Null Space and Nullity of a Matrix

 

Definition of Null Space of a Matrix

The null space of an $m \times n$ matrix $A$ is the set of all the solutions $\mathbf x$ to the homogeneous equation

$A \mathbf x = \mathbf 0$ where $\mathbf x$ is a column vector with $n$ rows and $\mathbf 0$ is a zero column vector with $n$ rows.
The null space of matrix $A$ is denoted as "Null $A$".
Null $A$ is a subspace of $\mathbb{R}^n$ and vectors $x$ are in $\mathbb{R}^n$.
Using set notation we write:       Null $A = \{ \mathbf x : \mathbf x \in \mathbb{R}^n | A \mathbf x = \mathbf 0 \}$
The nullity of matrix $A$ is the dimension of Null $A$ which equal to the number of vectors in Null $A$.

Properties of the Null Space

Let $A$ be an $m \times n$ matrix.

1.    The null space of a given matrix $A$ is never empty since $\mathbf x = \mathbf 0$ is a trivial solution to the homogeneous equation $A \mathbf x = \mathbf 0$.
2.    Null $A$ is a subspace of $\mathbb{R}^n$
3.    All elements of Null $A$ are vectors in $\mathbb{R}^n$.

Examples with Solutions

Example 1
Which of the vectors $\mathbf v = \begin{bmatrix} 3\\ -3\\ -1 \end{bmatrix} , \mathbf u = \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix}$ is an element of Null $A$ given that $A = \begin{bmatrix} 1 & 2 & -3\\ -1 & -1 & 0\\ -2 & -3 & 3 \end{bmatrix}$?
Solution to Example 1
From the above definition, any element of Null A must be a solution to the homogeneous equation $A \mathbf x = \mathbf 0$. Therefore, in order to find out if a given vector is an elelment of Null A, we just check whether it is a solution to the homogeneous equation.

Check vector $\mathbf v$
Evaluate the product $A \mathbf v$
$\begin{bmatrix} 1 & 2 & -3\\ -1 & -1 & 0\\ -2 & -3 & 3 \end{bmatrix} \begin{bmatrix} 3\\ -3\\ -1 \end{bmatrix} = \begin{bmatrix} (1)(3) + (2)(-3) + (-3)(-1)\\\ (-1)(3) + (-1)(-3) + (0)(-1) \\ (-2)(3) + (-3)(-3) + 3(-1) \end{bmatrix}$
Simplify
$= \begin{bmatrix} 0\\\ 0 \\ 0 \end{bmatrix}$
Hence since $\; A \mathbf v \;$ is equal to $\mathbf 0$, vector $\mathbf v = \begin{bmatrix} 3\\ -3\\ -1 \end{bmatrix}$ is a solution to the homogeneous equation $A \mathbf x = \mathbf 0$ and is therefore an element of Null $A$.

Check vector $\mathbf u$
Evaluate the product $A \mathbf u$
$\begin{bmatrix} 1 & 2 & -3\\ -1 & -1 & 0\\ -2 & -3 & 3 \end{bmatrix} \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(-2) + (-3)(1)\\\ (-1)(1) + (-1)(-2) + (0)(1) \\ (-2)(1) + (-3)(-2) + 3(1) \end{bmatrix}$
Simplify
$= \begin{bmatrix} -4\\\ 1\\ 7 \end{bmatrix}$
Hence since the $\; A \mathbf u \;$ is NOT equal to $\mathbf 0$, vector $\mathbf u = \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix}$ is a NOT solution to the homogeneous equation $A \mathbf x = \mathbf 0$ and therefore is NOT an element of Null $A$.

Example 2
a) Find Null A given that $A = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$
b) Find 3 elements that are in Null A.

Solution to Example 2
a)
To find the Null A, we need to solve the equation $A \mathbf x = \mathbf 0$ where $\mathbf x$ is a vector in $\mathbb{R}^2$
Let $\mathbf x = \begin{bmatrix} x_1\\ x_2 \end{bmatrix}$
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 $A \mathbf x = \mathbf 0$ $\begin{bmatrix} 1 & 2 & | & 0\\ 0 & 0 & | & 0 \end{bmatrix}$
Solve the above. $x_2$ is a free variable and
$x_1 = - 2 x_2$
Hence the solution set may be written of the form
$\mathbf x = x_2 \begin{bmatrix} -2\\ 1 \end{bmatrix}$ , where $x_2 \in \mathbb{R}$
Any real value of $x_3$ gives an element in Null $A$.
Hence Null $A$ is the subspace spanned by $\begin{bmatrix} -2 \\ 1 \end{bmatrix}$
b)
Note that for $x_2 = 0$, we have the trivial solution $\mathbf x = 0 \begin{bmatrix} -2\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$
More elements of Null $A$ may be obtained by setting the free variable $x_2$ to different values.

Let $x_2 = -1$, hence the corresponding element of Null $A$: $\mathbf x = -1 \begin{bmatrix} -2\\ 1 \end{bmatrix} = \begin{bmatrix} 2\\ -1 \end{bmatrix}$

Let $x_2 = 2$, hence the corresponding element of Null $A$: $\mathbf x = 2 \begin{bmatrix} -2\\ 1 \end{bmatrix} = \begin{bmatrix} -4\\ 2 \end{bmatrix}$

Let $x_2 = \dfrac{1}{2}$, hence the corresponding element of Null $A$: $\mathbf x = \dfrac{1}{2} \begin{bmatrix} -2\\ 1 \end{bmatrix} = \begin{bmatrix} -1\\ \dfrac{1}{2} \end{bmatrix}$

As an exercise, check that the three elements $\begin{bmatrix} 2\\ -1 \end{bmatrix} , \begin{bmatrix} -4\\ 2 \end{bmatrix} , \begin{bmatrix} -1\\ \dfrac{1}{2} \end{bmatrix}$ obtained above are solutions to the equation $A \mathbf x = \mathbf 0$.

Example 3
a) Find Null A and the nullity of $A$ given that $A = \begin{bmatrix} 1 & 2 & -1 \\ 1 & -1 & 1 \\ 3 & 0 & 1 \end{bmatrix}$

Solution to Example 3
Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}$ be the solution of $A \mathbf x = \mathbf 0$
We need to solve the system
$\begin{bmatrix} 1 & 2 & -1 \\ 1 & -1 & 1 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
Write the augmented matrix of the above equation.
$\begin{bmatrix} 1 & 2 & -1 & | & 0\\ 1 & -1 & 1 & | & 0\\ 3 & 0 & 1 & | & 0 \end{bmatrix}$
Row reduce using Gauss Jordan method
$\begin{bmatrix} 1 & 0 & \dfrac{1}{3} & | & 0\\ 0 & 1 & -\dfrac{2}{3} & | & 0\\ 0 & 0 & 0 & | & 0 \end{bmatrix}$
$x_3$ is the free variable.
$x_2 = \dfrac{2}{3} x_3$
$x_1 = -\dfrac{1}{3} x_3$
The solution $x$ may be written as
$\mathbf x = x_3 \begin{bmatrix} -\dfrac{1}{3} \\ \dfrac{2}{3} \\ 1 \end{bmatrix}$
Hence Null $A$ is the subspace spanned by $\begin{bmatrix} -\dfrac{1}{3} \\ \dfrac{2}{3} \\ 1 \end{bmatrix}$
Nullity of $A$ = the number of vector in Null $A$ = 1

Example 4

a) Find Null A and the nullity of $A$ given that $A = \begin{bmatrix} 3 & -2 & -1 & 0 & 2\\ 1 & -2 & 1 & -2 & 4\\ -4 & 4 & 0 & 2 & -6 \end{bmatrix}$

Solution to Example 4
The system to solve is
$A \mathbf x = \mathbf 0$
which may be written as
$\begin{bmatrix} 3 & -2 & -1 & 0 & 2\\ 1 & -2 & 1 & -2 & 4\\ -4 & 4 & 0 & 2 & -6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\\ x_3\\ x_4\\ x_5 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

We first write the augmented matrix of the above 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 above matrix
$\begin{bmatrix} 1 & 0 & -1 & 1 & -1 & | & 0\\ 0 & 1 & -1 & \dfrac{3}{2} & -\dfrac{5}{2} & | & 0 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$

Solve the system corresponding to the augmented matrix
The free variables are: $x_3, \;x_4, \;x_5$, hence
The second row gives: $x_2 = x_3 - \dfrac{3}{2} x_4 + \dfrac{5}{2} x_5$
The first row gives: $x_1 = x_3 - x_4 + x_5$
The solution vector $\mathbf x$ is given by
$\mathbf x = \begin{bmatrix} x_3 - x_4 + x_5 \\ x_3 - \dfrac{3}{2} x_4 + \dfrac{5}{2} x_5 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}$

which may be written as
$\mathbf x = x_3 \begin{bmatrix} 1\\ 1\\ 1\\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} - 1\\ - \dfrac{3}{2}\\ 0 \\ 1\\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 1\\ \dfrac{5}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix}$

Null A = Span $\left\{ \begin{bmatrix} 1\\ 1\\ 1\\ 0\\ 0 \end{bmatrix} , \begin{bmatrix} -1\\ -\dfrac{3}{2}\\ 0\\ 1\\ 0 \end{bmatrix} , \begin{bmatrix} 1\\ \dfrac{5}{2}\\ 0\\ 0\\ 1 \end{bmatrix} \right\}$

Nullity of $A$ is equal to the number of vectors in the basis of Null $A$ = 3.

As an exercise, select any real values for $x_3, \; x_4$ and $x_5$, calculate vector $\mathbf x = x_3 \begin{bmatrix} 1\\ 1\\ 1\\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} - 1\\ - \dfrac{3}{2}\\ 0 \\ 1\\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 1\\ \dfrac{5}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix}$ and verify that it is a solution to the system $A \mathbf x = \mathbf 0$.