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Math Help - Spanning

  1. #1
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    Spanning

    Let S=\{v_1,v_1,...,v_n\} be a spanning set of V.

    Prove that for every w\in V, the set \{w\}\cup S is a lin. dep spanning set of V.

    Seems trivially but not sure how to say it correctly.

    If S isn't a lin ind set of vectors, some vectors in S can be removed to form a minimal spanning set say S' of basis vectors of v. If S is lin. ind., then S is a set of basis vectors for V.

    By definition of span, any vector in V can be written as a linear combination of vectors in S or S' depending if S is lin ind. It seems to follow trivially. Is there a better way to say this?
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  2. #2
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    Re: Spanning

    Quote Originally Posted by dwsmith View Post
    Let S=\{v_1,v_1,...,v_n\} be a spanning set of V.

    Prove that for every w\in V, the set \{w\}\cup S is a lin. dep spanning set of V.

    Seems trivially but not sure how to say it correctly.

    If S isn't a lin ind set of vectors, some vectors in S can be removed to form a minimal spanning set say S' of basis vectors of v. If S is lin. ind., then S is a set of basis vectors for V.

    By definition of span, any vector in V can be written as a linear combination of vectors in S or S' depending if S is lin ind. It seems to follow trivially. Is there a better way to say this?
    Quote Originally Posted by dwsmith
    By definition of span, any vector in V can be written as a linear combination of vectors in S
    w can be written as a linear combination of vectors in S
    so, \{w\}\cup S is lin. dep.
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  3. #3
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    Re: Spanning

    to amplify: we can write (because {v1,...,vn} spans V) w = a1v1 + a2v2 +...+ anvn.

    therefore w - a1v1 - a2v2 - ... - anvn = 0. note that the coefficient of w is 1 ≠ 0, so we have found a linear combination

    c1w + c2v1 + ...+ c(n+1)vn = 0 with not all the cj = 0. so {w,v1,...,vn} is linearly dependent.
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  4. #4
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    Re: Spanning

    Quote Originally Posted by Deveno View Post
    to amplify: we can write (because {v1,...,vn} spans V) w = a1v1 + a2v2 +...+ anvn.

    therefore w - a1v1 - a2v2 - ... - anvn = 0. note that the coefficient of w is 1 ≠ 0, so we have found a linear combination

    c1w + c2v1 + ...+ c(n+1)vn = 0 with not all the cj = 0. so {w,v1,...,vn} is linearly dependent.
    But other than that, what I have is ok?

    I can add this along with what I have to make it more clear and stronger, correct?
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  5. #5
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    Re: Spanning

    you don't need to find the subset S'. all you need is the fact that S spans V, so you can express w as a linear combination of the vj.

    what you did is take S, create a basis S' and then show {w}U S' was linearly dependent. you don't need the middle part, adding anything to a spanning set S

    creates a linearly dependent set, whether or not S was linearly independent.
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