= Hermitian hyperbolic plane over a finite field
{c}
{title2=$\begin{pmatrix}0&1\\1&0\end{pmatrix}$}
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.
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