Null Space of a Matrix

The null space (or kernel) is the set of all solutions to the homogeneous equation. In other words, it is the set of all vectors x such that:

$$A\vec{x}=\vec{0}$$

Note: $\mathrm{Nul}(A)$ is a subspace of ${\mathbb{R}}^n$

Example

$$A=\left[\begin{array}{ccccc}-3& 6& -1& 1& -7\\ 1& -2& 2& 3& -1\\ 2& -4& 5& 8& -4\end{array}\right]$$

To find $\mathrm{Nul}(A)$, find the solution set to the augmented matrix of $A\vec{x}=\vec{0}$. This can be done using row reduction to RREF form.

$$\left[\begin{array}{cccccc}-3& 6& -1& 1& -7& 0\\ 1& -2& 2& 3& -1& 0\\ 2& -4& 5& 8& -4& 0\end{array}\right]\to \left[\begin{array}{cccccc}1& -2& 0& -1& 3& 0\\ 0& 0& 1& 2& -2& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right]$$

From this, we get 3 free variables which we can assign parameters to:

$$\{\begin{array}{l}{x}_2=s\\ x_4=t\\ x_5=u\end{array}$$

This generates the following system of equations:

$$\{\begin{array}{l}{x}_1=2s+t-3w\\ x_3=-2t+2w\end{array}$$

forming the general solution

$$\vec{x}=\left[\begin{array}{c}2s+t-3w\\ s\\ -2t+2w\\ t\\ w\end{array}\right]=\left[\begin{array}{c}2\\ 1\\ 0\\ 0\\ 0\end{array}\right]s+\left[\begin{array}{c}1\\ 0\\ -2\\ 1\\ 0\end{array}\right]t+\left[\begin{array}{c}-3\\ 0\\ 2\\ 0\\ 1\end{array}\right]w$$

it follows that the vector coefficients of the free variables form a spanning set of the null space.

$$\mathrm{Nul}(A)=\mathrm{span}\left\{\left[\begin{array}{c}2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 0\\ -2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 2\\ 0\\ 1\end{array}\right]\right\}$$