Span (linear algebra)

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Definition

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

$\text{Span}(\{v_\alpha\}_{\alpha\in I})$ for an arbitrary collection, or
$\text{Span}(v_1,\ldots,v_k)$ for a finite collection

is defined as follows:

$\text{Span}(\{v_\alpha\}_{\alpha\in I}):\eq\Bigg\{\underbrace{\sum_{\alpha\in I}\lambda_\alpha v_\alpha}_{\text{Linear combination} }\ \Bigg\vert\ \{\lambda_\alpha\}_{\alpha\in I}\in\underbrace{\big\{\{\lambda_\alpha\}_{\alpha\in I}\in\mathbb{F}^I\ \big\vert\ \overbrace{\vert\{\lambda_\alpha\ \vert\ \alpha\in I\wedge\lambda_\alpha\neq 0\}\vert\in\mathbb{N} }^{\text{There are only finitely many non-zero terms} }\big\} }_{\text{The set of }I\text{-indexed scalars such that}\{\lambda_\alpha\}_{\alpha\in I}\text{ only has finitely many non-zero terms} } \Bigg\}$ ^{[Note 1]}

For a finite collection, $\{v_1,\ldots,v_k\}$ this simplifies to:

$\text{Span}(v_1,\ldots,v_n):\eq\left\{\left.\sum_{i\eq 1}^k\lambda_i v_i\ \right\vert\ (\lambda_i)_{i\eq 1}^k\in\mathbb{F}\right\}$

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 $\{v_1,\ldots,v_n\}$ there is no order, we really mean $(v_i)_{i\eq 1}^n$ - 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

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

References

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

[2] 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 $0+v\eq v$ ) to "pretend" we included them, they have no effect on the summation's value.

[3] Notations like $\sum^\infty_{n\eq 0}v_n$ make no sense in a vector space. This definition of limit requires a topological space. That is where the "tending towards" comes from. $\mathbb{R}$ is a vector space (as is $\mathbb{R}^n$ and so forth) but also a metric space (infact an inner product space) and thus a topological space.

[Note 1]

Span (linear algebra)

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Definition

Caveats

Notes

References

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