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Skew Symmetric Determinant Proof

  1. Oct 7, 2009 #1
    Hi all! I was working on some homework for the linear algebra section of my "Math Methods for Physicists" class and was studying skew symmetric matrices. There was a proof I saw on Wikipedia that proves that the determinant of a skew symmetric matrix is zero if the number of rows is an odd number.

    [tex]det(A) = det(A^T) = det(-A) = (-1)^n*det(A)[/tex]

    This is followed up by, "Hence, det(A) = 0 when n is odd." The problem is that I don't understand the proof too well. I understand that the determinant of a matrix is equal to the determinant of its transpose. That means that the determinant of the negation of a matrix is equal to those as well (-A = A^T). Looks like the (-1)^n*det(A) means that multiplying each row by (-1) will produce the same result as the other derivations so far.

    If my logic is sound up to this point, then I get it all, until the big leap to, "Hence, det(A) = 0 when n is odd." Could someone point out either a flaw in my previous logic, or help me to understand how they get to the idea that det(A) must be zero when n is odd? Thank you! :)
     
  2. jcsd
  3. Oct 7, 2009 #2
    Okay, my math skills must be low tonight, because I think I got it. :P If someone would confirm what I'm thinking, that'd be nice!

    In the end we get that det(A) = (-1)^n*det(A). If n is odd, we get det(A) = -det(A), which is only possible when det(A) is zero. Does that sound right?
     
  4. Oct 14, 2009 #3
    Pretty much sums it up.
     
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