Editing Row and column spaces (section)

==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'')}}.