---
title: "span (linear algebra)"
type: concept
related: [Commutivity, Distributivity, Subspace, List, Vector]
source: https://www.jemoka.com/posts/kbhspan/
confidence: high
status: active
---

The span of a bunch of vectors is the set of all linear combinations of that bunch of vectors. We denote it as \(span(v_1, \dots v_{m)}\).
constituents for constructing a linear combination a list of vectors \(v_1,\dots,v_{m}\) and scalars \(a_1, a_2, \dots, a_{m} \in \mathbb{F}\) requirements \begin{equation} span(v_{1}..v_{m}) = \{a_1v_1+\dots +a_{m}v_{m}:a_1\dots a_{m} \in \mathbb{F}\} \end{equation}
additional information span is the smallest subspace containing all vectors in the list Part 1: that a span of a list of vectors is a subspace containing those vectors
By taking all \(a_{n}\) as \(0\), we show that the additive identity exists.
Taking two linear combinations and adding them (i.e. adding two members of the span) is still in the span by commutativity and distributivity (reorganize each constant \(a_{1}\) together)&mdash;creating another linear combination and therefore a member of the span.
Scaling a linear combination, by distributivity, just scales the scalars and create yet another linear combination.
Part 2: a subspace containing the list of vectors contain the span
subspaces are closed under scalar multiplication and addition. Therefore, we can just construct every linear combination.
By double-containment, a subspace is the smallest subspace containing all vectors. \(\blacksquare\)
spans If \(span(v_1, \dots v_{m})\) equals \(V\), we say that \(v_1, \dots, v_{m}\) spans \(V\).
NOTE! the two things have to be equal&mdash;if the span of a set of vectors is larger than \(V\), they do not span \(V\).
length of linearly-independent list \(\leq\) length of spanning list see here.
