Linear Algebra and Its Applications · Problem 3.1.18

The orthogonal complement $S^\perp$ contains all vectors perpendicular to every vector in $S$—think of it as the "orthogonal shadow" of the subspace. When $S$ is trivial (just the zero vector), nothing constrains perpendicularity, so $S^\perp$ is everything. When $S$ is one-dimensional, $S^\perp$ is the orthogonal hyperplane. The double complement $(S^\perp)^\perp$ recovers the original subspace because orthogonality is symmetric: if you're perpendicular to all perpendiculars, you're back where you started.

The orthogonal complement $S^\perp$ contains all vectors perpendicular to every vector in $S$—think of it as the "orthogonal shadow" of the subspace. When $S$ is trivial (just the zero vector), nothing constrains perpendicularity, so $S^\perp$ is everything. When $S$ is one-dimensional, $S^\perp$ is the orthogonal hyperplane. The double complement $(S^\perp)^\perp$ recovers the original subspace because orthogonality is symmetric: if you're perpendicular to all perpendiculars, you're back where you started.