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Number of subspaces of a vector space over a finite field

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove: If V is an n-dimensional vector space of a finite field, and if 0 <= m <= n, then the number of m-dimensional subspaces of V is the same as the number of (n-m)-dimensional subspaces.

    3. The attempt at a solution

    Well here's a sketch of my argument. Let U be an m-dimensional subspace of V, then the annihiator of U, U^0 is a (n-m)-dimensional subspace of V*, the dual space of V. Let W be the subspace of V whose dual space is U^0. I plan to show that W is in 1-to-1 correspondence with U, so there is an injection between the set of m-dimensional subspaces of V and the set of (n-m) dimensional subspaces. Since the situation is symmetric, it follows those sets have a bijection and therefore the same cardinality.

    Now before I work out the details, I wanna ask about one thing, this argument nowhere uses the fact that V is a vector space over a finite field (except perhaps at the very last, to substitute "cardinality" by "number of elements"). So is there something wrong with it? why is the problem specifically about vector spaces over finite fields if it works in the general case?

    Thanks.
     
  2. jcsd
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