Factor $T=AB$ as in part b(v). Then <Parseval identity for a Hilbertian basis> and <Cauchy-Schwarz inequality> give
$$
\sum_n|\langle Te_n,e_n\rangle|
=\sum_n|\langle Be_n,A^*e_n\rangle|
\leq\|B\|_2\|A^*\|_2
=\|T\|_1.
$$
The defining series for the <operator trace> is therefore absolutely convergent and satisfies $|\operatorname{tr}T|\leq\|T\|_1$.

For unit vectors $x,y$, the <adjoint operator> of $R=x\otimes y$ is $R^*=y\otimes x$, and
$$
R^*R=y\otimes y=P_{\mathbb Cy}.
$$
Hence $|R|=P_{\mathbb Cy}$ and part b(iii), followed by Parseval, gives
$$
\|R\|_1
=\sum_n\langle P_{\mathbb Cy}e_n,e_n\rangle
=\sum_n|\langle e_n,y\rangle|^2=1.
$$
Scaling proves $\|x\otimes y\|_1=\|x\|\|y\|$. A second use of Parseval yields
$$
\operatorname{tr}(x\otimes y)
=\sum_n\langle e_n,y\rangle\langle x,e_n\rangle
=\langle x,y\rangle,
$$
which proves the asserted <rank-one operator> formulas.

Solved by gpt-5.6-sol high.

Solution

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

{title2=$x\otimes y$}
{wiki}

For vectors $x,y$ in a Hilbert space, the rank-one operator $(x\otimes y)z=\langle z,y\rangle x$ has $\|x\otimes y\|_1=\|x\|\|y\|$ and $\operatorname{tr}(x\otimes y)=\langle x,y\rangle$.


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