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.
\(\begin{bmatrix} 1 & 2 & 3\\ \end{bmatrix}\).
\(\begin{bmatrix} 1 & 0 & 2 & 0\\ 0 & 1 & -1 & 1\\ 1 & 1 & 1 & 1 \end{bmatrix}\).