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

For a bilinear form on a finite-dimensional vector space, the displayed right bilinear orthogonal complement is a subspace. Nondegeneracy identifies the ambient space with its dual and gives $\dim W^{\perp }=\dim V-\dim W$. If the form is symmetric or alternating, orthogonality is reciprocal and $(W^{\perp }{)}^{\perp }=W$. For a general nonsymmetric form, left and right complements must be distinguished. A restricted symmetric or alternating form has radical of a bilinear form $W\cap W^{\perp }$; quotienting by that radical of a bilinear form gives a nondegenerate induced form.

For a bilinear form $B$ on a finite-dimensional vector space $V$, let $\Phi :V\to V^{\ast }$ send $u$ to $B(u,\cdot )$. Under the bidual evaluation map, $U^{\perp }$ is the annihilator of a vector subspace $\mathrm{im}(\Phi {|}_U)$, so $\dim U^{\perp }=\dim V-\mathrm{rank}(\Phi {|}_U)$. The rank-nullity theorem on $U$ now gives the displayed equality. In particular the sum of dimensions is at least $\dim V$. If $U\cap U^{\perp }=0$, the dimension formula for a sum of subspaces forces $V=U\oplus U^{\perp }$. The exact formula records the extra dimensions contributed by the left radical $\ker \Phi$, and does not assume an inner product.