# Linear span

In the mathematical subfield of linear algebra, the linear span of a set of vectors is the set of all linear combinations of the vectors. The linear span of a set of vectors is a therefore a vector space but unlike a basis the vectors need not be linearly independent.

The linear span is an example of a set-builder notation.

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## Definition

Given a vector space V over a field K and vectors v1,...,vn in V then

[itex] \mathrm{span}( \mathbf{v}_1 ,\ldots, \mathbf{v}_n) := \{ a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n : a_1 ,\ldots, a_n \in K \} \,[itex]

is a subspace S of V called the linear span of v1,...,vn.

The vectors are called spanning vectors and

[itex]\lbrace \mathbf{v}_1,\ldots,\mathbf{v}_n \rbrace[itex]

is called a spanning set or generating set of S.

## Notes

A spanning set is usually not a basis for S as the spanning vectors need not be linearly independent. On the other hand a minimal spanning set for a given vector space S is a basis. In other words a spanning set is a basis for S if and only if every vector in S can be written as a unique linear combination of elements in the spanning set.

## Examples

The real vector space R3 has {(1,0,0), (0,1,0), (0,0,1)} as a spanning set. This spanning set is actually a basis. Another spanning set for the same space is given by {(1,2,3), (0,1,2), (−1,1/2,3), (1,1,1)}, but this set is not a basis, because it is linearly dependent. The set {(1,0,0), (0,1,0), (1,1,0)} is not a spanning set of R3; instead its span is the space of all vectors in R3 whose last component is zero.

## Theorems

Theorem 1: span(v1,...,vn) is a subspace of V. Furthermore, this span is the smallest subspace of V that the vectors v1,...,vn all belong to.

This fact (which is proved later in this section) is one reason why the span is important.

Now let S be a subset of the vector space V. The linear span of S consists of all linear combinations of elements of S. In symbols,

[itex] \mathrm{span}( S ) = \{ a_1 v_1 + \cdots + a_k v_k : k \in \mathbf{N}, a_1 ,\ldots, a_k \in K , v_1 ,\ldots, v_k \in S \} \,[itex]

where N is the set of natural numbers (including zero). Notice that this time the number of vectors involved in the linear combination can vary, from zero on up, but it must still be finite each time.

Theorem 2: span(S) is also a subspace of V. Furthermore, this span is the smallest subspace of V that is a superset of S.

The rest of this section is a proof of Theorem 1. Theorem 2 is very similar, but a bit messier to write down, since the vectors involved in any given linear combination can vary.

Proof of Theorem 1:

Property 1:
The most general possible two elements of the span are x := a1v1 + ... + anvn and y := b1v1 + ... + bnvn. We have to show that x + y is also a linear combination. By using associativity and commutativity of addition and the distributive law, we can write

[itex] x + y = ( a_1 + b_1 ) v_1 + \cdots + ( a_n + b_n ) v_n \,[itex]

and since ai + bi is a scalar for every i, we see that x + y is indeed a linear combination of the given vectors.

Property 2:
Let c be a scalar and again take x := a1v1 + ... + anvn. We have to show that cx is also a linear combination. Now,

[itex] c x = ( c a_1 ) v_1 + \cdots + ( c a_n ) v_n \,[itex]

and since cai is a scalar for every i, we are done.

Property 3:
The zero element 0V of V is a linear combination because we can write

[itex] 0_V = 0_K v_1 + 0_K v_2 + \cdots + 0_K v_n \,[itex]

(Here, 0K is the zero element of the field K.) This equation is true because in every vector space we have 0Kv = 0V.

Minimality:
Suppose W is another subspace of V which contains the vectors v1,...,vn. Then W is closed under scalar multiplication and addition of vectors, so we can prove by mathematical induction that a1v1 + ... + anvn is an element of W for any scalars a1,...,an. Thus, span(v1,...,vn), the set of all such linear combinations, is a subset of W.de:Lineare Hlle he:פרישה ליניארית

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