Linear Independence of subsets

[topgear]

Homework Statement

Suppose {V1, V2, ..., Vp} form a linearly independent set of vectors. Show that any subset of this collection of vectors is also linearly independent. Is it necessarily true that is the vectors are dependent, that any subset is also dependent?

Homework Equations

The 10 axioms of subspaces
5 facts about subspaces

The Attempt at a Solution

Is a subset and a subspace the same thing? It makes since that if you have something in something that has finite answers the subset wouldn't have infinite answers. I am just having trouble proving it.

 

[micromass]

Is a subset and a subspace the same thing?

No, not at all! {2} is a subset of $$\mathbb{R}$$, but it isn't a subspace!

It makes since that if you have something in something that has finite answers the subset wouldn't have infinite answers. I am just having trouble proving it.

I'm having trouble seeing what you're getting at...

Like any problem, we need to start with the definitions. What does "linear independent" mean? How is it defines? Do you know equivalent properties for it?

 

[topgear]

Linear Independent: means it has only one or no answers
Would a subset be {v1,v2,v3}, only part of {v1,v2,...,vp}?

 

[micromass]

Linear Independent: means it has only one or no answers

Sorry, this makes no sense to me. Could you please copy the exact definition?

 

[topgear]

An indexed set of vectors {v1,v2,...vp} in R^n is said to be linearly independent if the vector equation x1v1+x2v2+...+xpvp=0 has only the trivial solution.

The zero vector solution is known as the trivial solution.

 

[micromass]

Indeed, so you know that $$x_1v_1+...+x_pv_p=0$$ only has the trivial solution. Now, we we take $$\{v_{i_1},...,v_{i_n}\}\subseteq \{v_1,...,v_p\}$$. And we need to show that $$x_{i_1}v_{i_1}+...+x_{i_p}v_{i_p}=0$$ only has a trivial solution. Maybe you should try contradiction here? Assume that the latter equation has a nontrivial solution, then make a solution for the first equation...

 

[topgear]

I'm sorry if I'm being slow I'm a graphic design major trying to take this class because my parents don't think art is a good major and I need math a math minor and it's killing me.

You're saying I should prove it by If P then not Q. So If x1v1+...xpvp=1 then xi1vi1+...+xipvip is non trivial? Doesn't that make it If not P then not Q?

 

[micromass]

No, you should prove "If not Q, then not P"...

I actually wonder why people consider math a good degree and graphic design a bad degree. That makes no sense to me. Graphic design is much more applied and real-worldy than math. Strange...