Finding the basis of a null space

The null space of $A$ is the set of solutions to $A{\bf x}={\bf 0}$. To find this, you may take the augmented matrix $[A|0]$ and row reduce to an echelon form. Note that every entry in the rightmost column of this matrix will always be 0 in the row reduction steps. So, we may as well just row reduce $A$, and when finding solutions to $A{\bf x}={\bf 0}$, just keep in mind that the missing column is all 0's.

Suppose after doing this, you obtain $$ \left[\matrix{1&0&0&0&-1 \cr 0&0&1&1&0 \cr 0&0&0&0&0 \cr 0&0&0&0&0 \cr }\right] $$

Now, look at the columns that do not contain any of the leading row entries. These columns correspond to the free variables of the solution set to $A{\bf x}={\bf 0}$ Note that at this point, we know the dimension of the null space is 3, since there are three free variables. That the null space has dimension 3 (and thus the solution set to $A{\bf x}={\bf 0}$ has three free variables) could have also been obtained by knowing that the dimension of the column space is 2 from the rank-nullity theorem.

The "free columns" in question are 2,4, and 5. We may assign any value to their corresponding variable.

So, we set $x_2=a$, $x_4=b$, and $x_5=c$, where $a$, $b$, and $c$ are arbitrary.

Now solve for $x_1$ and $x_3$:

The second row tells us $x_3=-x_4=-b$ and the first row tells us $x_1=x_5=c$.

So, the general solution to $A{\bf x}={\bf 0}$ is $$ {\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right] $$

Let's pause for a second. We know:

1) The null space of $A$ consists of all vectors of the form $\bf x $ above.

2) The dimension of the null space is 3.

3) We need three independent vectors for our basis for the null space.

So what we can do is take $\bf x$ and split it up as follows:

$$\eqalign{ {\bf x}=\left[\matrix{c\cr a\cr -b\cr b\cr c}\right] &=\left[ \matrix{0\cr a\cr 0\cr 0\cr 0}\right]+ \left[\matrix{c\cr 0\cr 0\cr 0\cr c}\right]+ \left[\matrix{0\cr 0\cr -b\cr b\cr 0}\right]\cr &= a\left[ \matrix{0\cr1\cr0\cr 0\cr 0}\right]+ c\left[ \matrix{1\cr 0\cr 0\cr 0\cr 1}\right]+ b\left[ \matrix{0\cr 0\cr -1\cr 1\cr 0}\right]\cr } $$ Each of the column vectors above are in the null space of $A$. Moreover, they are independent. Thus, they form a basis.

I'm not sure that this answers your question. I did a bit of "hand waving" here. What I glossed over were the facts:

1)The columns of the echelon form of $A$ that do not contain leading row entries correspond to the "free variables" to $A{\bf x}={\bf 0}$. If the number of these columns is $r$, then the dimension of the null space is $r$ (again, if you know the dimension of the column space, you can see that the dimension of the null space must be the number of these columns from the rank-nullity theorem).

2) If you split up the general solution to $A{\bf x}={\bf 0}$ as done above, then these vectors will be independent (and span of course since you'll have $r$ of them).

There is a mechanical part in the accepted (+1) answer that may possibly need a more step-by-step explanation, namely the process behind

what we can do is take $\mathbf x$ and split it up

Reducing the matrix $A$ to the reduced row echelon (rref) form results in:

$$A=\left(\begin{array}{rrrrrr} 1 & 3 & -2 & 0 & 2 & 0 \\ 2 & 6 & -5 & -2 & 4 & -3 \\ 0 & 0 & 5 & 10 & 0 & 15 \\ 2 & 6 & 0 & 8 & 4 & 18 \end{array}\right)\to\left(\begin{array}{rrrrrr} \bbox[5px,border:2px solid red]1 & 3 & 0 & 4 & 2 & 0 \\ 0 & 0 & \bbox[5px,border:2px solid red]1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \bbox[5px,border:2px solid red]1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$$

There are three pivot columns corresponding to the pivot $1$'s in red, and the rank-nullity theorem tells us that there are $n-r=3$ vectors in any basis of the $N(A)$, corresponding to the free variables.

At this point it is worth remembering where all this comes from: the homogeneous system of linear equations $A\vec x = \vec 0,$ which has now been reduced to:

$$\begin{align} 1x_1 + 3x_2+ 0x_3+4x_4+2x_5 +0x_6 &=0\\ 0x_1 + 0x_2+ 1x_3+2x_4+0x_5 +0x_6 &=0\\ 0x_1 + 0x_2+ 0x_3+0x_4+0x_5 +1x_6 &=0\\ 0x_1 + 0x_2+ 0x_3+0x_4+0x_5 +0x_6 &=0\\ \end{align}$$

Expressing the pivot variables in terms of the free variables:

$$\begin{align} 1x_1 &= - 3\;\color{blue}{x_2} - 4\;\color{red}{x_4} - 2\;\color{magenta}{x_5} \\ 1x_3 &= -2\;\color{red}{x_4}\\ 1x_6 &=\;0\\ \end{align}$$

immediately shows the way the basis vectors of the $N(A)$ will be filled in from their "skeleton" form simply indicating the column where the free variable in question is located (i.e. $x_2,x_4,x_5$):

$$\left\{\begin{bmatrix}0\\\color{blue}1\\0\\0\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\\color{red}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\0\\0\\\color{magenta}1\\0\end{bmatrix}\right\}\to \color{blue}{x_2}\,\begin{bmatrix}0\\1\\0\\0\\0\\0\end{bmatrix}+\color{red}{x_4}\,\begin{bmatrix}0\\0\\0\\1\\0\\0\end{bmatrix}+\color{magenta}{x_5}\,\begin{bmatrix}0\\0\\0\\0\\1\\0\end{bmatrix}$$

to the final form:

$$\color{blue}{x_2}\,\begin{bmatrix}-3\\1\\0\\0\\0\\0\end{bmatrix}+\color{red}{x_4}\,\begin{bmatrix}-4\\0\\-2\\1\\0\\0\end{bmatrix}+\color{magenta}{x_5}\,\begin{bmatrix}-2\\0\\0\\0\\1\\0\end{bmatrix}\to \text{basis }N(A)= \left\{\begin{bmatrix}-3\\\color{blue}1\\0\\0\\0\\0\end{bmatrix},\begin{bmatrix}-4\\0\\-2\\\color{red}1\\0\\0\end{bmatrix},\begin{bmatrix}-2\\0\\0\\0\\\color{magenta}1\\0\end{bmatrix}\right\}$$

So it boils down to changing the signs of the entries in the rref, and keeping track of the free variables.

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require("pracma")
A = matrix(c(1,3,-2,0,2,0,2,6,-5,-2,4,-3,0,0,5,10,0,15,2,6,0,8,4,18), nrow=4, byrow=T)
x2= c(-3,1,0,0,0,0); A %*% x2; x4=c(-4,0,-2,1,0,0); A %*% x4; x5=c(-2,0,0,0,1,0); A %*% x5