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Prove the property of skew symmetric matrix

  1. Sep 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Hi, I need to prove that if S is a skew-symmetric matrix with NXN dimension and B is any square real-valued matrix, therefore the product of transpose(B), S, and B is also askew symmetric matrix


    2. Relevant equations
    This is what I know so far.
    1.Transpose(S) = -S within R^N
    2. When N is odd number, S is invertible
    3. The sum of all diagonal elements of S is 0


    3. The attempt at a solution
    1. I try to multiply Transpose(B) with S and then with B to see if all the diagonal elements of the product is 0. Apparently, it does not turn out as I thought.
     
  2. jcsd
  3. Sep 8, 2008 #2

    Dick

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    What is Transpose(AB) in terms of Transpose(A) and Transpose(B) for any matrices A and B? What is Transpose(Transpose(B)*S*B)?
     
  4. Sep 8, 2008 #3
    Thank you for the quick post.

    Tranpose (AB) = Transpose(A) * Transpose(B) for any A and Bmatrices.
    Transpose( Transpose(B) * S * B) = B * Transpose(S) * Transpose(B).

    Am I correct? What would be the next step to approach the end goal?
     
  5. Sep 8, 2008 #4

    Dick

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    No. Transpose(AB)=Transpose(B)*Transpose(A). The order of the matrices is reversed.
     
  6. Sep 8, 2008 #5
    Oh my mistake... so anyway, that would make Transpose (Transpose(B) * S * B) = Transpose(B) *Transpose(S) * B, correct?

    How would this lead to the proof that Transpose(B) * S * B is a skew-symmetric matrix for S is a symmetric matrix and B is any square real-valued matrix?
     
  7. Sep 8, 2008 #6

    Dick

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    Now use Transpose(S)=(-S). A matrix is skew symmetric if that condition holds. So you just want to show that if C=Transpose(B)*S*B, then Tranpose(C)=-C.
     
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