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Vector Spaces, Subsets, and Subspaces

  1. Nov 8, 2007 #1
    1. The problem statement, all variables and given/known data

    What is an example of a subset of R^2 which is closed under vector addition and taking additive inverses which is not a subspace of R^2?

    R, in this question, is the real numbers.

    2. Relevant equations

    I know that, for example, V={(0,0)} is a subset for R^2 that is also a subspace, but I can't figure out how something can be a subset and not a subspace.

    3. The attempt at a solution

    Does this have anything to do with scalar multiplication being closed on the vector space?
  2. jcsd
  3. Nov 9, 2007 #2


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    Think about the set of all (x,y) where x and y are both integers.
  4. Nov 9, 2007 #3
    So for example, if we let the subset = (a,b) s.t. a,b are elements of Z. Then it is closed under addition but not under scalar multiplication. i.e. Let (a,b) = (1,3) and multiply by 1/2 for example (which is the example we used to figure it out). Then you get (1/2, 3/2), neither of which are in Z.
  5. Nov 9, 2007 #4


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    Sure, but it does have additive inverses.
  6. Nov 9, 2007 #5
    or both rational numbers
  7. Nov 9, 2007 #6


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    It's very easy to be a subset without being a subspace! Just look at any subset that does not satisfy the requirements for a subpace- what about { (1, 0)}?

    In order for a subset to be a subspace, it must be closed under addition, have additive inverses, and be closed under scalar multiplication. Since you are asked about a subset that IS closed under addition and has additive inverses, looks like there is only one thing left!
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