# basis for null space of matrix

• Mar 16th 2009, 05:28 PM
antman
basis for null space of matrix
How do you find a basis for the null space of the matrix:

$\begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}$
• Mar 16th 2009, 09:13 PM
ThePerfectHacker
Quote:

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

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

The first thing you should do is reduce this by Gaussian-Jordan elimination.
After you do this I tell you what to do next.
• Mar 17th 2009, 09:10 AM
antman
Reduced matrix
$
\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}
$
• Mar 17th 2009, 12:08 PM
ThePerfectHacker
Quote:

Originally Posted by antman
Reduced matrix
$
\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 $m\times n$ matrix $A$ the nullspace is the set of all $\bold{x}\in \mathbb{R}^n$ so that $A\bold{x} = \bold{0}$ where $\bold{0}$ is zero vector in $\mathbb{R}^m$. To find the nullspace you basically need to solve the homogenous equation $A\bold{x} = \bold{0}$. If you convert this problem in Gaussian-Jordan elimination, it means you need to solve the equation $(A|\bold{0})$ where $\bold{0}$ 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):
$
\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):
$-x_1 = 0$
$x_2 - 3x_4 + 6x_5 = 0$
$x_3 -x_4 + 3x_5 = 0$

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

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

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

Can you find a basis now?
• Mar 17th 2009, 07:47 PM
antman
Would the basis just be { $
\begin{pmatrix}0\\3\\1\\1\\0\end{pmatrix},\begin{p matrix}0\\-6\\-3\\0\\1\end{pmatrix}
$
}