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Math Help - Determinant of a 2n x 2n antisymmetric matrix

  1. #1
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    Determinant of a 2n x 2n antisymmetric matrix

    Compute the determinant of the 2n \times 2n matrix \left(\begin{array} {ccccc} 0 & 1 & 1 & ... & 1 \\ -1 & 0 & 1 & ... & 1 \\ -1 & -1 & 0 & ... & 1 \\ .. \\ -1 & -1 & ... & -1 & 0 \end{array}\right)
    (ie. all diagonal entries equal to zero, all above diagonal entries equal to 1, all below diagonal entries equal to -1)

    I am not sure how to go about finding this in general. I have found that the determinants when n is 1 and 2 are both 1, but that doesn't really help me in general, and I don't really have any idea what to start with for the general case. Any tips?
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  2. #2
    Super Member girdav's Avatar
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    Re: Determinant of a 2n x 2n antisymmetric matrix

    Hint: do c_k\rightarrow c_k-c_{k-1} for k=2,\ldots, 2n: you will have to compute, after expanding from the first line, the following (2n-1)\times (2n-1) determinant
    \begin{vmatrix}1&1&0&\ldots&0\\1&1&1&\ldots&0\\1&0  &1&\ldots&0\\\vdots&\vdots&\ddots&\ddots&\vdots\\1  &0&0&\ldots&1\end{vmatrix}.
    To compute this, notice that the first 2n-2 components of the vector which composes the first column can be written as linear combinaisons of the other columns, namely c_1 = \sum_{k=0}^{n-1}c_{2k+1}. After expanding, you only have to compute the determinant of a triangular matrix.
    Last edited by girdav; July 28th 2011 at 08:16 AM.
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  3. #3
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    Re: Determinant of a 2n x 2n antisymmetric matrix

    Thank you! I wasn't really sure how you ended up with that 2n-1 \times 2n-1 matrix, but via a similar procedure I did get the 2n \times 2n matrix  \left( \begin{array} {cccccc} 0 & 1 & 0 & 0 & ... & 0 \\ -1 & 1 & 1 & 0 & ... & 0 \\ -1 & 0 & 1 & 1 & ... & 0 \\ . & . & . & . & . & . \\ -1 & 0 & 0 & 0 & ... & 1 \end{array} \right) which, after adding \sum_{k=2}^{2n}c_k to c_1 gives  \left( \begin{array} {cccccc} 1 & 1 & 0 & 0 & ... & 0 \\ 1 & 1 & 1 & 0 & ... & 0 \\ 1 & 0 & 1 & 1 & ... & 0 \\ . & . & . & . & . & . \\ 1 & 0 & 0 & 0 & ... & 1  \\ 0 & 0 & 0 & 0 & ... & 1 \end{array} \right)
    Then, with expansion along the first row, you get the determinant to be 1, as it is of the form 1(det of triangular matrix with ones on the diagonals  ) -1( det of matrix which can be reduced to a triangular matrix with a zero on the diagonal by adding rows ), which is just 1(1)-1(0). Thanks!
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  4. #4
    Super Member girdav's Avatar
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    Re: Determinant of a 2n x 2n antisymmetric matrix

    Yes, I found 1 too.
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