Editing Linear subspace

{{Short description|In mathematics, vector subspace}}
In [[mathematics]], and more specifically in [[linear algebra]], a '''linear subspace''' or '''vector subspace'''<ref>{{Harvtxt|Halmos|1974}} pp. 16–17, § 10</ref><ref group="note">The term ''linear subspace'' is sometimes used for referring to [[flat (geometry)|flats]] and [[affine subspace]]s. In the case of vector spaces over the reals, linear subspaces, flats, and affine subspaces are also called ''linear manifolds'' for emphasizing that there are also [[manifold]]s.</ref> is a [[vector space]] that is a [[subset]] of some larger vector space. A linear subspace is usually simply called a ''subspace'' when the context serves to distinguish it from other types of [[subspace (mathematics)|subspaces]].

== Definition==
If ''V'' is a vector space over a [[Field (mathematics)|field]] ''K'', a subset ''W'' of ''V'' is a '''linear subspace''' of ''V'' if it is a [[vector space]] over ''K'' for the operations of ''V''. Equivalently, a linear subspace of ''V'' is a [[Empty set|nonempty]] subset ''W'' such that, whenever {{math|''w''<sub>1</sub>, ''w''<sub>2</sub>}} are elements of ''W'' and {{math|''α'', ''β''}} are elements of ''K'', it follows that {{math|''αw''<sub>1</sub> + ''βw''<sub>2</sub>}} is in ''W''.<ref>{{harvtxt|Anton|2005|p=155}}</ref><ref>{{harvtxt|Beauregard|Fraleigh|1973|p=176}}</ref><ref>{{harvtxt|Herstein|1964|p=132}}</ref><ref>{{harvtxt|Kreyszig|1972|p=200}}</ref><ref>{{harvtxt|Nering|1970|p=20}}</ref>

The [[singleton set]] consisting of the [[zero vector]] alone and the entire vector space itself are linear subspaces that are called the '''trivial subspaces''' of the vector space.<ref>{{harvtxt|Hefferon|2020}} p. 100, ch. 2, Definition 2.13</ref>

== Examples ==
=== Example I ===
In the vector space ''V'' = '''R'''<sup>3</sup> (the [[real coordinate space]] over the field '''R''' of [[real number]]s), take ''W'' to be the set of all vectors in ''V'' whose last component is 0.
Then ''W'' is a subspace of ''V''.

''Proof:''
#Given '''u''' and '''v''' in ''W'', then they can be expressed as {{nowrap|1='''u''' = (''u''<sub>1</sub>, ''u''<sub>2</sub>, 0)}} and {{nowrap|1='''v''' = (''v''<sub>1</sub>, ''v''<sub>2</sub>, 0)}}. Then {{nowrap|1='''u''' + '''v''' = (''u''<sub>1</sub>+''v''<sub>1</sub>, ''u''<sub>2</sub>+''v''<sub>2</sub>, 0+0) = (''u''<sub>1</sub>+''v''<sub>1</sub>, ''u''<sub>2</sub>+''v''<sub>2</sub>, 0)}}. Thus, '''u''' + '''v''' is an element of ''W'', too.
#Given '''u''' in ''W'' and a scalar ''c'' in '''R''', if {{nowrap|1='''u''' = (''u''<sub>1</sub>, ''u''<sub>2</sub>, 0)}} again, then {{nowrap|1=''c'''''u''' = (''cu''<sub>1</sub>, ''cu''<sub>2</sub>, ''c''0) = (''cu''<sub>1</sub>, ''cu''<sub>2</sub>,0)}}. Thus, ''c'''''u''' is an element of ''W'' too.

=== Example II ===
Let the field be '''R''' again, but now let the vector space ''V'' be the [[Cartesian plane]] '''R'''<sup>2</sup>.
Take ''W'' to be the set of points (''x'', ''y'') of '''R'''<sup>2</sup> such that ''x'' = ''y''.
Then ''W'' is a subspace of '''R'''<sup>2</sup>.

''Proof:''
#Let {{nowrap|1='''p''' = (''p''<sub>1</sub>, ''p''<sub>2</sub>)}} and {{nowrap|1='''q''' = (''q''<sub>1</sub>, ''q''<sub>2</sub>)}} be elements of ''W'', that is, points in the plane such that ''p''<sub>1</sub> = ''p''<sub>2</sub> and ''q''<sub>1</sub> = ''q''<sub>2</sub>. Then {{nowrap|1='''p''' + '''q''' = (''p''<sub>1</sub>+''q''<sub>1</sub>, ''p''<sub>2</sub>+''q''<sub>2</sub>)}}; since ''p''<sub>1</sub> = ''p''<sub>2</sub> and ''q''<sub>1</sub> = ''q''<sub>2</sub>, then ''p''<sub>1</sub> + ''q''<sub>1</sub> = ''p''<sub>2</sub> + ''q''<sub>2</sub>, so '''p''' + '''q''' is an element of ''W''.
#Let '''p''' = (''p''<sub>1</sub>, ''p''<sub>2</sub>) be an element of ''W'', that is, a point in the plane such that ''p''<sub>1</sub> = ''p''<sub>2</sub>, and let ''c'' be a scalar in '''R'''. Then {{nowrap|1=''c'''''p''' = (''cp''<sub>1</sub>, ''cp''<sub>2</sub>)}}; since ''p''<sub>1</sub> = ''p''<sub>2</sub>, then ''cp''<sub>1</sub> = ''cp''<sub>2</sub>, so ''c'''''p''' is an element of ''W''.

In general, any subset of the real coordinate space '''R'''<sup>''n''</sup> that is defined by a [[homogeneous system of linear equations]] will yield a subspace.
(The equation in example I was ''z''&nbsp;=&nbsp;0, and the equation in example II was ''x''&nbsp;=&nbsp;''y''.)

=== Example III ===
Again take the field to be '''R''', but now let the vector space ''V'' be the set '''R'''<sup>'''R'''</sup> of all [[function (mathematics)|function]]s from '''R''' to '''R'''.
Let C('''R''') be the subset consisting of [[continuous function]]s.
Then C('''R''') is a subspace of '''R'''<sup>'''R'''</sup>.

''Proof:''
#We know from calculus that {{nowrap|0 ∈ C('''R''') ⊂ '''R'''<sup>'''R'''</sup>}}.
#We know from calculus that the sum of continuous functions is continuous.
#Again, we know from calculus that the product of a continuous function and a number is continuous.

=== Example IV ===
Keep the same field and vector space as before, but now consider the set Diff('''R''') of all [[differentiable function]]s.
The same sort of argument as before shows that this is a subspace too.

Examples that extend these themes are common in [[functional analysis]].

== Properties of subspaces ==
From the definition of vector spaces, it follows that subspaces are nonempty, and are [[Closure (mathematics)|closed]] under sums and under scalar multiples.<ref>{{Harvtxt|MathWorld|2021}} Subspace.</ref> Equivalently, subspaces can be characterized by the property of being closed under linear combinations. That is, a nonempty set ''W'' is a subspace [[if and only if]] every linear combination of [[finite set|finite]]ly many elements of ''W'' also belongs to ''W''.
The equivalent definition states that it is also equivalent to consider linear combinations of two elements at a time.

In a [[topological vector space]] ''X'', a subspace ''W'' need not be topologically [[closed set|closed]], but a [[finite-dimensional]] subspace is always closed.<ref>{{harvtxt|DuChateau|2002}} Basic facts about Hilbert Space — class notes from Colorado State University on Partial Differential Equations (M645).</ref> The same is true for subspaces of finite [[codimension]] (i.e., subspaces determined by a finite number of continuous [[linear functional]]s).

==Descriptions==
Descriptions of subspaces include the solution set to a [[homogeneous system of linear equations]], the subset of Euclidean space described by a system of homogeneous linear [[parametric equations]], the [[linear span|span]] of a collection of vectors, and the [[null space]], [[column space]], and [[row space]] of a [[matrix (mathematics)|matrix]]. Geometrically (especially over the field of real numbers and its subfields), a subspace is a [[flat (geometry)|flat]] in an ''n''-space that passes through the origin.

A natural description of a 1-subspace is the [[scalar multiplication]] of one non-[[additive identity|zero]] vector '''v''' to all possible scalar values. 1-subspaces specified by two vectors are equal if and only if one vector can be obtained from another with scalar multiplication:
:<math>\exist c\in K: \mathbf{v}' = c\mathbf{v}\text{ (or }\mathbf{v} = \frac{1}{c}\mathbf{v}'\text{)}</math>
This idea is generalized for higher dimensions with [[linear span]], but criteria for [[equality (mathematics)|equality]] of ''k''-spaces specified by sets of ''k'' vectors are not so simple.

A [[duality (mathematics)|dual]] description is provided with [[linear functionals]] (usually implemented as linear equations). One non-[[additive identity|zero]] linear functional '''F''' specifies its [[kernel (linear algebra)|kernel]] subspace '''F'''&nbsp;=&nbsp;0 of codimension 1. Subspaces of codimension 1 specified by two linear functionals are equal, if and only if one functional can be obtained from another with scalar multiplication (in the [[dual space]]):
:<math>\exist c\in K: \mathbf{F}' = c\mathbf{F}\text{ (or }\mathbf{F} = \frac{1}{c}\mathbf{F}'\text{)}</math>
It is generalized for higher codimensions with a [[system of equations]]. The following two subsections will present this latter description in details, and [[#Span of vectors|the remaining]] four subsections further describe the idea of linear span.

===Systems of linear equations===
The solution set to any [[homogeneous system of linear equations]] with ''n'' variables is a subspace in the [[coordinate space]] ''K''<sup>''n''</sup>:
<math display="block">\left\{ \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right]  \in K^n : \begin{alignat}{6}
a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = 0&    \\
a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = 0&    \\
&&     &&            &&              &&                && \vdots\quad& \\
a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = 0&
\end{alignat} \right\}. </math>

For example, the set of all vectors {{math|(''x'', ''y'', ''z'')}} (over real or [[rational number]]s) satisfying the equations
<math display="block">x + 3y + 2z = 0 \quad\text{and}\quad 2x - 4y + 5z = 0</math>
is a one-dimensional subspace. More generally, that is to say that given a set of ''n'' independent functions, the dimension of the subspace in ''K''<sup>''k''</sup> will be the dimension of the [[null set]] of ''A'', the composite matrix of the ''n'' functions.

===Null space of a matrix===
{{main|Null space}}
In a finite-dimensional space, a homogeneous system of linear equations can be written as a single matrix equation:

:<math>A\mathbf{x} = \mathbf{0}.</math>

The set of solutions to this equation is known as the [[Null Space|null space]] of the matrix. For example, the subspace described above is the null space of the matrix

:<math>A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & -4 & 5 \end{bmatrix} .</math>

Every subspace of ''K''<sup>''n''</sup> can be described as the null space of some matrix (see {{slink||Algorithms}} below for more).

===Linear parametric equations===
The subset of ''K''<sup>''n''</sup> described by a system of homogeneous linear [[parametric equations]] is a subspace:

:<math>\left\{ \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right]  \in K^n : \begin{alignat}{7}
x_1 &&\; = \;&& a_{11} t_1 &&\; + \;&& a_{12} t_2 &&\; + \cdots + \;&& a_{1m} t_m &    \\
x_2 &&\; = \;&& a_{21} t_1 &&\; + \;&& a_{22} t_2 &&\; + \cdots + \;&& a_{2m} t_m &    \\
&& \vdots\;\; &&       &&       &&            &&                &&            &    \\
x_n &&\; = \;&& a_{n1} t_1 &&\; + \;&& a_{n2} t_2 &&\; + \cdots + \;&& a_{nm} t_m &    \\
\end{alignat} \text{ for some } t_1,\ldots,t_m\in K \right\}. </math>

For example, the set of all vectors (''x'',&nbsp;''y'',&nbsp;''z'') parameterized by the equations

:<math>x = 2t_1 + 3t_2,\;\;\;\;y = 5t_1 - 4t_2,\;\;\;\;\text{and}\;\;\;\;z = -t_1 + 2t_2</math>

is a two-dimensional subspace of ''K''<sup>3</sup>, if ''K'' is a [[number field]] (such as real or rational numbers).<ref name="fields" group="note">Generally, ''K'' can be any field of such [[characteristic (algebra)|characteristic]] that the given integer matrix has the appropriate [[rank (matrix theory)|rank]] in it. All fields include [[integer]]s, but some integers may equal to zero in some fields.</ref>

===Span of vectors===
{{main|Linear span}}
In linear algebra, the system of parametric equations can be written as a single vector equation:

:<math>\begin{bmatrix} x \\ y \\ z \end{bmatrix} \;=\; t_1 \!\begin{bmatrix} 2 \\ 5 \\ -1 \end{bmatrix} + t_2 \!\begin{bmatrix} 3 \\ -4 \\ 2 \end{bmatrix}.</math>

The expression on the right is called a linear combination of the vectors
(2,&nbsp;5,&nbsp;−1) and (3,&nbsp;−4,&nbsp;2). These two vectors are said to '''span''' the resulting subspace.

In general, a '''linear combination''' of vectors '''v'''<sub>1</sub>,&nbsp;'''v'''<sub>2</sub>,&nbsp;...&nbsp;,&nbsp;'''v'''<sub>''k''</sub> is any vector of the form

:<math>t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k.</math>

The set of all possible linear combinations is called the '''span''':

:<math>\text{Span} \{ \mathbf{v}_1, \ldots, \mathbf{v}_k \}
= \left\{ t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k : t_1,\ldots,t_k\in K \right\} .</math>

If the vectors '''v'''<sub>1</sub>,&nbsp;...&nbsp;,&nbsp;'''v'''<sub>''k''</sub> have ''n'' components, then their span is a subspace of ''K''<sup>''n''</sup>. Geometrically, the span is the flat through the origin in ''n''-dimensional space determined by the points '''v'''<sub>1</sub>,&nbsp;...&nbsp;,&nbsp;'''v'''<sub>''k''</sub>.

; Example
: The ''xz''-plane in '''R'''<sup>3</sup> can be parameterized by the equations
::<math>x = t_1, \;\;\; y = 0, \;\;\; z = t_2.</math>

:As a subspace, the ''xz''-plane is spanned by the vectors (1,&nbsp;0,&nbsp;0) and (0,&nbsp;0,&nbsp;1). Every vector in the ''xz''-plane can be written as a linear combination of these two:

::<math>(t_1, 0, t_2) = t_1(1,0,0) + t_2(0,0,1)\text{.}</math>

:Geometrically, this corresponds to the fact that every point on the ''xz''-plane can be reached from the origin by first moving some distance in the direction of (1,&nbsp;0,&nbsp;0) and then moving some distance in the direction of (0,&nbsp;0,&nbsp;1).

===Column space and row space===
{{main|Row and column spaces}}
A system of linear parametric equations in a finite-dimensional space can also be written as a single matrix equation:

:<math>\mathbf{x} = A\mathbf{t}\;\;\;\;\text{where}\;\;\;\;A = \left[ \begin{alignat}{2} 2 && 3 & \\ 5 && \;\;-4 & \\ -1 && 2 & \end{alignat} \,\right]\text{.}</math>

In this case, the subspace consists of all possible values of the vector '''x'''. In linear algebra, this subspace is known as the column space (or [[image (mathematics)|image]]) of the matrix ''A''. It is precisely the subspace of ''K''<sup>''n''</sup> spanned by the column vectors of ''A''.

The row space of a matrix is the subspace spanned by its row vectors. The row space is interesting because it is the [[orthogonal complement]] of the null space (see below).

===Independence, basis, and dimension===
{{main|Linear independence|Basis (linear algebra)|Dimension (vector space)}}
[[File:Basis for a plane.svg|thumb|280px|right|The vectors '''u''' and '''v''' are a basis for this two-dimensional subspace of '''R'''<sup>3</sup>.]]
In general, a subspace of ''K''<sup>''n''</sup> determined by ''k'' parameters (or spanned by ''k'' vectors) has dimension ''k''. However, there are exceptions to this rule. For example, the subspace of ''K''<sup>3</sup> spanned by the three vectors (1,&nbsp;0,&nbsp;0), (0,&nbsp;0,&nbsp;1), and (2,&nbsp;0,&nbsp;3) is just the ''xz''-plane, with each point on the plane described by infinitely many different values of {{nowrap| ''t''<sub>1</sub>, ''t''<sub>2</sub>, ''t''<sub>3</sub>}}.

In general, vectors '''v'''<sub>1</sub>,&nbsp;...&nbsp;,&nbsp;'''v'''<sub>''k''</sub> are called '''linearly independent''' if

:<math>t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k \;\ne\; u_1 \mathbf{v}_1 + \cdots + u_k \mathbf{v}_k</math>

for
(''t''<sub>1</sub>,&nbsp;''t''<sub>2</sub>,&nbsp;...&nbsp;,&nbsp;''t<sub>k</sub>'')&nbsp;≠ (''u''<sub>1</sub>,&nbsp;''u''<sub>2</sub>,&nbsp;...&nbsp;,&nbsp;''u<sub>k</sub>'').<ref group="note">This definition is often stated differently: vectors '''v'''<sub>1</sub>, ..., '''v'''<sub>''k''</sub> are linearly independent if
{{nowrap| ''t''<sub>1</sub>'''v'''<sub>1</sub> + ··· + ''t<sub>k</sub>'''''v'''<sub>''k''</sub> ≠ '''0'''}} for {{nowrap| (''t''<sub>1</sub>, ''t''<sub>2</sub>, ..., ''t<sub>k</sub>'') ≠ (0, 0, ..., 0)}}. The two definitions are equivalent.</ref>
If {{nowrap| '''v'''<sub>1</sub>, ..., '''v'''<sub>''k''</sub> }} are linearly independent, then the '''coordinates''' {{nowrap| ''t''<sub>1</sub>, ..., ''t<sub>k</sub>''}} for a vector in the span are uniquely determined.

A '''basis''' for a subspace ''S'' is a set of linearly independent vectors whose span is ''S''. The number of elements in a basis is always equal to the geometric dimension of the subspace. Any spanning set for a subspace can be changed into a basis by removing redundant vectors (see [[#Algorithms|§ Algorithms]] below for more).

; Example
: Let ''S'' be the subspace of '''R'''<sup>4</sup> defined by the equations
::<math>x_1 = 2 x_2\;\;\;\;\text{and}\;\;\;\;x_3 = 5x_4.</math>
:Then the vectors (2,&nbsp;1,&nbsp;0,&nbsp;0) and (0,&nbsp;0,&nbsp;5,&nbsp;1) are a basis for ''S''. In particular, every vector that satisfies the above equations can be written uniquely as a linear combination of the two basis vectors:

::<math>(2t_1, t_1, 5t_2, t_2) = t_1(2, 1, 0, 0) + t_2(0, 0, 5, 1).</math>

:The subspace ''S'' is two-dimensional. Geometrically, it is the plane in '''R'''<sup>4</sup> passing through the points (0,&nbsp;0,&nbsp;0,&nbsp;0), (2,&nbsp;1,&nbsp;0,&nbsp;0), and (0,&nbsp;0,&nbsp;5,&nbsp;1).

==Operations and relations on subspaces==

=== Inclusion ===
<!-- some illustration, please -->
The [[inclusion relation|set-theoretical inclusion]] binary relation specifies a [[partial order]] on the set of all subspaces (of any dimension).

A subspace cannot lie in any subspace of lesser dimension. If dim&nbsp;''U''&nbsp;=&nbsp;''k'', a finite number, and ''U''&nbsp;⊂&nbsp;''W'', then dim&nbsp;''W''&nbsp;=&nbsp;''k'' if and only if ''U''&nbsp;=&nbsp;''W''.

===Intersection===
[[File:Intersecting Planes 2.svg|thumb|right|In '''R'''<sup>3</sup>, the intersection of two distinct two-dimensional subspaces is one-dimensional]]
Given subspaces ''U'' and ''W'' of a vector space ''V'', then their [[intersection (set theory)|intersection]] ''U''&nbsp;∩&nbsp;''W''&nbsp;:= {'''v'''&nbsp;∈&nbsp;''V''&nbsp;: '''v'''&nbsp;is an element of both ''U'' and&nbsp;''W''} is also a subspace of ''V''.<ref>{{harvtxt|Nering|1970|p=21}}</ref>

''Proof:''
# Let '''v''' and '''w''' be elements of ''U''&nbsp;∩&nbsp;''W''. Then '''v''' and '''w''' belong to both ''U'' and ''W''. Because ''U'' is a subspace, then '''v'''&nbsp;+&nbsp;'''w''' belongs to ''U''. Similarly, since ''W'' is a subspace, then '''v'''&nbsp;+&nbsp;'''w''' belongs to ''W''. Thus, '''v'''&nbsp;+&nbsp;'''w''' belongs to ''U''&nbsp;∩&nbsp;''W''.
# Let '''v''' belong to ''U''&nbsp;∩&nbsp;''W'', and let ''c'' be a scalar. Then '''v''' belongs to both ''U'' and ''W''. Since ''U'' and ''W'' are subspaces, ''c'''''v''' belongs to both ''U'' and&nbsp;''W''.
# Since ''U'' and ''W'' are vector spaces, then '''0''' belongs to both sets. Thus, '''0''' belongs to ''U''&nbsp;∩&nbsp;''W''.

For every vector space ''V'', the [[zero vector space|set {'''0'''}]] and ''V'' itself are subspaces of ''V''.<ref>{{harvtxt|Hefferon|2020}} p. 100, ch. 2, Definition 2.13</ref><ref>{{harvtxt|Nering|1970|p=20}}</ref>

===Sum===
If ''U'' and ''W'' are subspaces, their '''sum''' is the subspace<ref>{{harvtxt|Nering|1970|p=21}}</ref><ref name=":1">Vector space related operators.</ref>
<math display="block">U + W = \left\{ \mathbf{u} + \mathbf{w} \colon \mathbf{u}\in U, \mathbf{w}\in W \right\}.</math>

For example, the sum of two lines is the plane that contains them both. The dimension of the sum satisfies the inequality
<math display="block">\max(\dim U,\dim W) \leq \dim(U + W) \leq \dim(U) + \dim(W).</math>

Here, the minimum only occurs if one subspace is contained in the other, while the maximum is the most general case. The dimension of the intersection and the sum are related by the following equation:<ref>{{harvtxt|Nering|1970|p=22}}</ref>
<math display="block">\dim(U+W) = \dim(U) + \dim(W) - \dim(U \cap W).</math>

A set of subspaces is '''independent''' when the only intersection between any pair of subspaces is the trivial subspace. The '''[[Direct sum of modules|direct sum]]''' is the sum of independent subspaces, written as <math>U \oplus W</math>. An equivalent restatement is that a direct sum is a subspace sum under the condition that every subspace contributes to the span of the sum.<ref>{{harvtxt|Hefferon|2020}} p. 148, ch. 2, §4.10</ref><ref>{{harvtxt|Axler|2015}} p. 21 § 1.40</ref><ref>{{harvtxt|Katznelson|Katznelson|2008}} pp. 10–11, § 1.2.5</ref><ref>{{harvtxt|Halmos|1974}} pp. 28–29, § 18</ref>

The dimension of a direct sum <math>U \oplus W</math> is the same as the sum of subspaces, but may be shortened because the dimension of the trivial subspace is zero.<ref>{{harvtxt|Halmos|1974}} pp. 30–31, § 19</ref>

<math display="block">\dim (U \oplus W) = \dim (U) + \dim (W)</math>

=== Lattice of subspaces ===
The operations [[#Intersection|intersection]] and [[#Sum|sum]] make the set of all subspaces a bounded [[modular lattice]], where the [[zero vector space|{0} subspace]], the [[least element]], is an [[identity element]] of the sum operation, and the identical subspace ''V'', the greatest element, is an identity element of the intersection operation.

=== 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}}

==Algorithms==
Most algorithms for dealing with subspaces involve [[row reduction]]. This is the process of applying [[elementary row operation]]s to a matrix, until it reaches either [[row echelon form]] or [[reduced row echelon form]]. Row reduction has the following important properties:
# The reduced matrix has the same null space as the original.
# Row reduction does not change the span of the row vectors, i.e. the reduced matrix has the same row space as the original.
# Row reduction does not affect the linear dependence of the column vectors.

===Basis for a row space===
:'''Input''' An ''m''&nbsp;×&nbsp;''n'' matrix ''A''.
:'''Output''' A basis for the row space of ''A''.
:# Use elementary row operations to put ''A'' into row echelon form.
:# The nonzero rows of the echelon form are a basis for the row space of ''A''.
See the article on [[row space]] for an [[Row and column spaces#Basis 2|example]].

If we instead put the matrix ''A'' into reduced row echelon form, then the resulting basis for the row space is uniquely determined. This provides an algorithm for checking whether two row spaces are equal and, by extension, whether two subspaces of ''K''<sup>''n''</sup> are equal.

===Subspace membership===
:'''Input''' A basis {'''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>} for a subspace ''S'' of ''K''<sup>''n''</sup>, and a vector '''v''' with ''n'' components.
:'''Output''' Determines whether '''v''' is an element of ''S''
:# Create a (''k''&nbsp;+&nbsp;1)&nbsp;×&nbsp;''n'' matrix ''A'' whose rows are the vectors '''b'''<sub>1</sub>,&nbsp;...&nbsp;,&nbsp;'''b'''<sub>''k''</sub> and '''v'''.
:# Use elementary row operations to put ''A'' into row echelon form.
:# If the echelon form has a row of zeroes, then the vectors {{nowrap| {'''b'''<sub>1</sub>, ..., '''b'''<sub>''k''</sub>, '''v'''} }} are linearly dependent, and therefore {{nowrap| '''v''' ∈ ''S''}}.

===Basis for a column space===
:'''Input''' An ''m''&nbsp;×&nbsp;''n'' matrix ''A''
:'''Output''' A basis for the column space of ''A''
:# Use elementary row operations to put ''A'' into row echelon form.
:# Determine which columns of the echelon form have [[Row echelon form|pivots]]. The corresponding columns of the original matrix are a basis for the column space.
See the article on column space for an [[Column space#Basis|example]].

This produces a basis for the column space that is a subset of the original column vectors. It works because the columns with pivots are a basis for the column space of the echelon form, and row reduction does not change the linear dependence relationships between the columns.

===Coordinates for a vector===
:'''Input''' A basis {'''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>} for a subspace ''S'' of ''K''<sup>''n''</sup>, and a vector {{nowrap| '''v''' ∈ ''S''}}
:'''Output''' Numbers ''t''<sub>1</sub>, ''t''<sub>2</sub>, ..., ''t''<sub>''k''</sub> such that {{nowrap|1= '''v''' = ''t''<sub>1</sub>'''b'''<sub>1</sub> + ··· + ''t''<sub>''k''</sub>'''b'''<sub>''k''</sub>}}
:# Create an [[augmented matrix]] ''A'' whose columns are '''b'''<sub>1</sub>,...,'''b'''<sub>''k''</sub> , with the last column being '''v'''.
:# Use elementary row operations to put ''A'' into reduced row echelon form.
:# Express the final column of the reduced echelon form as a linear combination of the first ''k'' columns. The coefficients used are the desired numbers {{nowrap| ''t''<sub>1</sub>, ''t''<sub>2</sub>, ..., ''t''<sub>''k''</sub>}}. (These should be precisely the first ''k'' entries in the final column of the reduced echelon form.)
If the final column of the reduced row echelon form contains a pivot, then the input vector '''v''' does not lie in ''S''.

===Basis for a null space===
:'''Input''' An ''m''&nbsp;×&nbsp;''n'' matrix ''A''.
:'''Output''' A basis for the null space of ''A''
:# Use elementary row operations to put ''A'' in reduced row echelon form.
:# Using the reduced row echelon form, determine which of the variables {{nowrap| ''x''<sub>1</sub>, ''x''<sub>2</sub>, ..., ''x<sub>n</sub>''}} are free. Write equations for the dependent variables in terms of the free variables.
:# For each free variable ''x<sub>i</sub>'', choose a vector in the null space for which {{nowrap|1= ''x<sub>i</sub>'' = 1}} and the remaining free variables are zero. The resulting collection of vectors is a basis for the null space of ''A''.
See the article on null space for an [[Kernel (matrix)#Basis|example]].

===Basis for the sum and intersection of two subspaces===
Given two subspaces {{mvar|U}} and {{mvar|W}} of {{mvar|V}}, a basis of the sum <math>U + W</math> and the intersection <math>U \cap W</math> can be calculated using the [[Zassenhaus algorithm]].

===Equations for a subspace===
:'''Input''' A basis {'''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>} for a subspace ''S'' of ''K''<sup>''n''</sup>
:'''Output''' An (''n''&nbsp;−&nbsp;''k'')&nbsp;×&nbsp;''n'' matrix whose null space is ''S''.
:# Create a matrix ''A'' whose rows are {{nowrap| '''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>}}.
:# Use elementary row operations to put ''A'' into reduced row echelon form.
:# Let {{nowrap| '''c'''<sub>1</sub>, '''c'''<sub>2</sub>, ..., '''c'''<sub>''n''</sub> }} be the columns of the reduced row echelon form. For each column without a pivot, write an equation expressing the column as a linear combination of the columns with pivots.
:# This results in a homogeneous system of ''n'' − ''k'' linear equations involving the variables '''c'''<sub>1</sub>,...,'''c'''<sub>''n''</sub>. The {{nowrap| (''n'' − ''k'') × ''n''}} matrix corresponding to this system is the desired matrix with nullspace ''S''.
; Example
:If the reduced row echelon form of ''A'' is

::<math>\left[ \begin{alignat}{6}
1 && 0 && -3 && 0 &&  2 && 0 \\
0 && 1 &&  5 && 0 && -1 && 4 \\
0 && 0 &&  0 && 1 &&  7 && -9 \\
0 && \;\;\;\;\;0 &&  \;\;\;\;\;0 && \;\;\;\;\;0 &&  \;\;\;\;\;0 && \;\;\;\;\;0 \end{alignat} \,\right] </math>

:then the column vectors {{nowrap| '''c'''<sub>1</sub>, ..., '''c'''<sub>6</sub>}} satisfy the equations

::<math> \begin{alignat}{1}
\mathbf{c}_3 &= -3\mathbf{c}_1 + 5\mathbf{c}_2 \\
\mathbf{c}_5 &= 2\mathbf{c}_1 - \mathbf{c}_2 + 7\mathbf{c}_4 \\
\mathbf{c}_6 &= 4\mathbf{c}_2 - 9\mathbf{c}_4
\end{alignat}</math>

:It follows that the row vectors of ''A'' satisfy the equations

::<math> \begin{alignat}{1}
x_3 &= -3x_1 + 5x_2 \\
x_5 &= 2x_1 - x_2 + 7x_4 \\
x_6 &= 4x_2 - 9x_4.
\end{alignat}</math>

:In particular, the row vectors of ''A'' are a basis for the null space of the corresponding matrix.

==See also==
* [[Cyclic subspace]]
* [[Invariant subspace]]
* [[Multilinear subspace learning]]
* [[Quotient space (linear algebra)]]
* [[Signal subspace]]
* [[Subspace topology]]

== Notes ==
<references group="note" />

==Citations==
{{Reflist}}

==Sources==

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=== Web ===

* {{Cite web|last=Weisstein|first=Eric Wolfgang|author-link=Eric W. Weisstein|title=Subspace|url=https://mathworld.wolfram.com/Subspace.html|access-date=16 Feb 2021|website=[[MathWorld]]|ref=CITEREFMathWorld2021}}
* {{Cite web|last=DuChateau|first=Paul|date=5 Sep 2002|title=Basic facts about Hilbert Space|url=https://www.math.colostate.edu/~pauld/M645/BasicHS.pdf|access-date=17 Feb 2021|website=[[Colorado State University]]}}

==External links==
*{{Cite web|last=Strang|first=Gilbert|author-link=Gilbert Strang|date=7 May 2009|title=The four fundamental subspaces|url=https://www.youtube.com/watch?v=nHlE7EgJFds| archive-url=https://ghostarchive.org/varchive/youtube/20211211/nHlE7EgJFds| archive-date=2021-12-11|url-status=live|access-date=17 Feb 2021|via=[[YouTube]]}}{{cbignore}}
*{{Cite web|last=Strang|first=Gilbert|author-link=Gilbert Strang|date=5 May 2020|title=The big picture of linear algebra|url=https://www.youtube.com/watch?v=rwLOfdfc4dw&list=PLUl4u3cNGP61iQEFiWLE21EJCxwmWvvek&index=3| archive-url=https://ghostarchive.org/varchive/youtube/20211211/rwLOfdfc4dw| archive-date=2021-12-11|url-status=live|access-date=17 Feb 2021|via=[[YouTube]]}}{{cbignore}}

{{Linear algebra}}

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