 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/34c/a/i/solution

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

Using

B^{2} = 0

and

B B^{†} = 1 - B^{†} B

H^{2} = B^{†} B B^{†} B = B^{†} (1 - B^{†} B) B = H .

Thus $H$ is a projection and its eigenvalues are $0$ and $1$ .

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