Example LINSB Linear independence of null space basis

Suppose that we are interested in the null space of a $3\times 7$ matrix, $A$, which row-reduces to \begin{equation*} B= \begin{bmatrix} \leading{1} & 0 & -2 & 4 & 0 & 3 & 9\\ 0 & \leading{1} & 5 & 6 & 0 & 7 & 1\\ 0 & 0 & 0 & 0 & \leading{1} & 8 & -5 \end{bmatrix} \end{equation*}

The set $F=\set{3,\,4,\,6,\,7}$ is the set of indices for our four free variables that would be used in a description of the solution set for the homogeneous system $\homosystem{A}$. Applying Theorem SSNS we can begin to construct a set of four vectors whose span is the null space of $A$, a set of vectors we will reference as $T$. \begin{equation*} \nsp{A}=\spn{T}=\spn{\set{\vect{z}_1,\,\vect{z}_2,\,\vect{z}_3,\,\vect{z}_4}}=\spn{\set{ \colvector{ \\ \\1\\0\\ \\0\\0},\, \colvector{ \\ \\0\\1\\ \\0\\0},\, \colvector{ \\ \\0\\0\\ \\1\\0},\, \colvector{ \\ \\0\\0\\ \\0\\1} }} \end{equation*}

So far, we have constructed as much of these individual vectors as we can, based just on the knowledge of the contents of the set $F$. This has allowed us to determine the entries in slots 3, 4, 6 and 7, while we have left slots 1, 2 and 5 blank. Without doing any more, let us ask if $T$ is linearly independent? Begin with a relation of linear dependence on $T$, and see what we can learn about the scalars, \begin{align*} \zerovector&=\alpha_1\vect{z}_1+\alpha_2\vect{z}_2+\alpha_3\vect{z}_3+\alpha_4\vect{z}_4\\ \colvector{0\\0\\0\\0\\0\\0\\0} &= \alpha_1\colvector{ \\ \\1\\0\\ \\0\\0}+ \alpha_2\colvector{ \\ \\0\\1\\ \\0\\0}+ \alpha_3\colvector{ \\ \\0\\0\\ \\1\\0}+ \alpha_4\colvector{ \\ \\0\\0\\ \\0\\1}\\ &= \colvector{ \\ \\\alpha_1\\0\\ \\0\\0}+ \colvector{ \\ \\0\\\alpha_2\\ \\0\\0}+ \colvector{ \\ \\0\\0\\ \\\alpha_3\\0}+ \colvector{ \\ \\0\\0\\ \\0\\\alpha_4} = \colvector{ \\ \\\alpha_1\\\alpha_2\\ \\\alpha_3\\\alpha_4} \end{align*}

Applying Definition CVE to the two ends of this chain of equalities, we see that $\alpha_1=\alpha_2=\alpha_3=\alpha_4=0$. So the only relation of linear dependence on the set $T$ is a trivial one. By Definition LICV the set $T$ is linearly independent. The important feature of this example is how the “pattern of zeros and ones” in the four vectors led to the conclusion of linear independence.