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| [[File:Matrix Rows.svg|thumb|right|The row vectors of a [[matrix (mathematics)|matrix]]]]
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| In [[linear algebra]], the '''row space''' of a [[matrix (mathematics)|matrix]] is the set of all possible [[linear combination]]s of its [[row vector]]s. Let ''K'' be a [[field (mathematics)|field]] (such as [[real number|real]] or [[complex number|complex]] numbers). The row space of an ''m'' × ''n'' matrix with components from ''K'' is a [[linear subspace]] of the [[Examples of vector spaces #Coordinate space|''n''-space]] ''K''<sup>''n''</sup>. The [[dimension (linear algebra)|dimension]] of the row space is called the '''[[rank (linear algebra)|row rank]]''' of the matrix.<ref>Linear algebra, as discussed in this article, is a very well established mathematical discipline for which there are many sources. Almost all of the material in this article can be found in Lay 2005, Meyer 2001, and Strang 2005.</ref>
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| A definition for matrices over a [[ring (mathematics)|ring]] ''K'' (such as [[integer]]s) is also possible.<ref>A definition and certain properties for rings are the same with replacement of the "[[vector space|vector ''n''-space]]" ''K''<sup>''n''</sup> with "left [[free module]]" and "linear subspace" with "[[submodule]]". For non-commutative rings this row space is sometimes disambiguated as ''left'' row space.</ref>
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| ==Definition==
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| Let ''K'' be a [[field (mathematics)|field]] of [[scalar (mathematics)|scalars]]. Let ''A'' be an ''m'' × ''n'' matrix, with row vectors '''r'''<sub>1</sub>, '''r'''<sub>2</sub>, ... , '''r'''<sub>''m''</sub>. A [[linear combination]] of these vectors is any vector of the form
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| :<math>c_1 \mathbf{r}_1 + c_2 \mathbf{r}_2 + \cdots + c_m \mathbf{r}_m,</math>
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| where ''c''<sub>1</sub>, ''c''<sub>2</sub>, ... , ''c<sub>m</sub>'' are scalars. The set of all possible linear combinations of '''r'''<sub>1</sub>, ... , '''r'''<sub>''m''</sub> is called the '''row space''' of ''A''. That is, the row space of ''A'' is the [[linear span|span]] of the vectors '''r'''<sub>1</sub>, ... , '''r'''<sub>''m''</sub>.
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| For example, if
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| :<math>A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \end{bmatrix},</math>
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| then the row vectors are '''r'''<sub>1</sub> = (1, 0, 2) and '''r'''<sub>2</sub> = (0, 1, 0). A linear combination of '''r'''<sub>1</sub> and '''r'''<sub>2</sub> is any vector of the form
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| :<math>c_1 (1,0,2) + c_2 (0,1,0) = (c_1,c_2,2c_1).\,</math>
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| The set of all such vectors is the row space of ''A''. In this case, the row space is precisely the set of vectors (''x'', ''y'', ''z'') ∈ ''K''<sup>3</sup> satisfying the equation ''z'' = 2''x'' (using [[Cartesian coordinates]], this set is a [[plane (mathematics)|plane]] through the origin in [[three-dimensional space]]).
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| For a matrix that represents a homogeneous [[system of linear equations]], the row space consists of all linear equations that follow from those in the system.
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| The column space of ''A'' is equal to the row space of ''A''<sup>T</sup>.
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| ==Basis==
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| The row space is not affected by [[elementary row operations]]. This makes it possible to use [[row reduction]] to find a [[basis (linear algebra)|basis]] for the row space.
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| For example, consider the matrix
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| :<math>A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2\end{bmatrix}.</math>
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| The rows of this matrix span the row space, but they may not be [[linearly independent]], in which case the rows will not be a basis. To find a basis, we reduce ''A'' to [[row echelon form]]:
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| '''r<sub>1</sub>''', '''r<sub>2</sub>''', '''r<sub>3</sub>''' represents the rows.
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| :<math>
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| \begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2\end{bmatrix} \underbrace{\sim}_{r_2-2r_1}
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| \begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 1 & 5 & 2\end{bmatrix} \underbrace{\sim}_{r_3-r_1}
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| \begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 0\end{bmatrix} \underbrace{\sim}_{r_3-2r_2}
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| \begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix} \underbrace{\sim}_{r_1-3r_2}
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| \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}.
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| </math>
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| Once the matrix is in echelon form, the nonzero rows are a basis for the row space. In this case, the basis is { (1, 3, 2), (2, 7, 4) }. Another possible basis { (1, 0, 2), (0, 1, 0) } comes from a further reduction.<ref name="example">The example is valid over real, [[rational number]]s, and other [[number field]]s. It is not necessarily correct over fields and rings with non-zero [[characteristic (algebra)|characteristic]].</ref>
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| This algorithm can be used in general to find a basis for the span of a set of vectors. If the matrix is further simplified to [[reduced row echelon form]], then the resulting basis is uniquely determined by the row space.
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| ==Dimension==
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| {{main|Rank (linear algebra)}}
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| The [[dimension (linear algebra)|dimension]] of the row space is called the '''[[rank (linear algebra)|rank]]''' of the matrix. This is the same as the maximum number of linearly independent rows that can be chosen from the matrix. For example, the 3 × 3 matrix in the example above has rank two.<ref name="example"/>
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| The rank of a matrix is also equal to the dimension of the [[column space]]. The dimension of the [[null space]] is called the '''nullity''' of the matrix, and is related to the rank by the following equation:
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| :<math>\operatorname{rank}(A) + \operatorname{nullity}(A) = n,</math>
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| where ''n'' is the number of columns of the matrix ''A''. The equation above is known as the [[rank-nullity theorem]].
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| ==Relation to the null space==
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| The [[null space]] of matrix ''A'' is the set of all vectors '''x''' for which ''A'''''x''' = '''0'''. The product of the matrix ''A'' and the vector '''x''' can be written in terms of the [[dot product]] of vectors:
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| :<math>A\mathbf{x} = \begin{bmatrix} \mathbf{r}_1 \cdot \mathbf{x} \\ \mathbf{r}_2 \cdot \mathbf{x} \\ \vdots \\ \mathbf{r}_m \cdot \mathbf{x} \end{bmatrix},</math>
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| where '''r'''<sub>1</sub>, ... , '''r'''<sub>''m''</sub> are the row vectors of ''A''. Thus ''A'''''x''' = '''0''' if and only if '''x''' is [[orthogonal]] (perpendicular) to each of the row vectors of ''A''.
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| It follows that the null space of ''A'' is the [[orthogonal complement]] to the row space. For example, if the row space is a plane through the origin in three dimensions, then the null space will be the perpendicular line through the origin. This provides a proof of the [[rank-nullity theorem]] (see [[#Dimension|dimension]] above).
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| The row space and null space are two of the [[four fundamental subspaces]] associated with a matrix ''A'' (the other two being the [[column space]] and [[left null space]]).
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| ==Relation to coimage==
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| If ''V'' and ''W'' are [[vector spaces]], then the [[kernel (linear algebra)|kernel]] of a [[linear transformation]] ''T'': ''V'' → ''W'' is the set of vectors '''v''' ∈ ''V'' for which ''T''('''v''') = '''0'''. The kernel of a linear transformation is analogous to the null space of a matrix.
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| If ''V'' is an [[inner product space]], then the orthogonal complement to the kernel can be thought of as a generalization of the row space. This is sometimes called the [[coimage]] of ''T''. The transformation ''T'' is one-to-one on its coimage, and the coimage maps [[isomorphism|isomorphically]] onto the [[image (mathematics)|image]] of ''T''.
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| When ''V'' is not an inner product space, the coimage of ''T'' can be defined as the [[quotient space (linear algebra)|quotient space]] ''V'' / ker(''T'').
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| ==Notes==
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| {{reflist}}
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| ==References==
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| {{see also|Linear algebra#Further reading}}
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| ===Textbooks===
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| * {{Citation
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| | last = Axler
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| | first = Sheldon Jay
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| | date = 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
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| | last = Lay
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| | first = David C.
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| | 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
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| | last = Meyer
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| | first = Carl D.
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| | 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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| }}
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| * {{Citation
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| | last = Poole
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| | first = David
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| | date = 2006
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| | 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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| }}
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| * {{Citation
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| | last = Anton
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| | first = Howard
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| | date = 2005
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| | title = Elementary Linear Algebra (Applications Version)
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| | publisher = Wiley International
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| | edition = 9th
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| }}
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| * {{Citation
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| | last = Leon
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| | first = Steven J.
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| | date = 2006
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| | title = Linear Algebra With Applications
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| | publisher = Pearson Prentice Hall
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| | edition = 7th
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| }}
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| ==External links==
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| {{wikibooks|Linear Algebra/Column and Row Spaces}}
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| * {{MathWorld |title=Row Space |urlname=RowSpace}}
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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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| {{linear algebra}}
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| [[Category:Linear algebra]]
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| [[Category:Matrices]]
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| [[it:Spazi delle righe e delle colonne]]
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| [[nl:Kolom- en rijruimte]]
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| [[ur:قطار اور ستون فضا]]
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| [[zh:行空间与列空间]]
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