Proof of Theorem ELIS

Suppose $S=\set{\vectorlist{v}{m}}$ and begin with a relation of linear dependence on $S^\prime$, \begin{equation*} \lincombo{a}{v}{m}+a_{m+1}\vect{w}=\zerovector. \end{equation*}

There are two cases to consider. First suppose that $a_{m+1}=0$. Then the relation of linear dependence on $S^\prime$ becomes \begin{equation*} \lincombo{a}{v}{m}=\zerovector. \end{equation*} and by the linear independence of the set $S$, we conclude that $a_1=a_2=a_3=\cdots=a_m=0$. So all of the scalars in the relation of linear dependence on $S^\prime$ are zero.

In the second case, suppose that $a_{m+1}\neq 0$. Then the relation of linear dependence on $S^\prime$ becomes \begin{align*} a_{m+1}\vect{w}&=-a_1\vect{v}_1-a_2\vect{v}_2-a_3\vect{v}_3-\cdots-a_m\vect{v}_m\\ \vect{w}&=-\frac{a_1}{a_{m+1}}\vect{v}_1-\frac{a_2}{a_{m+1}}\vect{v}_2-\frac{a_3}{a_{m+1}}\vect{v}_3-\cdots-\frac{a_m}{a_{m+1}}\vect{v}_m \end{align*}

This equation expresses $\vect{w}$ as a linear combination of the vectors in $S$, contrary to the assumption that $\vect{w}\not\in\spn{S}$, so this case leads to a contradiction.

The first case yielded only a trivial relation of linear dependence on $S^\prime$ and the second case led to a contradiction. So $S^\prime$ is a linearly independent set since any relation of linear dependence is trivial.