Linear Algebra: Vector Space proof

[Rocket254]

Linear Algebra: Vector Space proof...

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) = $\left\{ \sum_{i=1}^n \alpha_{i} v_{i}: n \in \mathbf{N}, v_{i} \in S, \alpha_{i} \in \mathbf{F} \right\}$. 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 $\vec{u}$ and $\vec{v}$ are "linear combinations of V1, V2,... , Vk" then $a\vec{u}+ b\vec{v}$ is also a linear combination?

 

[Rocket254]

Do you see that if $\vec{u}$ and $\vec{v}$ are "linear combinations of V1, V2,... , Vk" then $a\vec{u}+ b\vec{v}$ 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]

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]

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]

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.)