Vector Spaces

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  1. JBKProductions's Avatar
    • Overlord in Training
    • Posts: 2,105
    Vector Spaces
    Let V be a finite dimensional vector space. If C is a finite spanning set for V and if I is a linearly independent subset of V such that  I \subseteq C prove that there is a basis B of V such that  I \subseteq B \subseteq C . Bit stuck on this proof, any help would be appreciated, thanks.
  2. Narev's Avatar
    • Adored and Respected Member
    • Posts: 538
    Re: Vector Spaces
    I suppose one way would be to let the subset  I be comprised of linearly independent vectors  \utilde{c}_1, \utilde{c}_2, \hdots, \utilde{c}_i say, and  C be comprised of  \utilde{c}_1, \utilde{c}_2, \hdots, \utilde{c}_i, \utilde{c}_{i+1}, \hdots, \utilde{c}_{i+m} say.

    With judicious use of sifting, you should be able to prove this.
    Last edited by Narev; 21-05-2012 at 22:27.
  3. JBKProductions's Avatar
    • Overlord in Training
    • Posts: 2,105
    Re: Vector Spaces
    Cheers, I'll give it a go.
  4. nuodai's Avatar
    • PS Helper
    • TSR Legend
    Re: Vector Spaces
    Any linearly independent set is contained in a basis. (Standard proof: take the elements of I and then add to them by repeatedly adding vectors not contained in the linear span of what you already have, and check that it works.) You need to show that you can modify this proof by adding vectors which, not only are in the complement of the linear span, but which also lie in C, and Narev's post is the simplest way forward. (Namely, whereas in the 'standard proof' you'd pick "any element of the complement of the linear span of the vectors already picked", you instead pick specifically an element of C. You just have to argue that such elements exist.)
    Last edited by nuodai; 21-05-2012 at 23:59.
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