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Help me with these proofs (vector spaces over finite fields)

  1. Aug 13, 2007 #1
    It's hard to find the proofs of these theorems. Please help me... Thanks!

    Theorem 1: Let V be a vector space over GF(q). If dim(V)=k, then V has [tex]\frac{1}{k!}[/tex] [tex]\prod^{k-1}_{i=0}[/tex] (q[tex]^{k}[/tex]-q[tex]^{i}[/tex]) different bases.

    Theorem 2: Let S be a subset of F[tex]^{n}_{q}[/tex], then we have dim(<S>)+dim(S[tex]^{\bot}[/tex])=n.
     
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  3. Aug 13, 2007 #2

    mathwonk

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    think about the steps involved in choosing a basis. then the numbet of bases is the number of ways to carry out these steps.

    first choose any non zero vector. how many ways?

    then afterwrads chooise avector that is not on the line through that one - hiw many ways?............
     
  4. Aug 13, 2007 #3

    morphism

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    What is F[itex]^{n}_{q}[/itex]? An n-dimensional innerproduct space over GF(q)? And is <S> the span of S?

    My thoughts: get an orthonormal basis for <S>, and extend this to one for F[itex]^{n}_{q}[/itex]. The new vectors you add will probably be an o.n. basis for S[itex]^{\perp}[/itex].
     
  5. Aug 13, 2007 #4

    mathwonk

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    thm 2 is true in any vector space of finite dimn. over any field.
     
  6. Aug 13, 2007 #5

    matt grime

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    This doesn't make sense over a finite field (or any field of positive characteristic).
     
  7. Aug 13, 2007 #6

    morphism

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    I was wondering about this also. What is S[itex]^\perp[/itex] over a finite field?
     
  8. Aug 13, 2007 #7

    matt grime

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    Just pick any complementary subspace, or take the quotient space.
     
  9. Aug 13, 2007 #8

    mathwonk

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    Sperp is the subspace of the dual space that annihilates S. equivalently, it is the dual space of the quotient by <S>, which is why matt's hint gives the right dimension.
     
  10. Aug 14, 2007 #9

    morphism

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    Ah, my bad. I wasn't familiar with that notation for the annihilator of S. I thought it was the space of vectors orthogonal to S.
     
  11. Jan 30, 2011 #10
    I need the proof for theorem2,too.

    I know that theorem2 is an important property in coding theory.

    (Fq)^n is the direct product over (Fq)^n (similar with R^n over R)


    the S-perp in (Fq)^n have the same definition.

    But the defition of inner space (over finite field) may not need <x,x> > 0.
    I think the most important point is that the intersecton of C and C-perp may not be zero.
    (i.e. it is possible that <x,x>=0 for some x =/= 0 in (Fq)^n)

    Actually, in the language of coding. If C is a subspace of (Fq)^n with dim(C)=k,
    C-perp = {x in C| x(G)' = 0 in (Fq)^k} where G is the generator matrix,(G)' : transpose of G
    but I don't which is useful or not!!:)
     
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