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Math Help - Vector space and its subspace

  1. #1
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    Vector space and its subspace

    Let V be a vector space, and let W \subset V be its subspace. Is it true, that there exists a space W' such that V = W \oplus W'?

    It is clear for finite-dimensional spaces (you can work with bases, and matrices)! However i have no clue how to do this in the general case? Is it really true, or is there any counterexapmle?
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  2. #2
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    Re: Vector space and its subspace

    do you believe in the axiom of choice?

    if so, then every vector space has a basis. this includes W. so we may assert we have a basis B for W, and can extend this to a basis C for V.

    define W' = span(C\B).

    (there are some vector spaces for which such bases have never been adequately described. the usual example is the real numbers as a vector space over Q. R is uncountable, so any basis has to be uncountable, so explicitly listing one is out of the question).
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  3. #3
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    Re: Vector space and its subspace

    Quote Originally Posted by Deveno View Post
    do you believe in the axiom of choice?

    if so, then every vector space has a basis. this includes W. so we may assert we have a basis B for W, and can extend this to a basis C for V.

    define W' = span(C\B).

    (there are some vector spaces for which such bases have never been adequately described. the usual example is the real numbers as a vector space over Q. R is uncountable, so any basis has to be uncountable, so explicitly listing one is out of the question).
    Yes, I know, that when assuming axiom of choice every v. s. has a base. But I haven't seen in any book, the theorem allowing me to extend the base of the subspace to some base of space. Naturally this would allow us to prove the above theorem.
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    Re: Vector space and its subspace

    well, it's clear we can pick a basis B for W. consider the set L of all linearly independent sets of V that contain B. B is in L, so L is non-empty. L is also partially ordered by inclusion.

    for each chain C in L, define C' = UC. clearly, C' is an upper bound for C. let {x1,x2,...,xn} be any finite set of vectors in C'.

    since each xj is in C, there is some element Cj in C with xj in Cj, thus some finite union in C, C1 U C2 U....U Cn containing {x1,x2,...xn}
    such that xj is in Cj for 1 ≤ j ≤ n.

    since C is totally ordered, there is some Ck where 1 ≤ k ≤ n, such that Ck = C1 U C2 U...U Cn, so {x1,x2,...,xn} ⊆ Ck, and since Ck is in L, {x1,x2,...,xn} is linearly independent.

    but this means C' is in L (since any finite subset of it is linearly independent).

    so by the axiom of choice (ok, zorn's lemma, whatever) L has a maximal element, M, which is a basis for some subspace of V.

    suppose y is in V - span(M). let {v1,v2,....,vn} be any finite set of vectors in span(M), and suppose that:

    a1v1 + a2v2 +....+anvn - by = 0. if b = 0, for every such set, (and choice of coefficients) then a1 = a2 =...= an = 0, by the linear independence of M,
    but then M U {y} is linearly independent, contradicting its maximality.

    but if for some such set b ≠ 0, then y = (a1/b)v1 + (a2/b)v2 + ...+ (an/b)vn,

    contradicting our choice of y. so there must not be any such y, so V - span(M) = , so M is a basis for V, that contains B as a subset.
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  5. #5
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    Re: Vector space and its subspace

    Thank you, that is great
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