Linear Algebra: Vector Space proof....

Rocket254

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.

radou

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

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

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

What's asked here is very clear, and you simply have to read definitions, that's all.

HallsofIvy

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

Ah, It just clicked.

Thank you both very much.

Rocket254

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

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.)