Definition. Let \( A \) be an \( m\times n \) matrix. The null space of \( A \), written \( \Nul A \), is the set of all vectors \( \bbm x \in \mathbb R^n \) such that \( A \bbm x= \bbm 0 \).
In set-builder notation, we write \( \Nul A = \{ \bbm x\in \mathbb R^n : A\bbm x = \bbm 0 \} \). The elements of \( \Nul A \) are the vectors \( \bbm x\) with the property that \( A \bbm x = \bbm 0 \).
Example 1. Let \( A = \begin{bmatrix} 1 & -3 & -2 \\ -5 & 9 & 1 \end{bmatrix} \). Is \( \bbm u = \vecthree 5 3 {-2} \) in \( \Nul A \)?
We compute \( A \bbm u = \begin{bmatrix} 1 & -3 & -2 \\ -5 & 9 & 1 \end{bmatrix} \vecthree 5 3 {-2} = \vectwo 00 \). Since \( A \bbm u = \bbm 0\), yes, \( \bbm u \) is in \( \Nul A \). \( \Box \)
Example 2. Let \( B = \begin{bmatrix} -3 & 6 \\ 1 & -2 \\ -2 & 4 \end{bmatrix} \). For what values of \( h \) is \( \bbm v = \vectwo {-4} h \) in \( \Nul B \)?
We compute \( B\bbm v = \begin{bmatrix} -3 & 6 \\ 1 & -2 \\ -2 & 4 \end{bmatrix}\vectwo {-4} h = \vecthree {12+6h} {-4-2h} {8+4h} \). In order for \( B\bbm v = \bbm 0\), we must have \( 12+6h = 0\), \(-4-2h = 0\), and \( 8+4h =0\) for the same value of \( h \). The only solution is \( h = -2 \). \( \Box \)
Definition. Let \( T : \mathbb R^n \to \mathbb R^m \) be a linear transformation. The kernel of \( T \), written \( \mbox{ker } T \), is the set of all vectors \( \bbm x \in \mathbb R^n \) such that \( T(\bbm x) = \bbm 0 \).
Note that, if \( A \) is the standard matrix for \( T \), then \( \mbox{ker } T \) is the same as \( \Nul A \) since \( T(\bbm x)=A\bbm x \).
Theorem (Null Space). Let \( A \) be an \( m\times n\) matrix. Then \( \Nul A \) is a subspace of \( \mathbb R^n \).
Using the subspace defintion from Lecture 27, we must show that \( \Nul A \) contains the zero vector, is closed under addition, and is closed under scalar multiplication.
Our definition for null space is said to be "implicit," since it is define by a condition that must be checked. There is no obvious way to generate elements of \( \Nul A \) just from the definition. We would like to find a spanning set for \( \Nul A \):
Defintion. Given a subspace \( H \) of \( \mathbb R^n \), a spanning set for \( H \) is a finite set of vectors \( \{ \bbm v_1, \ldots, \bbm v_p \} \) such that \( \mbox{Span} \{ \bbm v_1, \ldots, \bbm v_p \} = H \).
Once you have a spanning set for a subspace, it is easy to generate new elements of that subspace: just form linear combinations of vectors in the spanning set. To find a spanning set for \( \Nul A \), we start by solving the matrix equation \( A \bbm x = \bbm 0 \) in the normal way.
Example 3. Find a spanning set for \( \Nul A \), where \( A = \begin{bmatrix} -3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1 \\ 2 & -4 & 5 & 8 & -4 \end{bmatrix} \).
We form the augmented matrix corresponding to \( A \bbm x = \bbm 0 \) and row-reduce: \[ \begin{bmatrix} -3 & 6 & -1 & 1 & -7 & 0 \\ 1 & -2 & 2 & 3 & -1 & 0 \\ 2 & -4 & 5 & 8 & -4 & 0 \end{bmatrix} \longrightarrow \begin{bmatrix} 1 & -2 & 0 & -1 & 3 & 0 \\ 0 & 0 & 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]
Now, write the solution in parametric vector form as we learned in Lecture 10: \[ \bbm x = \vecfive {x_1} {x_2} {x_3} {x_4} {x_5} = \vecfive {2x_2+x_4-3x_5} {x_2} {-2x_4+2x_5} {x_4} {x_5} = x_2 \vecfive 21000 + x_4 \vecfive 10{-2}10 + x_5 \vecfive {-3}0201. \]
From this, we can see that every solution of \( A \bbm x = \bbm 0 \) can be written as a linear combination of \( \bbm v_1 = \vecfive 21000, \bbm v_2 = \vecfive 10{-2}10 \), and \( v_3 = \vecfive {-3}0201 \). So, \( \{ \bbm v_1, \bbm v_2, \bbm v_3 \}\) is a spanning set for \( \Nul A \). \( \Box \)
Example 4. Suppose that \( B = \begin{bmatrix} -4 & 0 & 1 & 4 \\ 2 & 0 & 3 & 5 \\ -6 & 0 & -2 & -1 \end{bmatrix} \), which row-reduces to \( \begin{bmatrix} 1 & 0 & 0 & -1/2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \). Find a spanning set for \( \Nul B \).
Since the row reduction has been done for us, we can now write the parametric solution to \( B \bbm x = \bbm 0 \): \[ \bbm x = \vecfour {x_1} {x_2} {x_3} {x_4} = x_2 \vecfour 0 1 0 0 + x_4 \vecfour {1/2} 0 {-2} 1. \]
So, a spanning set for \( \Nul B \) is \( \left\{ \vecfour 0 1 0 0, \vecfour {1/2} 0 {-2} 1 \right\} \). \( \Box \)
From these examples, you should notice that the spanning set we construct for \( \Nul A \) contains one vector for each free variable in the equation \( A \bbm x = \bbm 0\). The Linearly Independent Columns Theorem implies that \( \Nul A = \{ \bbm 0 \} \) exactly when \( A \) has a pivot in every column.
Challenge Question. What can you say about an \( m\times n\) matrix \( A \) if \( \Nul A = \mathbb R^n \) ?
Sometimes we have limited information about a matrix, but we can still sometimes draw conclusions about its null space.
Example 5. Suppose that \( A \) is a \( 3\times 4\) matrix whose third column is all zeroes. Find a nonzero vector in \( \Nul A \).
We are looking for a nonzero vector \( \bbm x \) for which \( A \bbm x = \bbm 0 \). Remember that \( A \bbm x \) is just a linear combination of the columns of \( A\). Now, \( A \vecfour 0010 \) will equal the third column of \( A \), which we know is all zeroes. So, \( \bbm x= \vecfour 0010 \) (or any nonzero multiple of this vector) is a nonzero vector that is guaranteed to be in \( \Nul A \). \( \Box \)
Example 6. Suppose that \( A \) is a \( 2\times 5 \) matrix whose second column equals 3 times its fourth column. Find a nonzero vector in \( \Nul A \).
Write \( \bbm a_1, \ldots, \bbm a_5 \) for the columns of \( A \). We are given that \( \bbm a_2 = 3 \bbm a_4 \). We can rewrite this as \( \bbm a_2 - 3\bbm a_4 = \bbm 0 \), or \( A \vecfive 0 1 0 {-3} 0 = \bbm 0 \). Thus \( \vecfive 0 1 0 {-3} 0 \in \Nul A \). \( \Box \)
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