View source for Basis (linear algebra)

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==Definition==
Let {{M|(V,\mathbb{F})}} be a [[vector space]] and let {{M|(b_i)_{i\eq 1}^n\subseteq V }} be a finite set of vectors of {{M|V}}. Then {{M|(b_i)_{i\eq 1}^n}} is a ''basis'' for {{M|V}} if:
* {{M|\{b_1,\ldots,b_n\} }} is a [[linearly independent set]] (over {{M|\mathbb{F} }}), and
* [[Span (linear algebra)|{{M|\text{Span} }}]]{{M|(b_1,\ldots,b_n)\eq V}}
** That is to say every {{M|v\in V}} is a [[linear combination]] of elements of {{M|\{b_i\}_{i\eq 1}^n }}, that is {{M|v\eq\sum_{j\eq 1}^k \lambda_jb_{\alpha_j} }} where {{M|(\alpha_j)_{j\eq 1}^k}} are indices for basis elements and {{M|(\lambda_j)_{j\eq 1}^k\subseteq\mathbb{F} }} are scalars.
{{Caveat|The infinite case}} - we're actually able to write {{M|\sum_{\alpha\in I}\lambda_\alpha x_\alpha}} for an arbitrary set of vectors {{M|\{x_\alpha\}_{\alpha\in I}\subseteq V}} for a linear combination, provided that {{M|\vert\{\lambda_\alpha\in\mathbb{F}\ \vert\ \alpha\in I\wedge \lambda_\alpha\neq 0\}\vert\in\mathbb{N} }} - that is only finitely many of the {{M|\lambda_\alpha}} are non-zero, this way all the (possibly infinitely many) zero terms vanish.
* - See [[Hamal basis]]
==Important theorems==
==See also==
==References==
<references/>
{{Definition|Linear Algebra|Abstract Algebra}}