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.
Best Answer
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.