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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.

For a Hermitian form or a Hermitian form over a quadratic finite field, the displayed subspace is the orthogonal complement of $W$. If the form restricts nondegenerately to a finite-dimensional $W$, its invertible Gram matrix supplies, for each $v$, a unique $w_0\in W$ with $B(v-w_0,W)=0$. Consequently $V=W\oplus W^{\perp }$. When the form on $V$ is nondegenerate, its restriction to $W^{\perp }$ is too: a radical vector there is orthogonal to both summands and hence is zero.

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.

Take $B$ to be linear in its first argument and conjugate-linear in its second, with $\bar{a}=a^q$. For a Hermitian form over a quadratic finite field, $B(v,v)\in {\mathbb{F}}_q$. We first establish the quadratic finite-field norm and trace surjectivity needed for orthonormalization of a finite-field Hermitian form. The field norm $N(a)=a^{q+1}$ maps ${\mathbb{F}}_{q^2}^{\ast }$ onto ${\mathbb{F}}_q^{\ast }$ because the multiplicative group of a finite field is cyclic of order $q^2-1$. The field trace $\mathrm{Tr}(a)=a+a^q$ is an ${\mathbb{F}}_q$-linear map into ${\mathbb{F}}_q$. It is nonzero: a nonzero polynomial of degree $q$ cannot vanish at all $q^2$ elements. It is therefore onto, including in characteristic two.

There must be $v$ with $B(v,v)\ne 0$. Otherwise, for any $v,w,a$, expanding the diagonal value givesIf $B(v,w)\ne 0$, varying $a$ contradicts the surjectivity of the field trace. Thus all pairings would vanish, contrary to nondegeneracy. Choose $a$ with $N(a)=B(v,v{)}^{-1}$; then $v_1=av$ has norm one. Every $w$ decomposes uniquely asThe restricted Hermitian form over a quadratic finite field on the orthogonal complement for a sesquilinear form is nondegenerate: a vector there orthogonal to that complement is also orthogonal to $v_1$, hence to all of $V$, and is zero. Induction produces an orthonormal basis $v_1,\dots ,v_n$.

The unitary group over a finite field is the group of invertible ${\mathbb{F}}_{q^2}$-linear maps preserving $B$. In the orthonormal basis just constructed, its concise description isThe notation here uses the size of the matrix field; it is also frequently written $U_n(q)$. There is no determinant-one condition in this definition.

The group of invertible linear maps preserving a nondegenerate Hermitian form over a quadratic finite field is, in an orthonormal basis, the displayed matrix group. Here $q^2$ denotes the size of the field on which the matrices act; some authors denote this same group by $U_n(q)$. Imposing determinant one gives the special unitary subgroup, and subsequently quotienting its scalar center gives the projective special unitary group over a finite field.