= 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 $\operatorname{Tr}(\bar a B(v,w))=0$ 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>.
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