View source for Span (linear algebra)

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==Definition==
Let {{M|(V,\mathbb{F})}} be a [[vector space]] over a [[field]] {{M|\mathbb{F} }} and let {{M|\{v_\alpha\}_{\alpha\in I}\subseteq V}} be an ''[[arbitrary]]'' collection of vectors of {{M|V}}. The ''span'' of {{M|\{v_\alpha\}_{\alpha\in I} }}, denoted:
* {{M|\text{Span}(\{v_\alpha\}_{\alpha\in I})}} for an arbitrary collection, or
* {{M|\text{Span}(v_1,\ldots,v_k)}} for a finite collection
is defined as follows:
* {{MM|\text{Span}(\{v_\alpha\}_{\alpha\in I}):\eq\Bigg\{\underbrace{\sum_{\alpha\in I}\lambda_\alpha v_\alpha}_{\text{Linear combination} }\ \Bigg\vert\ \{\lambda_\alpha\}_{\alpha\in I}\in\underbrace{\big\{\{\lambda_\alpha\}_{\alpha\in I}\in\mathbb{F}^I\ \big\vert\ \overbrace{\vert\{\lambda_\alpha\ \vert\ \alpha\in I\wedge\lambda_\alpha\neq 0\}\vert\in\mathbb{N} }^{\text{There are only finitely many non-zero terms} }\big\} }_{\text{The set of }I\text{-indexed scalars such that}\{\lambda_\alpha\}_{\alpha\in I}\text{ only has finitely many non-zero terms} } \Bigg\} }}<ref group="Note">Remember that the "vector addition" is a [[binary function]] on {{M|V}}. It's also associative so {{M|(u+v)+w\eq u+(v+w)}} which makes things easier. However we can only do this finitely many times ultimately. So we use the following [[abuse of notation]]:
* We may define arbitrary sums on the condition that it only has ''finitely many'' non zero terms. We use the fact that zero vector is the additive identity (and thus {{M|0+v\eq v}}) to "pretend" we included them, they have no effect on the summation's value. 
We can still only sum finitely many non-zero terms however. Lastly:
* Notations like {{M|\sum^\infty_{n\eq 0}v_n}} make no sense in a vector space. This definition of [[limit]] requires a [[topological space]]. That is where the "tending towards" comes from. {{M|\mathbb{R} }} is a vector space (as is {{M|\mathbb{R}^n}} and so forth) but also a [[metric space]] (infact an [[inner product space]]) and thus a [[topological space]].</ref>
For a [[finite]] collection, {{M|\{v_1,\ldots,v_k\} }} this simplifies to:
* {{M|\text{Span}(v_1,\ldots,v_n):\eq\left\{\left.\sum_{i\eq 1}^k\lambda_i v_i\ \right\vert\ (\lambda_i)_{i\eq 1}^k\in\mathbb{F}\right\} }}
See [[linear combination]] for details of why we need the "finitely many part"
===Caveats===
{{Caveat|There are a few problems here}}
# Ordered basis - in the set {{M|\{v_1,\ldots,v_n\} }} there is no order, we really mean {{M|(v_i)_{i\eq 1}^n}} - this implies order. Also for an arbitrary collection, tuple notation doesn't make sense unless there's an ordering in play. This needs to be "united"
==Notes==
<references group="Note"/>
==References==
<references/>
{{Definition|Linear Algebra|Abstract Algebra}}