 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/12k/b/solution

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

M (x_{1}, \dots, x_{d}) = (x_{2}, \dots, x_{d}, \sum_{i} a_{i} x_{i}) .

For the stated matrix

H

c = (c_{0}, \dots, c_{k - 1})

lies in

C

exactly when

\sum_{j} c_{j} M^{j} x = 0

. If

σ c

is its cyclic shift, periodicity

M^{k} x = x

gives

H (σ c) = M (Hc) = 0.

Thus $ker H$ is shift-invariant and is a binary cyclic code.

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