 Orthonormalization of a finite-field Hermitian form

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.