Recipe

Find the orthogonal complement of a subspace under a bilinear form

Use when: you have a bilinear form's Gram matrix and a subspace W, and need W^{\perp_B}=\{\mathbf v: B(\mathbf v,\mathbf w)=0 \text{ for every } \mathbf w\in W\}

Goal: a spanning set for W^{\perp_B}, with its dimension confirmed against \dim V-\dim W

Taught in Nondegenerate forms and orthogonal complements

1. 1

   Compute \det(A) for the form's Gram matrix A; a genuine orthogonal complement with the dimension formula needs \det(A)\ne0.

   Example

   For A=\begin{pmatrix}1&2\\2&4\end{pmatrix}, \det(A)=(1)(4)-(2)(2)=0, so this particular form is degenerate and the steps below don't apply to it as stated.

   Why: the dimension formula \dim(W^{\perp_B})=\dim V-\dim W is only guaranteed when B is nondegenerate; section 8's degenerate case needs its own separate treatment.

2. 2

   For each vector \mathbf w spanning W, write B(\mathbf v,\mathbf w)=0 using a general vector \mathbf v=(x,y,\dots).

   Example

   For A=\operatorname{diag}(1,-1) and W=\operatorname{span}\{(1,1)\}: B((x,y),(1,1))=x-y=0.

   Why: \mathbf v belongs to W^{\perp_B} exactly when it pairs to zero against every spanning vector of W, and a spanning set is enough to check.

3. 3

   Solve the equation(s) from Step 2 the same way as any homogeneous linear system.

   Example

   x-y=0 gives x=y, so the solution set is \operatorname{span}\{(1,1)\}.

   • the form is positive definite — W\cap W^{\perp_B} always comes out \{\mathbf 0\}, matching an ordinary inner product's behavior.
   • the form is indefinite — W can sit partly or entirely inside its own W^{\perp_B}, since a nonzero vector can be self-orthogonal.

   Why: solving the system is the only genuinely new computation here; everything before it is just setting the system up correctly.

4. 4

   Report W^{\perp_B} as a span, then check \dim(W^{\perp_B})=\dim V-\dim W holds regardless of any overlap with W itself.

   Example

   W^{\perp_B}=\operatorname{span}\{(1,1)\} is 1-dimensional, matching \dim V-\dim W=2-1=1 exactly, even though W\subseteq W^{\perp_B} here.

   Why: the dimension formula only tracks dimension, never disjointness, so it survives even the overlap an indefinite form can produce.