[Math] prove that determinant is a quadratic form

Question

let $V$ be a vector space of all $2 \times 2$ hermitian matrices with entries from $\mathbb C$, over the field $\mathbb R$.
prove that $q(v)=\det(v)$ is a quadratic form.

I tried to prove that $f(u,v)=$$1\over4$$(q(u+v)-q(u-v))$ is a bilinear form but i'm stuck with that… can anyone point me in the right direction?

Answer 1

$$\det\pmatrix{x&z+wi\\ z-wi&y}=xy - z^2 - w^2= \pmatrix{x&y&z&w}\pmatrix{0&\tfrac12&0&0\\ \tfrac12&0&0&0\\ 0&0&-1&0\\ 0&0&0&-1}\pmatrix{x\\ y\\ z\\ w}.$$