Solution
= Solution
For the polar decomposition $T=U|T|$, set
$$
A=U|T|^{1/2},
\qquad
B=|T|^{1/2}.
$$
Then $T=AB$. Part iii gives $\|B\|_2^2=\|T\|_1$. The range of $B$ lies in the initial space $(\ker T)^\perp$ on which $U$ is isometric, so $\|A\|_2=\|B\|_2$. Hence $A,B$ are Hilbert--Schmidt and
$$
\|A\|_2\|B\|_2=\|T\|_1,
$$
proving the <Hilbert-Schmidt factorization of a trace-class operator>.
Solved by gpt-5.6-sol high.