Row and column spaces

The row vectors of a matrix. The row space of this matrix is the vector space generated by linear combinations of the row vectors.

The column vectors of a matrix. The column space of this matrix is the vector space generated by linear combinations of the column vectors.

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 $\mathbb {F}$ be a field. The column space of an m × n matrix with components from $\mathbb {F}$ is a linear subspace of the m-space $\mathbb {F} ^{m}$ . 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 $\mathbb {K}$ is also possible.

The row space is defined similarly.

This article considers matrices of real numbers. The row and column spaces are subspaces of the real spaces Rⁿ and R^m respectively.

Overview

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

rank(A) = dim(rowsp(A)) = dim(colsp(A)),
rank(A) = number of pivots in any echelon form of A,
rank(A) = the maximum number of linearly independent rows or columns of A.

If one considers the matrix as a linear transformation from Rⁿ to 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 = [a₁, ...., a_n], then colsp(A) = span {a₁, ...., a_n}.

The concept of row space generalizes to matrices over 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 = A x must reside in the column space of A. See singular value decomposition for more details on this second interpretation.

Example

Given a matrix J:

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 r₁ = (2,4,1,3,2), r₂ = (−1,−2,1,0,5), r₃ = (1,6,2,2,2), r₄ = (3,6,2,5,1). Consequently, the row space of J is the subspace of R⁵ 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 = (6,−1,4,−4,0), so it can be deduced that the row space consists of all vectors in R⁵ 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

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:

{\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 &\vdots &\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}}

Therefore, the column space of A consists of all possible products Ax, for x ∈ Cⁿ. This is the same as the image (or range) of the corresponding matrix transformation.

Example

A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 0 \end{bmatrix}

, then the column vectors are v₁ = (1, 0, 2)^T and v₂ = (0, 1, 0)^T.

A linear combination of v₁ and v₂ is any vector of the form

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}\,

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) ∈ R³ satisfying the equation z = 2x (using Cartesian coordinates, this set is a plane through the origin in three-dimensional space).

Basis

The columns of A span the column space, but they may not form a 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 for the column space.

For example, consider the matrix

A = \begin{bmatrix} 1 & 3 & 1 & 4 \\ 2 & 7 & 3 & 9 \\ 1 & 5 & 3 & 1 \\ 1 & 2 & 0 & 8 \end{bmatrix}\text{.}

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:

\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}\text{.}

^[2]

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, v₃ = –2v₁ + v₂.) Therefore, the first, second, and fourth columns of the original matrix are a basis for the column space:

\begin{bmatrix} 1 \\ 2 \\ 1 \\ 1\end{bmatrix},\;\; \begin{bmatrix} 3 \\ 7 \\ 5 \\ 2\end{bmatrix},\;\; \begin{bmatrix} 4 \\ 9 \\ 1 \\ 8\end{bmatrix}\text{.}

Note that the independent columns of the reduced row echelon form are precisely the columns with pivots. This makes it possible to determine which columns are linearly independent by reducing only to 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. 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^T.

Dimension

The dimension of the column space is called the 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.

Because the column space is the image of the corresponding matrix transformation, the rank of a matrix is the same as the dimension of the image. For example, the transformation R⁴ → R⁴ described by the matrix above maps all of R⁴ to some three-dimensional subspace.

The nullity of a matrix is the dimension of the null space, and is equal to the number of columns in the reduced row echelon form that do not have pivots.^[3] The rank and nullity of a matrix A with n columns are related by the equation:

\text{rank}(A) + \text{nullity}(A) = n.\,

This is known as the rank–nullity theorem.

Relation to the left null space

The left null space of A is the set of all vectors x such that x^TA = 0^T. It is the same as the null space of the transpose of A. The product of the matrix A^T and the vector x can be written in terms of the dot product of vectors:

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},

because row vectors of A^T are transposes of column vectors v_k of A. Thus A^Tx = 0 if and only if x is orthogonal (perpendicular) to each of the column vectors of A.

It follows that the left null space (the null space of A^T) is the orthogonal complement to the column space of A.

For a matrix 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 K as

\sum\limits_{k=1}^n \mathbf{v}_k c_k

for any c₁, ..., c_n, with replacement of the vector m-space with "right free module", which changes the order of scalar multiplication of the vector v_k to the scalar c_k such that it is written in an unusual order vector–scalar.^[4]