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Math Help - [SOLVED] Vector space

  1. #1
    Neoxl
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    [SOLVED] Vector space

    I am sure this is easy but I have been trying all day and just don't understand the question.

    Determine whether the set, together with the indicated operations, is a vector space.

    1. The set {(x, -x) : x is a real number} with the standard operations.

    2. The set {(x, y) : x, y are a real numbers such that x > y} with the standard operations.

    Any help would be appreciated. Thanks
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  2. #2
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    Quote Originally Posted by Neoxl View Post

    1. The set {(x, -x) : x is a real number} with the standard operations.
    Let S = \{(x,-x)\}. Now if a\in S, \ b\in S then a=(x,-x) and b=(y,-y) so a+b = (x+y,-(x+y)) so a+b in S. Similarly ka in S where k\in \mathbb{R}. So it is a vector space.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let S = \{(x,-x)\}. Now if a\in S, \ b\in S then a=(x,-x) and b=(y,-y) so a+b = (x+y,-(x+y)) so a+b in S. Similarly ka in S where k\in \mathbb{R}. So it is a vector space.
    don't you have to verify a list of like 10 axioms or something. that's what we had to do in my linear algebra class. i guess if the standard operations hold, then verifying that it is closed under addition and scalar multiplication is fine, but i don't know
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    Quote Originally Posted by Jhevon View Post
    don't you have to verify a list of like 10 axioms or something. that's what we had to do in my linear algebra class. i guess if the standard operations hold, then verifying that it is closed under addition and scalar multiplication is fine, but i don't know
    You do not need to. For example, since \mathbb{R}^2 is commutative then certainly any subset! So it just comes down to checking the closure of addition and scalar multiplication.
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