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

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If S1 is a subset of S2 in a vector space V, then the span of S1 is a subset of the span of S2. This is established by showing that any linear combination of elements from S1 can also be expressed using elements from S2. The discussion emphasizes that the span is defined through finite linear combinations, which applies regardless of whether the subsets are finite or infinite. Additionally, it clarifies that any vector from span(S1) can be represented in span(S2) due to the properties of vector spaces. The proof hinges on the closure properties of spans and the definition of linear combinations.
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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.

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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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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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First of all, I didn't say that the set is a finite set of linear combinations. I said it is a set of finite linear combinations, which is something completely different.

Also, z = a1x1 + ... + anxn is not a set, and it needn't contain all the elements of s1. In any case, if you want to prove that span(S1) is a subset of span(S2), you will need to make clear that you are taking any element z from span(S1). You can use that if you pick such a z, you can write it in the form you gave; rather than: if you write something in the form you gave, it is in span(S1).

Finally, a sentence like
"Since x1...xn are all elements of span(s2) for some a1...an"
is weird: why are x1 ... xn elements of span(S2) and what does a1...an have to do with them? I don't see a1...an in x1...xn?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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