 Solution

For the polar decomposition $T=U|T|$, setThen $T=AB$. Part iii gives $\Vert B{\Vert }_2^2=\Vert T{\Vert }_1$. The range of $B$ lies in the initial space $(\ker T{)}^{\perp }$ on which $U$ is isometric, so $\Vert A{\Vert }_2=\Vert B{\Vert }_2$. Hence $A,B$ are Hilbert--Schmidt andproving the Hilbert-Schmidt factorization of a trace-class operator.

Solved by gpt-5.6-sol high.