> [!NOTE] Lemma (Span of linearly independent vectors) > Let $V$ be a [[Real Vector Space|real vector space]]. Let $B \subset V$ a [[Linearly Independent Subset of Real Vector Space|linearly independent subset]]. Then for all $v\in V,$ $v\in \langle B \rangle \iff B \cup \{v\}$ is linearly dependent; where $\langle B \rangle$ denotes the [[Span of Subset of Vector Space|span]] of $B$. *Proof*. If $v\in \langle B \rangle$, then $v= \lambda_{1}w_{1}+\dots+\lambda_{s}w_{s}$ for some $w_{1},\dots,w_{s}\in B$ and $\lambda_{1},\dots\lambda_{s}\in \mathbb{R}$. This is a nontrivial linear dependence relation among $B \cup \{ v \}$ since the coefficient of $v$ is not zero, so $B \cup \{ v \}$ is linearly dependent. Suppose $v \not \in \langle B \rangle$. Consider a linear combination $\lambda v + \mu_{1} w_{1} +\dots+\mu_{s}w_{s}=0_{V}\tag{1}$for some $w_{1},\dots,w_{s}\in B$ and $\lambda_{1},\dots,\lambda_{s}\in \mathbb{R}$. Now if $\lambda \neq 0$, then $v= - \left( \frac{\mu_{1}}{\lambda} \right)w_{1} + \dots + \left( \frac{\mu_{s}}{\lambda} \right)w_{s}$But $v \not \in \langle B \rangle$ so we must have $\lambda=0$. Therefore $(1)$ reads $\sum \mu_{1}v_{i} = 0_{V}$. Since $B$ is linearly independent, it follows that $\mu_{i}=0$ so all coefficients of $(1)$ are zero which shows that $B \cup \{ v \}$ is linearly independent.