The following theorem is from my textbook:
Theorem 1.7. Let $S$ be a linearly independent subset of a vector space $V$, and let $v$ be a vector in $V$ that is not in $S$. Then $S \cup \{v\}$ is linearly dependent if and only if $v \in\operatorname{span}(S)$.
I'm having trouble understanding the proof in the reverse direction. Here's the text:
Conversely, let $v \in\operatorname{span}(S)$. Then there exist vectors $v_1, v_2,\dots, v_m$ in $S$ and scalars $b_1, b_2,\dots, b_m$ such that $v = b_1v_1 + b_2v_2+\dots+b_mv_m.$ Hence $0 = b_1v_1 + b_2v_2 + \dots+ b_mv_m + (-1)v$.
Since $v \neq v_i$ for $i = 1, 2,\dots, m$, the coefficient of $v$ in this linear combination is nonzero, and so the set $\{v_1, v_2,\dots, v_m, v\}$ is linearly dependent. Therefore $S \cup \{v\}$ is linearly dependent.
What does the proof mean when it says "Since $v \neq v_i$ for $i = 1, 2,\dots, m$"? Wasn't it enough to show already that there is a non-trivial representation of $S \cup \{v\} = 0$ to prove that it is linearly dependent?
I'm just unsure why that qualification is necessary. In the case that $v = v_i$ for $i = 1, 2,\dots, m$, then wouldn't it be easy to find non-zero scalars for $v$ and that $v_i$ to prove that it is linearly dependent?