> [!NOTE] Lemma (Span is a vector subspace of $\mathbb{R}^{n}$) > Let $\underline{v}_{1},\underline{v}_{2},\dots,\underline{v}_{s}\in \mathbb{R}^{n}$ be a [[Finite Set|finite set]] of [[Real n-Space|real vectors]]. Then their [[Span of Subset of Real n-Space|span]] is a [[Subspace of Real n-Space|subspace of the real n-space]]. **Proof**: Let $W=\langle\underline{v}_{1},\dots,\underline{v}_{s} \rangle$. Suppose $\underline{v},\underline{w}\in W$. Hence there are scalars $\lambda_{i},\mu_{j}\in \mathbb{R}$ for which $\underline{v} = \lambda_{1}v_{1}+\dots\lambda_{s} \underline{v}_{s} \quad \text{and} \quad \underline{w}=\mu_{1}v_{1}+\dots \mu_{s} \underline{v}_{s}$Therefore $\underline{v}+\underline{w}=(\lambda_{1}+\mu_{1})\underline{v}_{1}+\dots+(\lambda_{s}+\mu_{s})\underline{v}_{s}$which is a linear combination of $v_{1},\dots,\underline{v}_{s}$ and so $\underline{v}+\underline{w}\in W$. Similarly, if $\alpha\in \mathbb{R}$, then $\alpha \underline{v}=(\alpha\lambda_{1})\underline{v}_{1}+\dots+(\alpha\lambda_{s})\underline{v}_{s}$which is a linear combination of $\underline{v}_{1},\dots,\underline{v}_{s}$ so $\alpha \underline{v}\in W$. Thus the span is a subspace of $\mathbb{R}^{n}.$