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).
Why not just reduce the matrix ? its not more work and often makes things easier.
\begin{bmatrix}
1&0&0&0
\\0&1&0&0
\\0&0&1&0
\end{bmatrix}
1) We see that the row and column rank (they are always equal) are $3$.
The nullity $=n-$rank$=4-3=1$.
To find the null space we have to solve $A\mathbf{x}=0$, and this is easy now in row reduced form $\mathbf{x}=\begin{bmatrix}
0\\0\\0\\a
\end{bmatrix}$.
2) The row space has dimension $3$ as mentioned, for the basis one can take:
$$(1,0,0,0)$$
$$(0,1,0,0)$$
$$(0,0,1,0)$$
Or one could take the rows of the original matrix, since the rank is $3$.
3) The column rank is also $3$ row reduction has not changed the column vectors, just expressed them in a different basis so a basis for the column space will be the first $3$ vectors of the original matrix (corresponding to the pivot position of the reduced matrix):
$$\begin{bmatrix}
1\\3\\-1
\end{bmatrix},
\begin{bmatrix}
1\\-1\\2
\end{bmatrix},
\begin{bmatrix}
-1\\2\\-4\\
\end{bmatrix}. $$
Best Answer
The basic principal is that elementary row options do not change the rowspace of a matrix. To prove that, you could demonstrate that the processes of scaling a basis element, interchanging two basis elements, and adding multiples of one basis element to another, all result in a new basis.
So, if you perform elementary row operations and change it entirely to row-echelon form, you are left with a matrix that has some zero rows and some nonzero rows. The row-echelon shape makes it obvious that the nonzero rows are linearly independent. Can you see why?
Since this is the case, the remaining rows are a basis for the rowspace of the matrix.
Basic row operations do not change the (right) nullspace either, so you can use the new matrix to work with the nullspace. Using that matrix instead of the original matrix, it will be particularly easy to solve the equation $A'x=0$ with back substitution to discover what is in the nullspace.