The null space of a matrix, denoted \(\text{Nul }A\), is the set of all solutions to the homogeneous equation \(A\vec{x}=\vec{0}\). Since the homogeneous equation always has the trivial solution (\(\vec{x} = \vec{0}\)), we know the zero vector is always in the null space of a matrix. Let’s use an example to explore what other vectors are in the null space.

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## Example of describing the null space of a matrix

Describe the null space of the matrix: \(A = \left[\begin{array}{cccc} 1 & 3 & 2 & 1\\ 4 & 1 & 0 & 1\\ 2 & 1 & 1 & 4\\ \end{array}\right]\)

It may seem strange to use the word “describe”, but “find” doesn’t really make sense here. What we are going to do is describe the null space of matrix A as the span of a set of vectors. This is similar to the column space of a matrix.

Every matrix equation can be written as a vector equation or an augmented matrix. Here, an augmented matrix is the most useful. So we will write \(A\vec{x}=\vec{0}\) as an augmented matrix.

\(\left[\begin{array}{cccc|c} 1 & 3 & 2 & 1 & 0\\ 4 & 1 & 0 & 1 & 0\\ 2 & 1 & 1 & 4 & 0\\ \end{array}\right]\)

Row reducing this matrix using our calculators, we get:

\(\left[\begin{array}{cccc|c} 1 & 3 & 2 & 1 & 0\\ 4 & 1 & 0 & 1 & 0\\ 2 & 1 & 1 & 4 & 0\\ \end{array}\right] \sim \left[\begin{array}{cccc|c} 1 & 0 & 0 & \frac{8}{7} & 0\\ 0 & 1 & 0 & -\frac{25}{7} & 0\\ 0 & 0 & 1 & \frac{37}{7} & 0\\ \end{array}\right]\)

Let’s write a kind of general solution to this since the system has free variables. What we will do is write a vector in the null space and solve for those vectors in the pivot columns and write the solution of each as an entry in the vector. Free variables will remain themselves.

If \(\vec{x} \in \text{Nul }A\), then:

\(\vec{x} = \left[\begin{array}{c} -\frac{8}{7}x_4\\ \frac{25}{7}x_4\\ -\frac{37}{7}x_4\\ x_4 \end{array}\right]\)

If you aren’t sure about where these values came from, write out the general solution to the system of equations. You will see that that is the same as this, this just written in vector form.

Now \(x_4\) is a constant, so we can factor it out.

\(\vec{x} = x_4\left[\begin{array}{c} -\frac{8}{7}\\ \frac{25}{7}\\ -\frac{37}{7}\\ 1 \end{array}\right]\)

The free variable \(x_4\) can take on any real number value. Pick a value, and you have found a solution to the equation we started with. In other words, pick a value and you have found a vector in \(\text{Nul }A\). When you have multiples of a single vector, you are looking at linear combinations of that vector. By definition, that is the span of a set of vectors, or a vector in this case.

Therefore:

\(\text{Nul }A = \text{Span}\{\left[\begin{array}{c} -\frac{8}{7}\\ \frac{25}{7}\\ -\frac{37}{7}\\ 1 \end{array}\right]\}\)

This is what we mean by a description of the null space of the matrix.

### Some vectors in the null space

There are infinitely many vectors in the null space of this matrix since the span of a set of vectors includes ALL linear combinations. So, you can find vectors in the null space simply by finding multiples of this vector. For example:

\((7)\left[\begin{array}{c} -\frac{8}{7}\\ \frac{25}{7}\\ -\frac{37}{7}\\ 1 \end{array}\right] = \left[\begin{array}{c} -8\\ 25\\ -37\\ 7 \end{array}\right] \in \text{Nul }A\)

You already know that \(\vec{0} \in \text{Nul }A\) since not only is it a multiple of this vector but it is also the trivial solution to the homogeneous equation \(A\vec{x}=\vec{0}\). Remember all solutions to this equation defines the null space.

### Other spaces associated with a matrix

You can also talk about the column space or the row space of a matrix.