Let
A be an
m by
n matrix. The space spanned by the rows of
A is called the
row space of
A, denoted
RS(A); it is a subspace of
R
n
. The space spanned by the columns of
A is called the
column space of
A, denoted
CS(A); it is a subspace of
R
m
.
The collection {
r1,
r2, …,
r
m
} consisting of the rows of
A may not form a basis for
RS(A), because the collection may not be linearly independent. However, a maximal linearly independent subset of {
r1,
r2, …,
r
m
}
does give a basis for the row space. Since the maximum number of linearly independent rows of
A is equal to the rank of
A,
Similarly, if
c1,
c2, …,
c
n
denote the columns of
A, then a maximal linearly independent subset of {
c1,
c2, …,
c
n
} gives a basis for the column space of
A. But the maximum number of linearly independent columns is also equal to the rank of the matrix, so
Therefore, although
RS(A) is a subspace of
R
n
and
CS(A) is a subspace of
R
m
, equations (*) and (**) imply that
even if
m ≠ n.
Example 1: Determine the dimension of, and a basis for, the row space of the matrix
A sequence of elementary row operations reduces this matrix to the echelon matrix
The rank of
B is 3, so dim
RS(B) = 3. A basis for
RS(B) consists of the nonzero rows in the reduced matrix:
Another basis for
RS(B), one consisting of some of the original rows of
B, is
Note that since the row space is a 3-dimensional subspace of
R3, it must be all of
R3.
Criteria for membership in the column space. If
A is an
m x n matrix and
x is an
n-vector, written as a column matrix, then the product
A
x is equal to a linear combination of the columns of
A:
By definition, a vector
b in
R
m
is in the column space of
A if it can be written as a linear combination of the columns of
A. That is,
b ∈
CS(A) precisely when there exist scalars
x1,
x2, …,
x
n
such that
Combining (*) and (**), then, leads to the following conclusion:
Example 2: For what value of
b is the vector
b = (1, 2, 3,
b)T in the column space of the following matrix?
Form the augmented matrix [
A/
b] and reduce:
Because of the bottom row of zeros in
A′ (the reduced form of
A), the bottom entry in the last column must also be 0—giving a complete row of zeros at the bottom of [
A′/
b′]—in order for the system
A
x =
b to have a solution. Setting (6 − 8
b) − (17/27)(6 − 12
b) equal to 0 and solving for
b yields
Therefore,
b = (1, 2, 3,
b)T is in
CS(A) if and only if
b = 5.
Since elementary row operations do not change the rank of a matrix, it is clear that in the calculation above, rank
A = rank
A′ and rank [
A/
b] = rank [
A′/
b′]. (Since the bottom row of
A′ consisted entirely of zeros, rank
A′ = 3, implying rank
A = 3 also.) With
b = 5, the bottom row of [
A′/
b′] also consists entirely of zeros, giving rank [
A′/
b′] = 3. However, if
b were not equal to 5, then the bottom row of [
A′/
b′] would not consist entirely of zeros, and the rank of [
A′/
b′] would have been 4, not 3. This example illustrates the following general fact: When
b is in
CS(A), the rank of [
A/
b] is the same as the rank of
A; and, conversely, when
b is not in
CS(A), the rank of [
A/
b] is not the same as (it's strictly greater than) the rank of
A. Therefore, an equivalent criterion for membership in the column space of a matrix reads as follows:
Example 3: Determine the dimension of, and a basis for, the column space of the matrix
from Example 1 above.
Because the dimension of the column space of a matrix always equals the dimension of its row space,
CS(B) must also have dimension 3:
CS(B) is a 3-dimensional subspace of
R4. Since
B contains only 3 columns, these columns must be linearly independent and therefore form a basis:
Example 4: Find a basis for the column space of the matrix
Since the column space of
A consists precisely of those vectors
b such that
A
x =
b is a solvable system, one way to determine a basis for
CS(A) would be to first find the space of all vectors
b such that
A
x =
b is consistent, then constructing a basis for this space. However, an elementary observation suggests a simpler approach:
Since the columns of A are the rows of AT, finding a basis for CS(A) is equivalent to finding a basis for RS(AT)
. Row-reducing
AT yields
Since there are two nonzero rows left in the reduced form of
AT, the rank of
AT is 2, so
Furthermore, since {
v1,
v2} = {(1, 2, −3), (0, −4, 7)} is a basis for
RS(AT), the collection
is a basis for
CS(A), a 2-dimensional subspace of
R3.
Vector Algebra
Real Euclidean Vector Spaces
