Consider the matrix
$$\begin{pmatrix}1&5\\2&10\end{pmatrix}.$$
Its column space is one-dimensional, spanned by $\binom12$ (or equivalently by $\binom5{10}$ since these vectors are proportional). After a row operation, subtracting twice the first row from the second, the matrix becomes
$$\begin{pmatrix}1&5\\0&0\end{pmatrix}.$$
The column space of this is still one-dimensional, but it's a quite different one-dimensional space from before; the new one is spanned by $\binom10$. (Edit to correct an error in a comment: The column space of the original matrix was a line of slope $2$ (not $5$); the column space of the new matrix has slope $0$.)
Remember, you have no solution to a system if say, you have a system
$$
\left[
\begin{array}{ccc c|c}
1 & 0 & 0 & 0 & a \\
0 & 1 & 0 & 0 & b \\
0 & 0 & 0 & 0 & c\\
\end{array}
\right]$$
where $c\neq 0$. This is because the system essentially says $0=c$.
The nullspace is the subspace of all solutions to $A\mathbf{x}=\mathbf{0}$, and the column space is the span of the columns.
To find the null space, you simply want to determine a basis for the solution vectors of the homogeneous system. For the reduced row-echelon form of your homogeneous system, you have
$$
\left[
\begin{array}{ccc c|c}
1 & 0 & 0 & 2 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0\\
\end{array}
\right]$$
Let's say each respective column corresponds to the variables $x_{1}$, $x_{2}$, $x_{3}$, and $x_{4}$. Since the column for $x_{4}$ does not have a leading $1$, we let $x_{4}$ be the free variable.
Then the RREF gives us the following solution to the homogeneous system:
$$x_{1}+2x_{4}=0\ \mathrm{or}\ x_{1}=-2x_{4}$$
$$x_{2} =0$$
$$x_{3}+x_{4}=0\ \mathrm{or}\ x_{3}=-x_{4}$$
In vector form, we have
$$\begin{bmatrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{bmatrix}=\begin{bmatrix}-2x_{4} \\ 0 \\ -x_{4} \\ x_{4}\end{bmatrix}=\begin{bmatrix}-2x_{4} \\ 0x_{4} \\ -x_{4} \\ x_{4}\end{bmatrix}=\begin{bmatrix}-2 \\ 0 \\ -1 \\ 1\end{bmatrix}x_{4}.$$
So the nullspace of your original matrix is the span of the vector $\begin{bmatrix}-2 \\ 0 \\ -1 \\ 1\end{bmatrix}$, since $x_{4}$ can take any value.
For the column space, you need to look at the columns in the RREF that have leading $1$'s. the column space will be the span of the columns from your original matrix which have a leading $1$ in the RREF (i.e. the column space is the span of columns $1$, $2$, and $3$).
You should post the original matrix that you're working with if you are going to ask this type of question, so that we can fully answer your question and ensure you did not make any mistakes after applying row operations.
Best Answer
A matrix has a $0$-dimensional null-space if and only if its columns are linearly independent. This follows immediately from the definitions of null-space and linear independence.
If the matrix is square, then this is equivalent to its columns being linearly independent (and to the matrix having maximal rank, and a non-zero determinant, etc.)
If the matrix has more rows than columns, then its rows cannot possibly be linearly independent, and so linear independence of the rows tells you nothing about anything.
If the matrix has more columns than rows, then its rows may or may not be linearly independent, but the matrix cannot possibly have a non-zero null-space.