Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 17G / c / Solution

Taking traces in the defining equation gives $\mathrm{tr}C+\overline{\mathrm{tr}C}=2\mathrm{tr}C$, so the trace is real. Put $a=\mathrm{tr}C/2$. Then $K=C-aI$ is traceless and satisfies $K^{\ast }=-K$, so $-iK$ is a traceless Hermitian matrix. Hence every element has a unique representationConversely 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 givesIts 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 $aI+{\sum }_j{u}_j{A}_j$ to $aI-i{\sum }_j{u}_j{A}_j$.