Proof of Theorem SSNS

Consider the homogeneous system with $A$ as a coefficient matrix, $\homosystem{A}$. Its set of solutions, $S$, is by Definition NSM, the null space of $A$, $\nsp{A}$. Let $B^{\prime}$ denote the result of row-reducing the augmented matrix of this homogeneous system. Since the system is homogeneous, the final column of the augmented matrix will be all zeros, and after any number of row operations (Definition RO), the column will still be all zeros. So $B^{\prime}$ has a final column that is totally zeros.

Now apply Theorem VFSLS to $B^{\prime}$, after noting that our homogeneous system must be consistent (Theorem HSC). The vector $\vect{c}$ has zeros for each entry that has an index in $F$. For entries with their index in $D$, the value is $-\matrixentry{B^{\prime}}{k,{n+1}}$, but for $B^{\prime}$ any entry in the final column (index $n+1$) is zero. So $\vect{c}=\zerovector$. The vectors $\vect{z}_j$, $1\leq j\leq n-r$ are identical to the vectors $\vect{u}_j$, $1\leq j\leq n-r$ described in Theorem VFSLS. Putting it all together and applying Definition SSCV in the final step, \begin{align*} \nsp{A} &=S\\ &=\setparts{ \vect{c}+\alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3+\cdots+\alpha_{n-r}\vect{u}_{n-r} }{ \alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_{n-r}\in\complex{\null} }\\ &=\setparts{ \alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3+\cdots+\alpha_{n-r}\vect{u}_{n-r} }{ \alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_{n-r}\in\complex{\null} }\\ &=\spn{\set{\vectorlist{z}{n-r}}} \end{align*}