How to use MATLAB to find a basis Nullspace of A

Enter your matrix $A$ in MATLAB.

1. Type

$A= \left[ \begin{array}{rrrrrrr} 1 & 3 & 0 & 2 & 6 & 3 & 1\\ -2 & -6 & 0 & ... ...& 9 & 0 & 0 & 6 & 6 & 2\\ -1 & -3 & 0 & 1 & 0 & 9 & 3\\ \end{array} \right]$

2. Find $rref(A)$ you will get

$A= \left[ \begin{array}{rrrrrrr} 1 & 3 & 0 & 0 & 2 & 0& 0 \\ 0 & 0& 0 & 1 & ... ...0& 0& 1 & { {1} \over {3} } \\ 0& 0 &0 &0 & 0 & 0 & 0 \\ \end{array} \right]$

cm

3. Solve for leading variables:

$\begin{array}{rrrrr} x_1= & -3x_2 & & -2x_5 \\ x_4= & -2x_5 & & \\ x_6= & -{ {1} \over {3} } x_7 & & \\ \end{array}$

cm

Set $x_2 = r, x_3 = s , x_5 = t$ and $x_7=w$ we obtain

$\left[ \begin{array}{rrrrrr} x_1= & &-3r & & -2t & \\ x_4= & & & -2t & & \\ x_6= & & & & &-{ {1} \over {3} } w \\ \end{array} \right]$

In matrix form we can write this as

$\left[ \begin{array}{r} x_1 \ x_2 \ x_3 \ x_4 \ x_5 \ x_6 \ x_7 \\ \end... ...}{r} 0 \ 0 \ 0 \ 0 \ 0 \ - { {1} \over {3} } \ 1 \\ \end{array} \right]$

So the vectors

$v_1 = \left[ \begin{array}{r} -3 \ 1 \ 0 \ 0 \ 0 \ 0 \ 0 \ \end{array}\... ...ay}{r} 0 \ 0 \ 0 \ 0 \ 0 \ { {1} \over {3} } \ 1 \ \end{array}\right]$

form a basis for the solution space of $AX=0$. The solution space of $AX=0$ is called the nullspace of $A$ and denoted by $null(A)$.

Example 6 of section 1.2, Example 10 of section 5.4, Example 4 of section 5.5 show how to solve $AX=0$ and how to find a basis for the nullspace of $A$.

Dimension of the nullspace of $A$ is called nullity of A and denoted by $nullity(A)$.

NOTE: in the above example $rank(A) = 3$ and $nullity(A) = 4.$ Recall the dimension theorem : Is $A$ is an $m \times n$ matrix then $rank(A) + nullity(A) = n$.