Basis (linear algebra)

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Let [ilmath](V,\mathbb{F})[/ilmath] be a vector space and let [ilmath](b_i)_{i\eq 1}^n\subseteq V [/ilmath] be a finite set of vectors of [ilmath]V[/ilmath]. Then [ilmath](b_i)_{i\eq 1}^n[/ilmath] is a basis for [ilmath]V[/ilmath] if:

  • [ilmath]\{b_1,\ldots,b_n\} [/ilmath] is a linearly independent set (over [ilmath]\mathbb{F} [/ilmath]), and
  • [ilmath]\text{Span} [/ilmath][ilmath](b_1,\ldots,b_n)\eq V[/ilmath]
    • That is to say every [ilmath]v\in V[/ilmath] is a linear combination of elements of [ilmath]\{b_i\}_{i\eq 1}^n [/ilmath], that is [ilmath]v\eq\sum_{j\eq 1}^k \lambda_jb_{\alpha_j} [/ilmath] where [ilmath](\alpha_j)_{j\eq 1}^k[/ilmath] are indices for basis elements and [ilmath](\lambda_j)_{j\eq 1}^k\subseteq\mathbb{F} [/ilmath] are scalars.

Caveat:The infinite case - we're actually able to write [ilmath]\sum_{\alpha\in I}\lambda_\alpha x_\alpha[/ilmath] for an arbitrary set of vectors [ilmath]\{x_\alpha\}_{\alpha\in I}\subseteq V[/ilmath] for a linear combination, provided that [ilmath]\vert\{\lambda_\alpha\in\mathbb{F}\ \vert\ \alpha\in I\wedge \lambda_\alpha\neq 0\}\vert\in\mathbb{N} [/ilmath] - that is only finitely many of the [ilmath]\lambda_\alpha[/ilmath] are non-zero, this way all the (possibly infinitely many) zero terms vanish.

Important theorems

See also