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Linear Algebra: Vector Space proof...

by Rocket254
Tags: algebra, linear, proof, space, vector
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Rocket254
#1
Oct7-07, 05:10 PM
P: 33
I'm really having trouble comprehending this problem. This is not exactly a "homework problem" but I need a good, formal definition of this to help with some other problems.

Let (Vectors) V1, V2,......,Vk be vectors in vector space V. Then the set W of all linear combinations of Vectors V1, V2,.....Vk is a "subspace" of V.

Exactly how do you prove this?

After setting up two vectors to find the subspace, I'm lost.

Gladly appreciate any help.
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radou
#2
Oct7-07, 05:40 PM
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Well, for some vector space V over a field F, and for the set S = {v1, ..., vk} of vectors from V, we define the span of S as the set of all linear combinations of vectors from S, i.e. span(S) = [itex]\left\{ \sum_{i=1}^n \alpha_{i} v_{i}: n \in \mathbf{N}, v_{i} \in S, \alpha_{i} \in \mathbf{F} \right\}[/itex]. Now, span(S) is obviously a subset of V, right? What simple condition must be satisfied in order for span(S) to be a subspace of V?
HallsofIvy
#3
Oct7-07, 05:57 PM
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Do you see that if [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] are "linear combinations of V1, V2,... , Vk" then [itex]a\vec{u}+ b\vec{v}[/itex] is also a linear combination?

Rocket254
#4
Oct7-07, 06:04 PM
P: 33
Linear Algebra: Vector Space proof...

Quote Quote by HallsofIvy View Post
Do you see that if [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] are "linear combinations of V1, V2,... , Vk" then [itex]a\vec{u}+ b\vec{v}[/itex] is also a linear combination?
Yes, I see that. I'm sorry. I am just not grasping this well at all today.

So, this would be an acceptable proof?

I'm not even exactly sure what is being asked here...

doh!!
radou
#5
Oct7-07, 06:07 PM
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Quote Quote by Rocket254 View Post
I'm not even exactly sure what is being asked here...
What's asked here is very clear, and you simply have to read definitions, that's all.
HallsofIvy
#6
Oct7-07, 06:11 PM
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Surely you've seen examples of subspaces. What do you need to prove to show that a subset of a vector space is a subspace?
Rocket254
#7
Oct7-07, 06:11 PM
P: 33
Ah, It just clicked.

Thank you both very much.
Rocket254
#8
Oct7-07, 06:25 PM
P: 33
Just making sure I have this down pat...
Could I say:

A vector V that is a subspace of "V" is a linear combination of Vectors V1,V2,.....,Vk if
Vector V= C1V1 + C2V2 + C3V3 + .......+ CkVk if C1,.....,Ck is all real numbers?

Would that be acceptable or do I still need to set up vectors "u and v" and write this equation for both vectors?
radou
#9
Oct8-07, 04:34 AM
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Quote Quote by Rocket254 View Post
A vector V that is a subspace of "V" is a linear combination of Vectors V1,V2,.....,Vk if Vector V= C1V1 + C2V2 + C3V3 + .......+ CkVk if C1,.....,Ck is all real numbers?
What?

Look, if you have a vector space V and some *subset* M of V, then this very subset M is a *subspace* of V if and only if, for every two scalars a, b and vectors u, v from M, au+bv is in M, too. (One can easily show that all the other vector space axioms follow from this condition, which makes this condition efficient on an operative level.)


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