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Math Help - Skew-symmetric matrix determinant question

  1. #1
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    Skew-symmetric matrix determinant question

    How do I prove that the determinant of an nxn, skew-symmetric matrix is always zero if n is odd?
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  2. #2
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    Well, you know that \det(-A)=(-1)^{n}\det(A). Does that help?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Well, you know that \det(-A)=(-1)^{n}\det(A). Does that help?
    I'm sure it does. I'll let you know when my brain starts working again. Lol.
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  4. #4
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    And what's the definition of a skew-symmetric matrix? Can you tie that in with the equation I just gave you?
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  5. #5
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    I got it already. Thanks alot for the help. I just proved that an inverse retains the upper triangularity of the original matrix, and I the determinant scalar can't affect the zero entries.
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  6. #6
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    There's a problem: not all skew-symmetric matrices are invertible.
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  7. #7
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    Re: Skew-symmetric matrix determinant question

    Basic facts:
    (*) det(A) = det(A^T)
    (**) det(kA) = k^n det(A) where A is nxn matric and k is scalar

    Let n be odd and let A be a skew-symmetric nxn matrix. Then A^T = -A. Then det(A^T) = det(-A). Note that by (**) where k = -1, we see that det(-A) = (-1)^n det(A). Then since n is odd, (-1)^n = -1. Then det(-A) = -det(A). Then by (*) we see that det(A) = det(A^T) = det(-A) = -det(A). Thus, det(A) = -det(A). Then 2*det(A) = 0. Then det(A) = 0.
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