# A Matrix Nullspace (Kernel) Tutorial - Finding the Nullspace

· 2018/03/01 · 7 minute read

# 1. Introduction

In this article we describe what the nullspace is. We go on to explain how to find the nullspace.

# 2. Nullspace, Null Space or Kernel?

The nullspace (or null space) of a matrix is also known as the kernel of a matrix. These terms are interchangeable.

The nullspace is a set of vectors that when multiplied by a matrix returns $0$ (well, the zero vector).

# 3. How To Find the Nullspace

Have a look at this matrix:

$A % = % \begin{bmatrix} 1 & 1 & 2 & 3 \newline 2 & 2 & 8 & 10 \newline 3 & 3 & 10 & 13 \end{bmatrix}$

We’re interested in finding its nullspace: that is, all vectors such that when multiplied by our matrix $A$, return the zero vector, $\begin{bmatrix} 0 & 0 & 0 \end{bmatrix} ^ {T}$.

I happen to know that $\begin{bmatrix} -3 & 1 & -2 & 2 \end{bmatrix} ^ {T}$ is one such vector:

$\begin{bmatrix} 1 & 1 & 2 & 3 \newline 2 & 2 & 8 & 10 \newline 3 & 3 & 10 & 13 \end{bmatrix} % \begin{bmatrix} -3 \newline 1 \newline -2 \newline 2 \end{bmatrix} % = % % \begin{bmatrix} -3 + 1 - 4 + 6 \newline -6 + 2 - 16 + 20 \newline -9 + 3 - 20 + 26 \end{bmatrix} % = % \begin{bmatrix} 0 \newline 0 \newline 0 \end{bmatrix}$

If we set $x = \begin{bmatrix} -3 & 1 & -2 & 2 \end{bmatrix} ^ {T}$ , then we can write out the above equation more succinctly: $Ax = 0$

We say that $x$ is in the nullspace of $A$.

We’re going to look at a systematic way to find all the vectors in the nullspace. We’ll do this is by:

1. Using elimination to find our pivots
2. Using back substitution to find our solutions

We can think of this as our two-step recipe to find our nullspace.

## 3.1 Finding The Pivots

Suppose we performed elimination on our matrix $A$, which results in our new $U$:

$U % = % \begin{bmatrix} 1 & 1 & 2 & 3 \newline 0 & 0 & 4 & 4 \newline 0 & 0 & 0 & 0 \end{bmatrix}$

The cool thing is that $A$ and $U$ have the same nullspace. So, if we know that $Ux = 0$, we also know that $Ax = 0$.

Let’s suppose that we didn’t know that $x = \begin{bmatrix} -3 & 1 & -2 & 2 \end{bmatrix} ^ {T}$ was in the nullspace of the matrix $A$. We know that there is some vector $x = \begin{bmatrix} x_1 & x_2 & x_3 & x_4 \end{bmatrix} ^ {T}$ such that $Ux = 0$ (even if that vector is just $\begin{bmatrix} 0 & 0 & 0 & 0 \end{bmatrix} ^ {T}$).

Writing this out more explicitly:

$\tag{1} \label{1} \begin{bmatrix} 1 & 1 & 2 & 3 \newline 0 & 0 & 4 & 4 \newline 0 & 0 & 0 & 0 \end{bmatrix} % \begin{bmatrix} x_1 \newline x_2 \newline x_3 \newline x_4 \end{bmatrix} % = % \begin{bmatrix} 0 \newline 0 \newline 0 \end{bmatrix}$

According to our recipe for finding the nullspace, we need to find our pivots. A pivot is simply the first non-zero element in each row of the matrix.

Looking at $U$, the first row has a pivot in the first column, with value $1$. The second row has a pivot in the third column: $4$. The third row does not have a pivot.

We call $x_1$ and $x_3$ pivot variables since columns 1 and 3 contain pivots. $x_2$ and $x_4$ are called free variables since columns 2 and 4 have no pivots.

Note that there are 4 variables in total - 2 free variables, and 2 pivot variables. This is because the nullspace vectors are in $\mathbb{R}^4$ in this case, since $A$ (and $U$) are $3 \times 4$ matrices.

## 3.2 Using Back Substitution to Find our Nullspace

Now that we know what our pivots are, we want to find our special solutions. Special solutions are vectors in our nullspace which will eventually help us find all the vectors in our nullspace.

We will have as many special solutions as we have free variables. In this case, we have 2 free variables, so we will have 2 special solutions. The way we find the special solutions is by back substitution.

Before we get to what back substition is, let’s rewrite equation $(\ref{1})$ as a set of equations, instead of its matrix form:

$x_1 + x_2 + 2x_3 + 3x_4 = 0$

$4x_3 + 4x_4 = 0$

We want to find values for $x_1$, $x_2$, $x_3$ and $x_4$ which satisfy the above set of equations.

Since $x_2$ and $x_4$ are free variables, we can give them any values we wish - they are free variables, after all! The simples choice for our free variables is ones or zeroes.

Let’s set $x_2 = 1$ and $x_4 = 0$:

$x_1 + (1) + 2x_3 + 3(0) = 0$

$4x_3 + 4(0) = 0$

which gives:

$\label{2} \tag{2} x_1 + 1 + 2x_3 = 0$

$\label{3} \tag{3} x_3 = 0$

Great! We now know that $x_3 = 0$ (and we have values for $x_2$ and $x_4$ ). All that’s left is for us to find $x_1$

Now, let’s use back substitution to plug $x_3 = 0$ into equation ($\ref{2}$):

$x_1 + 1 + 2(0) = 0$

so

$x_1 = -1$

We now have our first special solution:

$\begin{bmatrix} x_1 \newline x_2 \newline x_3 \newline x_4 \end{bmatrix} % = % \begin{bmatrix} -1 \newline 1 \newline 0 \newline 0 \end{bmatrix}$

Setting $x_2 = 1$ and $x_4 = 0$ gave us our first special solution. We need to find our second special solution. The easiest way to do this is to set $x_2 = 0$ and $x_4 = 1$. Using the same procedure as we did for the first special solution we find our second special solution:

$\begin{bmatrix} -1 \newline 0 \newline -1 \newline 1 \end{bmatrix}$

Since we had 2 free variables, we get 2 special solutions:

$\begin{bmatrix} -1 \newline 1 \newline 0 \newline 0 \end{bmatrix}, \begin{bmatrix} -1 \newline 0 \newline -1 \newline 1 \end{bmatrix}$

These two special solutions are in the nullspace of $U$ and therefore also $A$. This means that if you multiply the matrix $A$ or $U$ by these vectors, you will get the zero vector.

Crucially - every combination of our special solutions is in the nullspace.

So, the complete solution is:

$x % = % x_2 % \begin{bmatrix} -1 \newline 1 \newline 0 \newline 0 \end{bmatrix} % + % x_4 % \begin{bmatrix} -1 \newline 0 \newline -1 \newline 1 \end{bmatrix} % = % \begin{bmatrix} -x_2 - x_4 \newline x_2 \newline -x_4 \newline x_4 \end{bmatrix}$

What does this actually mean? It means that we can pick our free variables $x_2$ and $x_4$ however we wish, and we will get a valid solution in response.

For example let’s pick: $x_2=1$ and $x_4 = 2$. So:

$\begin{bmatrix} -x_2 - x_4 \newline x_2 \newline -x_4 \newline x_4 \end{bmatrix} % = % \begin{bmatrix} -3 \newline 1 \newline -2 \newline 2 \end{bmatrix}$

which was my example at the beginning of the article.

## 4 Another Example

Say now I wish to find the nullspace of

$U % = % \begin{bmatrix} 1 & 5 & 7 \newline 0 & 0 & 9 \newline \end{bmatrix}$

The second column contains our free variable. Since there is only one free variable, there will only be one special solution.

We need to back substitute $x_2 = 1$ (no need to set zeroes this time, as there is only one free variable).

Then $9x_3 = 0$

gives

$x_3 = 0$

and

$x_1 + 5x_2 = 0$

so

$x_1 = -5$

since

$x_2 = 1$

So, $x % = % x_2 % \begin{bmatrix} -5 \newline 1 \newline 0 \end{bmatrix}$

The nullspace in this case is a line in $\mathbb{R}^3$. It contains multiples of the special solution.