Editing Linear subspace (section)

=== Orthogonal complements ===

If <math>V</math> is an [[inner product space]] and <math>N</math> is a subset of <math>V</math>, then the [[orthogonal complement]] of <math>N</math>, denoted <math>N^{\perp}</math>, is again a subspace.<ref>{{harvtxt|Axler|2015}} p. 193, § 6.46</ref> If <math>V</math> is finite-dimensional and <math>N</math> is a subspace, then the dimensions of <math>N</math> and <math>N^{\perp}</math> satisfy the complementary relationship <math>\dim (N) + \dim (N^{\perp}) = \dim (V) </math>.<ref>{{harvtxt|Axler|2015}} p. 195, § 6.50</ref>  Moreover, no vector is orthogonal to itself, so <math> N \cap N^\perp = \{ 0 \}</math> and <math>V</math> is the [[direct sum]] of <math>N</math> and <math>N^{\perp}</math>.<ref>{{harvtxt|Axler|2015}} p. 194, § 6.47</ref>  Applying orthogonal complements twice returns the original subspace: <math>(N^{\perp})^{\perp} = N</math> for every subspace <math>N</math>.<ref>{{harvtxt|Axler|2015}} p. 195, § 6.51</ref>

This operation, understood as [[negation]] (<math>\neg</math>), makes the lattice of subspaces a (possibly [[infinite set|infinite]]) orthocomplemented lattice (although not a distributive lattice).{{citation needed|date=January 2019}}

In spaces with other [[bilinear form]]s, some but not all of these results still hold. In [[pseudo-Euclidean space]]s and [[symplectic vector space]]s, for example, orthogonal complements exist. However, these spaces may have [[null vector]]s that are orthogonal to themselves, and consequently there exist subspaces <math>N</math> such that <math>N \cap N^{\perp} \ne \{ 0 \}</math>. As a result, this operation does not turn the lattice of subspaces into a Boolean algebra (nor a [[Heyting algebra]]).{{citation needed|date=January 2019}}