Let $A$ be a $3 \times 4$ matrix and it has a particular solution of $\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= \begin{bmatrix} 1\\ 0\\ -1\\ 0 \end{bmatrix}$
The nullspace of $A$ is $\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}= t\begin{bmatrix} 1\\ 1\\ 0\\ 1 \end{bmatrix}+ s\begin{bmatrix} -2\\ 1\\ 1\\ 0 \end{bmatrix}$, where $t,s \in \mathbb{R}$.
From this, do I have enough information to find the column space of $A$? I am thinking that I could because I can find out the reduced row echelon form of $A$ by working backwards from the nullspace. And I could get the $rref(A)=\begin{bmatrix} 1 & 0 & 2 & -1\\ 0 & 1& -1 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix}$. From here, isn't the column space of $A$ just the $span(\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix})$?
We could also determine that the left-nullspace is going to be of rank 1.
But from the book that I am reading, it says that I should not have enough information to find the column space of $A$. Why is it so? Isn't the span I wrote above the column space of $A$ already?