> [!NOTE] Theorem (Linearly Independent Subset of Finite Dimensional Real Vector Space is Finite) > Let $V$ be a [[Finite Dimensional Real Vector Space|finite dimensional real vector space]]. Let $L \subset V$ be [[Linearly Independent Subset of Real Vector Space|linearly independent subset]]. Then $L$ is [[Finite Set|finite]]. **Proof**. Since $V$ is finite dimensional, there is a spanning set $w_{1},\dots,w_{n}\in V$. Suppose there are $n+1$ distinct elements $v_{1},\dots,v_{n+1}\in S$. Since $w_{1},\dots,w_{n}$ span $V$, there are scalars $a_{ij}\in\mathbb{F}$ so that $\underline{v}_j=\sum_{i=1}^na_{ij}\underline{e}_i$for each $j=1,\ldots,n+1$. Consider the [[Matrix|matrix]] $A=(a_{ij})\in\mathbb{F}^{n\times n+1}$. Note that the vectors $\underline{v}_{j}$ are the columns of $A$. Since $A$ has fewer rows than columns, there is a nonzero vector $\underline{k}=$ $(k_1,\ldots,k_{n+1})^{T}$ such that $A\underline{k}=0_{V}$ (since $(A \mid 0_{V})$ [[Linear System of Equations#^fc36e2|has a at least one column without a pivot in its RREF]]). Since the $\underline{v}_j$ are the columns of $A$, this is exactly saying that $k_1\underline{v}_1+\ldots+k_{n+1}\underline{v}_{n+1}=\underline{0}$which is a linear dependence relation for $\underline{v}_1,\ldots,\underline{v}_{n+1}$hence a contradiction. $\square$