> [!NOTE] Proposition (Spanning set of $\mathbb{F}^n$ contains at least $n$ elements) > Let $\mathbb{R}^{n}$ be a [[Real n-Space|real n-space]]. Suppose $S \subset \mathbb{R}^{n}$ is a [[Spanning Set of Real n-Space|spanning set]] of $\mathbb{R}^{n}$. Then $S$ contains at least $n$ elements. **Proof**: Suppose $S$ has $m<n$ elements. Let $S=\{ \underline{v}_{1}, \dots, \underline{v}_{m} \}$Let $\underline{e}_{1},\dots,\underline{e}_{n}\in \mathbb{R}^{n}$ denote the [[Standard basis of real n-space|standard basis]] of $\mathbb{R}^{n}.$ Since $S$ spans $\mathbb{R}^{n},$ each $\underline{e}_{i}$ is a linear combination of the elements of $S,$ that is, there exist $a_{ij} \in \mathbb{R}$ so that we have the following [[Linear System of Equations|linear system]] $\forall i \in \{ 1,2,\dots,n \}: \quad\underline{e}_{i} = \sum_{j=1}^{m} a_{ij}\underline{v}_{j} \tag{1}.$ Consider $A=(a_{ij})\in \mathbb{R}^{n\times m}$. We can rewrite $(1)$ as $I_{n}=AV\tag{2}$where the rows of $V\in \mathbb{R}^{m\times n}$ are $v_{1},\dots v_{m}$. Let $EA$ denote the [[Reduced Row Echelon Form for Real Matrix|RREF]] of $A$ where $E\in\mathbb{R}^{n\times n}$ is [[Inverse of Real Square Matrix|invertible]] so its rows are non zero. Pre-multiplying both sides of $(2)$ by $E$ gives $E=(EA)V$ Since $n>m$, the last row of $EA$ is $\underline{0}$ so the last row of $E$ must also be $\underline{0}$ which leads to contradiction.