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| |[[Image:Projectivisation F5P^1.svg|120px]][[Image:Projectivisation F5P^1.svg|120px]]<br/>[[Image:Projectivisation F5P^1.svg|120px]][[Image:Projectivisation F5P^1.svg|120px]]
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| | style="font-size:87%" |One-dimensional subspaces in the two-dimensional vector space over the [[finite field]] '''F'''<sub>5</sub>. The [[origin (mathematics)|origin]] (0, 0), marked with green circles, belongs to any of six 1-subspaces, while each of 24 remaining points belongs to exactly one; a property which holds for 1-subspaces over any field and in all dimensions. All '''F'''<sub>5</sub><sup>2</sup> (i.e. a 5 × 5 square) is pictured four times for a better visualization
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| |}
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| The concept of a '''linear subspace''' (or '''vector subspace''') is important in [[linear algebra]] and related fields of [[mathematics]].
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| A linear subspace is usually called simply a ''subspace'' when the context serves to distinguish it from other kinds of [[subspace (disambiguation)|subspace]]s.
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| == Definition and useful characterization of subspace ==
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| Let ''K'' be a [[field (mathematics)|field]] (such as the field of [[real number]]s), and let ''V'' be a [[vector space]] over ''K''.
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| As usual, we call elements of ''V'' ''vectors'' and call elements of ''K'' ''[[scalar (mathematics)|scalars]]''. Ignoring the full extent of mathematical generalization, scalars can be understood simply as [[number]]s.
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| Suppose that ''W'' is a [[subset]] of ''V''.
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| If ''W'' is a vector space itself (which means that it is [[closure (mathematics)|closed]] under operations of [[addition]] and [[scalar multiplication]]), with the same vector space operations as ''V'' has, then ''W'' is a '''subspace''' of ''V''.
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| To use this definition, we don't have to prove that all the properties of a vector space hold for ''W''.
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| Instead, we can prove a theorem that gives us an easier way to show that a subset of a vector space is a subspace.
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| | |
| '''Theorem:'''
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| Let ''V'' be a vector space over the field ''K'', and let ''W'' be a subset of ''V''.
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| Then ''W'' is a subspace [[if and only if]] ''W'' satisfies the following three conditions:
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| #The zero vector, '''0''', is in ''W''.
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| #If '''u''' and '''v''' are elements of ''W'', then the sum '''u''' + '''v''' is an element of ''W'';
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| #If '''u''' is an element of ''W'' and ''c'' is a scalar from ''K'', then the product ''c'''''u''' is an element of ''W'';
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| '''Proof:'''
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| Firstly, property 1 ensures ''W'' is nonempty. Looking at the definition of a [[vector space]], we see that properties 2 and 3 above assure closure of ''W'' under addition and scalar multiplication, so the vector space operations are well defined. Since elements of ''W'' are necessarily elements of ''V'', axioms 1, 2 and 5–8 of a vector space are satisfied. By the closure of ''W'' under scalar multiplication (specifically by 0 and -1), vector space's definitional axiom [[identity element]] of [[vector_addition#Addition_and_subtraction|addition]] and axiom [[inverse element]] of addition are satisfied.
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| Conversely, if ''W'' is subspace of ''V'', then ''W'' is itself a vector space under the operations induced by
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| ''V'', so properties 2 and 3 are satisfied. By property 3, −'''w''' is in ''W'' whenever '''w''' is, and it follows that
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| ''W'' is closed under subtraction as well. Since | |
| ''W'' is nonempty, there is an element '''x''' in ''W'', and <math> \bold x - \bold x = {\bold 0} </math>
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| is in ''W'', so property 1 is satisfied. One can also argue that since ''W'' is nonempty, there is an element '''x''' in ''W'', and 0 is in the field ''K'' so <math> 0 \bold x = {\bold 0} </math> and therefore property 1 is satisfied. | |
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| == Examples ==
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| '''Example I:'''
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| Let the field ''K'' be the set '''R''' of [[real number]]s, and let the vector space ''V'' be the [[real coordinate space]] '''R'''<sup>3</sup>.
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| Take ''W'' to be the set of all vectors in ''V'' whose last component is 0.
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| Then ''W'' is a subspace of ''V''.
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| ''Proof:''
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| #Given '''u''' and '''v''' in ''W'', then they can be expressed as '''u''' = (''u''<sub>1</sub>, ''u''<sub>2</sub>, 0) and '''v''' = (''v''<sub>1</sub>, ''v''<sub>2</sub>, 0). Then '''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.
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| #Given '''u''' in ''W'' and a scalar ''c'' in '''R''', if '''u''' = (''u''<sub>1</sub>, ''u''<sub>2</sub>, 0) again, then ''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.
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| '''Example II:'''
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| Let the field be '''R''' again, but now let the vector space be the [[Cartesian plane]] '''R'''<sup>2</sup>.
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| Take ''W'' to be the set of points (''x'', ''y'') of '''R'''<sup>2</sup> such that ''x'' = ''y''.
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| Then ''W'' is a subspace of '''R'''<sup>2</sup>.
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| ''Proof:''
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| #Let '''p''' = (''p''<sub>1</sub>, ''p''<sub>2</sub>) and '''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 '''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''.
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| #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 ''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''.
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| In general, any subset of the [[real coordinate space]] '''R'''<sup>''n''</sup> that is defined by a system of homogeneous [[linear equation]]s will yield a subspace.
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| (The equation in example I was ''z'' = 0, and the equation in example II was ''x'' = ''y''.)
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| Geometrically, these subspaces are points, lines, planes, and so on, that pass through the point '''0'''.
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| === Examples related to calculus ===
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| '''Example III:'''
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| 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'''.
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| Let C('''R''') be the subset consisting of [[continuous function|continuous]] functions.
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| Then C('''R''') is a subspace of '''R'''<sup>'''R'''</sup>.
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| ''Proof:''
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| #We know from calculus that {{nowrap|0 ∈ C('''R''') ⊂ '''R'''<sup>'''R'''</sup>}}.
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| #We know from calculus the sum of continuous functions is continuous.
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| #Again, we know from calculus that the product of a continuous function and a number is continuous.
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| '''Example IV:'''
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| Keep the same field and vector space as before, but now consider the set Diff('''R''') of all [[differentiable]] functions.
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| The same sort of argument as before shows that this is a subspace too.
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| Examples that extend these themes are common in [[functional analysis]].
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| == Properties of subspaces ==
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| A way to characterize subspaces is that they are [[Closure (mathematics)|closed]] under [[linear combination]]s.
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| 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''.
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| Conditions 2 and 3 for a subspace are simply the most basic kinds of linear combinations.
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| In a [[topological vector space]] ''X'', a subspace ''W'' need not be [[closed set|closed]] in general, but a [[finite-dimensional]] subspace is always closed.<ref>See {{cite url|title=Basic Facts About Hilbert Space|url=http://www.math.colostate.edu/~pauld/M645/BasicHS.pdf|format=pdf|accessdate=September 17, 2012|author=Paul DuChateau}} for [[Hilbert space]]s</ref> The same is true for subspaces of finite [[codimension]], i.e. determined by a finite number of continuous [[linear functional]]s.
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| ==Descriptions==
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| 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 number]]s and its subfields), a subspace is a [[flat (geometry)|flat]] in an ''n''-space that passes through the origin.
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| A natural description of an 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:
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| :<math>\exist c\in K: \mathbf{v}' = c\mathbf{v}\text{ (or }\mathbf{v} = \frac{1}{c}\mathbf{v}'\text{)}</math>
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| 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.
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| A [[duality (mathematics)|dual]] description is provided with [[linear functional]]s (usually implemented as [[linear equation]]s). One non-[[additive identity|zero]] linear functional '''F''' specifies its [[kernel (linear algebra)|kernel]] subspace '''F''' = 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]]):
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| :<math>\exist c\in K: \mathbf{F}' = c\mathbf{F}\text{ (or }\mathbf{F} = \frac{1}{c}\mathbf{F}'\text{)}</math>
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| 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 liner span.
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| ===Systems of linear equations===
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| The solution set to any homogeneous [[system of linear equations]] with ''n'' variables is a subspace in the [[coordinate space]] ''K''<sup>''n''</sup>:
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| :<math>\left\{ \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right] \in K^n : \begin{alignat}{6}
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| a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = 0& \\
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| a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = 0& \\
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| \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && \vdots\,& \\
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| a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = 0&
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| \end{alignat} \right\}. </math>
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| For example (over real or [[rational number]]s), the set of all vectors (''x'', ''y'', ''z'') satisfying the equations
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| :<math>x + 3y + 2z = 0 \;\;\;\;\text{and}\;\;\;\; 2x - 4y + 5z = 0</math>
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| 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.
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| ===Null space of a matrix===
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| {{main|Null space}}
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| In a finite-dimensional space, a homogeneous system of linear equations can be written as a single [[matrix (mathematics)|matrix]] equation:
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| :<math>A\mathbf{x} = \mathbf{0}.</math>
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| The set of solutions to this equation is known as the [[null space]] of the matrix. For example, the subspace described above is the null space of the matrix
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| :<math>A = \left[ \begin{alignat}{3} 1 && 3 && 2 &\\ 2 && \;\;-4 && \;\;\;\;5 &\end{alignat} \,\right]\text{.}</math>
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| Every subspace of ''K''<sup>''n''</sup> can be described as the null space of some matrix (see [[#Algorithms|algorithms]], below).
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| ===Linear parametric equations===
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| The subset of ''K''<sup>''n''</sup> described by a system of homogeneous linear [[parametric equations]] is a subspace:
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| :<math>\left\{ \left[\!\! \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \!\!\right] \in K^n : \begin{alignat}{7}
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| x_1 &&\; = \;&& a_{11} t_1 &&\; + \;&& a_{12} t_2 &&\; + \cdots + \;&& a_{1m} t_m & \\
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| x_2 &&\; = \;&& a_{21} t_1 &&\; + \;&& a_{22} t_2 &&\; + \cdots + \;&& a_{2m} t_m & \\
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| \vdots \,&& && \vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; & \\
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| x_n &&\; = \;&& a_{n1} t_1 &&\; + \;&& a_{n2} t_2 &&\; + \cdots + \;&& a_{nm} t_m & \\
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| \end{alignat} \text{ for some } t_1,\ldots,t_m\in K \right\}. </math>
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| For example, the set of all vectors (''x'', ''y'', ''z'') parameterized by the equations
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| :<math>x = 2t_1 + 3t_2,\;\;\;\;y = 5t_1 - 4t_2,\;\;\;\;\text{and}\;\;\;\;z = -t_1 + 2t_2</math>
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| 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">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>
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| ===Span of vectors===
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| {{main|Linear span}}
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| In linear algebra, the system of parametric equations can be written as a single vector equation:
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| :<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>
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| The expression on the right is called a [[linear combination]] of the vectors
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| (2, 5, −1) and (3, −4, 2). These two vectors are said to '''span''' the resulting subspace.
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| In general, a '''linear combination''' of vectors '''v'''<sub>1</sub>, '''v'''<sub>2</sub>, ... , '''v'''<sub>''k''</sub> is any vector of the form
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| :<math>t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k.</math>
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| The set of all possible linear combinations is called the '''span''':
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| :<math>\text{Span} \{ \mathbf{v}_1, \ldots, \mathbf{v}_k \}
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| = \left\{ t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k : t_1,\ldots,t_k\in K \right\} .</math>
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| If the vectors '''v'''<sub>1</sub>, ... , '''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>, ... , '''v'''<sub>''k''</sub>.
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| ; Example
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| : The ''xz''-plane in '''R'''<sup>3</sup> can be parameterized by the equations
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| ::<math>x = t_1, \;\;\; y = 0, \;\;\; z = t_2.</math>
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| :As a subspace, the ''xz''-plane is spanned by the vectors (1, 0, 0) and (0, 0, 1). Every vector in the ''xz''-plane can be written as a linear combination of these two:
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| ::<math>(t_1, 0, t_2) = t_1(1,0,0) + t_2(0,0,1)\text{.}\,</math>
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| :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, 0, 0) and then moving some distance in the direction of (0, 0, 1).
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| ===Column space and row space===
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| {{main|Row and column spaces}}
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| A system of linear parametric equations in a finite-dimensional space can also be written as a single matrix equation:
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| :<math>\mathbf{x} = A\mathbf{t}\;\;\;\;\text{where}\;\;\;\;A = \left[ \begin{alignat}{2} 2 && 3 & \\ 5 && \;\;-4 & \\ -1 && 2 & \end{alignat} \,\right]\text{.}</math>
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| 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''.
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| 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).
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| ===Independence, basis, and dimension===
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| {{main|Linear independence|Basis (linear algebra)|Dimension (vector space)}}
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| [[File:Basis for a Plane.PNG|thumb|280px|right|The vectors '''u''' and '''v''' are a basis for this two-dimensional subspace of '''R'''<sup>3</sup>.]]
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| 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, 0, 0), (0, 0, 1), and
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| (2, 0, 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>}}.
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| In general, vectors '''v'''<sub>1</sub>, ... , '''v'''<sub>''k''</sub> are called '''linearly independent''' if
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| :<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>
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| for
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| (''t''<sub>1</sub>, ''t''<sub>2</sub>, ... , ''t<sub>k</sub>'') ≠ (''u''<sub>1</sub>, ''u''<sub>2</sub>, ... , ''u<sub>k</sub>'').<ref>This definition is often stated differently: vectors '''v'''<sub>1</sub>, ..., '''v'''<sub>''k''</sub> are linearly independent if
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| {{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>
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| 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.
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| 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).
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| ; Example
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| : Let ''S'' be the subspace of '''R'''<sup>4</sup> defined by the equations
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| ::<math>x_1 = 2 x_2\;\;\;\;\text{and}\;\;\;\;x_3 = 5x_4.</math>
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| :Then the vectors (2, 1, 0, 0) and (0, 0, 5, 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:
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| ::<math>(2t_1, t_1, 5t_2, t_2) = t_1(2, 1, 0, 0) + t_2(0, 0, 5, 1).\,</math>
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| :The subspace ''S'' is two-dimensional. Geometrically, it is the plane in '''R'''<sup>4</sup> passing through the points (0, 0, 0, 0), (2, 1, 0, 0), and (0, 0, 5, 1).
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| ==Operations and relations on subspaces==
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| === Inclusion ===
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| <!-- some illustration, please -->
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| The [[inclusion relation|set-theoretical inclusion]] binary relation specified a [[partial order]] on the set of all subspaces (of any dimension).
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| A subspace cannot lie in any subspace of lesser dimension. If dim ''U'' = ''k'', a finite number, and ''U'' ⊂ ''W'', then dim ''W'' = ''k'' if and only if ''U'' = ''W''.
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| ===Intersection===
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| [[File:IntersectingPlanes.png|thumb|right|In '''R'''<sup>3</sup>, the intersection of two-dimensional subspaces is one-dimensional]]
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| Given subspaces ''U'' and ''W'' of a vector space ''V'', then their [[intersection (set theory)|intersection]] ''U'' ∩ ''W'' := {'''v''' ∈ ''V'' : '''v''' is an element of both ''U'' and ''W''} is also a subspace of ''V''.
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| ''Proof:''
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| # Let '''v''' and '''w''' be elements of ''U'' ∩ ''W''. Then '''v''' and '''w''' belong to both ''U'' and ''W''. Because ''U'' is a subspace, then '''v''' + '''w''' belongs to ''U''. Similarly, since ''W'' is a subspace, then '''v''' + '''w''' belongs to ''W''. Thus, '''v''' + '''w''' belongs to ''U'' ∩ ''W''.
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| # Let '''v''' belong to ''U'' ∩ ''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 ''W''.
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| # Since ''U'' and ''W'' are vector spaces, then '''0''' belongs to both sets. Thus, '''0''' belongs to ''U'' ∩ ''W''.
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| For every vector space ''V'', the [[zero vector space|set {'''0'''}]] and ''V'' itself are subspaces of ''V''.
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| ===Sum===
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| If ''U'' and ''W'' are subspaces, their '''sum''' is the subspace
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| :<math>U + W = \left\{ \mathbf{u} + \mathbf{w} \colon \mathbf{u}\in U, \mathbf{w}\in W \right\}.</math>
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| For example, the sum of two lines is the plane that contains them both. The dimension of the sum satisfies the inequality
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| :<math>\max(\dim U,\dim W) \leq \dim(U + W) \leq \dim(U) + \dim(W).</math>
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| 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:
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| :<math>\dim(U+W) = \dim(U) + \dim(W) - \dim(U \cap W).</math>
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| === Lattice of subspaces ===
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| Aforementioned two operations make the set of all subspaces a bounded [[distributive 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.
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| === Other ===
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| {{unreferenced section|date=April 2013}}
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| If ''V'' is an [[inner product space]], then the [[orthogonal complement]] ⊥ of any subspace of ''V'' is again a subspace. This operation, understood as [[negation]] (¬), makes the lattice of subspaces a (possibly [[infinite set|infinite]]) [[Boolean algebra (structure)|Boolean algebra]].
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| In a [[pseudo-Euclidean space]] there are orthogonal complements too, but such operation does not form a Boolean algebra (nor a [[Heyting algebra]]) because of [[null vector|null]] subspaces, for which ''N'' ∩ ''N''<sup>⊥</sup> = ''N'' ≠ {0}. The same case presents the <sup>⊥</sup> operation in [[symplectic vector space]]s.
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| ==Algorithms==
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| 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:
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| # The reduced matrix has the same null space as the original.
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| # Row reduction does not change the span of the row vectors, i.e. the reduced matrix has the same row space as the original.
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| # Row reduction does not affect the linear dependence of the column vectors.
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| ===Basis for a row space===
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| :'''Input''' An ''m'' × ''n'' matrix ''A''.
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| :'''Output''' A basis for the [[row space]] of ''A''.
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| :# Use elementary row operations to put ''A'' into row echelon form.
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| :# The nonzero rows of the echelon form are a basis for the row space of ''A''.
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| See the article on [[row space]] for an [[Row space#Basis|example]].
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| 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.
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| ===Subspace membership===
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| :'''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.
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| :'''Output''' Determines whether '''v''' is an element of ''S''
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| :# Create a (''k'' + 1) × ''n'' matrix ''A'' whose rows are the vectors '''b'''<sub>1</sub>, ... , '''b'''<sub>''k''</sub> and '''v'''.
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| :# Use elementary row operations to put ''A'' into row echelon form.
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| :# 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'' }}.
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| ===Basis for a column space===
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| :'''Input''' An ''m'' × ''n'' matrix ''A''
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| :'''Output''' A basis for the [[column space]] of ''A''
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| :# Use elementary row operations to put ''A'' into row echelon form.
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| :# 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.
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| See the article on [[column space]] for an [[Column space#Basis|example]].
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| 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.
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| ===Coordinates for a vector===
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| :'''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''}}
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| :'''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>}}
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| :# Create an [[augmented matrix]] ''A'' whose columns are '''b'''<sub>1</sub>,...,'''b'''<sub>''k''</sub> , with the last column being '''v'''.
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| :# Use elementary row operations to put ''A'' into reduced row echelon form.
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| :# 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.)
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| If the final column of the reduced row echelon form contains a pivot, then the input vector '''v''' does not lie in ''S''.
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| ===Basis for a null space===
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| :'''Input''' An ''m'' × ''n'' matrix ''A''.
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| :'''Output''' A basis for the null space of ''A''
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| :# Use elementary row operations to put ''A'' in reduced row echelon form.
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| :# 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.
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| :# 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''.
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| See the article on [[null space]] for an [[Kernel_(matrix)#Basis|example]].
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| ===Equations for a subspace===
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| :'''Input''' A basis {'''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>} for a subspace ''S'' of ''K''<sup>''n''</sup>
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| :'''Output''' An (''n'' − ''k'') × ''n'' matrix whose null space is ''S''.
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| :# Create a matrix ''A'' whose rows are {{nowrap| '''b'''<sub>1</sub>, '''b'''<sub>2</sub>, ..., '''b'''<sub>''k''</sub>}}.
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| :# Use elementary row operations to put ''A'' into reduced row echelon form.
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| :# 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.
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| :# 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''.
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| ; Example
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| :If the reduced row echelon form of ''A'' is
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| ::<math>\left[ \begin{alignat}{6}
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| 1 && 0 && -3 && 0 && 2 && 0 \\
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| 0 && 1 && 5 && 0 && -1 && 4 \\
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| 0 && 0 && 0 && 1 && 7 && -9 \\
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| 0 && \;\;\;\;\;0 && \;\;\;\;\;0 && \;\;\;\;\;0 && \;\;\;\;\;0 && \;\;\;\;\;0 \end{alignat} \,\right] </math>
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| :then the column vectors {{nowrap| '''c'''<sub>1</sub>, ..., '''c'''<sub>6</sub>}} satisfy the equations
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| ::<math> \begin{alignat}{1}
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| \mathbf{c}_3 &= -3\mathbf{c}_1 + 5\mathbf{c}_2 \\
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| \mathbf{c}_5 &= 2\mathbf{c}_1 - \mathbf{c}_2 + 7\mathbf{c}_3 \\
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| \mathbf{c}_6 &= 4\mathbf{c}_2 - 9\mathbf{c}_3.
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| \end{alignat}</math>
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| :It follows that the row vectors of ''A'' satisfy the equations
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| ::<math> \begin{alignat}{1}
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| x_3 &= -3x_1 + 5x_2 \\
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| x_5 &= 2x_1 - x_2 + 7x_3 \\
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| x_6 &= 4x_2 - 9x_3.
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| \end{alignat}</math>
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| :In particular, the row vectors of ''A'' are a basis for the null space of the corresponding matrix.
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| ==See also==
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| * [[Signal subspace]]
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| * [[Multilinear subspace learning]]
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| * [[Cyclic subspace]]
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| ==Textbooks==
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| * {{Citation
| |
| | last = Axler
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| | first = Sheldon Jay
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| | year = 1997
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| | title = Linear Algebra Done Right
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| | publisher = Springer-Verlag
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| | edition = 2nd
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| | isbn = 0-387-98259-0
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| }}
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| * {{Citation
| |
| | last = Lay
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| | first = David C.
| |
| | date = August 22, 2005
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| | title = Linear Algebra and Its Applications
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| | publisher = Addison Wesley
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| | edition = 3rd
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| | isbn = 978-0-321-28713-7
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| }}
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| * {{Citation
| |
| | last = Meyer
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| | first = Carl D.
| |
| | date = February 15, 2001
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| | title = Matrix Analysis and Applied Linear Algebra
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| | publisher = Society for Industrial and Applied Mathematics (SIAM)
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| | isbn = 978-0-89871-454-8
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| | url = http://www.matrixanalysis.com/DownloadChapters.html
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| }}
| |
| * {{Citation
| |
| | last = Poole
| |
| | first = David
| |
| | year = 2006
| |
| | title = Linear Algebra: A Modern Introduction
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| | publisher = Brooks/Cole
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| | edition = 2nd
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| | isbn = 0-534-99845-3
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| }}
| |
| * {{Citation
| |
| | last = Anton
| |
| | first = Howard
| |
| | year = 2005
| |
| | title = Elementary Linear Algebra (Applications Version)
| |
| | publisher = Wiley International
| |
| | edition = 9th
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| }}
| |
| * {{Citation
| |
| | last = Leon
| |
| | first = Steven J.
| |
| | year = 2006
| |
| | title = Linear Algebra With Applications
| |
| | publisher = Pearson Prentice Hall
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| | edition = 7th
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| }}
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| ==External links==
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| * {{planetmath reference|id=624|title=Vector subspace}}.
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| * {{aut|[[Gilbert Strang]]}}, [http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture10.htm MIT Linear Algebra Lecture on the Four Fundamental Subspaces] at Google Video, from [[MIT OpenCourseWare]]
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| ==References==
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| {{Reflist}}
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| {{DEFAULTSORT:Linear Subspace}}
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| [[Category:Linear algebra]]
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| [[Category:Articles containing proofs]]
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| [[Category:Operator theory]]
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| [[Category:Functional analysis]]
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| [[ru:Векторное подпространство]]
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