Recipe

Find the orthogonal complement of a subspace

Use when: you have a subspace $W$ given as a span of vectors and need $W^\perp$

Goal: $W^\perp$ described as the span of a basis, found by solving one homogeneous system

Taught in Orthogonal sets and orthogonal complements

Quick recipe

Identify a spanning set for $W$
Write one equation per spanning vector
Solve the resulting homogeneous system
Write $W^\perp$ as the span of the solution

1
Write down the vectors $\mathbf w_1,\dots,\mathbf w_m$ that span $W$ — testing orthogonality against just these is enough, never mind the infinitely many other vectors $W$ contains.

Example

Let $W=\operatorname{span}\{(1,0,-2),\ (0,1,3)\}$ , a plane through the origin in $\mathbb R^3$ .

Why: orthogonality to a spanning set forces orthogonality to every linear combination of it, so checking finitely many vectors already checks the whole subspace.
2
For $\mathbf x=(x,y,z)$ , set $\mathbf x\cdot\mathbf w_i=0$ for each spanning vector $\mathbf w_i$ in turn.

Example

$\mathbf x\cdot(1,0,-2)=x-2z=0$ and $\mathbf x\cdot(0,1,3)=y+3z=0$ .

Why: this is exactly the membership condition $W^\perp=\{\mathbf x:\langle\mathbf x,\mathbf w\rangle=0\text{ for all }\mathbf w\in W\}$ , restricted to a finite, checkable list.
3
Treat the equations as an ordinary linear system in the unknown entries of $\mathbf x$ ; let any leftover unknown be a free parameter and back-substitute through the other equations.

Example

From $x-2z=0$ : $x=2z$ . From $y+3z=0$ : $y=-3z$ . Let $z=t$ : $x=2t,\ y=-3t,\ z=t$ .

Why: $W^\perp$ is always the solution set of a homogeneous system built from $W$ 's spanning vectors, so solving that system finds every vector in it at once.
4
Factor the free parameter out of the parametrized solution to read off a basis for $W^\perp$ .

Example

$(2t,-3t,t)=t(2,-3,1)$ , so $W^\perp=\operatorname{span}\{(2,-3,1)\}$ — check: $(2,-3,1)\cdot(1,0,-2)=2-2=0$ and $(2,-3,1)\cdot(0,1,3)=-3+3=0$ .

Why: the solution set of a homogeneous system is automatically a subspace, and writing it as the span of a basis is the standard way to describe any subspace.

Common mistakes

Testing orthogonality against only one spanning vector instead of every one of them.
Treating a vector orthogonal to some vector in $W$ as automatically in $W^\perp$ — it has to be orthogonal to all of $W$ .
Reporting a single vector when the solution actually needs more than one free parameter (a $W^\perp$ can be a plane, not just a line).
Sign errors moving a term across the equals sign while solving the homogeneous system.