 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-106/3/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

∥ T ∥_{2}^{2} = \sum_{n} \sum_{m} ∣ ⟨ T e_{n}, e_{m} ⟩ ∣^{2} = \sum_{m} \sum_{n} ∣ ⟨ e_{n}, T^{*} e_{m} ⟩ ∣^{2} = ∥ T^{*} ∥_{2}^{2} .

Thus $∥ T ∥_{2} = ∥ T^{*} ∥_{2}$ .

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