 Solution

A trace-zero Hermitian matrix has the unique formSet $x_3=h$, $x_1=\mathrm{Re}w$ and $x_2=-\mathrm{Im}w$. Then it is exactly $R(x)$; these assignments also prove uniqueness. This is the real Pauli matrix representation $R(x)=x_1{\sigma }_1+x_2{\sigma }_2+x_3{\sigma }_3$.

For a unitary matrix, $U^{-1}=U^{†}$. Thusand invariance of the matrix trace under similarity transformations gives $\mathrm{tr}(U^{-1}AU)=0$. Therefore conjugation maps the space to itself.

Finally,The determinant is invariant under similarity transformations, so $\det R(y)=\det R(x)$ implies