Definition. Let \( A \) be an \( m\times n \) matrix. Write \( \bbm a_1, \bbm a_2, \ldots, \bbm a_n \in \mathbb R^m \) for the columns of \( A \). The column space of \( A \), written \( \Col A \), is the span of the columns of \( A \): \[ \Col A = \mbox{Span} \{ \bbm a_1, \bbm a_2, \ldots, \bbm a_n \}. \]
In set-builder notation, we write \( \Col A = \{ c_1 \bbm a_1 + c_2 \bbm a_2 + \cdots + c_n \bbm a_n : c_1, c_2, \ldots, c_n \in \mathbb R \} \). The elements of \( \Col A \) are the vectors \( \bbm x\) of the form \( \bbm x = c_1 \bbm a_1 + c_2 \bbm a_2 + \cdots + c_n \bbm a_n \).
Example 1. Let \( A = \begin{bmatrix} -4 & -2 & -9 \\ 3 & 4 & 8 \\ 2 & 0 & 4 \end{bmatrix} \). Generate three elements of \( \Col A \).
We need only construct linear combinations of the columns of \( A \): \[ 1 \vecthree {-4} 3 2 + 2 \vecthree {-2} 4 0 + 3 \vecthree {-9} 8 4 = \vecthree {-35} {35} {14} \] \[ 0 \vecthree {-4} 3 2 + 0 \vecthree {-2} 4 0 + 0 \vecthree {-9} 8 4 = \vecthree 000 \] \[ 1 \vecthree {-4} 3 2 + 0 \vecthree {-2} 4 0 + (-1) \vecthree {-9} 8 4 = \vecthree 5 {-5} {-2}.\ \Box \]
Example 2. Let \( B = \begin{bmatrix} -1 & 1 & 0 \\ 5 & -3 & 2 \\ -2 & 2 & 0 \end{bmatrix} \). Is \( \bbm w = \vecthree 5 0 4 \) in \( \Col B \)?
This question is asking whether \( x_1 \vecthree {-1} 5 {-2} + x_2 \vecthree 1 {-3} 2 + x_3 \vecthree 0 2 0 = \vecthree 5 0 4 \) for some values of \( x_1 \), \( x_2 \), and \( x_3 \). We solve this vector equation in the normal way by row-reducing: \[ \begin{bmatrix} -1 & 1 & 0 & 5 \\ 5 & -3 & 2 & 0 \\ -2 & 2 & 0 & 4 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
Since the augmented matrix has a pivot in the last column, the vector equation has no solutions. We conclude that, no, \( \bbm w \) is not in \( \Col B \). \( \Box \)
Definition. Let \( T : \mathbb R^n \to \mathbb R^m \) be a linear transformation. The image of \( T \), written \( \mbox{Im } T \), is the set of all vectors \( T(\bbm x) \).
For any \( \bbm x \in \mathbb R^n \), recall that \( A\bbm x \) is a linear combination of the columns of \( A \). So, another way to write the definition of the column space is \( \Col A = \{ A\bbm x : \bbm x\in \mathbb R^n \} \). If \( A \) is the standard matrix for \( T \), then the image of \( T \) is the same as \( \Col A \) since \( T(\bbm x)=A\bbm x \).
Finding a spanning set for \( \Col A \) is easy: the columns of \( A \) themselves span \( \Col A \). This directly leads to the following theorem:
Theorem (Column Space). Let \( A \) be an \( m\times n\) matrix. Then \( \Col A \) is a subspace of \( \mathbb R^n \).
Proof. This follows immediately from the Subspace Spanned By a Set Theorem from Lecture 27. \( \Box \)
From the Spanning Columns Theorem, we know that the columns of \( A \) span \( \mathbb R^m \) exactly when \( A \) has a pivot in every row. In this case, \( \Col A = \mathbb R^m \).
Challenge Question. What can you say about an \( m\times n\) matrix \( A \) if \( \Col A = \{ \bbm 0 \} \) ?
Now that we have studied both the null space and the column space, we can compare different questions that we can ask about these spaces for a given matrix \( A \). Some of these questions are easier to answer for \( \Nul A \), and others are easier for \( \Col A \).
| Null Space \( \Nul A \) | \( \qquad \) | Column Space \( \Col A \) |
|---|---|---|
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To illustrate the relative difficulty of these questions, we will work through one more example. Here we are talking about the computational difficulty of the problem. That is, the number of arithmetic operations that are required to answer the question. Any question that simply requires us to multiply a matrix by a vector is "easy," since this is a relatively low number of arithmetic operations. However, any question that requires row-reduction is "difficult" since, as we have seen, row-reduction can require a large number of steps.
Example 3. Let \( A = \begin{bmatrix} -1 & 3 & 2 \\ 3 & 1 & -1 \\ 0 & 2 & 1 \\ -4 & -6 & -1 \end{bmatrix} \).
For question 1, we just multiply \( A \bbm u \) and see if we get the zero vector: \[ A \bbm u = \begin{bmatrix} -1 & 3 & 2 \\ 3 & 1 & -1 \\ 0 & 2 & 1 \\ -4 & -6 & -1 \end{bmatrix} \vecthree {-1} 2 0 = \vecfour 7 {-1} 4 {-8}. \] Since \( A\bbm u \ne 0 \), no, \( \bbm u \) is not in \( \Nul A \).
For question 2, we have to solve \( A\bbm x = \bbm 0 \), which requires row-reducing \( A \): \[ \begin{bmatrix} -1 & 3 & 2 \\ 3 & 1 & -1 \\ 0 & 2 & 1 \\ -4 & -6 & -1 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & -1/2 \\ 0 & 1 & 1/2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \] The parametric form of the solution is \( \bbm x = x_3 \vecthree {1/2} {-1/2} 1 \), and so one example of an element of \( \Nul A \) is to take \( x_3 = 1 \), which gives \( \vecthree {1/2} {-1/2} 1 \).
For question 3, we have to solve \( A \bbm x = \bbm v \), which again requires row-reduction. Unfortunately, we can't re-use the work from question 2, since we don't know what will happen in the augmented column. \[ \begin{bmatrix} -1 & 3 & 2 & 5 \\ 3 & 1 & -1 & 0 \\ 0 & 2 & 1 & 3 \\ -4 & -6 & -1 & -7 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & 0 & -1/2 & -1/2 \\ 0 & 1 & 1/2 & 3/2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \] Since there is no pivot in the last column, the equation \( A \bbm x = \bbm v \) is consistent. We conclude that, yes, \( \bbm v\) is in \( \Col A \).
For question 4, we can construct any linear combination of the columns of \( A\). Equivalently, we can multiply \( A \) by any vector in \( \mathbb R^3 \). For example, \[ A \vecthree 111 = \begin{bmatrix} -1 & 3 & 2 \\ 3 & 1 & -1 \\ 0 & 2 & 1 \\ -4 & -6 & -1 \end{bmatrix} \vecthree 111 = \vecfour 4 3 3 {-11} \in \Col A.\ \Box \]
From the Spanning Columns Theorem, we know that the columns of \( A \) span \( \mathbb R^m \) exactly when \( A \) has a pivot in every row. In this case, \( \Col A = \mathbb R^m \). We can use this fact to draw conclusions about matrices even when we have limited information.
Example 4. Suppose that \( A \) is a \( 3\times 3\) matrix whose first and third columns are equal. Is \( \Col A = \mathbb R^3 \)?
From the given information, we can tell that the columns of \( A \) are not linearly independent. This tells us that \( A \) does not have a pivot in every column, so \( A \) has at most 2 pivots. But \( A \) has 3 rows, so \( A \) cannot have a pivot in every row. By the Spanning Columns Theorem, the columns of \( A \) cannot span \( \mathbb R^3 \). We conclude that, no, \( \Col A \) does not equal \( \mathbb R^3 \).
Example 5. Suppose that \( A \) is a \( 3\times 4\) matrix whose first and third columns are equal. Is \( \Col A = \mathbb R^3 \)?
While at first glance this may seem to be the same as Example 4, the situation is significantly different. As in that example, we can conclude that \( A \) does not have a pivot in every column. However, this time \( A \) has 4 columns, so the best we can conclude is that \( A \) has at most 3 pivots. It is possible that \( A \) has exactly 3 pivots, in which case \( \Col A \) would equal \( \mathbb R^3 \). However, without additional information, we cannot know for sure. In this case, there is not enough information to answer this question.
Challenge Question. What can you say about an \( m\times n\) matrix \( A \) if \( \Col A = \{ \bbm 0 \} \) ?
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