We saw in Example VFSAL that when a system of equations is homogeneous the solution set can be expressed in the form described by Theorem VFSLS where the vector $\vect{c}$ is the zero vector. We can essentially ignore this vector, so that the remainder of the typical expression for a solution looks like an arbitrary linear combination, where the scalars are the free variables and the vectors are $\vectorlist{u}{n-r}$. Which sounds a lot like a span. This is the substance of the next theorem.
Suppose that $A$ is an $m\times n$ matrix, and $B$ is a row-equivalent matrix in reduced row-echelon form. Suppose that $B$ has $r$ pivot columns, with indices given by $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$, while the $n-r$ non-pivot columns have indices $F=\set{f_1,\,f_2,\,f_3,\,\ldots,\,f_{n-r},\,n+1}$. Construct the $n-r$ vectors $\vect{z}_j$, $1\leq j\leq n-r$ of size $n$, \begin{equation*} \vectorentry{\vect{z}_j}{i}= \begin{cases} 1&\text{if $i\in F$, $i=f_j$}\\ 0&\text{if $i\in F$, $i\neq f_j$}\\ -\matrixentry{B}{k,f_j}&\text{if $i\in D$, $i=d_k$} \end{cases} \end{equation*}
Then the null space of $A$ is given by \begin{equation*} \nsp{A}=\spn{\left\{\vectorlist{z}{n-r}\right\}} \end{equation*}
Notice that the hypotheses of Theorem VFSLS and Theorem SSNS are slightly different. In the former, $B$ is the row-reduced version of an augmented matrix of a linear system, while in the latter, $B$ is the row-reduced version of an arbitrary matrix. Understanding this subtlety now will avoid confusion later.
Here is an example that will simultaneously exercise the span construction and Theorem SSNS, while also pointing the way to the next section.