Begin with the set of four vectors of size $3$ \begin{equation*} T=\set{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3,\,\vect{w}_4} =\set{ \colvector{2\\-3\\1},\, \colvector{1\\4\\1},\, \colvector{7\\-5\\4},\, \colvector{-7\\-6\\-5} } \end{equation*} and consider the infinite set $W=\spn{T}$. The vectors of $T$ have been chosen as the four columns of the coefficient matrix in Archetype D. Check that the vector \begin{equation*} \vect{z}_2=\colvector{2\\3\\0\\1} \end{equation*} is a solution to the homogeneous system $\homosystem{D}$ (it is the vector $\vect{z}_2$ provided by the description of the null space of the coefficient matrix $D$ from Theorem SSNS).
Applying Theorem SLSLC, we can write the linear combination, \begin{equation*} 2\vect{w}_1+3\vect{w}_2+0\vect{w}_3+1\vect{w}_4=\zerovector \end{equation*} which we can solve for $\vect{w}_4$, \begin{equation*} \vect{w}_4=(-2)\vect{w}_1+(-3)\vect{w}_2. \end{equation*}
This equation says that whenever we encounter the vector $\vect{w}_4$, we can replace it with a specific linear combination of the vectors $\vect{w}_1$ and $\vect{w}_2$. So using $\vect{w}_4$ in the set $T$, along with $\vect{w}_1$ and $\vect{w}_2$, is excessive. An example of what we mean here can be illustrated by the computation, \begin{align*} 5\vect{w}_1&+(-4)\vect{w}_2+6\vect{w}_3+(-3)\vect{w}_4\\ &=5\vect{w}_1+(-4)\vect{w}_2+6\vect{w}_3+(-3)\left((-2)\vect{w}_1+(-3)\vect{w}_2\right)\\ &=5\vect{w}_1+(-4)\vect{w}_2+6\vect{w}_3+\left(6\vect{w}_1+9\vect{w}_2\right)\\ &=11\vect{w}_1+5\vect{w}_2+6\vect{w}_3 \end{align*}
So what began as a linear combination of the vectors $\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3,\,\vect{w}_4$ has been reduced to a linear combination of the vectors $\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3$. A careful proof using our definition of set equality (Definition SE) would now allow us to conclude that this reduction is possible for any vector in $W$, so \begin{equation*} W=\spn{\left\{\vect{w}_1,\,\vect{w}_2,\,\vect{w}_3\right\}} \end{equation*} So the span of our set of vectors, $W$, has not changed, but we have described it by the span of a set of three vectors, rather than four. Furthermore, we can achieve yet another, similar, reduction.
Check that the vector \begin{equation*} \vect{z}_1=\colvector{-3\\-1\\1\\0} \end{equation*} is a solution to the homogeneous system $\linearsystem{D}{\zerovector}$ (it is the vector $\vect{z}_1$ provided by the description of the null space of the coefficient matrix $D$ from Theorem SSNS). Applying Theorem SLSLC, we can write the linear combination, \begin{equation*} (-3)\vect{w}_1+(-1)\vect{w}_2+1\vect{w}_3=\zerovector \end{equation*} which we can solve for $\vect{w}_3$, \begin{equation*} \vect{w}_3=3\vect{w}_1+1\vect{w}_2 \end{equation*}
This equation says that whenever we encounter the vector $\vect{w}_3$, we can replace it with a specific linear combination of the vectors $\vect{w}_1$ and $\vect{w}_2$. So, as before, the vector $\vect{w}_3$ is not needed in the description of $W$, provided we have $\vect{w}_1$ and $\vect{w}_2$ available. In particular, a careful proof (such as is done in Example RSC5) would show that \begin{equation*} W=\spn{\left\{\vect{w}_1,\,\vect{w}_2\right\}} \end{equation*}
So $W$ began life as the span of a set of four vectors, and we have now shown (utilizing solutions to a homogeneous system) that $W$ can also be described as the span of a set of just two vectors. Convince yourself that we cannot go any further. In other words, it is not possible to dismiss either $\vect{w}_1$ or $\vect{w}_2$ in a similar fashion and winnow the set down to just one vector.
What was it about the original set of four vectors that allowed us to declare certain vectors as surplus? And just which vectors were we able to dismiss? And why did we have to stop once we had two vectors remaining? The answers to these questions motivate “linear independence,” our next section and next definition, and so are worth considering carefully now.