# Part 2Find a Subset that is not a Subspace

1. Jul 8, 2009

Here we go again

Give an Example of a subset U of R^2 that is closed under scalar multiplication, but is not a subspace of R^2

I am thinking let U={(x1,x2) : x1=-x2}

If x=(x1,x2) and y=(y1, y2) then x+y= (-x2-y2), x2+y2) = (-(x2+y2), x2+y2) okaaayy so that does not work

Let me think.

2. Jul 8, 2009

### Dick

Here's a hint. The intersection of two subspaces is a subspace. The union of two subspaces may not be a subspace.

3. Jul 9, 2009

Okay. Can I do like.... the subspace of R^2 whose elements are all even numbers? Since (0) is not in there? But it is closed under scalar mult over R.

Edit: I don't know if I used your hint, though. :/

4. Jul 9, 2009

### Dick

You mean even integers? Like (n1,n2) where n1 and n2 are even? i) Most people would say 0 is even. ii) That's not closed under scalar multiplication. (n1,n2)*sqrt(2) isn't in the set. Suggest you use the hint.

5. Jul 9, 2009

Yeah. I wasn't sure about 0. And I guess I don't know how to use your hint Are you saying that I should join two subsets of R^2, such that they do not form a subspace? Of course you, are.... why am I asking... that's clearly what he is saying.

6. Jul 9, 2009

### Dick

Give me any two different nontrivial subspaces of R^2. Any two. That's a good exercise.

7. Jul 9, 2009

Z^2 and Q^2 I think these are the notations for the subspaces that are integers and rational numbers. Maybe.

Q^2 is a subspace right? I think it is.

8. Jul 9, 2009

### Dick

I don't think either one is, if you are talking about a vector space over the field R. They aren't closed under scalar multiplication by an element of R. Give me an example.

9. Jul 9, 2009

Oh jeez... you are right How about all (x1,x2) whose sum is (0) and all (x1,x2) for which x1=2*x2.

10. Jul 9, 2009

### Dick

Much better. So the union (as in set theory) of those two (call it U) is closed under scalar multiplication, right? Then take a vector in the first one and a vector in the second one and add them. Is the sum in U?

11. Jul 9, 2009

### HallsofIvy

Staff Emeritus
Actually, I was thinking even simpler. {(x, 0)} is a subspace of R2 and so is {(0, y)}. Is their union, the set of all pairs (x, y) such that at least one is 0, a subspace? Is it closed under scalar multiplication? What about vector addition?

12. Jul 9, 2009

I am confused now. I thought the definition of 'union' was that if A has elements a1,...,an and B has b1,...,bn then the union of A&B is a set that contains a1,...,an & b1,...,bn.

Why are you saying that the union of {(x,0)} and {(0,y)} is {(x,y)}. What did I miss? It looks like you added the two sets.

Shouldn't the union of {(x,0)} & {(0,y)} be the set that includes only all vectors whose y coordinate is 0 OR whose x coordinate is 0 ?

Isn't {(x,y)} by definition R^2 ?

13. Jul 9, 2009

### Staff: Mentor

HallsOfIvy didn't say that the union of {(x,0)} and {(0,y)} is {(x,y)}. What you missed is the "such that" part.

14. Jul 9, 2009