 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-4/1e/ii/solution

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

For

X = (a c b d),

ϕ_{A} (X) = A X - X A = (c 0 d - a - c) .

(E_{11}, E_{12}, E_{21}, E_{22})

, its matrix is therefore

[ϕ_{A}] = 0 - 1 00 0000 100 - 1 0100 .

ϕ_{A}^{3} = 0

, while

ϕ_{A}^{2} \neq = 0

. Moreover,

rank ϕ_{A} = 2, rank ϕ_{A}^{2} = 1.

These ranks determine one nilpotent block of size three and one of size one. Hence its Jordan normal form is

J_{3} (0) \oplus J_{1} (0) .

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