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Math Help - Span of a Set

  1. #1
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    Span of a Set

    The question is.......

    Let u and v be any vectors in R^n. Prove that the spans of {u,v} and {u+v, u-v} are equal.

    I'm thinking if they u and v were inverses of each other then this would work, but i'm not sure.

    I'm not sure how to go about this problem. TIA
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  2. #2
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    Quote Originally Posted by TreeMoney View Post
    The question is.......

    Let u and v be any vectors in R^n. Prove that the spans of {u,v} and {u+v, u-v} are equal.

    I'm thinking if they u and v were inverses of each other then this would work, but i'm not sure.

    I'm not sure how to go about this problem. TIA
    Let x in Span(u,v), then there exists a, b in R, such that:

    x=au+bv.

    Now put a'=(a+b)/2 and b'=(a-b)/2, then x=a'(u+v) +b'(u-v) so x is in
    Span(u+v, u-v).

    A similar argument (left to the reader) will show that if y in Span(u+v, u-v)
    then y is also in Span(u,v).

    RonL
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  3. #3
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    Quote Originally Posted by TreeMoney View Post
    The question is.......

    Let u and v be any vectors in R^n. Prove that the spans of {u,v} and {u+v, u-v} are equal.

    I'm thinking if they u and v were inverses of each other then this would work, but i'm not sure.

    I'm not sure how to go about this problem. TIA
    We need to show that the sets,
    S={au+bv|a,b in R and u,v in R^n}
    T={a(u+v)+b(u-v)|a,b in R}
    Are equal.

    Equivalently, S is subset of T and T is subset of S.

    In set T if a=1/2 and b=1/2 then, u is in S
    In set T if a=1/2 and b=-1/2 then v is in S.
    Thus, their linear combination is in T.
    Thus, T contains S.

    In set S if a=1 and b=1 then u+v is in T.
    In set S if a=1 and b=-1 then u-v is in T.
    Thus, their linear combination is in T.
    Thus, S contains T.

    Thus, T=S
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