Let $\vec v_{1}, \dots, \vec v_{n} \in \R^{n}$, The [[Span]] $\op{span}(\vec v_{1}, \dots, \vec v_{n})$ is the [[Set]] of all [[Linear Combination|Linear Combinations]] of the [[Vector|Vectors]] $\set{\vec v_{1}, \dots, \vec v_{n}}$. A [[Vector]] $\vec b$ is [[Existential Quantifier|included]] in the [[Set]] $\op{span}(\vec v_{1}, \dots, \vec v_{n})$ if there is a [[Linear Combination]] in the set that adds to $\vec b$. >[!example]- Span of $\hat{i}$ and $\hat{j}$ > $\huge \op{span}\pa{\hat{i}, \hat{j}} = \R^{2}$ >[!example] Span with a single point- > $\op{span}\left(\vec 0, \vec 0\right) = \set{\vec 0} $ >[!example]- Is some specific vector $\vec b\in\op{span}\pa{\dots}$? >$ >\let \vec v_{1} = \mat{1\\0\\-1\\3}, \vec v_{2} = \mat{2\\1\\-1\\0}, \vec b = \mat{0\\1\\1\\-6} \\ >$ >Is $\vec b \in \op{span}{(\vec v_{1}, \vec v_{2})}$? > >$\huge\begin{align} >x_{1}\vec v_{1}+x_{2} \vec v_{2} &= \vec b >\sim \augmented{cc|c}{ >1&2&0\\0&1&1\\-1&-1&1\\3&0&-6 } >\\&\vdots >\\\sim& >\underbrace{\augmented{cc|c}{ >1&0&-2\\0&1&1\\0&0&0\\0&0&0 >}}_{\text{{Consistent }} } \\ >\therefore& \,\vec b\, \in \op{span}\pa{\vec v_{1}, \vec v_{2}} >\end{align}$ > #### Span covering all of $\Rn n$ $\huge \forall \vec v_{1}, \dots, \vec v_{m} \in \R^{n} $ $ \huge \begin{align} \op{rref}(\mat{\vec v_{1}, \dots, \vec v_{m}}) \text{has a leading term in every row} \iff \op{span}(\vec v_{1}, \dots) = \R^{n} \end{align}$ [[Universal Quantifier|For All]] $\vec b \in \R^{n}$ and some [[Ordered Set]] $\set{\vec v_{1},\dots,\vec v_{n}}$, if the [[Reduced Row Echelon Form]] of the [[Augmented Matrix]] either (or both): - contains a [[Free Parameter]] - contains less [[Vector|Vectors]] [[Existential Quantifier|in]] then the dimension of each - $\vec v_{i} \in \R^{n}$ $ \lvert \set{\vec v_{1}, \dots, \vec v_{m}} \rvert < n \implies \ba{\op{span}\pa{\vec v_{1}, \dots, \vec v_{m}} \ne \R^{n} } $ then the [[Span]] of the set is smaller than the [[Domain]] $\Rn n$, $x \in\Rn{n} , \exists x \ba{ \vec x \notin \op{span}(\vec v_{1}, \dots, \vec v_{n})}$.