Hermitian hyperbolic plane over a finite field ($\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$)

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.