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3.6 Row Space and Column Space

If A is an $m \times n$ $m \times n$ matrix, each row of A is an n-tuple of real numbers and hence can be considered as a vector in $ℝ^{1 \times n}$ $ℝ^{1 \times n}$ . The m vectors corresponding to the rows of A will be referred to as the row vectors of A. Similarly, each column of A can be considered as a vector in $ℝ^{m}$ $ℝ^{m}$ , and we can associate n column vectors with the matrix A.

Example 1

Let

A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}]

$A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}]$

The row space of A is the set of all 3-tuples of the form

α (1, 0, 0) + β (0, 1, 0) = (α, β, 0)

$α (1, 0, 0) + β (0, 1, 0) = (α, β, 0)$

The column space of A is the set of all vectors of the form

α [\begin{matrix} 1 \\ 0 \end{matrix}] + β [\begin{matrix} 0 \\ 1 \end{matrix}] + γ [\begin{matrix} 0 \\ 0 \end{matrix}] = [\begin{matrix} α \\ β \end{matrix}]

$α [\begin{matrix} 1 \\ 0 \end{matrix}] + β [\begin{matrix} 0 \\ 1 \end{matrix}] + γ [\begin{matrix} 0 \\ 0 \end{matrix}] = [\begin{matrix} α \\ β \end{matrix}]$

Thus, the row space of A is a two-dimensional subspace of $ℝ^{1 \times 1}$ $ℝ^{1 \times 1}$ , and the column space of A is $ℝ^{2}$ $ℝ^{2}$ .

Theorem 3.6.1

Two row equivalent matrices have the same row space.

Proof

If B is row equivalent to A, then B can be formed from A by a finite sequence of row operations. Thus, the row vectors of B must be linear combinations of the row vectors of A. Consequently, the row space of B must be a subspace of the row space of A. Since A is row equivalent to B, by the same reasoning, the row space of A is a subspace of the row space of B.

∎

To determine the rank of a matrix, we can reduce the matrix to row echelon form. The nonzero rows of the row echelon matrix will form a basis for the row space.

Example 2

Let

A = [\begin{matrix} 1 & - 2 & 3 \\ 2 & - 5 & 1 \\ 1 & - 4 & - 7 \end{matrix}]

$A = [\begin{matrix} 1 & - 2 & 3 \\ 2 & - 5 & 1 \\ 1 & - 4 & - 7 \end{matrix}]$

Reducing A to row echelon form, we obtain the matrix

U = [\begin{matrix} 1 & - 2 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 0 \end{matrix}]

$U = [\begin{matrix} 1 & - 2 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 0 \end{matrix}]$

Clearly, $(1, - 2, 3)$ $(1, - 2, 3)$ and $(0, 1, 5)$ $(0, 1, 5)$ form a basis for the row space of U. Since U and A are row equivalent, they have the same row space, and hence the rank of A is 2.

Linear Systems

The concepts of row space and column space are useful in the study of linear systems. A system $A x = b$ $A x = b$ can be written in the form

x_{1} [\begin{matrix} a_{11} \\ a_{21} \\ ⋮ \\ a_{m 1} \end{matrix}] + x_{2} [\begin{matrix} a_{12} \\ a_{22} \\ ⋮ \\ a_{m 2} \end{matrix}] + \dots + x_{n} [\begin{matrix} a_{1 n} \\ a_{2 n} \\ ⋮ \\ a_{m n} \end{matrix}] = [\begin{matrix} b_{1} \\ b_{2} \\ ⋮ \\ b_{m} \end{matrix}]

$x_{1} [\begin{matrix} a_{11} \\ a_{21} \\ ⋮ \\ a_{m 1} \end{matrix}] + x_{2} [\begin{matrix} a_{12} \\ a_{22} \\ ⋮ \\ a_{m 2} \end{matrix}] + \dots + x_{n} [\begin{matrix} a_{1 n} \\ a_{2 n} \\ ⋮ \\ a_{m n} \end{matrix}] = [\begin{matrix} b_{1} \\ b_{2} \\ ⋮ \\ b_{m} \end{matrix}]$ (1)

In Chapter 1 we used this representation to characterize when a linear system will be consistent. The result, Theorem 1.3.1, can now be restated in terms of the column space of the matrix.

Theorem 3.6.2 Consistency Theorem for Linear Systems

A linear system $A x = b$ $A x = b$ is consistent if and only if b is in the column space of A.

If b is replaced by the zero vector, then (1) becomes

x_{1} a_{1} + x_{2} a_{2} + \dots + x_{n} a_{n} = 0

$x_{1} a_{1} + x_{2} a_{2} + \dots + x_{n} a_{n} = 0$ (2)

It follows from (2) that the system $A x = 0$ $A x = 0$ will have only the trivial solution $x = 0$ $x = 0$ if and only if the column vectors of A are linearly independent.

Theorem 3.6.3

Let A be an m×n matrix. The linear system $A x = b$ $A x = b$ is consistent for every $b \in ℝ^{m}$ $b \in ℝ^{m}$ if and only if the column vectors of A span $ℝ^{m}$ $ℝ^{m}$ . The system $A x = b$ $A x = b$ has at most one solution for every $b \in ℝ^{m}$ $b \in ℝ^{m}$ if and only if the column vectors of A are linearly independent.

Proof

We have seen that the system $A x = b$ $A x = b$ is consistent if and only if b is in the column space of A. It follows that $A x = b$ $A x = b$ will be consistent for every $b \in ℝ^{m}$ $b \in ℝ^{m}$ if and only if the column vectors of A span $ℝ^{m}$ $ℝ^{m}$ . To prove the second statement, note that, if $A x = b$ $A x = b$ has at most one solution for every b, then in particular the system $A x = 0$ $A x = 0$ can have only the trivial solution, and hence the column vectors of A must be linearly independent. Conversely, if the column vectors of A are linearly independent, $A x = 0$ $A x = 0$ has only the trivial solution. Now, if $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ were both solutions of $A x = b$ $A x = b$ , then $x_{1}$ $x_{1}$ − $x_{2}$ $x_{2}$ would be a solution of $A x = 0$ $A x = 0$ :

A (x_{1} - x_{2}) = A x_{1} - A x_{2} = b - b = 0

$A (x_{1} - x_{2}) = A x_{1} - A x_{2} = b - b = 0$

It follows that $x_{1} - x_{2} = 0$ $x_{1} - x_{2} = 0$ , and hence $x_{1}$ $x_{1}$ must equal $x_{2}$ $x_{2}$ .

∎

Let A be an $m \times n$ $m \times n$ matrix. If the column vectors of A span $ℝ^{m}$ $ℝ^{m}$ , then n must be greater than or equal to m, since no set of fewer than m vectors could span $ℝ^{m}$ $ℝ^{m}$ . If the columns of A are linearly independent, then n must be less than or equal to m, since every set of more than m vectors in $ℝ^{m}$ $ℝ^{m}$ is linearly dependent. Thus, if the column vectors of A form a basis for $ℝ^{m}$ $ℝ^{m}$ , then n must equal m.

Corollary 3.6.4

An $n \times n$ $n \times n$ matrix A is nonsingular if and only if the column vectors of A form a basis for $ℝ^{n}$ $ℝ^{n}$ .

In general, the rank and the dimension of the null space always add up to the number of columns of the matrix. The dimension of the null space of a matrix is called the nullity of the matrix.

Theorem 3.6.5 The Rank—Nullity Theorem

If A is an $m \times n$ $m \times n$ matrix, then the rank of A plus the nullity of A equals n.

Proof

Let U be the reduced row echelon form of A. The system $A x = 0$ $A x = 0$ is equivalent to the system $U x = 0$ $U x = 0$ . If A has rank r, then U will have r nonzero rows, and consequently, the system $U x = 0$ $U x = 0$ will involve r lead variables and $n - r$ $n - r$ free variables. The dimension of $N (A)$ $N (A)$ will equal the number of free variables.

∎

Example 3

Let

A = [\begin{matrix} 1 & 2 & - 1 & 1 \\ 2 & 4 & - 3 & 0 \\ 1 & 2 & 1 & 5 \end{matrix}]

$A = [\begin{matrix} 1 & 2 & - 1 & 1 \\ 2 & 4 & - 3 & 0 \\ 1 & 2 & 1 & 5 \end{matrix}]$

Find a basis for the row space of A and a basis for $N (A)$ $N (A)$ . Verify that dim $N (A) = n - r$ $N (A) = n - r$ .

SOLUTION

The reduced row echelon form of A is given by

U = [\begin{matrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{matrix}]

$U = [\begin{matrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{matrix}]$

Thus, ${(1, 2, 0, 3), (0, 0, 1, 2)}$ ${(1, 2, 0, 3), (0, 0, 1, 2)}$ is a basis for the row space of A, and A has rank 2. Since the systems $A x = 0$ $A x = 0$ and $U x = 0$ $U x = 0$ are equivalent, it follows that x is in $N (A)$ $N (A)$ if and only if

\begin{array}{r} x_{1} + 2 x_{2} + 3 x_{4} = 0 \\ x_{3} + 2 x_{4} = 0 \end{array}

$\begin{array}{r} x_{1} + 2 x_{2} + 3 x_{4} = 0 \\ x_{3} + 2 x_{4} = 0 \end{array}$

The lead variables $x_{1}$ $x_{1}$ and $x_{3}$ $x_{3}$ can be solved for in terms of the free variables $x_{2}$ $x_{2}$ and x₄:

\begin{array}{l} x_{1} = - 2 x_{2} - {3 x}_{4} \\ x_{3} = - 2 x_{4} \end{array}

$\begin{array}{l} x_{1} = - 2 x_{2} - {3 x}_{4} \\ x_{3} = - 2 x_{4} \end{array}$

Let $x_{2} = α$ $x_{2} = α$ and $x_{4} = β$ $x_{4} = β$ . It follows that $N (A)$ $N (A)$ consists of all vectors of the form

[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} - 2 α - 3 β \\ α \\ - 2 β \\ β \end{matrix}] = α [\begin{array}{r} - 2 \\ 1 \\ 0 \\ 0 \end{array}] + β [\begin{array}{r} - 3 \\ 0 \\ - 2 \\ 1 \end{array}]

$[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix} - 2 α - 3 β \\ α \\ - 2 β \\ β \end{matrix}] = α [\begin{array}{r} - 2 \\ 1 \\ 0 \\ 0 \end{array}] + β [\begin{array}{r} - 3 \\ 0 \\ - 2 \\ 1 \end{array}]$

The vectors $(- 2, 1, 0, 0)^{T}$ $(- 2, 1, 0, 0)^{T}$ and $(- 3, 0, - 2, 1)^{T}$ $(- 3, 0, - 2, 1)^{T}$ form a basis for $N (A)$ $N (A)$ . Note that

n - r = 4 - 2 = 2 = dim N (A)

$n - r = 4 - 2 = 2 = dim N (A)$

The Column Space

The matrices A and U in Example 3 have different column spaces; however, their column vectors satisfy the same dependency relations. For the matrix U, the column vectors $u_{1}$ $u_{1}$ and $u_{3}$ $u_{3}$ are linearly independent, while

\begin{array}{l} u_{2} = 2 u_{1} \\ u_{4} = 3 u_{1} + 2 u_{3} \end{array}

$\begin{array}{l} u_{2} = 2 u_{1} \\ u_{4} = 3 u_{1} + 2 u_{3} \end{array}$

The same relations hold for the columns of A: The vectors $a_{1}$ $a_{1}$ and $a_{3}$ $a_{3}$ are linearly independent, while

\begin{array}{l} a_{2} = 2 a_{1} \\ a_{4} = 3 a_{1} + 2 a_{3} \end{array}

$\begin{array}{l} a_{2} = 2 a_{1} \\ a_{4} = 3 a_{1} + 2 a_{3} \end{array}$

In general, if A is an $m \times n$ $m \times n$ matrix and U is the row echelon form of A, then, since $A x = 0$ $A x = 0$ if and only if $U x = 0$ $U x = 0$ , their column vectors satisfy the same dependency relations. We will use this property to prove that the dimension of the column space of A is equal to the dimension of the row space of A.

Theorem 3.6.6

If A is an $m \times n$ $m \times n$ matrix, the dimension of the row space of A equals the dimension of the column space of A.

Proof

If A is an $m \times n$ $m \times n$ matrix of rank r, the row echelon form U of A will have r leading 1’s. The columns of U corresponding to the leading 1’s will be linearly independent. They do not, however, form a basis for the column space of A, since, in general, A and U will have different column spaces. Let $U_{L}$ $U_{L}$ denote the matrix obtained from U by deleting all the columns corresponding to the free variables. Delete the same columns from A and denote the new matrix by $A_{L}$ $A_{L}$ . The matrices $A_{L}$ $A_{L}$ and $U_{L}$ $U_{L}$ are row equivalent. Thus, if x is a solution of $A_{L} x = 0$ $A_{L} x = 0$ , then x must also be a solution of $U_{L} x = 0$ $U_{L} x = 0$ . Since the columns of $U_{L}$ $U_{L}$ are linearly independent, x must equal 0. It follows from the remarks preceding Theorem 3.6.3 that the columns of $A_{L}$ $A_{L}$ are linearly independent. Since $A_{L}$ $A_{L}$ has r columns, the dimension of the column space of A is at least r.

We have proved that, for any matrix, the dimension of the column space is greater than or equal to the dimension of the row space. Applying this result to the matrix $A^{T}$ $A^{T}$ , we see that

\begin{matrix} dim (row space of A) & = & dim (column space of A^{T}) \\ \geq & dim (row space of A^{T}) \\ = & dim (column space of A) \end{matrix}

$\begin{matrix} dim (row space of A) & = & dim (column space of A^{T}) \\ \geq & dim (row space of A^{T}) \\ = & dim (column space of A) \end{matrix}$

Thus, for any matrix A, the dimension of the row space must equal the dimension of the column space.

∎

We can use the row echelon form U of A to find a basis for the column space of A. We need only determine the columns of U that correspond to the leading 1’s. These same columns of A will be linearly independent and form a basis for the column space of A.

Note

The row echelon form U tells us only which columns of A to use to form a basis. We cannot use the column vectors from U, since, in general, U and A have different column spaces.

Example 4

Let

A = [\begin{matrix} 1 & - 2 & 1 & 1 & 2 \\ - 1 & 3 & 0 & 2 & - 2 \\ 0 & 1 & 1 & 3 & 4 \\ 1 & 2 & 5 & 13 & 5 \end{matrix}]

$A = [\begin{matrix} 1 & - 2 & 1 & 1 & 2 \\ - 1 & 3 & 0 & 2 & - 2 \\ 0 & 1 & 1 & 3 & 4 \\ 1 & 2 & 5 & 13 & 5 \end{matrix}]$

The row echelon form of A is given by

U = [\begin{matrix} 1 & - 2 & 1 & 1 & 2 \\ 0 & 1 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]

$U = [\begin{matrix} 1 & - 2 & 1 & 1 & 2 \\ 0 & 1 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]$

The leading 1’s occur in the first, second, and fifth columns. Thus,

\begin{matrix} a_{1} = [\begin{array}{r} 1 \\ - 1 \\ 0 \\ 1 \end{array}], & a_{2} = [\begin{array}{r} - 2 \\ 3 \\ 1 \\ 2 \end{array}], & a_{5} = [\begin{array}{r} 2 \\ - 2 \\ 4 \\ 5 \end{array}] \end{matrix}

$\begin{matrix} a_{1} = [\begin{array}{r} 1 \\ - 1 \\ 0 \\ 1 \end{array}], & a_{2} = [\begin{array}{r} - 2 \\ 3 \\ 1 \\ 2 \end{array}], & a_{5} = [\begin{array}{r} 2 \\ - 2 \\ 4 \\ 5 \end{array}] \end{matrix}$

form a basis for the column space of A.

Example 5

Find the dimension of the subspace of $ℝ^{4}$ $ℝ^{4}$ spanned by

\begin{matrix} \begin{matrix} x_{1} = [\begin{array}{r} 1 \\ 2 \\ - 1 \\ 0 \end{array}], & x_{2} = [\begin{array}{r} 2 \\ 5 \\ - 3 \\ 2 \end{array}], & x_{3} = [\begin{array}{r} 2 \\ 4 \\ - 2 \\ 0 \end{array}], \end{matrix} & x_{4} = [\begin{array}{r} 3 \\ 8 \\ - 5 \\ 4 \end{array}] \end{matrix}

$\begin{matrix} \begin{matrix} x_{1} = [\begin{array}{r} 1 \\ 2 \\ - 1 \\ 0 \end{array}], & x_{2} = [\begin{array}{r} 2 \\ 5 \\ - 3 \\ 2 \end{array}], & x_{3} = [\begin{array}{r} 2 \\ 4 \\ - 2 \\ 0 \end{array}], \end{matrix} & x_{4} = [\begin{array}{r} 3 \\ 8 \\ - 5 \\ 4 \end{array}] \end{matrix}$

SOLUTION

The subspace $Span (x_{1}, x_{2}, x_{3}, x_{4})$ $Span (x_{1}, x_{2}, x_{3}, x_{4})$ is the same as the column space of the matrix

X = [\begin{matrix} 1 & 2 & 2 & 3 \\ 2 & 5 & 4 & 8 \\ - 1 & - 3 & - 2 & - 5 \\ 0 & 2 & 0 & 4 \end{matrix}]

$X = [\begin{matrix} 1 & 2 & 2 & 3 \\ 2 & 5 & 4 & 8 \\ - 1 & - 3 & - 2 & - 5 \\ 0 & 2 & 0 & 4 \end{matrix}]$

The row echelon form of X is

[\begin{matrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]

$[\begin{matrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix}]$

The first two columns $x_{1}$ $x_{1}$ , $x_{2}$ $x_{2}$ of X will form a basis for the column space of X. Thus, dim $Span (x_{1}, x_{2}, x_{3}, x_{4}) = 2$ $Span (x_{1}, x_{2}, x_{3}, x_{4}) = 2$ .

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Table of Contents for 3.6 Row Space and Column Space

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Table of Contents for
3.6 Row Space and Column Space