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Math Help - Sylvester Matrix

  1. #1
    Senior Member slevvio's Avatar
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    Sylvester Matrix

    Hello I am in the process of proving that the determinant of the sylvester matrix gives the resultant


    f(x) = a_0 x^m + ... + a_{m-1} x + a_m
    g(x) = b_0 x^n +... + b_{n-1} x + b_n

    (can't get latex to work)

    I am supposed to show that the determinant of the Sylvester matrix associated to f,g has degree less than or equal to mn. (see PlanetMath: proof that Sylvester's matrix equals the resultant) I don't understand how it is clear that the degree is less than or equal to mn. I do understand that each a_k or b_k has degree k.

    Any help would be appreciated , thank you
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  2. #2
    A Plied Mathematician
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    You have to show that the determinant of the Sylvester matrix associated with f and g has degree less than mn, or m + n?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Ackbeet View Post
    You have to show that the determinant of the Sylvester matrix associated with f and g has degree less than mn, or m + n?
    It has degree less than mn.

    Hint : remember that the determinant of the matrix (a_{ij}) can be written \sum_{\sigma \in S_n} (\mbox{sgn }\sigma) a_{1\sigma(1)}\dots a_{n\sigma(n)}, where the sum is taken over the symmetric group S_n and  \mbox{sgn } is the sign.
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  4. #4
    Senior Member slevvio's Avatar
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    thank you, where i can find information about this determinant result?
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    MHF Contributor Bruno J.'s Avatar
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  6. #6
    Senior Member slevvio's Avatar
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    thank you. I would thank your posts but that facility seems to have gone since i was last here
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    MHF Contributor Bruno J.'s Avatar
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    You're welcome!

    There is a little star in the lower left corner, if you must. But you've thanked me twice already!
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  8. #8
    Senior Member slevvio's Avatar
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    I understand now why this is true. It's a bit like a sudoko you pick an element in each row and then pick an element in the next row but not in the column of the first one you picked, and multiply them and keep going down. In some simple examples i see that the highest possible degree after multiplying the entries is mn, but how do i show this for the general case of an (m+n)x(m+n) matrix?
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  9. #9
    Senior Member slevvio's Avatar
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    i know its something to do with the fact there are m entries of a_i polynomials, and n entries of b_j, and for some reason when you add up the i's and j's it can't be bigger than mn, because the zeros in the matrix force you to pick certain values , but i can't work out how to make this rigorous!
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