# Prove the property of skew symmetric matrix

1. Sep 8, 2008

### frostshoxx

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. Sep 8, 2008

### Dick

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)?

3. Sep 8, 2008

### frostshoxx

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?

4. Sep 8, 2008

### Dick

No. Transpose(AB)=Transpose(B)*Transpose(A). The order of the matrices is reversed.

5. Sep 8, 2008

### frostshoxx

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?

6. Sep 8, 2008

### Dick

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.