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Math Help - Subspace of a vector space

  1. #1
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    Subspace of a vector space

    The following statement is false. Could someone give a counterexample?

    If V is a real vector space and W is a (non-empty) subset of V such that for all vectors
    w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V .
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    Re: Subspace of a vector space

    Quote Originally Posted by Lotte1990 View Post
    The following statement is false. Could someone give a counterexample?

    If V is a real vector space and W is a (non-empty) subset of V such that for all vectors
    w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V .
    This is not true. Consider \mathbb{Z}\subseteq\mathbb{R}. Said differently, (if you had changed w_1+w_2 to w_1-w_2) being a subgroup of a vector space's underlying abelian group is not equivalent to being a subspace of the vector space.
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    Re: Subspace of a vector space

    Quote Originally Posted by Drexel28 View Post
    This is not true. Consider \mathbb{Z}\subseteq\mathbb{R}. Said differently, (if you had changed w_1+w_2 to w_1-w_2) being a subgroup of a vector space's underlying abelian group is not equivalent to being a subspace of the vector space.
    What do you mean? Could you be more specific? Could you define two specific vectors for w1 and w2? Or is this not possible?
    I need some help in understanding this...
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    MHF Contributor Drexel28's Avatar
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    Re: Subspace of a vector space

    Quote Originally Posted by Lotte1990 View Post
    What do you mean? Could you be more specific? Could you define two specific vectors for w1 and w2? Or is this not possible?
    I need some help in understanding this...
    Well, I'll give you a hint. As I said, the problem is that W being a subgroup of V does not imply that W is a subspace. Said more concretely, we have that w_1,w_2\in W implies w_1+w_2\in W DOES NOT imply that \alpha w_1\in W for all \alpha\in\mathbb{R}. To see this, look at my example. For all x,y\in\mathbb{Z} one has that x+y\in\mathbb{Z} so that \mathbb{Z} is closed under addition. But, look at where the problem comes in, is \mathbb{Z} closed under arbitrary real number multiplication?
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    Re: Subspace of a vector space

    Quote Originally Posted by Lotte1990 View Post
    The following statement is false. Could someone give a counterexample?

    If V is a real vector space and W is a (non-empty) subset of V such that for all vectors
    w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V .

    Set of all n-dimensional vectors with positive (non-zero) real coordinates.
    Last edited by Hartlw; December 1st 2011 at 01:38 PM.
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