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Linear Algebra: Vector Space proof... 
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#1
Oct707, 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. 


#2
Oct707, 05:40 PM

HW Helper
P: 3,220

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?



#3
Oct707, 05:57 PM

Math
Emeritus
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Thanks
PF Gold
P: 39,503

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?



#4
Oct707, 06:04 PM

P: 33

Linear Algebra: Vector Space proof...
So, this would be an acceptable proof? I'm not even exactly sure what is being asked here... doh!! 


#5
Oct707, 06:07 PM

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P: 3,220




#6
Oct707, 06:11 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,503

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?



#7
Oct707, 06:11 PM

P: 33

Ah, It just clicked.
Thank you both very much. 


#8
Oct707, 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? 


#9
Oct807, 04:34 AM

HW Helper
P: 3,220

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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