I have to get a basis and dimension for the null space of a matrix A:
$$
\begin{pmatrix}
1 & 2 & 0 & 0 & 2 & -3 \\
2 & 2 & 2 & 2 & 2 & 2 \\
1 & 2 & 1 & 4 & 3 & 6 \\
3 & 4 & 3 & 6 & 5 & 8 \\
\end{pmatrix}
$$
I attained the reduced row echelon form of the matrix:
$$
\begin{pmatrix}
1 & 0 & 0 & -6 & -2 & -13 \\
0 & 1 & 0 & 3 & 2 & 5 \\
0 & 0 & 1 & 4 & 1 & 9 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}
$$
I know that the dimension of the null space is equal to the number of free variables, which in the case is $3$ (the last three columns). So, the dimension of the null space of $A$ is $3$.
I'm aware of this method for finding the null space: Finding the basis of a null space
However, I'm finding it hard to wrap my head around as to how I can apply it here as the $RREF(A)$ is a little more complex.
Best Answer
It is always the same procedure. Consider the pivots in red.
$$ \begin{pmatrix} \color{red}{1} & 0 & 0 & -6 & -2 & -13 \\ 0 & \color{red}{1} & 0 & 3 & 2 & 5 \\ 0 & 0 & \color{red}{1} & 4 & 1 & 9 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} $$
We always write stuff based on the pivots. The first row would become: $$x_1-6x_4-2x_5-13x_6=0\to x_1=6x_4+2x_5+13x_6$$
Similarly for row 2 and 3,
$$x_2+3x_4+2x_5+5x_6=0\to x_2=-3x_4-2x_5-5x_6$$
$$x_3+4x_4+x_5+9x_6=0\to x_3=-4x_4-x_5-9x_6$$
For row 4, there are no pivots, so we simply have $x_4=x_4$.
Notice that since we have $6$ columns, we do indeed have a $x_5$ and $x_6$, but similarly they are not pivots, and so $x_5 = x_5$, $x_6=x_6$.
Now, we can rewrite the system as the following:
$$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{pmatrix}= \begin{pmatrix} 6x_4+2x_5+13x_6 \\ -3x_4-2x_5-5x_6 \\ -4x_4-x_5-9x_6 \\ x_4 \\ x_5 \\ x_6 \\ \end{pmatrix} = x_4\begin{pmatrix} 6 \\ -3 \\ -4 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} + x_5 \begin{pmatrix} 2 \\ -2 \\ -1 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} + x_6 \begin{pmatrix} 13 \\ -5 \\ -9 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix} $$
So your basis for null space would be:
$$\{\begin{pmatrix} 6 \\ -3 \\ -4 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} ,\begin{pmatrix} 2 \\ -2 \\ -1 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} ,\begin{pmatrix} 13 \\ -5 \\ -9 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}\}$$