>[!quote] Its a big one - [[../Digipen/Alexander Young|Alexander Young]] A big [[Thereom]] of bijections between statements about [[Linear Independence]] and the [[Span]] of a [[Set]] of [[Vector|Vectors]]. #### [[Linear Independence]] Let $A$ be an $n\times m$ [[Matrix]] with column [[Vector|Vectors]]: $\vec v_{1},\vec v_{2}, \dots, \vec v_{m} \in \R^{n}$ *All* of the following are [[Biconditional|If and Only If]]: - $\op{rref}(A)$ has a [[Reduced Row Echelon Form|leading term]] for every column. ($\op{rank}(A) = m$) - For any $\vec b \in \R^n$, if $A\vec x = \vec b$ is [[Consistent]], then there is only one possible value of $\vec x$. - If $\vec b \in \op{span}\pa{\vec v_{1}, \dots, \vec v_{m}}$, then $\vec b$ is equal to a *unique* [[Linear Combination]] $\vec b = \sum^m_{i=0} x_{i}\vec v_{i}$ - $A\vec x = \vec 0$ has only the [[Trivial Solution]] where $\vec x=\vec 0$. - The [[Ordered Set]] $\set{\vec v_{1}, \dots, \vec v_{m}}$ is [[Linear Independence|Linearly Independent]] - The corresponding [[Linear Transformation]] is [[Injective|one-to-one]]. - For any $\forall x_{1},\dots,x_{m} \in \R$ (where they are not *all* equal to 0), $\sum_{i=0}^m x_{i}\vec v_{i} \ne \vec 0$. --- #### [[Span]] inside $\R^n$ Let $A$ be an $n\times m$ [[Matrix]] with column [[Vector|Vectors]]: $\vec v_{1},\vec v_{2}, \dots, \vec v_{m} \in \R^{n}$ *All* of the following are [[Biconditional|If and Only If]]: - $\op{rref}(A)$ has a leading term in every *row* - For every $\vec b \in \R^n$, $A\vec x = \vec b$ is [[Consistent]] - Every $\vec b\in \R^n$ is equal to *at least one* [[Linear Combination]] $\sum^m_{i=0} x_{i}\vec v_{i}$. - $\op{span}(\vec v_{1}, \dots, \vec v_{m}) = \R^n$ - The corresponding [[Linear Transformation]] is [[Surjective|Onto]]