 Solution

The displayed matrices are the three Pauli matrices, in the order ${\sigma }_3,{\sigma }_1,{\sigma }_2$. They satisfy $\mathrm{tr}{A}_i=0$ and $\mathrm{tr}(A_i{A}_j)=2{\delta }_{ij}$. Every Hermitian matrix has a unique expression $aI+{\sum }_i{v}_i{A}_i$ with all four coefficients real. Therefore they form a basis together with $I$.

The preceding trace identities give $b(I,I)=-1$, $b(I,A_i)=0$, and $b(A_i,A_j)={\delta }_{ij}$. The Gram matrix is thusConsequently rank $=4$ and inertia signature $(3,1)$. If signature is defined as the number of positive squares minus the number of negative squares, its value is $2$. This is the Hermitian matrix representation of Minkowski four-vectors, with the negative-determinant sign convention.