# Null Space and Nullity of a Matrix

## Definition of Null Space of a Matrix

The null space of an m × n matrix A is the set of all the solutions x to the homogeneous equation

A x = 0

where x is a column vector with n rows and 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 n and vectors x are in n .
Using set notation we write:

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 × n matrix.

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

## Examples with Solutions

Example 1
Which of the vectors is an element of Null A given that
Solution to Example 1
From the above definition, any element of Null A must be a solution to the homogeneous equation A x = 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 v
Evaluate the product A v

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

Check vector $$u$$
Evaluate the product $$A u$$
$$\begin{bmatrix} 1 & 2 & -3\\ -1 & -1 & 0\\ -2 & -3 & 3 \end{bmatrix} \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix} = \begin{bmatrix} (1)(1) + (2)(-2) + (-3)(1)\\\ (-1)(1) + (-1)(-2) + (0)(1) \\ (-2)(1) + (-3)(-2) + 3(1) \end{bmatrix}$$
Simplify
$$= \begin{bmatrix} -4\\\ 1\\ 7 \end{bmatrix}$$
Hence since the $$\; A u \;$$ is NOT equal to $$0$$, vector $$u = \begin{bmatrix} 1\\ -2\\ 1 \end{bmatrix}$$ is a NOT solution to the homogeneous equation $$A x = 0$$ 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 $$A x = 0$$ 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 $$A x = 0$$ $$\begin{bmatrix} 1 & 2 & | & 0\\ 0 & 0 & | & 0 \end{bmatrix}$$
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
$$x = x_2 \begin{bmatrix} -2\\ 1 \end{bmatrix}$$ , 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 $$x = 0 \begin{bmatrix} -2\\ 1 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$
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 $$A$$: $$x = -1 \begin{bmatrix} -2\\ 1 \end{bmatrix} = \begin{bmatrix} 2\\ -1 \end{bmatrix}$$

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

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

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 $$A x = 0$$.

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 $$x = \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix}$$ be the solution of $$A x = 0$$
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
$$x = x_3 \begin{bmatrix} -\dfrac{1}{3} \\ \dfrac{2}{3} \\ 1 \end{bmatrix}$$
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
$$A x = 0$$
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 $$x$$ is given by
$$x = \begin{bmatrix} x_3 - x_4 + x_5 \\ x_3 - \dfrac{3}{2} x_4 + \dfrac{5}{2} x_5 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}$$

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

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 $$x = x_3 \begin{bmatrix} 1\\ 1\\ 1\\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} - 1\\ - \dfrac{3}{2}\\ 0 \\ 1\\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 1\\ \dfrac{5}{2} \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ and verify that it is a solution to the system $$A x = 0$$.