Find a set of vectors, $S$, so that the null space of the matrix $A$ below is the span of $S$, that is, $\spn{S}=\nsp{A}$. \begin{equation*} A= \begin{bmatrix} 1 & 3 & 3 & -1 & -5\\ 2 & 5 & 7 & 1 & 1\\ 1 & 1 & 5 & 1 & 5\\ -1 & -4 & -2 & 0 & 4 \end{bmatrix} \end{equation*}
The null space of $A$ is the set of all solutions to the homogeneous system $\homosystem{A}$. If we find the vector form of the solutions to this homogeneous system (Theorem VFSLS) then the vectors $\vect{u}_j$, $1\leq j\leq n-r$ in the linear combination are exactly the vectors $\vect{z}_j$, $1\leq j\leq n-r$ described in Theorem SSNS. So we can mimic Example VFSAL to arrive at these vectors (rather than being a slave to the formulas in the statement of the theorem).
Begin by row-reducing $A$. The result is \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 6 & 0 & 4\\ 0 & \leading{1} & -1 & 0 & -2\\ 0 & 0 & 0 & \leading{1} & 3\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
With $D=\set{1,\,2,\,4}$ and $F=\set{3,\,5}$ we recognize that $x_3$ and $x_5$ are free variables and we can interpret each nonzero row as an expression for the dependent variables $x_1$, $x_2$, $x_4$ (respectively) in the free variables $x_3$ and $x_5$. With this we can write the vector form of a solution vector as \begin{equation*} \colvector{x_1\\x_2\\x_3\\x_4\\x_5}= \colvector{-6x_3-4x_5\\x_3+2x_5\\x_3\\-3x_5\\x_5}= x_3\colvector{-6\\1\\1\\0\\0}+ x_5\colvector{-4\\2\\0\\-3\\1} \end{equation*}
Then in the notation of Theorem SSNS, \begin{align*} \vect{z}_1&=\colvector{-6\\1\\1\\0\\0}& \vect{z}_2&=\colvector{-4\\2\\0\\-3\\1} \end{align*} and \begin{equation*} \nsp{A} =\spn{\set{\vect{z}_1,\,\vect{z}_2}} =\spn{\set{\colvector{-6\\1\\1\\0\\0},\,\colvector{-4\\2\\0\\-3\\1}}} \end{equation*}