> [!NOTE] Lemma > Let $\underline{f}_{1},\underline{f}_{2},\dots,\underline{f}_{n}$ be a [[Basis of Real n-Space|basis]] of $\mathbb{R}^{n}.$ Then for all $\underline{w}\in \mathbb{R}^{n}$ elements of a [[Real n-Space|real n-space]], there exists unique scalars $\mu_{1},\dots,\mu_{n}\in \mathbb{R}$ so that $\underline{w}=\mu_{1}\underline{f}_{1}+\dots+\mu\underline{f}_{n}.$ **Proof**: By definition, the [[Span of Subset of Real n-Space|span]] of basis equals $\mathbb{R}^{n}$ thus there exists $\mu_{1},\mu_{2},\dots,\mu_{n}\in \mathbb{R}$ so that $\underline{w}=\mu_{1}\underline{f}_{1}+\dots+\mu\underline{f}_{n}.$Now suppose there exists $\nu_{1}, \nu_{2},\dots,\nu_{n}\in \mathbb{R}$ so that $\underline{w}=\nu_{1}\underline{f}_{1}+\dots+\nu\underline{f}_{n}.$Then $(\mu_{1}-\nu_{1})\underline{f}_{1}+\dots+(\mu_{n}-\nu_{n})\underline{f}_{n}=\underline{0}$which gives $\mu_{1}-\nu_{1}=\dots=\mu_{n}-\nu_{n}=0$since the basis is also [[Linearly Independent Subset of Real n-Space|linearly independent]]: that is, for all $i=1,\dots,n$ $\mu_{i}=\nu_{i}.$