The concept of the null space of a matrix is a foundational topic in linear algebra. Below, you will find definitions, mathematical properties, and practice examples with interactive step-by-step solutions to help you understand null spaces and nullity.
The null space of an \( m \times n \) matrix \( A \) is the set of all solutions \( x \) to the homogeneous equation:
\[ Ax = 0 \]where \( x \) is a column vector with \( n \) rows and \( 0 \) is a zero column vector with \( m \) rows.
Let \( A \) be an \( m \times n \) matrix. The following properties hold true:
Which of the vectors \( v \) and \( u \) below is an element of Null \( A \), given that:
\[ v = \begin{bmatrix} 3 \\ -3 \\ -1 \end{bmatrix}, \quad u = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 0 \\ -2 & -3 & 3 \end{bmatrix} \]From the definition, any element of Null \( A \) must be a solution to the homogeneous equation \( Ax = 0 \). To determine if a given vector is an element of Null \( A \), we simply multiply the matrix by the vector to check if the result is the zero vector.
Checking vector \( v \):
Evaluate the product \( A \cdot v \):
\[ A v = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 0 \\ -2 & -3 & 3 \end{bmatrix} \begin{bmatrix} 3 \\ -3 \\ -1 \end{bmatrix} = \begin{bmatrix} (1)(3) + (2)(-3) + (-3)(-1) \\ (-1)(3) + (-1)(-3) + (0)(-1) \\ (-2)(3) + (-3)(-3) + (3)(-1) \end{bmatrix} \] \[ = \begin{bmatrix} 3 - 6 + 3 \\ -3 + 3 + 0 \\ -6 + 9 - 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]Since \( A \cdot v = 0 \), vector \( v \) is a solution to the homogeneous equation \( Ax = 0 \) and is therefore an element of Null \( A \).
Checking vector \( u \):
Evaluate the product \( A \cdot u \):
\[ A u = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 0 \\ -2 & -3 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(-2) + (-3)(1) \\ (-1)(1) + (-1)(-2) + (0)(1) \\ (-2)(1) + (-3)(-2) + (3)(1) \end{bmatrix} \] \[ = \begin{bmatrix} 1 - 4 - 3 \\ -1 + 2 + 0 \\ -2 + 6 + 3 \end{bmatrix} = \begin{bmatrix} -6 \\ 1 \\ 7 \end{bmatrix} \]Since \( A \cdot u \neq 0 \), vector \( u \) is not a solution to the homogeneous equation \( Ax = 0 \) and therefore is not an element of Null \( A \).
Part a)
To find Null \( A \), we need to solve the equation \( Ax = 0 \) where \( x \) is a vector in \( \mathbb{R}^2 \). Let \( x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \).
We need to solve:
\[ \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]Write the augmented matrix of the equation \( Ax = 0 \):
\[ \begin{bmatrix} 1 & 2 & | & 0 \\ 0 & 0 & | & 0 \end{bmatrix} \]Solving this reveals that \( x_2 \) is a free variable and \( x_1 = -2x_2 \).
Hence, the solution set may be written in the form:
\[ x = x_2 \begin{bmatrix} -2 \\ 1 \end{bmatrix} , \quad \text{where } x_2 \in \mathbb{R} \]Any real value of \( x_2 \) gives an element in Null \( A \). Therefore, Null \( A \) is the subspace spanned by \( \begin{bmatrix} -2 \\ 1 \end{bmatrix} \).
Part b)
More elements of Null \( A \) can be obtained by setting the free variable \( x_2 \) to different values:
Find Null \( A \) and the nullity of \( A \) given that:
\[ A = \begin{bmatrix} 1 & 2 & -1 \\ 1 & -1 & 1 \\ 3 & 0 & 1 \end{bmatrix} \]Let \( x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \) be the solution of \( Ax = 0 \).
We write the augmented matrix for the system \( Ax = 0 \):
\[ \begin{bmatrix} 1 & 2 & -1 & | & 0 \\ 1 & -1 & 1 & | & 0 \\ 3 & 0 & 1 & | & 0 \end{bmatrix} \]Row reduce using the Gauss-Jordan method:
\[ \begin{bmatrix} 1 & 0 & \frac{1}{3} & | & 0 \\ 0 & 1 & -\frac{2}{3} & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix} \]Here, \( x_3 \) is the free variable. We extract the equations:
The solution \( x \) may be written as:
\[ x = x_3 \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ 1 \end{bmatrix} \]Hence, Null \( A \) is the subspace spanned by \( \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ 1 \end{bmatrix} \).
Nullity of \( A \) is equal to the number of basis vectors in Null \( A \), which is 1.
Find Null \( A \) and the nullity of \( A \) given that:
\[ A = \begin{bmatrix} 3 & -2 & -1 & 0 & 2 \\ 1 & -2 & 1 & -2 & 4 \\ -4 & 4 & 0 & 2 & -6 \end{bmatrix} \]We need to solve the system \( Ax = 0 \), which maps out to 5 variables (\(x_1\) through \(x_5\)). First, we write the augmented matrix of the system:
\[ \begin{bmatrix} 3 & -2 & -1 & 0 & 2 & | & 0 \\ 1 & -2 & 1 & -2 & 4 & | & 0 \\ -4 & 4 & 0 & 2 & -6 & | & 0 \end{bmatrix} \]We row reduce the augmented matrix to Reduced Row Echelon Form (RREF):
\[ \begin{bmatrix} 1 & 0 & -1 & 1 & -1 & | & 0 \\ 0 & 1 & -1 & \frac{3}{2} & -\frac{5}{2} & | & 0 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \end{bmatrix} \]Solving the system corresponding to the reduced matrix gives us three free variables: \( x_3, x_4, x_5 \).
The solution vector \( x \) is parameterized as:
\[ x = \begin{bmatrix} x_3 - x_4 + x_5 \\ x_3 - \frac{3}{2}x_4 + \frac{5}{2}x_5 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} \]Which may be written by separating the variables as a linear combination:
\[ x = x_3 \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -1 \\ -\frac{3}{2} \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 1 \\ \frac{5}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix} \]Therefore, Null \( A \) is the span of these three vectors:
\[ \text{Null } A = \text{Span} \left\{ \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} -1 \\ -\frac{3}{2} \\ 0 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ \frac{5}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} \]Nullity of \( A \) is equal to the number of vectors in the basis of Null \( A \), which is 3.