Let \(V\) be the real vector space of all \(2 \times 2\) (complex) Hermitian matrices, that is, \(2 \times 2\) complex matrices \(A\) which satisfy \(A_{ij} = \overline{A_{ji}}\).
Show that the equation \(q(A) = \det A\) defines a quadratic form \(q\) on \(V\).
Let \(W\) be the subspace of \(V\) of matrices of trace 0. Show that the bilinear form \(f\) determined by \(q\) is negative definite on the subspace \(W\).
Solution
Every Hermitian matrix in \(V\) has a unique expression \[ A= \begin{pmatrix} a&z\\ \overline z&d \end{pmatrix}, \qquad a,d\in\mathbb R,\quad z=x+iy\in\mathbb C. \]
1. The determinant is a quadratic form on the real vector space \(V\).
In the real coordinates \((a,d,x,y)\), \[ q(A)=\det A=ad-|z|^2=ad-x^2-y^2. \] Thus \(q\) is a homogeneous polynomial of degree two in the real coordinates of \(V\). Equivalently, its polarization \[ f(A,B)=\frac12\bigl(q(A+B)-q(A)-q(B)\bigr) \] is bilinear over \(\mathbb R\) and satisfies \(q(A)=f(A,A)\). Hence \(q\) is a quadratic form on \(V\).
More explicitly, if \[ B= \begin{pmatrix} b&w\\ \overline w&e \end{pmatrix}, \] then direct expansion gives \[ \boxed{ f(A,B)=\frac12(ae+db)-\operatorname{Re}(z\overline w).} \] This formula is visibly real-bilinear and symmetric.
2. On the trace-zero subspace, the quadratic form is strictly negative away from zero.
A matrix \(A\in W\) has trace zero, so \(d=-a\). Therefore \[ q(A)=a(-a)-|z|^2=-a^2-|z|^2. \] If \(A\ne0\), then either \(a\ne0\) or \(z\ne0\), and hence \[ q(A)=-a^2-|z|^2<0. \] Since \(q(A)=f(A,A)\), it follows that \[ \boxed{f(A,A)<0\qquad\text{for every }0\ne A\in W.} \] Thus the restriction of \(f\) to \(W\) is negative definite.