Hermitian form over a quadratic finite field ($B(v,w)=\overline{B(w,v)},\bar{a}=a^q$)

Over the finite field $K={\mathbb{F}}_{q^2}$, a sesquilinear form relative to $a\mapsto a^q$ is linear in its first argument and conjugate-linear in its second. It is Hermitian when $B(w,v)=\overline{B(v,w)}$. Its diagonal values belong to ${\mathbb{F}}_q$; it is nondegenerate when $B(v,w)=0$ for every $w$ forces $v=0$. This definition applies in characteristic two as well.

In a two-dimensional nondegenerate Hermitian form over a quadratic finite field, choose an orthonormal basis $v,w$ and $a$ with $a\bar{a}=-1$. Then $e=v+aw$ is an isotropic vector. Choose $g$ with $B(e,g)=1$, and choose $c$ with $c+\bar{c}=B(g,g)$. The vector $f=g-ce$ satisfies $B(e,f)=1$ and $B(f,f)=0$. Thus $e,f$ give the displayed matrix, including in characteristic two. Iterating on nondegenerate orthogonal complement for a sesquilinear form leaves only a unit-norm line in odd dimension.

A nondegenerate Hermitian form over a quadratic finite field has an orthonormal basis. If all diagonal values vanished, evaluating $B(v+aw,v+aw)$ would give $\mathrm{Tr}(\bar{a}B(v,w))=0$ for every $a$, contradicting quadratic finite-field norm and trace surjectivity and nondegeneracy. Thus some $v$ has nonzero norm; rescaling using the field norm makes it one. Its orthogonal complement for a sesquilinear form is nondegenerate, so induction completes the basis.