Span (linear algebra)

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Let [ilmath](V,\mathbb{F})[/ilmath] be a vector space over a field [ilmath]\mathbb{F} [/ilmath] and let [ilmath]\{v_\alpha\}_{\alpha\in I}\subseteq V[/ilmath] be an arbitrary collection of vectors of [ilmath]V[/ilmath]. The span of [ilmath]\{v_\alpha\}_{\alpha\in I} [/ilmath], denoted:

  • [ilmath]\text{Span}(\{v_\alpha\}_{\alpha\in I})[/ilmath] for an arbitrary collection, or
  • [ilmath]\text{Span}(v_1,\ldots,v_k)[/ilmath] for a finite collection

is defined as follows:

  • [math]\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\} [/math][Note 1]

For a finite collection, [ilmath]\{v_1,\ldots,v_k\} [/ilmath] this simplifies to:

  • [ilmath]\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\} [/ilmath]

See linear combination for details of why we need the "finitely many part"


Caveat:There are a few problems here

  1. Ordered basis - in the set [ilmath]\{v_1,\ldots,v_n\} [/ilmath] there is no order, we really mean [ilmath](v_i)_{i\eq 1}^n[/ilmath] - 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"


  1. Remember that the "vector addition" is a binary function on [ilmath]V[/ilmath]. It's also associative so [ilmath](u+v)+w\eq u+(v+w)[/ilmath] 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 [ilmath]0+v\eq v[/ilmath]) 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 [ilmath]\sum^\infty_{n\eq 0}v_n[/ilmath] make no sense in a vector space. This definition of limit requires a topological space. That is where the "tending towards" comes from. [ilmath]\mathbb{R} [/ilmath] is a vector space (as is [ilmath]\mathbb{R}^n[/ilmath] and so forth) but also a metric space (infact an inner product space) and thus a topological space.