In linear algebra, the column space (also called the range or image) of a matrix A is the span (set of all possible linear combinations) of its column vectors. The column space of a matrix is the image or range of the corresponding matrix transformation.

Let ${\displaystyle F}$ be a field. The column space of an m × n matrix with components from ${\displaystyle F}$ is a linear subspace of the m-space ${\displaystyle F^{m}}$. The dimension of the column space is called the rank of the matrix and is at most min(m, n).^[1] A definition for matrices over a ring ${\displaystyle R}$ is also possible.

The row space is defined similarly.

The row space and the column space of a matrix A are sometimes denoted as C(A^T) and C(A) respectively.^[2]

This article considers matrices of real numbers. The row and column spaces are subspaces of the real spaces ${\displaystyle \mathbb {R} ^{n}}$ and ${\displaystyle \mathbb {R} ^{m}}$ respectively.^[3]

Overview

Let A be an m-by-n matrix. Then

1. rank(A) = dim(rowsp(A)) = dim(colsp(A)),^[4]
2. rank(A) = number of pivots in any echelon form of A,
3. rank(A) = the maximum number of linearly independent rows or columns of A.^[5]

If one considers the matrix as a linear transformation from ${\displaystyle \mathbb {R} ^{n}}$ to ${\displaystyle \mathbb {R} ^{m}}$, then the column space of the matrix equals the image of this linear transformation.

The column space of a matrix A is the set of all linear combinations of the columns in A. If A = , then colsp(A) = span({a₁, ..., a_n}).

The concept of row space generalizes to matrices over ${\displaystyle \mathbb {C} }$, the field of complex numbers, or over any field.

Intuitively, given a matrix A, the action of the matrix A on a vector x will return a linear combination of the columns of A weighted by the coordinates of x as coefficients. Another way to look at this is that it will (1) first project x into the row space of A, (2) perform an invertible transformation, and (3) place the resulting vector y in the column space of A. Thus the result y = Ax must reside in the column space of A. See singular value decomposition for more details on this second interpretation.^{[clarification needed]}

Example

Given a matrix J:

${\displaystyle J={\begin{bmatrix}2&4&1&3&2\\-1&-2&1&0&5\\1&6&2&2&2\\3&6&2&5&1\end{bmatrix}}}$

the rows are ${\displaystyle \mathbf {r} _{1}={\begin{bmatrix}2&4&1&3&2\end{bmatrix}}}$, ${\displaystyle \mathbf {r} _{2}={\begin{bmatrix}-1&-2&1&0&5\end{bmatrix}}}$, ${\displaystyle \mathbf {r} _{3}={\begin{bmatrix}1&6&2&2&2\end{bmatrix}}}$, ${\displaystyle \mathbf {r} _{4}={\begin{bmatrix}3&6&2&5&1\end{bmatrix}}}$. Consequently, the row space of J is the subspace of ${\displaystyle \mathbb {R} ^{5}}$ spanned by { r₁, r₂, r₃, r₄ }. Since these four row vectors are linearly independent, the row space is 4-dimensional. Moreover, in this case it can be seen that they are all orthogonal to the vector n = , so it can be deduced that the row space consists of all vectors in ${\displaystyle \mathbb {R} ^{5}}$ that are orthogonal to n.

Column space

Definition

Let K be a field of scalars. Let A be an m × n matrix, with column vectors v₁, v₂, ..., v_n. A linear combination of these vectors is any vector of the form

${\displaystyle c_{1}\mathbf {v} _{1}+c_{2}\mathbf {v} _{2}+\cdots +c_{n}\mathbf {v} _{n},}$

where c₁, c₂, ..., c_n are scalars. The set of all possible linear combinations of v₁, ..., v_n is called the column space of A. That is, the column space of A is the span of the vectors v₁, ..., v_n.

Any linear combination of the column vectors of a matrix A can be written as the product of A with a column vector: