 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-2/17g/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Taking traces in the defining equation gives

tr C + \overline{tr C} = 2 tr C

, so the trace is real. Put

a = tr C /2

. Then

K = C - a I

is traceless and satisfies

K^{*} = - K

, so

- i K

is a traceless Hermitian matrix. Hence every element has a unique representation

C = a I + i \sum_{j} v_{j} A_{j}, a, v_{j} \in R .

Conversely each such matrix satisfies the defining equation. This proves $Q$ is a real four-dimensional vector space, with basis

I, i A_{1}, i A_{2}, i A_{3}

Substitution into the given map gives

Φ (a I + i j \sum v_{j} A_{j}) = a I - j \sum v_{j} A_{j} .

Its values are Hermitian, and it acts on real coordinates by

(a, v_{1}, v_{2}, v_{3}) \mapsto (a, - v_{1}, - v_{2}, - v_{3})

. This invertible real linear coordinate transformation proves the required isomorphism. The inverse sends

a I + \sum_{j} u_{j} A_{j}

a I - i \sum_{j} u_{j} A_{j}

 New to topics? Read the docs here!