# Prove that if s1 and s2 are subsets of a vectorspaceV such that

Prove that if s1 and s2 are subsets of a vectorspaceV such that....

## Homework Statement

Prove that if s1 and s2 are subsets of a vector space V such that S1 is a subset of S2,then span(S1) is a subset of span(s2). In particular, if s1 is a subset of s2 and span(s1)=V, deduce that span(s2)=V.

## The Attempt at a Solution

I came up with this, but I doubt its right. Particularly, it only applies to a finite subset. I don't know how I'd modify it to fit any subset.

Let s1,s2 be subsets of V such that s1 is a subset of s2. In cases1=s2, it is clear that span(s1)=span(s2). In case s1 does not equal s2, let x1...xn be the elements of s1. Then a1x1+a2x2+....anxn for all scalars A are in span(s1). We can write x1....xn....xk as the elements of s2. Then by definition a1x1+...anxn+....akxk are in span(s2). So span(s1) is a subset of span(s2).

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CompuChip
Homework Helper

There is a theorem (or perhaps in your text it is even used as a definition) that the span of a set (possibly infinite) consists of finite linear combinations. So if you take any vector v from span(S1) then you can write it as a1x1+a2x2+....anxn for some scalars ai and finitely many vectors xi in S1. Now you only have to show that v is then also in span(S2). That's almost trivial though, if you note that span(S2) is a vector space by definition of span.

I just checked my text, and for some reason it doesn't mention that the span is a finite set of linear combinations. But now I know.

Let s1,s2 be subsets of V such that s1 is a subset of s2. In case s1=s2, it is clear that span(s1)=span(s2). In case s1 does not equal s2, let z=a1x1+a2x2+....anxn be the set of linear combinations of the elements of s1. By definition, z is an element of span(s1). Since the span of any set is a subspace, span(s2) is closed under addition and scalar multiplication. Since x1....xn are all elements of span(s2) for some a1...an, by the definition of a subspace, x=a1x1+a2x2+....anxn is also in span(s2).

Does this work?

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CompuChip