 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/7b/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Over

C

, triangularize

B

β_{1}, \dots, β_{n}

. Then

χ_{B^{m}} (z^{m}) = \prod_{j = 1}^{n} (β_{j}^{m} - z^{m}) .

ω_{l} = e^{2 πi l / m}

are all the root of unity solutions of

w^{m} = 1

, so

\prod_{l = 1}^{m} (t - ω_{l} z) = t^{m} - z^{m} .

Consequently

l = 1 \prod m χ_{B} (ω_{l} z) = j = 1 \prod n l = 1 \prod m (β_{j} - ω_{l} z) = j = 1 \prod n (β_{j}^{m} - z^{m}) = χ_{B^{m}} (z^{m}) .

λ

is an eigenvalue of

B^{m}

, choose

z

with

z^{m} = λ

. The displayed identity makes at least one factor

χ_{B} (ω_{l} z)

zero. Thus

μ = ω_{l} z

is an eigenvalue of

B

and

μ^{m} = λ .

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