Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$

Let $V$ be a vector space over a field $F$ (Vector space over a field).

Finite sums of vectors

By axiom (V1) the triple $(V, +, 0_V)$ is an abelian group (Group and abelian group), hence in particular a commutative monoid (Semigroup and monoid). So the finite products of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity are available in it, and we write them additively: for $n \in \mathbb{N}$ and a finite list $u : n \to V$, that is a function on the von Neumann natural $n = \{0, \dots, n-1\}$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$),

$\sum_{i<n} u_i \;:=\; \prod_{i<n} u_i \quad \text{computed in } (V,+,0_V),$

so that $\sum_{i<0} u_i = 0_V$ and $\sum_{i<\sigma(n)} u_i = \bigl(\sum_{i<n} u_i\bigr) + u_n$, and the value depends only on $u_0, \dots, u_{n-1}$.

Linear combinations

A linear combination in $V$ is a vector of the form

$\sum_{i<n} \lambda_i v_i$

where $n \in \mathbb{N}$, $\lambda : n \to F$ is a finite list of scalars and $v : n \to V$ is a finite list of vectors; the sum is the finite sum just described, of the list $i \mapsto \lambda_i v_i$. For $S \subseteq V$, a vector $w \in V$ is a linear combination of elements of $S$ when there are $n \in \mathbb{N}$, $\lambda : n \to F$ and $v : n \to S$ with $w = \sum_{i<n} \lambda_i v_i$.

The empty case is a real case. $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)), and at $n = 0$ the sum is the empty sum, which is $0_V$. So $0_V$ is a linear combination of elements of every subset of $V$, including $S = \varnothing$. The lists are indexed from $0$, so a linear combination of length $n$ is $\lambda_0 v_0 + \dots + \lambda_{n-1}v_{n-1}$; no statement here is restricted to $n \ge 1$.

The span

Let $S \subseteq V$. The set of linear subspaces of $V$ containing $S$ is nonempty, since $V$ itself is one, so its intersection is a linear subspace of $V$ by The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$. That intersection is the span of $S$,

$\operatorname{span}(S) \;:=\; \bigcap \{\, W \;:\; W \text{ is a linear subspace of } V \text{ and } S \subseteq W \,\} .$

It contains $S$, being an intersection of sets each of which contains $S$, and it is contained in every linear subspace of $V$ that contains $S$. So it is the smallest linear subspace of $V$ containing $S$, and those two properties determine it uniquely: if $W$ and $W'$ both contain $S$ and are each contained in every linear subspace containing $S$, then each is contained in the other. This is what licenses the definite article.

A subset $S \subseteq V$ spans $V$, or is a spanning set of $V$, when $\operatorname{span}(S) = V$.

• The definition is the one already used for subgroups. The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups defines $\langle S \rangle$ as the intersection of all subgroups containing $S$, licensed by The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$. Its Remarks also record a description from inside, as a set of products, proved there only for a single generator ($\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian) with the general case deferred to a later page. The span is defined here in exactly that outside shape, and the identification from inside, that $\operatorname{span}(S)$ is precisely the set of linear combinations of elements of $S$, is proved in full as $\operatorname{span}(S)$ is exactly the set of linear combinations of finite lists of elements of $S$, and $\operatorname{span}(\varnothing) = \{0_V\}$. In particular $\operatorname{span}(\varnothing) = \{0_V\}$ is proved there, as a consequence of the definition, and is not stipulated here.

• Why the finite sum is The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity and not Finite sums and finite products, by recursion. The latter is stated for sequences into the complete ordered field, so it cannot carry a sum of vectors in an arbitrary vector space over an arbitrary field. The monoid finite product is defined by recursion in any monoid, its empty value is the identity, and Generalised associativity: in a monoid the product of a finite list does not depend on the bracketing, and in a commutative monoid it does not depend on the order of the factors either supplies the splitting, regrouping and reordering laws for it. Reading it additively in $(V,+,0_V)$ costs nothing and is the only sum of vectors this page uses.

• A linear combination is a value, not an expression. Two different lists may produce the same vector, and nothing above asserts otherwise. Repetitions are allowed in the list $v$, and so are coefficients equal to $0_F$; asking when a vector is a linear combination of a set in only one way is the question of linear independence, which belongs to a later page and is not raised here.