 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/b/solution

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

C

C^{- 1} = C^{†}

. Let

M = C^{†} A C

M^{†} = C^{†} A^{†} C = M

, so

M

M^{n}

More explicitly, if

A v_{j} = a_{j} v_{j}

with

a_{j}

real, then

C^{- 1} v_{j}

form an eigenbasis and

M^{n} (C^{- 1} v_{j}) = C^{- 1} A^{n} v_{j} = a_{j}^{n} (C^{- 1} v_{j}) .

Therefore

spec ((C^{- 1} A C)^{n}) = {a_{j}^{n} : j = 1, \dots, n_{A}} \subset R,

where

n_{A}

denotes the matrix size, to distinguish it from the power

n

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