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

Solution

Factor

T = A B

\sum_{n} ∣ ⟨ T e_{n}, e_{n} ⟩ ∣ = \sum_{n} ∣ ⟨ B e_{n}, A^{*} e_{n} ⟩ ∣ \leq ∥ B ∥_{2} ∥ A^{*} ∥_{2} = ∥ T ∥_{1} .

The defining series for the operator trace is therefore absolutely convergent and satisfies

∣ tr T ∣ \leq ∥ T ∥_{1}

x, y

R = x \otimes y

R^{*} = y \otimes x

, and

R^{*} R = y \otimes y = P_{C y} .

Hence

∣ R ∣ = P_{C y}

and part b(iii), followed by Parseval, gives

∥ R ∥_{1} = \sum_{n} ⟨ P_{C y} e_{n}, e_{n} ⟩ = \sum_{n} ∣ ⟨ e_{n}, y ⟩ ∣^{2} = 1.

Scaling proves

∥ x \otimes y ∥_{1} = ∥ x ∥∥ y ∥

. A second use of Parseval yields

tr (x \otimes y) = \sum_{n} ⟨ e_{n}, y ⟩ ⟨ x, e_{n} ⟩ = ⟨ x, y ⟩,

which proves the asserted rank-one operator formulas.

Solved by gpt-5.6-sol high.

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