# Thread: basis for null space of matrix

1. ## basis for null space of matrix

How do you find a basis for the null space of the matrix:

$\displaystyle \begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}$

2. Originally Posted by antman
How do you find a basis for the null space of the matrix:

$\displaystyle \begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}$
The first thing you should do is reduce this by Gaussian-Jordan elimination.
After you do this I tell you what to do next.

3. Reduced matrix
$\displaystyle \begin{pmatrix} -1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & -3 & 6\\0 & 0 & 1 & -1 & 3\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{pmatrix}$

4. Originally Posted by antman
Reduced matrix
$\displaystyle \begin{pmatrix} -1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & -3 & 6\\0 & 0 & 1 & -1 & 3\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{pmatrix}$
Given a $\displaystyle m\times n$ matrix $\displaystyle A$ the nullspace is the set of all $\displaystyle \bold{x}\in \mathbb{R}^n$ so that $\displaystyle A\bold{x} = \bold{0}$ where $\displaystyle \bold{0}$ is zero vector in $\displaystyle \mathbb{R}^m$. To find the nullspace you basically need to solve the homogenous equation $\displaystyle A\bold{x} = \bold{0}$. If you convert this problem in Gaussian-Jordan elimination, it means you need to solve the equation $\displaystyle (A|\bold{0})$ where $\displaystyle \bold{0}$ was an attached coloumn of zeros (corresponding to the fact that we are solving a homogenous system). Therefore, you end up with (notice the new coloum of zeros):
$\displaystyle \begin{pmatrix} -1 & 0 & 0 & 0 & 0&0\\0 & 1 & 0 & -3 & 6&0\\0 & 0 & 1 & -1 & 3&0\\0 & 0 & 0 & 0 & 0&0\\0 & 0 & 0 & 0 & 0&0\end{pmatrix}$
This matrix is telling us that (the last two coloums do not tell us anything):
$\displaystyle -x_1 = 0$
$\displaystyle x_2 - 3x_4 + 6x_5 = 0$
$\displaystyle x_3 -x_4 + 3x_5 = 0$

Thus,
$\displaystyle x_1=0$
$\displaystyle x_2 = 3x_4 - 6x_5$
$\displaystyle x_3 = x_4 - 3x_5$.

If we let $\displaystyle r,s\in \mathbb{R}$ then:
$\displaystyle x_1=0$
$\displaystyle x_2 = 3r - 6s$
$\displaystyle x_3 = r - 3s$
Are always in the nullspace.

Thus, the nullspace is, $\displaystyle \left\{ \begin{bmatrix}0\\3r-6s\\r-3s\\r\\s \end{bmatrix} \right\}$

Can you find a basis now?

5. Would the basis just be {$\displaystyle \begin{pmatrix}0\\3\\1\\1\\0\end{pmatrix},\begin{p matrix}0\\-6\\-3\\0\\1\end{pmatrix}$}