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Part 2Find a Subset that is not a Subspace

  1. Jul 8, 2009 #1
    Here we go again:biggrin:

    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 :redface:

    Let me think.
     
  2. jcsd
  3. Jul 8, 2009 #2

    Dick

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    Here's a hint. The intersection of two subspaces is a subspace. The union of two subspaces may not be a subspace.
     
  4. Jul 9, 2009 #3
    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. :/
     
  5. Jul 9, 2009 #4

    Dick

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    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.
     
  6. Jul 9, 2009 #5
    Yeah. I wasn't sure about 0. And I guess I don't know how to use your hint :redface: 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.
     
  7. Jul 9, 2009 #6

    Dick

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    Give me any two different nontrivial subspaces of R^2. Any two. That's a good exercise.
     
  8. Jul 9, 2009 #7
    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.
     
  9. Jul 9, 2009 #8

    Dick

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    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.
     
  10. Jul 9, 2009 #9
    Oh jeez... you are right How about all (x1,x2) whose sum is (0) and all (x1,x2) for which x1=2*x2.
     
  11. Jul 9, 2009 #10

    Dick

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    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?
     
  12. Jul 9, 2009 #11

    HallsofIvy

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    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?
     
  13. Jul 9, 2009 #12
    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 ?
     
  14. Jul 9, 2009 #13

    Mark44

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    HallsOfIvy didn't say that the union of {(x,0)} and {(0,y)} is {(x,y)}. What you missed is the "such that" part.
     
  15. Jul 9, 2009 #14
    Oh.:smile: That makes way more sense. :biggrin:
     
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