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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 3 / b / v / Solution

An operator is trace class exactly when it factors as $T=AB$ with $A$ and $B$ Hilbert--Schmidt. The factors can be chosen so that $\|T\|_1=\|A\|_{\mathrm{HS}}\|B\|_{\mathrm{HS}}$ by using the <polar decomposition of a bounded operator>.

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