Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 17G (Quadratic Mathematics)

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 $\mathrm{tr}(AI)=\mathrm{tr}A$,

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.

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 $aI-i{\sum }_j{u}_j{A}_j$.

Let $r=\mathrm{tr}C$, $s=\mathrm{tr}D$, and $\beta =(1-i)/2$. Their traces are real by part c, and $\Phi (C)=\beta rI+iC$. Thus $\mathrm{tr}\Phi (C)=(2\beta +i)r=r$. Expanding the product and using $2{\beta }^2+2i\beta =1$ givesThereforeIn particular the trace expression is real and symmetric on $Q$. Under the coordinates $C=aI+i\mathbf{v}\cdot \mathbf{A}$, $D=dI+i\mathbf{w}\cdot \mathbf{A}$, it is $-ad+\mathbf{v}\cdot \mathbf{w}$. Thus the trace form on scalar-plus-skew-Hermitian two-by-two matrices has rank $4$, inertia $(3,1)$, or numerical signature $2$, agreeing with $b$ under the real isomorphism.