_i\rangle < \infty$$ Let $|A| = V\big||A|\big| $ be the polar decomposition of $|A|$. Then $$|A| = \sqrt{|A|^* U^* U |A|} = \sqrt{|A|^*|A|} = \sqrt{\sqrt{AA^*}\sqrt{A^*A}}.$$ But since $|A| \neq |A^*|$ in general I would need to show that $\sqrt{AA^*}\sqrt{A^*A}$ is positive, so I am not certain if this is correct.
Any hints?
Since $|A| \geq 0$ and $|A|^2 = |A|^*\,|A| = \big||A|\big|^2$, by uniqueness of the square root of a positive operator, $$ \big||A|\big| = |A|. $$ Hence, $A$ is trace class iff $|A|$ is trace class and $$ \mathrm{Tr}\,\big||A|\big| = \mathrm{Tr}\,|A|. $$