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| [[Image:Matrix Columns.svg|thumb|right|The column vectors of a [[matrix (mathematics)|matrix]]]]
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| In [[linear algebra]], the '''column space''' C(''A'') of a [[matrix (mathematics)|matrix]] ''A'' (sometimes called the '''range''' of a matrix) is the set of all possible [[linear combination]]s of its [[column vector]]s.
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| Let ''K'' be a [[field (mathematics)|field]] (such as [[real number|real]] or [[complex number|complex]] numbers). The column space of an ''m'' × ''n'' matrix with components from ''K'' is a [[linear subspace]] of the [[Examples of vector spaces #Coordinate space|''m''-space]] ''K''<sup>''m''</sup>. The [[dimension (linear algebra)|dimension]] of the column space is called the [[rank (linear algebra)|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> A definition for matrices over a [[ring (mathematics)|ring]] ''K'' (such as [[integer]]s) [[#For matrices over a ring|is also possible]].
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| The column space of a matrix is the [[image (mathematics)|image]] or [[range (mathematics)|range]] of the corresponding [[matrix transformation]].
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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 column vectors '''v'''<sub>1</sub>, '''v'''<sub>2</sub>, ..., '''v'''<sub>''n''</sub>. A [[linear combination]] of these vectors is any vector of the form
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| :<math>c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n,</math>
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| where ''c''<sub>1</sub>, ''c''<sub>2</sub>, ..., ''c<sub>n</sub>'' are scalars. The set of all possible linear combinations of '''v'''<sub>1</sub>, ... ,'''v'''<sub>''n''</sub> is called the '''column space''' of ''A''. That is, the column space of ''A'' is the [[linear span|span]] of the vectors '''v'''<sub>1</sub>, ... , '''v'''<sub>''n''</sub>.
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| Any linear combination of the column vectors of a matrix ''A'' can be written as the product of ''A'' with a column vector:
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| :<math>\begin{array} {rcl}
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| A \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}
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| & = & \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}
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| = \begin{bmatrix} c_1 a_{11} + & \cdots & + c_{n} a_{1n} \\ \vdots & \vdots & \vdots \\ c_{1} a_{n1} + & \cdots & + c_{n} a_{nn} \end{bmatrix} = c_1 \begin{bmatrix} a_{11} \\ \vdots \\ a_{n1} \end{bmatrix} + \cdots + c_n \begin{bmatrix} a_{1n} \\ \vdots \\ a_{nn} \end{bmatrix} \\
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| & = & c_1 \mathbf{v}_1 + \cdots + c_n \mathbf{v}_n
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| \end{array}</math>
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| Therefore, the column space of ''A'' consists of all possible products ''A'''''x''', for '''x''' ∈ '''C'''<sup>''n''</sup>. This is the same as the [[image (mathematics)|image]] (or [[range (mathematics)|range]]) of the corresponding [[matrix transformation]].
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| ;Example
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| :If <math>A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 0 \end{bmatrix}</math>, then the column vectors are '''v'''<sub>1</sub> = (1, 0, 2)<sup>T</sup> and '''v'''<sub>2</sub> = (0, 1, 0)<sup>T</sup>.
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| :A linear combination of '''v'''<sub>1</sub> and '''v'''<sub>2</sub> is any vector of the form
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| ::<math>c_1 \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ 2c_1 \end{bmatrix}\,</math>
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| :The set of all such vectors is the column space of ''A''. In this case, the column space is precisely the set of vectors (''x'', ''y'', ''z'') ∈ '''R'''<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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| ==Basis==
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| The columns of ''A'' span the column space, but they may not form a [[basis (linear algebra)|basis]] if the column vectors are not [[linearly independent]]. Fortunately, [[elementary row operations]] do not affect the dependence relations between the column vectors. This makes it possible to use [[row reduction]] to find a [[basis (linear algebra)|basis]] for the column space.
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| For example, consider the matrix
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| :<math>A = \begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}\text{.}</math>
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| The columns of this matrix span the column space, but they may not be [[linearly independent]], in which case some subset of them will form a basis. To find this basis, we reduce ''A'' to [[reduced row echelon form]]:
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| :<math>\begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}
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| \sim \begin{bmatrix} 1 & 3 & 1 & 4 \\ 0 & 1 & 1 & 1 \\ 0 & 2 & 2 & -3 \\ 0 & -1 & -1 & 4 \end{bmatrix}
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| \sim \begin{bmatrix} 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & 5 \end{bmatrix}
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| \sim \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}\text{.}</math><ref>This computation uses the [[Gaussian elimination|Gauss–Jordan]] row-reduction algorithm. Each of the shown steps involves multiple elementary row operations.</ref>
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| At this point, it is clear that the first, second, and fourth columns are linearly independent, while the third column is a linear combination of the first two. (Specifically, '''v'''<sub>3</sub> = –2'''v'''<sub>1</sub> + '''v'''<sub>2</sub>.) Therefore, the first, second, and fourth columns of the original matrix are a basis for the column space:
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| :<math>\begin{bmatrix} 1 \\ 2 \\ 1 \\ 1\end{bmatrix},\;\;
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| \begin{bmatrix} 3 \\ 7 \\ 5 \\ 2\end{bmatrix},\;\;
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| \begin{bmatrix} 4 \\ 9 \\ 1 \\ 8\end{bmatrix}\text{.}</math>
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| Note that the independent columns of the reduced row echelon form are precisely the columns with [[Pivot_element|pivots]]. This makes it possible to determine which columns are linearly independent by reducing only to [[row echelon form|echelon form]].
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| The above algorithm can be used in general to find the dependence relations between any set of vectors, and to pick out a basis from any spanning set. A different algorithm for finding a basis from a spanning set is given in the [[row space]] article; finding a basis for the column space of ''A'' is equivalent to finding a basis for the row space of the [[transpose]] matrix ''A''<sup>T</sup>.
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| ==Dimension==
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| {{main|Rank (linear algebra)}}
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| The [[dimension (linear algebra)|dimension]] of the column space is called the '''[[rank (linear algebra)|rank]]''' of the matrix. The rank is equal to the number of pivots in the [[reduced row echelon form]], and is the maximum number of linearly independent columns that can be chosen from the matrix. For example, the 4 × 4 matrix in the example above has rank three.
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| Because the column space is the [[image (mathematics)|image]] of the corresponding [[matrix transformation]], the rank of a matrix is the same as the dimension of the image. For example, the transformation '''R'''<sup>4</sup> → '''R'''<sup>4</sup> described by the matrix above maps all of '''R'''<sup>4</sup> to some four-dimensional [[Euclidean subspace|subspace]].
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| The '''nullity''' of a matrix is the dimension of the [[kernel (matrix)|null space]], and is equal to the number of columns in the reduced row echelon form that do not have pivots.<ref>Columns without pivots represent free variables in the associated homogeneous [[system of linear equations]].</ref> The rank and nullity of a matrix ''A'' with ''n'' columns are related by the equation:
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| :<math>\text{rank}(A) + \text{nullity}(A) = n.\,</math>
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| This is known as the [[rank-nullity theorem]].
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| ==Relation to the left null space==
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| The [[left null space]] of ''A'' is the set of all vectors '''x''' such that '''x'''<sup>T</sup>''A'' = '''0'''<sup>T</sup>. It is the same as the [[kernel (matrix)|null space]] of the [[transpose]] of ''A''. The product of the matrix ''A''<sup>T</sup> and the vector '''x''' can be written in terms of the [[dot product]] of vectors:
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| :<math>A^\mathsf{T}\mathbf{x} = \begin{bmatrix} \mathbf{v}_1 \cdot \mathbf{x} \\ \mathbf{v}_2 \cdot \mathbf{x} \\ \vdots \\ \mathbf{v}_n \cdot \mathbf{x} \end{bmatrix},</math>
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| because [[row vector]]s of ''A''<sup>T</sup> are transposes of column vectors '''v'''<sub>''k''</sub> of ''A''. Thus ''A''<sup>T</sup>'''x''' = '''0''' if and only if '''x''' is [[orthogonal]] (perpendicular) to each of the column vectors of ''A''.
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| It follows that the left null space (the null space of ''A''<sup>T</sup>) is the [[orthogonal complement]] to the column space of A.
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| For a matrix ''A'', the column space, row space, null space, and left null space are sometimes referred to as the [[four fundamental subspaces]].
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| ==For matrices over a ring==
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| Similarly the column space (sometimes disambiguated as ''right'' column space) can be defined for matrices over a [[ring (mathematics)|ring]] ''K'' as
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| :<math>\sum\limits_{k=1}^n \mathbf{v}_k c_k</math>
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| for any ''c''<sub>1</sub>, ..., ''c<sub>n</sub>'', with replacement of the vector ''m''-space with "[[left and right (algebra)|right]] [[free module]]", which changes the order of [[scalar multiplication]] of the vector '''v'''<sub>''k''</sub> to the scalar ''c<sub>k</sub>'' such that it is written in an unusual order ''vector''–''scalar''.<ref>Important only if ''K'' is not [[commutative ring|commutative]]. Actually, this form is merely a [[matrix multiplication|product]] ''A'''''c''' of the matrix ''A'' to the column vector '''c''' from ''K''<sup>''n''</sup> where the order of factors is ''preserved'', unlike [[#Definition|the formula above]].</ref>
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| ==See also==
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| * [[Euclidean subspace]]
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| * [[Kernel (matrix)]]
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| * [[Row space]]
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| * [[Four fundamental subspaces]]
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| * [[Rank (linear algebra)]]
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| * [[Linear span]]
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| * [[Matrix (mathematics)]]
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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 = Strang
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| | first = Gilbert
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| | date = July 19, 2005
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| | title = Linear Algebra and Its Applications
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| | publisher = Brooks Cole
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| | edition = 4th
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| | isbn = 978-0-03-010567-8
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| }}
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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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| * [http://www.khanacademy.org/video/column-space-of-a-matrix?playlist=Linear+Algebra Khan Academy video tutorial]
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| * {{MathWorld |title=Column Space |urlname=ColumnSpace}}
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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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