Orthogonal complement for a sesquilinear form ($W^{\perp }=\{v:B(v,w)=0\text{ for all }w\in W\}$)

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.