 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/19i/b/solution

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

ρ (A) X = A_{1} X A_{1}^{- 1}

, we have

ρ (A) ρ (B) X = A_{1} B_{1} X B_{1}^{- 1} A_{1}^{- 1} = (A B)_{1} X (A B)_{1}^{- 1} = ρ (A B) X,

and

ρ (I) = I

The underlying three-dimensional module is

W = C^{3} ≅ V_{1} \oplus V_{0} .

Using

M_{3} (C) = End (W) ≅ W \otimes W^{*}

V_{1} \otimes V_{1} ≅ V_{2} \oplus V_{0}

, we obtain

M_{3} (C) ≅ (V_{1} \oplus V_{0}) \otimes (V_{1} \oplus V_{0}) ≅ (V_{2} \oplus V_{0}) \oplus V_{1} \oplus V_{1} \oplus V_{0} .

Therefore the irreducible decomposition is

M_{3} (C) ≅ V_{2} \oplus 2 V_{1} \oplus 2 V_{0} .

 New to topics? Read the docs here!