> [!NOTE] Lemma > Let $S\subset \mathbb{R}^{n}$ be a [[Linearly Independent Subset of Real n-Space|linearly independent subset of a real n-space]]. Then its $S$ is [[Finite Set|finite]] and its [[Cardinality|cardinality]] satisfies $|S|\leq n.$ **Proof**: Suppose $S$ has strictly more than $n$ elements. Choose $n+1$ distinct elements $\underline{v}_{1},\underline{v}_{2},\dots,\underline{v}_{n+1}\in S.$ These must be linearly independent. 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 they span $\mathbb{R}^{n},$ there are scalars $a_{ij}\in \mathbb{R}$ so that $\forall j\in \{ 1,2,\dots,n+1 \}: \quad \underline{v}_{j}=\sum_{i=1}^{n} a_{ij}\underline{e}_{i}\tag{1}$Let $A=(a_{ij})$ be the [[Real Matrices|real matrix]] of order $n\times n+1.$ Then we can rewrite $(1)$ as $V=A$ where $V$ is the matrix whose columns are $V$: that is, the vectors $\underline{v}_{j}$ are the columns of $A.$ By [[Infinitely Many Solutions to Homogeneous Linear System of Equations with More Variables Than Equations]], since $A$ has fewer rows than columns, there exists a nonzero vector $\underline{k}=(k_{1},\dots,k_{n+1})$ so that $A\underline{k}=\underline{0}$: that is, $k_{1}\underline{v}_{1}+\dots+k_{n+1}\underline{v}_{n+1}=\underline{0}.$This gives a linear dependence relation which contradicts the fact the vectors are linearly independent. **Proof**: Follows directly from [[Real n-Space is Finite Dimensional Real Vector Space]] and [[Linearly Independent Subset of Finite Dimensional Real Vector Space contains at Most n Elements]].