In Subsection SS.SSNS we proved Theorem SSNS which provided $n-r$ vectors that could be used with the span construction to build the entire null space of a matrix. As we have hinted in Example SCAD, and as we will see again going forward, linearly dependent sets carry redundant vectors with them when used in building a set as a span. Our aim now is to show that the vectors provided by Theorem SSNS form a linearly independent set, so in one sense they are as efficient as possible a way to describe the null space. Notice that the vectors $\vect{z}_j$, $1\leq j\leq n-r$ first appear in the vector form of solutions to arbitrary linear systems (Theorem VFSLS). The exact same vectors appear again in the span construction in the conclusion of Theorem SSNS. Since this second theorem specializes to homogeneous systems the only real difference is that the vector $\vect{c}$ in Theorem VFSLS is the zero vector for a homogeneous system. Finally, Theorem BNS will now show that these same vectors are a linearly independent set. We will set the stage for the proof of this theorem with a moderately large example. Study the example carefully, as it will make it easier to understand the proof.
The proof of Theorem BNS is really quite straightforward, and relies on the “pattern of zeros and ones” that arise in the vectors $\vect{z}_i$, $1\leq i\leq n-r$ in the entries that arise with the locations of the non-pivot columns. Play along with Example LINSB as you study the proof. Also, take a look at Example VFSAD, Example VFSAI and Example VFSAL, especially at the conclusion of Step 2 (temporarily ignore the construction of the constant vector, $\vect{c}$). This proof is also a good first example of how to prove a conclusion that states a set is linearly independent.
Suppose that $A$ is an $m\times n$ matrix, and $B$ is a row-equivalent matrix in reduced row-echelon form with $r$ pivot columns. Let $D=\{d_1,\,d_2,\,d_3,\,\ldots,\,d_r\}$ and $F=\{f_1,\,f_2,\,f_3,\,\ldots,\,f_{n-r}\}$ be the sets of column indices where $B$ does and does not (respectively) have pivot columns. Construct the $n-r$ vectors $\vect{z}_j$, $1\leq j\leq n-r$ of size $n$ as \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*}
Define the set $S=\set{\vectorlist{z}{n-r}}$.Then