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

A Hermitian two-by-two matrix has the formAddition and real scalar multiplication preserve this condition, and the two diagonal entries plus the real and imaginary parts of $z$ give four independent real coordinates. Thus $S$ is a real four-dimensional vector space. The trace product is real because $\overline{\mathrm{tr}(AB)}=\mathrm{tr}(BA)=\mathrm{tr}(AB)$; this also proves symmetry of $b$.

Directly, $\mathrm{tr}(A^2)={\alpha }^2+{\delta }^2+2|z{|}^2$ and $(\mathrm{tr}A{)}^2={\alpha }^2+2\alpha \delta +{\delta }^2$. HenceSince $\mathrm{tr}I=2$ and $\mathrm{tr}(AI)=\mathrm{tr}A$,