A Set is a Basis iff it is a Maximal Linearly Independent Set
# Statement
Let $V$ be a Vector Space Then $S \subset V$ is a Vector Space Basis for $V$ If and Only If it is a Maximal Linearly Independent subset of $V$. That is, there does not exist any Linearly Independent $R \subset V$ so that $S \subsetneq R$.
# Proof
($\Leftarrow$) Suppose not. Since $S$ is Linearly Independent, we must have that $\text{span} S \subsetneq V$. But then there exists $\mathbf{v} \in V \setminus \text{span} S$. Consider $R:= S \cup {\mathbf{v}}$. By Growing a Linearly Independent Set, we have that $R$ is Linearly Independent, contradicting our assumption that $S$ is a Maximal Linearly Independent subset of $V$ $\unicode{x21af}$. Therefore $\text{span} S = V$ and $S$ is a Vector Space Basis for $V$. $\checkmark$
($\Rightarrow$) Let $\mathbf{v} \in V \setminus S$. Let $R \subset V$ is such that $R \supset S \cup {\mathbf{v}} \supsetneq S$. Since $S$ is a Vector Space Basis for $V$, we have that $\mathbf{v} \in \text{span} S$. Because The Subspace Span is the Set of all Linear Combinations, there exists $c_{1}, \dots, c_{n} \in F$, $\mathbf{a}{1}, \dots, \mathbf{a}{n} \in S$ for some $n \in \mathbb{Z}{\geq 0}$ so that $$\begin{align*} &\mathbf{v} = c{1} \mathbf{a}{1} + \cdots + c{n} \mathbf{a}{n}\\ \Rightarrow &\mathbf{0} = c{1} \mathbf{a}{1} + \cdots + c{n} \mathbf{a}_{n} - \mathbf{v}, \end{align*}$$ so $R$ is Linearly Dependent. Since $\mathbf{v}$ was arbitrary, $S$ must be Maximal. $\checkmark$ $\blacksquare$
# Remarks
- Note that we account for $S = \emptyset$ in the $(\Rightarrow)$ direction by letting $n \in \mathbb{Z}_{\geq 0}$.