 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-1/7b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

x \neq = 0

x^{†} G x = ∣ A x ∣^{2} > 0,

so the Hermitian matrix

G

has positive real eigenvalues. If

G e_{i} = λ_{i} e_{i}

, then

H (A e_{i}) = A A^{†} A e_{i} = A (G e_{i}) = λ_{i} A e_{i} .

Moreover

∣ A e_{i} ∣^{2} = λ_{i} ∣ e_{i} ∣^{2}

, so

∣ f_{i} ∣/∣ e_{i} ∣ = λ_{i}

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