Finding a spanning set for a null space I have a matrix $A$ that is like this:
\begin{equation}
A = \pmatrix{
1 & 2 & -3 & 1 & 5 \\
1 & 3 & -1 & 4 & -2 \\
1 & 1 & -5 & -2 & 12 \\
1 & 4 & 1 & 7 & -7
}
\end{equation}
The question is: Find a set of $5\times 1$ matrices whose linear span is the null space of $A$.
I did Gauss-Jordan and I got the matrix down to:
\begin{equation}
\pmatrix{
1 & 2 & -3 & 1 & 5 \\
0 & -1 & -2 & -3 & -7 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -2
}
\end{equation}
But doesn't that mean the matrix is inconsistent therefore there is no linear span? Any ideas? Thanks

Your reduced matrix is correct. First you need to characterize the set of vectors $x$ that satisfy $A x = 0$. This set is called the null space or kernel, and I use the standard notation $\ker A$.
The reduction process above corresponds to pre-multiplying $A$ by an invertible matrix $G$ such that $G A = \tilde{A}$, where $\tilde{A}$ is the reduced matrix above. Since $G$ is invertible, you have $Ax = 0 $ iff $\tilde{A}x = 0$, or in other words, $\ker A = \ker \tilde{A}$. So we can focus on finding $\ker \tilde{A}$, since the matrix has a nicer form.
Suppose $ \tilde{A}\pmatrix{x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5} = 0$. Then you can immediately see that we must have $x_5 = 0$. The third row tells us nothing. And the first two rows can be re-written as
$\pmatrix{ 1 && 2 \\ 0 && 1} \pmatrix{x_1 \\ x_2}+\pmatrix{ -3 && 1 \\ 2 && 3} \pmatrix{x_3 \\ x_4} = 0$. Since $\pmatrix{ 1 && 2 \\ 0 && 1}^{-1} = \pmatrix{ 1 && -2 \\ 0 && 1}$, we get
$\pmatrix{x_1 \\ x_2} = \pmatrix{ 7 && 5 \\ -2 && -3} \pmatrix{x_3 \\ x_4}$.
This tells us that if we select $x_3, x_4$, then $x_1, x_2$ are completely determined. Thus $ \tilde{A}x = 0$ iff $x_5 = 0$, $x_3, x_4$ are arbitrary, and $x_1,x_2$ given by the above formula, or in other words $\ker A = \{\pmatrix{7 x_3+5 x_4 \\ -2x_3-3 x_4 \\ x_3 \\ x_4 \\ 0} | x_3, x_4 \text{arbitrary} \}$.
Now you want to find a more convenient way of expressing this. Note that we can write $\pmatrix{7 x_3+5 x_4 \\ -2x_3-3 x_4 \\ x_3 \\ x_4 \\ 0} = x_3 \pmatrix{7  \\ -2 \\ 1 \\ 0 \\ 0} + x_4 \pmatrix{5 \\ -3 \\ 0 \\ 1 \\ 0}$. So we can write $\ker A = \text{sp} \{\pmatrix{7  \\ -2 \\ 1 \\ 0 \\ 0}, \pmatrix{5 \\ -3 \\ 0 \\ 1 \\ 0} \}$. That is, the null space of $A$ is the span of these two vectors.
