Orthogonal complement

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In the mathematical fields of linear algebra and functional analysis, the orthogonal complement W^⊥ of a subspace W of an inner product space V is the set of all vectors in V that are orthogonal to every vector in W, i.e., it is

$W^\bot=\left\{x\in V : \langle x, y \rangle = 0 \mbox{ for all } y\in W \right\}.\,$

Informally, it is called the perp, short for perpendicular complement.

[ Properties

The orthogonal complement is always closed in the metric topology. In finite-dimensional spaces, that is merely an instance of the fact that all subspaces of a vector space are closed. In infinite-dimensional Hilbert spaces, some subspaces are not closed, but all orthogonal complements are closed. In such spaces, the orthogonal complement of the orthogonal complement of W is the closure of W, i.e.,

$W^{\bot\,\bot}=\overline{W}.$

The orthogonal complement generalizes to the annihilator, and gives a Galois connection on subsets of the inner product space, with associated closure operator the topological closure of the span.

For a finite dimensional inner product space of dimension n, the orthogonal complement of a k-dimensional subspace is an (n − k)-dimensional subspace, and the double orthogonal complement is the original subspace:

$W^{\bot\,\bot}=W.$

If A is an m × n matrix, where Row A, Col A, and Null A refer to the row space, column space, and null space of A (respectively), we have

$\begin{align} (\mbox{Row}\,A)^\bot &= \mbox{Null}\,A\\ (\mbox{Col}\,A)^\bot &= \mbox{Null}\,A^T. \end{align}$

[ Banach spaces

There is a natural analog of this notion in general Banach spaces. In this case one defines the orthogonal complement of W to be a subspace of the dual of V defined similarly by

$W^\bot = \left\{\,x\in V^* : \forall y\in W, x(y) = 0 \, \right\}.\,$

It is always a closed subspace of V^∗. There is also an analog of the double complement property. W^⊥⊥ is now a subspace of V^∗∗ (which is not identical to V). However, the reflexive spaces have a natural isomorphism i between V and V^∗∗. In this case we have