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Math Help - vector space

  1. #1
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    vector space

    Hello

    How can I prove that the set

    W={(t+s, t-s, 2t+3s, t, s);t and s belong to R} is a vector space under the standard operations?

    do i have to prove all the axioms ?

    thank you
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  2. #2
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    Quote Originally Posted by qwerty321 View Post
    do i have to prove all the axioms ?
    No. This is a subset of \mathbb{R}^5 which is a vector space. So immediately most of the definitions for a vector space are immediately verified. All you need to do is show if that if \bold{w}_1,\bold{w}_2\in W then \bold{w}_1+\bold{w}_2\in W and that if k\in \mathbb{R} then k\bold{w}\in W for any \bold{w}\in W.
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  3. #3
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    Hey! thanks for your reply
    Ok but how did you know that it is a subset of R^5?
    I also have to prove that 0 belongs to W no?which is evident

    Anyway can you help me prove the other ones?
    for example for the first one how do i begin with it
    thank you
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    No. This is a subset of \mathbb{R}^5 which is a vector space. So immediately most of the definitions for a vector space are immediately verified.
    I am curious: The empty set {} is also a subset of \mathbb{R}^5, but it is not a vector space. If S is a subset of vector space V, does that really imply that S is itself a vector space?

    Ah, nevermind, I see that your statement is true for any non-empty subset.
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  5. #5
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    this is a theorem scorpion:
    if V is a subset of W(which is a vector space) and you can prove that V is a subspace of W, then V is a subset itself.Now can you help me prove the properties?
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  6. #6
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    Quote Originally Posted by qwerty321 View Post
    this is a theorem scorpion:
    if V is a subset of W(which is a vector space) and you can prove that V is a subspace of W, then V is a subset itself.Now can you help me prove the properties?
    Indeed.

    Ok but how did you know that it is a subset of R^5?
    Because each element of W is a 5-tuple of reals (a_1, a_2, a_3, a_4, a_5), a_i \in \mathbb{R}.

    I also have to prove that 0 belongs to W no?which is evident
    No, because if you can prove that W is closed under scalar multiplication, then the fact that is has a 0 element is implied, by \forall \vec{x} \in W, 0\vec{x} = \vec{0} \in W.(Corrected 30/3/09)

    So as long as you can show that it is not empty, and is closed under scalar multiplication and vector addition, that will suffice to prove that it is a subspace of R^5.

    Anyway can you help me prove the other ones?
    for example for the first one how do i begin with it
    thank you
    For closure under vector addition, you must show that for any \bold{x}, \bold{y} \in W, that \bold{x} + \bold{y} \in W.

    E.g. Let t_1, t_2, s_1, s_2 \in \mathbb{R} and \bold{x}, \bold{y} be elements of W such that,

    \bold{x} = (t_1 + s_1, t_1 - s_1, 2t_1 + 3s_1, t_1, s_1) and
    \bold{y} = (t_2 + s_2, t_2 - s_2, 2t_2 + 3s_2, t_2, s_2).

    Then,

    \bold{x} + \bold{y} = ((t_1 + t_2) + (s_1 + s_2), (t_1 + t_2) - (s_1 + s_2), 2(t_1 + t_2) + 3(s_1 + s_2), t_1 + t_2, s_1 + s_2) which is also in W. (I grouped the terms in the same step, but you can show it separately first.)

    Hence, W is closed under vector addition.

    Scalar multiplication can be shown similarly.
    Last edited by scorpion007; March 29th 2009 at 09:54 PM.
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  7. #7
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    ok that's what i did and it worked
    now suppose i have the same set except that instead of 2t+3s, i have 2t+3s+1..will it be a vector space?
    no right?
    because w1+w2 is not in W(because of the 1 we will have 2)
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  8. #8
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    you said:
    No, because any non-empty subset of a vector space is itself a vector space, and so most of the axioms are inherited and subsequently verified, as ThePerfectHacker pointed out.


    but when we want to prove that w is a vector subspace of v we have to prove 3 properties:

    the scalar multiplication, the addition and the fact that 0 belongs to W, that is, that w is a non empty set
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  9. #9
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    I'm a bit confused
    For example, when we want to prove that thet set W={3x+2y=0, for (x,y)belong to R^2}, we prove that (x1,y1)+(x2,y2) belong to W.

    here, u proved that w1+w2 belong to W.Don't we have to pove that for al s and t, (t1,s1)+(t2,s2) belong to W?

    or because it is a set we can't say that?

    thank yu for your cooperation
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  10. #10
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    I said that because for the scalar multiplication it works for (t,s) not for (w1,w2)
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  11. #11
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    please i need help for the scalar proof
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