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 { **r** _{1}, **r** _{2}, …, **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 { **r** _{1}, **r** _{2}, …, **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 **c** _{1}, **c** _{2}, …, **c** _{n }denote the columns of *A*, then a maximal linearly independent subset of { **c** _{1}, **c** _{2}, …, **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 **R** ^{3}, it must be all of **R** ^{3}.

**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 *x* _{1}, *x* _{2}, …, *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 **R** ^{4}. 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 A ^{T}, finding a basis for CS(A) is equivalent to finding a basis for RS(A ^{T}) *. Row‐reducing

*A*

^{T}yields

Since there are two nonzero rows left in the reduced form of *A* ^{T}, the rank of *A* ^{T} is 2, so

Furthermore, since { **v** _{1}, **v** _{2}} = {(1, 2, −3), (0, −4, 7)} is a basis for *RS(A* ^{T}), the collection

is a basis for *CS(A)*, a 2‐dimensional subspace of **R** ^{3}.