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I'm stuck on a proof that's probably trivial, any insights?

  1. Jul 9, 2015 #1
    Given polynomial P(a,b,c,d,e,f........n) = (a^0)(b^1)(c^2).......(n^n)

    show that the sum of P(a,b,c,d,e,f........x) when acted upon by the symmetric group of order n and each time multipled by the sgn function (1 if even and -1 if odd); that this sum is equal to the vandermonde determinant of these variables. Any insights? I'm quite lost.
  2. jcsd
  3. Jul 9, 2015 #2
    What is your definition of the determinant? Do you know of the definition which uses the sign function?
  4. Jul 9, 2015 #3
    Well, I believe that I'm using the standard definition for the determinant of the matrix (the resulting polynomial from co-factor expansion). However, the resulting determinant from a vandermonde matrix can be given by a simpler formula; it's given on this page under 'properties' https://en.wikipedia.org/wiki/Vandermonde_matrix.
  5. Jul 9, 2015 #4
    So yes, I'm using the definition that involves the sign function. I'm still having trouble getting from the sum of alternating polynomials acted on by the symmetric group and the definition of the determinant of the vandermonde matrix.
  6. Jul 9, 2015 #5
  7. Jul 9, 2015 #6
    Wow, so this is like exactly what I'm looking for?
  8. Jul 9, 2015 #7
  9. Jul 9, 2015 #8
    u da boss :D
  10. Jul 10, 2015 #9
    So just to clarify, what you're saying is that the map that i'm trying to prove is equivalent to the vandermonde determinant is really equivalent to determinants in general, and thus equivalent to the vandermonde determinant.
  11. Jul 10, 2015 #10
    Yes. The identity you are trying to prove is a special case of a more general identity.
  12. Jul 10, 2015 #11


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    If you want to go through the details of this standard permutation approach to determinants, including the proof that it agrees with the expansion by rows you mention, it is explained on pages 62-67 of these class notes:

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