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{{Short description|Vector spaces associated to a matrix}}
[[File:Matrix Rows.svg|thumb|right|The row vectors of a [[matrix (mathematics)|matrix]]. The row space of this matrix is the vector space spanned by the row vectors.]]

[[Image:Matrix Columns.svg|thumb|right|The column vectors of a [[matrix (mathematics)|matrix]]. The column space of this matrix is the vector space spanned by the column vectors.]]

In [[linear algebra]], the '''column space''' (also called the '''range''' or [[Image (mathematics)|'''image''']]) of a [[matrix (mathematics)|matrix]] ''A'' is the [[Linear span|span]] (set of all possible [[linear combination]]s) of its [[column vector]]s. The column space of a matrix is the [[image (mathematics)|image]] or [[range of a function|range]] of the corresponding [[matrix transformation]].

Let <math>F</math> be a [[field (mathematics)|field]]. The column space of an {{math|''m'' × ''n''}} matrix with components from <math>F</math> is a [[linear subspace]] of the [[Examples of vector spaces#Coordinate space|''m''-space]] <math>F^m</math>.  The [[dimension (linear algebra)|dimension]] of the column space is called the [[rank (linear algebra)|rank]] of the matrix and is at most {{math|min(''m'', ''n'')}}.<ref name="ReferenceA">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]] <math>R</math> [[#For matrices over a ring|is also possible]].

The '''row space''' is defined similarly.

The row space and the column space of a matrix {{mvar|A}} are sometimes denoted as {{math|'''''C'''''(''A''<sup>T</sup>)}} and {{math|'''''C'''''(''A'')}} respectively.<ref>{{Cite book|last=Strang|first=Gilbert|title=Introduction to linear algebra|publisher=Wellesley-Cambridge Press|year=2016|isbn=978-0-9802327-7-6|edition=Fifth|location=Wellesley, MA|pages=128,168|oclc=956503593}}</ref>

This article considers matrices of [[real number]]s.  The row and column spaces are subspaces of the [[real coordinate space|real spaces]] <math>\R^n</math> and <math>\R^m</math> respectively.<ref>{{harvtxt|Anton|1987|p=179}}</ref>

==Overview==
Let {{mvar|A}} be an {{mvar|m}}-by-{{mvar|n}} matrix. Then
* {{math|1=rank(''A'') = dim(rowsp(''A'')) = dim(colsp(''A''))}},<ref>{{harvtxt|Anton|1987|p=183}}</ref>
* {{math|rank(''A'')}} = number of [[Pivot element|pivots]] in any echelon form of {{mvar|A}},
* {{math|rank(''A'')}} = the maximum number of linearly independent rows or columns of {{mvar|A}}.<ref>{{harvtxt|Beauregard|Fraleigh|1973|p=254}}</ref>

If the matrix represents a [[linear transformation]], the column space of the matrix equals the [[image (mathematics)|image]] of this linear transformation.

The column space of a matrix {{mvar|A}} is the set of all linear combinations of the columns in {{mvar|A}}.  If {{math|1=''A'' = ['''a'''<sub>1</sub> ⋯ '''a'''<sub>''n''</sub>]}}, then {{math|1=colsp(''A'') = span({{mset|'''a'''<sub>1</sub>, ..., '''a'''<sub>''n''</sub>}})}}.

Given a matrix {{mvar|A}}, the action of the matrix {{mvar|A}} on a vector {{math|'''x'''}} returns a linear combination of the columns of {{mvar|A}} with the coordinates of {{math|'''x'''}} as coefficients; that is, the columns of the matrix generate the column space.

===Example===
Given a matrix {{mvar|J}}:
:<math>
  J =
  \begin{bmatrix}
    2 & 4 & 1 & 3 & 2\\
    -1 & -2 & 1 & 0 & 5\\
    1 & 6 & 2 & 2 & 2\\
    3 & 6 & 2 & 5 & 1
  \end{bmatrix}
</math>
the rows are
<math>\mathbf{r}_1 = \begin{bmatrix} 2 & 4 & 1 & 3 & 2 \end{bmatrix}</math>,
<math>\mathbf{r}_2 = \begin{bmatrix} -1 & -2 & 1 & 0 & 5 \end{bmatrix}</math>,
<math>\mathbf{r}_3 = \begin{bmatrix} 1 & 6 & 2 & 2 & 2 \end{bmatrix}</math>,
<math>\mathbf{r}_4 = \begin{bmatrix} 3 & 6 & 2 & 5 & 1 \end{bmatrix}</math>.
Consequently, the row space of {{mvar|J}} is the subspace of <math>\R^5</math> [[linear span|spanned]] by {{math|{{mset| '''r'''<sub>1</sub>, '''r'''<sub>2</sub>, '''r'''<sub>3</sub>, '''r'''<sub>4</sub> }}}}.   
Since these four row vectors are [[Linear independence|linearly independent]], the row space is 4-dimensional.  Moreover, in this case it can be seen that they are all [[orthogonality|orthogonal]] to the vector {{math|1='''n''' = [6, −1, 4, −4, 0]}} ({{math|1='''n'''}} is an element of the [[Kernel (linear algebra)|kernel]] of {{mvar|J}} ), so it can be deduced that the row space consists of all vectors in <math>\R^5</math> that are orthogonal to {{math|'''n'''}}.

==Column space==

===Definition===

Let {{mvar|K}} be a [[field (mathematics)|field]] of [[scalar (mathematics)|scalars]]. Let {{math|A}} be an {{math|''m'' × ''n''}} matrix, with column vectors {{math|'''v'''<sub>1</sub>, '''v'''<sub>2</sub>, ..., '''v'''<sub>''n''</sub>}}.  A [[linear combination]] of these vectors is any vector of the form
:<math>c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_n \mathbf{v}_n,</math>
where {{math|''c''<sub>1</sub>, ''c''<sub>2</sub>, ..., ''c<sub>n</sub>''}} are scalars.  The set of all possible linear combinations of {{math|'''v'''<sub>1</sub>, ..., '''v'''<sub>''n''</sub>}} is called the '''column space''' of {{mvar|A}}.  That is, the column space of {{mvar|A}} is the [[linear span|span]] of the vectors {{math|'''v'''<sub>1</sub>, ..., '''v'''<sub>''n''</sub>}}.

Any linear combination of the column vectors of a matrix {{mvar|A}} can be written as the product of {{mvar|A}} with a column vector:
:<math>\begin{array} {rcl}
A \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} 
& = & \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix}
= \begin{bmatrix} c_1 a_{11} + \cdots + c_{n} a_{1n} \\ \vdots \\ c_{1} a_{m1} + \cdots + c_{n} a_{mn} \end{bmatrix} = c_1 \begin{bmatrix} a_{11} \\ \vdots \\ a_{m1} \end{bmatrix} + \cdots + c_n \begin{bmatrix} a_{1n} \\ \vdots \\ a_{mn} \end{bmatrix} \\
& = & c_1 \mathbf{v}_1 + \cdots + c_n \mathbf{v}_n
\end{array}</math>

Therefore, the column space of {{mvar|A}} consists of all possible products {{math|''A'''''x'''}}, for {{math|'''x''' ∈ ''K''<sup>''n''</sup>}}.  This is the same as the [[image (mathematics)|image]] (or [[range of a function|range]]) of the corresponding [[matrix transformation]].

==== Example ====
If <math>A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 0 \end{bmatrix}</math>, then the column vectors are {{math|1='''v'''<sub>1</sub> = [1, 0, 2]<sup>T</sup>}} and {{math|1='''v'''<sub>2</sub> = [0, 1, 0]<sup>T</sup>}}.
A linear combination of '''v'''<sub>1</sub> and '''v'''<sub>2</sub> is any vector of the form
<math display="block">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>
The set of all such vectors is the column space of {{mvar|A}}. In this case, the column space is precisely the set of vectors {{math|(''x'', ''y'', ''z'') ∈ '''R'''<sup>3</sup>}} satisfying the equation {{math|1=''z'' = 2''x''}} (using [[Cartesian coordinates]], this set is a [[plane (mathematics)|plane]] through the origin in [[three-dimensional space]]).

===Basis===
The columns of {{mvar|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.

For example, consider the matrix
:<math>A = \begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}.</math>
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 {{mvar|A}} to [[reduced row echelon form]]:
:<math>\begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}
\sim \begin{bmatrix} 1 & 3 & 1 & 4 \\ 0 & 1 & 1 & 1 \\ 0 & 2 & 2 & -3 \\ 0 & -1 & -1 & 4 \end{bmatrix}
\sim  \begin{bmatrix} 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & 5 \end{bmatrix}
\sim  \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.</math><ref>This computation uses the [[Gaussian elimination|Gauss–Jordan]] row-reduction algorithm.  Each of the shown steps involves multiple elementary row operations.</ref>
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, {{math|1='''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:
:<math>\begin{bmatrix} 1 \\ 2 \\ 1 \\ 1\end{bmatrix},\;\;
\begin{bmatrix} 3 \\ 7 \\ 5 \\ 2\end{bmatrix},\;\;
\begin{bmatrix} 4 \\ 9 \\ 1 \\ 8\end{bmatrix}.</math>
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]].

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.  Also finding a basis for the column space of {{mvar|A}} is equivalent to finding a basis for the row space of the [[transpose]] matrix&nbsp;{{math|''A''<sup>T</sup>}}.

To find the basis in a practical setting (e.g., for large matrices), the [[singular-value decomposition]] is typically used.

===Dimension===
{{main|Rank (linear algebra)}}
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&nbsp;×&nbsp;4 matrix in the example above has rank three.

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 <math>\R^4 \to \R^4</math> described by the matrix above maps all of <math>\R^4</math> to some three-dimensional [[Euclidean subspace|subspace]].

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 {{mvar|A}} with {{mvar|n}} columns are related by the equation:
:<math>\operatorname{rank}(A) + \operatorname{nullity}(A) = n.\,</math>
This is known as the [[rank–nullity theorem]].

===Relation to the left null space===
The [[left null space]] of {{mvar|A}} is the set of all vectors {{math|'''x'''}} such that {{math|1='''x'''<sup>T</sup>''A'' = '''0'''<sup>T</sup>}}.  It is the same as the [[kernel (matrix)|null space]] of the [[transpose]] of {{mvar|A}}. The product of the matrix {{math|''A''<sup>T</sup>}} and the vector {{math|'''x'''}} can be written in terms of the [[dot product]] of vectors:
:<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>
because [[row vector]]s of {{math|''A''<sup>T</sup>}} are transposes of column vectors {{math|'''v'''<sub>''k''</sub>}} of {{mvar|A}}.  Thus {{math|1=''A''<sup>T</sup>'''x''' = '''0'''}} if and only if {{math|'''x'''}} is [[orthogonal]] (perpendicular) to each of the column vectors of {{mvar|A}}.

It follows that the left null space (the null space of {{math|''A''<sup>T</sup>}}) is the [[orthogonal complement]] to the column space of {{mvar|A}}.

For a matrix {{mvar|A}}, the column space, row space, null space, and left null space are sometimes referred to as the ''four fundamental subspaces''.

===For matrices over a ring===
Similarly the column space (sometimes disambiguated as ''right'' column space) can be defined for matrices over a [[ring (mathematics)|ring]] {{mvar|K}} as
:<math>\sum\limits_{k=1}^n \mathbf{v}_k c_k</math>
for any {{math|''c''<sub>1</sub>, ..., ''c<sub>n</sub>''}}, with replacement of the vector {{mvar|m}}-space with "[[left and right (algebra)|right]] [[free module]]", which changes the order of [[scalar multiplication]] of the vector {{math|'''v'''<sub>''k''</sub>}} to the scalar {{math|''c<sub>k</sub>''}} such that it is written in an unusual order ''vector''–''scalar''.<ref>Important only if {{mvar|K}} is not [[commutative ring|commutative]]. Actually, this form is merely a [[matrix multiplication|product]] {{math|''A'''''c'''}} of the matrix {{mvar|A}} to the column vector {{math|'''c'''}} from {{math|''K''<sup>''n''</sup>}} where the order of factors is ''preserved'', unlike [[#Definition|the formula above]].</ref>

==Row space==

===Definition===
Let {{mvar|K}} be a [[field (mathematics)|field]] of [[scalar (mathematics)|scalars]]. Let {{mvar|A}} be an {{math|''m'' × ''n''}} matrix, with row vectors {{math|'''r'''<sub>1</sub>, '''r'''<sub>2</sub>, ..., '''r'''<sub>''m''</sub>}}.  A [[linear combination]] of these vectors is any vector of the form
:<math>c_1 \mathbf{r}_1 + c_2 \mathbf{r}_2 + \cdots + c_m \mathbf{r}_m,</math>
where {{math|''c''<sub>1</sub>, ''c''<sub>2</sub>, ..., ''c<sub>m</sub>''}} are scalars.  The set of all possible linear combinations of {{math|'''r'''<sub>1</sub>, ..., '''r'''<sub>''m''</sub>}} is called the '''row space''' of {{mvar|A}}.  That is, the row space of {{mvar|A}} is the [[linear span|span]] of the vectors {{math|'''r'''<sub>1</sub>, ..., '''r'''<sub>''m''</sub>}}.

For example, if
:<math>A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \end{bmatrix},</math>
then the row vectors are {{math|1='''r'''<sub>1</sub> = [1, 0, 2]}} and {{math|1='''r'''<sub>2</sub> = [0, 1, 0]}}.  A linear combination of {{math|'''r'''<sub>1</sub>}} and {{math|'''r'''<sub>2</sub>}} is any vector of the form
:<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>
The set of all such vectors is the row space of {{mvar|A}}.  In this case, the row space is precisely the set of vectors {{math|(''x'', ''y'', ''z'') ∈ ''K''<sup>3</sup>}} satisfying the equation {{math|1=''z'' = 2''x''}} (using [[Cartesian coordinates]], this set is a [[plane (mathematics)|plane]] through the origin in [[three-dimensional space]]).

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.

The column space of {{mvar|A}} is equal to the row space of {{math|''A''<sup>T</sup>}}.

===Basis===
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.

For example, consider the matrix
:<math>A = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2\end{bmatrix}.</math>
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 {{mvar|A}} to [[row echelon form]]:

{{math|'''r'''<sub>1</sub>}}, {{math|'''r'''<sub>2</sub>}}, {{math|'''r'''<sub>3</sub>}} represents the rows.
:<math>
\begin{align}
\begin{bmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2\end{bmatrix}
&\xrightarrow{\mathbf{r}_2-2\mathbf{r}_1 \to \mathbf{r}_2}
\begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 1 & 5 & 2\end{bmatrix}
\xrightarrow{\mathbf{r}_3-\,\,\mathbf{r}_1 \to \mathbf{r}_3}
\begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 2 & 0\end{bmatrix} \\
&\xrightarrow{\mathbf{r}_3-2\mathbf{r}_2 \to \mathbf{r}_3}
\begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}
\xrightarrow{\mathbf{r}_1-3\mathbf{r}_2 \to \mathbf{r}_1}
\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}.
\end{align}
</math>
Once the matrix is in echelon form, the nonzero rows are a basis for the row space.  In this case, the basis is {{math|{{mset| [1, 3, 2], [2, 7, 4] }}}}. Another possible basis {{math|{{mset| [1, 0, 2], [0, 1, 0] }}}} comes from a further reduction.<ref name="example">The example is valid over the [[real number]]s, the [[rational number]]s, and other [[number field]]s. It is not necessarily correct over fields and rings with non-zero [[characteristic (algebra)|characteristic]].</ref>

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.

It is sometimes convenient to find a basis for the row space from among the rows of the original matrix instead (for example, this result is useful in giving an elementary proof that the [[Rank (linear algebra)#Alternative definitions|determinantal rank]] of a matrix is equal to its rank). Since row operations can affect linear dependence relations of the row vectors, such a basis is instead found indirectly using the fact that the column space of {{math|''A''<sup>T</sup>}} is equal to the row space of {{mvar|A}}. Using the example matrix {{mvar|A}} above, find {{math|''A''<sup>T</sup>}} and reduce it to row echelon form:

:<math>
A^{\mathrm{T}} = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 7 & 5 \\ 2 & 4 & 2\end{bmatrix} \sim 
\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0\end{bmatrix}. 
</math>

The pivots indicate that the first two columns of {{math|''A''<sup>T</sup>}} form a basis of the column space of {{math|''A''<sup>T</sup>}}. Therefore, the first two rows of {{mvar|A}} (before any row reductions) also form a basis of the row space of {{mvar|A}}.

===Dimension===
{{main|Rank (linear algebra)}}
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, or equivalently the number of pivots.  For example, the 3&nbsp;×&nbsp;3 matrix in the example above has rank two.<ref name="example"/>

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:
:<math>\operatorname{rank}(A) + \operatorname{nullity}(A) = n,</math>
where {{mvar|n}} is the number of columns of the matrix {{mvar|A}}.  The equation above is known as the [[rank–nullity theorem]].

===Relation to the null space===
The [[null space]] of matrix {{mvar|A}} is the set of all vectors {{math|'''x'''}} for which {{math|1=''A'''''x''' = '''0'''}}.  The product of the matrix {{mvar|A}} and the vector {{math|'''x'''}} can be written in terms of the [[dot product]] of vectors:
:<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>
where {{math|'''r'''<sub>1</sub>, ..., '''r'''<sub>''m''</sub>}} are the row vectors of {{mvar|A}}.  Thus {{math|1=''A'''''x''' = '''0'''}} if and only if {{math|'''x'''}} is [[orthogonal]] (perpendicular) to each of the row vectors of {{mvar|A}}.

It follows that the null space of {{mvar|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).

The row space and null space are two of the [[four fundamental subspaces]] associated with a matrix {{mvar|A}} (the other two being the [[column space]] and [[left null space]]).

===Relation to coimage===
If {{mvar|V}} and {{mvar|W}} are [[vector spaces]], then the [[kernel (linear algebra)|kernel]] of a [[linear transformation]] {{math|''T'': ''V'' → ''W''}} is the set of vectors {{math|'''v''' ∈ ''V''}} for which {{math|1=''T''('''v''') = '''0'''}}.  The kernel of a linear transformation is analogous to the null space of a matrix.

If {{mvar|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 {{mvar|T}}.  The transformation {{mvar|T}} is one-to-one on its coimage, and the coimage maps [[isomorphism|isomorphically]] onto the [[image (mathematics)|image]] of {{mvar|T}}.

When {{mvar|V}} is not an inner product space, the coimage of {{mvar|T}} can be defined as the [[quotient space (linear algebra)|quotient space]] {{math|''V'' / ker(''T'')}}.

==See also==
* [[Euclidean subspace]]

==References & Notes==
{{reflist}}

{{see also|Linear algebra#Further reading}}

==Further reading==
* {{Citation
 | last = Anton
 | first = Howard
 | date = 1987
 | title = Elementary Linear Algebra
 | location = New York
 | publisher = [[John Wiley & Sons|Wiley]]
 | edition = 5th
 | isbn = 0-471-84819-0
}}
* {{Citation
 | last = Axler
 | first = Sheldon Jay
 | date = 1997
 | title = Linear Algebra Done Right
 | publisher = Springer-Verlag
 | edition = 2nd
 | isbn = 0-387-98259-0
}}
* {{Citation 
| last1 = Banerjee 
| first1 = Sudipto
| last2 = Roy
| first2 = Anindya 
| date = June 6, 2014 
| title = Linear Algebra and Matrix Analysis for Statistics 
| publisher = CRC Press 
| edition = 1st
| isbn = 978-1-42-009538-8
}}
* {{citation 
| last1 = Beauregard 
| first1 = Raymond A. 
| last2 = Fraleigh 
| first2 = John B. 
| title = A First Course In Linear Algebra: with Optional Introduction to Groups, Rings, and Fields 
| location = Boston 
| publisher = [[Houghton Mifflin Company]] 
| year = 1973 
| isbn = 0-395-14017-X 
| url-access = registration 
| url = https://archive.org/details/firstcourseinlin0000beau 
}}
* {{Citation
 | last = Lay
 | first = David C.
 | date = August 22, 2005
 | title = Linear Algebra and Its Applications
 | publisher = Addison Wesley
 | edition = 3rd
 | isbn = 978-0-321-28713-7
}}
* {{Citation
 | last = Leon
 | first = Steven J.
 | date = 2006
 | title = Linear Algebra With Applications
 | publisher = Pearson Prentice Hall
 | edition = 7th
}}
* {{Citation
 |last        = Meyer
 |first       = Carl D.
 |date        = February 15, 2001
 |title       = Matrix Analysis and Applied Linear Algebra
 |publisher   = Society for Industrial and Applied Mathematics (SIAM)
 |isbn        = 978-0-89871-454-8
 |url         = http://www.matrixanalysis.com/DownloadChapters.html
 |url-status     = dead
 |archive-url  = https://web.archive.org/web/20010301161440/http://matrixanalysis.com/DownloadChapters.html
 |archive-date = March 1, 2001
}}
* {{Citation
 | last = Poole
 | first = David
 | date = 2006
 | title = Linear Algebra: A Modern Introduction
 | publisher = Brooks/Cole
 | edition = 2nd
 | isbn = 0-534-99845-3
}}
* {{Citation 
| last = Strang 
| first = Gilbert 
| date = July 19, 2005 
| title = Linear Algebra and Its Applications 
| publisher = Brooks Cole 
| edition = 4th 
| isbn = 978-0-03-010567-8
}}

==External links==

{{wikibooks|Linear Algebra/Column and Row Spaces}}

*{{MathWorld |title=Row Space |urlname=RowSpace}}
*{{MathWorld |title=Column Space |urlname=ColumnSpace}}
*{{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]]
*[http://www.khanacademy.org/video/column-space-of-a-matrix?playlist=Linear+Algebra Khan Academy video tutorial]
*[http://mfile.akamai.com/7870/rm/mitstorage.download.akamai.com/7870/18/18.06/videolectures/strang-1806-lec06-20sep1999-80k.rm Lecture on column space and nullspace by Gilbert Strang of MIT]
*[http://wps.prenhall.com/am_leon_linearalg_7/53/13727/3514210.cw/content/index.html Row Space and Column Space]

{{linear algebra}}

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