Orthogonal complement

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In the mathematical fields of linear algebra and functional analysis, the orthogonal complement $W^\bot$ 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.

[edit] Properties

The orthogonal complement is always closed in the metric topology. In Hilbert 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.

[edit] Finite dimensions

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 $m \times n$ matrix, where $\mbox{Row}\,A$ , $\mbox{Col}\,A$ , and $\mbox{Nul}\,A$ refer to the row space, column space, and null space of A (respectively), we have

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

[edit] 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. $W^{\bot\,\bot}$ 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