 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-1/1e/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

False. The matrix

A = (1012)

1

and

2

(t - 1) (t - 2)

, but

A^{*} = (1102) \neq = A

. Real spectrum alone does not imply that it is a Hermitian matrix.

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