Let $U$ be a [[Linear Subspace]] of $\R^k$. If $U$ has a [[Basis]] of $\set{\vec v_{1}, \dots, \vec v_{k}}$, then every [[Basis]] of $U$ has $k$ [[Vector|Vectors]]. If $\dim U =k$ ([[Dimension]]), then every [[Set|Set]] of more than $k$ [[Vector|Vectors]] will be [[Linear Dependence|Linearly Dependent]], and any [[Set]] with *less* than $k$ [[Vector|Vectors]] will *not* [[Span]] $U$. For any $u_{1},\dots,u_{k}\in U$, *either*: - $\set{\vec u_{1}, \dots, \vec u_k} \neq U$ **and** $\set{\vec u_{1}, \dots,\vec u_{k}}$ is [[Linear Dependence|Linearly Dependent]] - $\op{span}(\set{\vec u_{1}, \dots, \vec u_{k}}) = U$ **and** $\set{\vec u_{1}, \dots,\vec u_{k}}$ is [[Linear Independence|Linearly Independent]], ie. the [[Set]] is a [[Basis]] of $U$. $\let U \subset \R^n$ be a [[Linear Subspace|Subspace]], and $\let \vec u_{1}, \dots, \vec u_{m} \in U$. - If $\set{\vec u_{1}, \dots, \vec u_{m}}$ is [[Linear Independence|Linearly Independent]], $\dim U \ge m$ - If $\op{span}{\vec u_{1}, \dots, \vec u_{m}=U}$, $\dim U \le m$