 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-1/6c/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

B^{k} = S^{- 1} A^{k} S

, so the preceding argument applied to

A^{k}, B^{k}

gives equal ranks.

 New to topics? Read the docs here!