Finding the column space from nullspace and particular solution 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?
 A: What you wrote is the column space of $rref(A)$, not the column space of $A$.  The column space changes when you perform row operations on the matrix.  You have enough information to find the dimension of the column space (since that doesn't change when you perform row operations), but not the column space itself.
A: *

*It makes no sense to talk about "particular solution of a $3\times 4$ matrix." Matrices don't have "particular solutions", systems of equations have particular solutions. If you are interpreting $A$ as the augmented matrix of a $3\times 3$ system of linear equations, then you need to say that this is what you mean. If you are interpreting $A$ as the coefficient matrix of a $3\times 4$ system of homogeneous linear equations, then you need to say so.  If you are interpreting $A$ as the coefficient matrix of a $3\times 4$ system of linear equations with unspecified "solution vector" $\mathbf{b}$, then you need to say so as well.
I am guessing that what you have is that $A$ is the coefficient matrix of a system $A\mathbf{x}=\mathbf{b}$, and that the vector you give happens to be a particular solution to that system.


*Elementary row operations do not preserve the columnspace. That is, the columnspace of the reduced row echelon form of $A$ is not the same as the columnspace of $A$. For example, if $A$ is a matrix that consists only of $1$s in every entry, then the reduced row-echelon form of $A$ is a row of $1$s and all the other rows $0$. The column space of $REF(A)$ is then the span of $(1,0,\ldots,0)^t$; but the column space of $A$ is the span of $(1,1,\ldots,1)^t$, which is different.


*However, say that $C$ is a matrix obtained from $A$ by the application of elementary row operations (for example, $C$ is the reduced row echelon form of $A$, or the row echelon form of $A$). If the $j_1,j_2,\ldots,j_k$ columns of $C$ are a basis for the columnspace of $C$, then the corresponding columns of $A$ are a basis for the columnspace of $A$. In your example, the first and second columns of the reduced row echelon form of $A$ are a basis for its columnspace. That means that the first and second columns of $A$  are a basis for the columnspace of $A$.
