A Linearly Dependent Set contains a vector in the Span of the others
# Statement
Let $V$ be a Vector Space on Field $F$ and let $S \subset V$ be Linearly Dependent. Then there exists $\mathbf{v} \in S$ so that $\mathbf{v} \in \text{span} (S \setminus {v})$.
# Proof
Since $S$ is Linearly Dependent, there exist distinct $\mathbf{a}{1}, \dots, \mathbf{a}{n} \in S$, $c_{1}, \dots, c_{n} \in F$ for some $n \in \mathbb{N}$ so that 1. $\exists i \in [n]$ $c_{i}\neq 0$ 2. $c_{1} \mathbf{a}{1} + \cdots + c{n} \mathbf{a}_{n} = \mathbf{0}$
If $n=1$, then $c_{1} \neq 0$, do we must have that $\mathbf{a}{1} = \mathbf{0}$. Then $\mathbf{a}{1} \in \text{span} \emptyset \subset \text{span} (S \setminus {\mathbf{a}1})$ since $\emptyset \subset S \setminus {\mathbf{a}{1}}$ and Span is Monotonic.
If $n > 1$, then Without Loss of Generality, let $c_{1} \neq 0$ (by Commutativity). Then
$$\begin{align*} &c_{1} \mathbf{a}{1} + \cdots + c{n} \mathbf{a}{n} = \mathbf{0}\\ \Rightarrow &c{1} \mathbf{a}{1} = -c{2} \mathbf{a}{2} - \cdots - c{n} \mathbf{a}{n}\\ \Rightarrow & \mathbf{a}{1} = -c_{1}^{-1}c_{2} \mathbf{a}{2} - \cdots - c{1}^{-1}c_{n} \mathbf{a}{n}. \end{align*}$$ Since The Subspace Span is the Set of all Linear Combinations, we have that $\mathbf{a}{1} \in \text{span} (S \setminus {\mathbf{a}_1})$. $\blacksquare$