1. Sep 10, 2009

### mat175

1. The problem statement, all variables and given/known data

1. let u and v be any vectors in Rn. Prove that the spans of {u,v,} and {u+v, u-v} are equal.

2. Let S1 and S2 be finite subsets of Rn such that S1 is contained in S2. Use only the definition of span s1 is contained in span s2.

2. Relevant equations
3. The attempt at a solution

1. w in the span (u+v, u-v) show that w is in the span (u,v)
w is a linear combination of (u+v, u-v)
w= c1 (u+v) + c2(u-v)------
w=(c1+c2)u+(c1-c2)v

How do you prove it going the other way though?

2. I am not sure how to start this one, other than having the definition of span there.

2. Sep 10, 2009

### Staff: Mentor

For 1, start with a vector w that is in the span of {u, v}, which means that w = c1u + c2v. See if you can get creative on the two constants to rewrite them as the sum of two numbers and the difference of two numbers, respectively. If haven't done this, but that's the direction I would take.

For 2, you are missing some words in the problem statement. I'm guessing that this is the actual problem statement: "Use only the definition of span to show that span s1 is contained in span s2." Please confirm that this is the correct interpretation.

3. Sep 10, 2009

### mat175

Yes that is correct.

For 1, I did have what you have then I dont know for sure where to go from there.

Thanks.

4. Sep 10, 2009

### Staff: Mentor

Show me what you've done for the 2nd half of this problem.

5. Sep 10, 2009

### mat175

All I have is let y be a linear combination of (u,v)
which is written y=c1u1+c2u2

6. Sep 10, 2009

### Staff: Mentor

Well, no. It would be y = c1u + c2v, right? You want y to be a linear combination of u and v, not u1 and u2.

Starting with y as I have written it, you want to end up with y (the same y) as a linear combination of u + v and u - v, right? What does that look like?

Now you know what you're starting with, and you sort of know what you want to end up with. Can you work backward from the end to the beginning and fill in the middle?