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
For the column space: note that the reduced row-echelon form of $A$ is given by $$\mathrm{RREF}(A)=\begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$
To find the column space of $A$, we want to look at the columns in $A$ which have leading $1$'s in $\mathrm{RREF}(A)$. Since columns one and two have leading $1$'s, then that tells us that $$\left\{\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\right\}$$ forms a basis for $\mathrm{col}(A)$.
Moreover, the entries in columns three and four tell us exactly how to write said columns as a linear combination of the basis:
$$\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}=1\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}+0\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix},$$ $$\begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix}=2\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}+1\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}.$$ Thus, columns three and four are indeed in $$\mathrm{span}\left\{\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\right\},$$ and $\left\{\begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}\right\}$ is a linearly independent set (if not, $\mathrm{RREF}(A)$ would not have a leading $1$ in both of the first two columns).