I'm supposed to find the column space, null space, and the special solution for a matrix. Changing to reduced row echelon form I get...
$ \begin{bmatrix} 1 & 0 & 0 & 2\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$
That is the furthest I get for reduced echelon row.
Would there be no solution because of that last row not being all zeros? Can someone elaborate for the column space, null space and special solution for this, thanks.
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.
By looking at the row reduced matrix, we can see that the rank of the matrix is $3$ because there are $3$ pivots. Therefore, the matrix has 3 linearly independent column vectors in $\mathbb{R}^3$ and so the column space of $A$ is $\mathbb{R}^3$. (Note, normally you would need the original matrix to find the column space).
Since the rank of the matrix is $r = 3$, the nullity of the matrix is $n-r = 1$, meaning the nullspace of the matrix is 1-dimensional. Furthermore, the nullspace of the original matrix is the same as the nullspace of the row reduction of the matrix. If $\begin{bmatrix} x_1 & x_2 & x_3 & x_4 \end{bmatrix}^T$ is a column vector in the nullspace of the row reduced matrix, then: $$ \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\x_4 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix} $$ \begin{align} x_1 + 2x_4 = 0 \\ x_2 = 0 \\ x_3 + x_4 = 0 \end{align} We can paramatrize the solutions as follows: \begin{align} x_1 &= -2t \\ x_2 &= 0 \\ x_3 &= -t \\ x_4 &= t \end{align} and see that the nullspace is spanned by the vector $\vec{v} = \langle 2,0,1, -1 \rangle$, that is, the nullspace is the set of all multiples of $\vec{v}$ $$ $$
