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Span and Subspace Question

by kbartlett
Tags: span, subspace
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kbartlett
#1
Oct13-08, 01:10 PM
P: 2
I got a small test tomorrow and i have been working throu exercises but i can't seem to solve this question:

Let V be a vector space over a field F, and let [tex]S\subset S'[/tex] be subsets of V.

a) Show that span(S) is a subspace of V.

b) Show that span(S) is a subset of Span(S').

c) Take [tex] V = R^3 [/tex] and give an example to show that it is possible that Span(S) = Span (S') even though [tex] S \subset S'[/tex] and [tex] S \neq S' [/tex].

d) Let U,W be subspaces of V. Prove that U+W is also a subspace of V.
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Tac-Tics
#2
Oct13-08, 03:10 PM
P: 810
Quote Quote by kbartlett View Post
I got a small test tomorrow and i have been working throu exercises but i can't seem to solve this question:

Let V be a vector space over a field F, and let [tex]S\subset S'[/tex] be subsets of V.

a) Show that span(S) is a subspace of V.

b) Show that span(S) is a subset of Span(S').

c) Take [tex] V = R^3 [/tex] and give an example to show that it is possible that Span(S) = Span (S') even though [tex] S \subset S'[/tex] and [tex] S \neq S' [/tex].

d) Let U,W be subspaces of V. Prove that U+W is also a subspace of V.
Go back to your definitions of subspace and span. These problems become pretty easy when you do.

A subspace is a subset of a vector space which is still closed under addition and scalar multiplication. Stated another way, any linear combination of vectors in S will be in S (or of S' in S').

The span is simply the set of linear combinations of a set of vectors. That is, you can take any finite number of vectors from S, make them as long or as short as you want, and add them together and the result is always in Span(S).

For the proof of (d), I'm not sure of your use of + in U+W. U and W are sets, which we don't usually have an addition operation for. Maybe you meant U union W or maybe you're using a notation I'm just not familiar with or something.

(Also, use the [tex ] tag, not [math ] on these boards. The latter doesn't work.
CompuChip
#3
Oct13-08, 03:19 PM
Sci Advisor
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P: 4,300
Welcome to PF kbartlett. Instead of "math", you may want to try "tex"

It seems to me that a) and b) are the same question, if you replace S by V and S' by S in b).
Try writing down an element from span(S) which is not in V, and use the axioms of a vector space to derive a contradiction.

For c: try to make S a set that spans the plane [tex]\mathbb R^2[/tex]. Do you see how to construct S'?

For d: check the axioms of a vector space.

If you need more specific advise, please post some workings.

CompuChip
#4
Oct13-08, 03:21 PM
Sci Advisor
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P: 4,300
Span and Subspace Question

Quote Quote by Tac-Tics View Post
For the proof of (d), I'm not sure of your use of + in U+W. U and W are sets, which we don't usually have an addition operation for. Maybe you meant U union W or maybe you're using a notation I'm just not familiar with or something.
As U and W are subspaces of a vector space (which has an addition [itex]+_V[/itex]), I think it is not unreasonable to assume that
[tex]U + W = \{ u +_V w \mid u \in U, w \in W \}[/tex]
kbartlett
#5
Oct14-08, 05:29 AM
P: 2
Thanks i think ive got the answer to part d), but im still stuck on the rest.
CompuChip
#6
Oct14-08, 06:31 AM
Sci Advisor
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P: 4,300
Suppose that I tell you that some vector, v, is in Span(S). What does this mean?


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