Proof of Subspace and Basis Relationship in R^n - Homework Help"

[simmonj7]

Homework Statement

Prove or disprove this with counter example:
Let U,V be subspaces of R^n and let B = {v1, v2,...,vr} be a basis of U. If B is a subset of V, then U is a subset of V.

Homework Equations

U and V are subspaces so
1. zero vector is contained in them
2. u1 + u2 is in U and v1 + v2 is in V when u1 and u2 are already in U and v1 and v2 are in V
3. au1 is in U and av1 is in V when a is in R

The Attempt at a Solution

Since B is a basis of U, B is a linearly independent spanning set of U and all of the elements of B are in U.

Since B is a subset of V, all of the elements of B are in V.

However this does no guarantee that all the elements of U are in V...Correct?

Well assuming that that is correct, I am having the hardest time finding a counter example.

At first I was just taking a set of vectors and calling that U. Finding the basis of that set of vectors and calling it B. And then creating another random set of vectors that contained the elements that were in B but left out the ones in U. But then I realized that U and V are subspaces so I got lost.

 

[Mark44]

I don't think your assumption here is correct

However this does no guarantee that all the elements of U are in V...Correct?

 

[simmonj7]

Could you elaborate a little more or hint to how that isn't true then...

 

[vela]

Take an arbitrary vector in U and express it in terms of basis B. Go from there.

 

[simmonj7]

Are you trying to say this:

Say B = {v1,v2,...,vp} is the basis for U where U is a subspace of R^n. Then if x is in U then x can be represented unique in the terms of the basis B. Meaning that there are unique scalars a1, a2,...,ap such that x = a1v1 + a2v2 +...+ apvp

And since V is a subspace, the properties av1 is in V and v1 + v2 is in V applies. So since B is in V, and U can be expressed in terms of B, then U is in V...?

 

[vela]

Well, I wasn't trying to say anything. I was trying to get you figure it out. :)

You should fill in some blanks when writing the actual proof, though. You have x=a1 v1+a2 v2+...+an vn, and because the v's are in V and V is a subspace, you can conclude that x is in V. So what you've shown is that $x \in U$ implies $x \in V$. By definition, this means that U is a subset of V.

 

[simmonj7]

Thank you so much! :)