Let \(A \in \mathbb{F}^{m \times n}\) where \(\mathbb{F}\) is a field. We have seen \(N(A)\), the nullspace of \(A\), is given by \( \{ x \in \mathbb{F}^n : Ax = 0\}\). There are two additional vector spaces associated with a matrix that we will now discuss.

The **column space** of \(A\), denoted by
\({\cal C}(A)\), is the
**span of the columns of \(A\)**. In other words, the
we treat the
columns of \(A\)
as vectors in \(\mathbb{F}^m\)
and take all possible linear combinations of these vectors
to form the span. So \({\cal C}(A)\) is a subspace of \(\mathbb{F}^m\).

If \(A\) is the matrix \(\begin{bmatrix} 1 & 0 & 2 \\ 0 & -1 & 3\end{bmatrix}\) defined over the real numbers, then \({\cal C}(A)\) consists of the vectors \(\alpha \begin{bmatrix} 1 \\ 0\end{bmatrix} + \beta \begin{bmatrix} 0 \\ -1\end{bmatrix} + \gamma \begin{bmatrix} 2 \\ 3\end{bmatrix}\) where \(\alpha, \beta, \gamma \in \mathbb{R}\). Writing \(\alpha \begin{bmatrix} 1 \\ 0\end{bmatrix} + \beta \begin{bmatrix} 0 \\ -1\end{bmatrix} + \gamma \begin{bmatrix} 2 \\ 3\end{bmatrix} \) as \( \begin{bmatrix} 1 & 0 & 2\\ 0 & -1 & 3\end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma\end{bmatrix}, \) we have \[{\cal C}(A) = \left \{ \begin{bmatrix} 1 & 0 & 2\\ 0 & -1 & 3\end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma\end{bmatrix} : \alpha,\beta,\gamma \in \mathbb{R} \right \}.\] As a result, \(\begin{bmatrix} a \\b\end{bmatrix}\) is in \({\cal C}(A)\) if and only if the system \(\begin{bmatrix} 1 & 0 & 2\\ 0 & -1 & 3\end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma\end{bmatrix} = \begin{bmatrix} a \\b\end{bmatrix}\) has a solution.

The **row space** of \(A\), denoted by \({\cal R}(A)\)
is given
by \({\cal C}(A^\mathsf{T})\).
So \({\cal R}(A)\),
is
a subspace of
\(\mathbb{F}^n\).

We are going to see how to find a basis for the row space and a basis for the column space. We first consider the case when the matrix is in reduced row-echelon form.

Let \(A = \begin{bmatrix} 1 & -2 & 0 & -1\\0 & 0 & 1 & 1\\ 0 & 0 & 0 &0\end{bmatrix}\). Say we want to find a basis for \({\cal R}(A)\).

Now, \(A^\mathsf{T} = \begin{bmatrix} 1 & 0 & 0\\-2 & 0 & 0\\ 0 & 1 &0 \\ 1 & 1 & 0\end{bmatrix}\). Clearly, the span of the first two columns is the same as the column space of \(A^\mathsf{T}\) since the third column, which has all entries equal to \(0\)'s, contributes nothing in a linear combination.

Note that each of the first two columns is needed to span \({\cal C}(A^\mathsf{T})\) since the first column is the only column with a nonzero in the first entry and the second column is the only column with a nonzero in the third entry. Hence, the first two columns of \(A^\mathsf{T}\) together form a basis for \({\cal R}(A)\).

Now, let's find a basis for \({\cal C}(A)\). Note that \(A_2\), the second column of \(A\), is a scalar multiple of \(A_1\). As well, \(A_4 = -A_1 + A_3.\) Thus, the span of \(\{A_1,A_3\}\) is still \({\cal C}(A)\). Clearly, \(\{A_1,A_3\}\) is a linearly independent set. So it gives a basis for \({\cal C}(A)\).

Given a matrix \(A\) in reduced row-echelon form, the nonzero columns in \(A^\mathsf{T}\) form a basis for \({\cal R}(A)\), and the pivot columns in \(A\) form a basis for \({\cal C}(A)\).

A useful fact concerning the nullspace and the row space of a matrix is the following:

Elementary row operations do not affect the nullspace or the row space of the matrix.

Hence, given a matrix \(A\), first transform it to a matrix \(R\) in reduced row-echelon form using elementary row operations. Then find a basis for the row space of \(R\). It will then be a basis for the row space of \(A\).

What about the column space? Clearly,
elementary row operations *do* affect the column space.
However, we can still make use of row reduction to help find a basis
for the column space.

The key is the following: If \(A\) has been row-reduced to a matrix \(R\) in reduced row-echelon form, then the columns of \(A\) that correspond to the pivot columns of \(R\) form a basis for \({\cal C}(A)\). In other words, if in \(R\), columns 1,3,7 are the pivot columns, then \(A_1,A_3,A_7\) form a basis for \({\cal C}(A)\).

Let \(A = \begin{bmatrix} 1 & -2 & 3 & 1\\ 0 & 1 & -1 & 0\\ -1 & 1 & -2 & -1 \end{bmatrix}\) be defined over the real numbers. Transforming \(A\) to a matrix in reduced row-echelon form gives \(\begin{bmatrix} 1 & 0 & 1 & 1\\ 0 & 1 & -1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\). Since the pivot columns are the first two columns, \(A_1\) and \(A_2\) will give a basis for \({\cal C}(A)\). Indeed, \(A_1\) and \(A_2\) are clearly linearly independent, \(A_3 = A_1 - A_2\) and \(A_4 = A_1\).

It can be shown that \(\dim({\cal C}(A)) = \dim({\cal R}(A)).\)
The dimension of the row space or column space of \(A\) is called
the **rank** of \(A\), denoted by \(\operatorname{rank}(A)\). Hence, for the matrix
\(A = \begin{bmatrix}
1 & -2 & 3 & 1\\
0 & 1 & -1 & 0\\
-1 & 1 & -2 & -1
\end{bmatrix}\) above, we have \(\operatorname{rank}(A) = 2\).

For each of the following matrices, find a basis for its nullspace, a basis for its row space, and a basis for its column space.

\(\begin{bmatrix} 1 & 2 & 3\\ \end{bmatrix}\).

\(\begin{bmatrix} 1 & 0 & 2 & 0\\ 0 & 1 & -1 & 1\\ 1 & 1 & 1 & 1 \end{bmatrix}\).