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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.

We construct a Hermitian hyperbolic plane over a finite field inside each successive two-dimensional piece. By the field norm surjectivity proved above, choose $a\in {\mathbb{F}}_{q^2}$ with $a\bar{a}=-1$. From two vectors in an orthonormal basis, form $e=v_1+a{v}_2$. Then $B(e,e)=1+a\bar{a}=0$, so $e$ is a nonzero isotropic vector. Nondegeneracy supplies $g$ with $B(e,g)=1$; rescaling the second argument achieves this normalization. In fact $g=v_1$ already works in this two-dimensional piece. By quadratic finite-field norm and trace surjectivity, choose $c$ with $c+\bar{c}=B(g,g)$ and put $f=g-ce$. Direct expansion givesThus $e,f$ are linearly independent, and their Gram matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, which is invertible even in characteristic two. Their span is nondegenerate, so its orthogonal complement for a sesquilinear form is again nondegenerate. Repeat the construction there. In even dimension this exhausts $V$; in odd dimension the final one-dimensional complement has a nonzero diagonal value, which can be normalized to one using the field norm. Calling its vector $d$, we obtainwith the $d$ terms present only in odd dimension. This proves all the requested pairings and the basis assertion.

The printed hint needs a qualification. If $X^2-X-1$ is irreducible over ${\mathbb{F}}_q$, its two roots are conjugate, so $\zeta +\bar{\zeta }=1$ and $\zeta \bar{\zeta }=-1$; the hinted vectors then do work. They need not work when it splits. For example, over ${\mathbb{F}}_{11}$ its roots are $4,8$, fixed by conjugation in ${\mathbb{F}}_{121}$. For either choice the proposed $e$ has norm $1+{\zeta }^2$, respectively $6$ or $10$, rather than zero. The norm-and-trace construction above proves the assertion without that restriction.

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.