 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/5/i/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

The conjugate transpose reverses products and conjugates scalars. Since

A, B

H^{†} = - i (A B - B A)^{†} = - i (B A - A B) = i (A B - B A) = H .

Therefore

H is Hermitian .

In particular the commutator of two Hermitian matrices is anti-Hermitian, and multiplication by

i

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