So I'm trying to understand the proof presented by Titu Adreescu for the converse of the following theorem:
Let $S$ be a set of vectors in some vector space $V$. Then $S$ is linearly dependent if and only if there is $v\in S$ such that $v\in\text{span}(S\setminus\{v\})$
The proof for the converse is as follows:
Suppose that there is $v\in S$ such that $s\in\texttt{span}(S\setminus \{v\})$. That means that we can find $v_1, v_2, \dots, v_n \in S\setminus v$ and scalars $a_1, \dots, a_n$ such that $v = a_1 v_1 + \dots + a_n v_n$ but then $1\cdot v + (-a_1) v_1 + \dots + (-a_n) v_n = 0$ and the vectors $v, v_1, \dots, v_n$ are linearly dependent. Since $v \not\in \{v_1, \dots, v_n \}$, it follows that $S$ has a finite subset which is linearly dependent and so $S$ is linearly dependent. The result follows.
Now, I get most of the proof but I think it should be enough to conclude that S is linearly dependent from $1\cdot v + (-a_1) v_1 + \dots + (-a_n) v_n = 0$ yet Titu goes and arguments that $S\setminus \{v\}$ is an linearly dependent subset of $S$ (which I don't understand how it follows from $v \not\in \{v_1, \dots, v_n \}$) and concludes by using that to prove that $S$ having a linearly dependent subset implies S is linearly dependent.
Please help me make sense out of this proof. Thank you.
Let $v\in S$ such that $v\in \text{span}(S \setminus \{{v}\})$ so exists $\alpha_1,\alpha_2,\ldots,\alpha_n$ such that
$$ v = \alpha_1 v_1 + \ldots + \alpha_n v_n, v_i \in S \setminus \{v\} $$
I think the fact that $v \notin \{v_1,v_2,\ldots,v_n\}$ serves to see that indeed $\{v,v_1,\ldots,v_n\}$ is linearly dependent. Why?, if some $v_i=v$ suppose $v_1$ you have
$$ (1-\alpha_1)v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n = 0 $$
And you couldn't say that $\{v_1,v_2,\ldots,v_n\}$ is linearly dependent.
However I think it's obvious that $\{v,v_1,\ldots,v_n\} \subseteq S$ is linearly dependent because they already tell you that $v_i \in S \setminus \{v\}$.
Notice that if $S$ have a subset linearly dependent, $S$ is linearly dependent.