Prove the property of skew symmetric matrix

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To prove that the product of transpose(B), S, and B is a skew-symmetric matrix when S is skew-symmetric, one must show that the transpose of the product equals its negative. The key properties of skew-symmetric matrices include that the transpose of S is equal to -S and that the sum of its diagonal elements is zero. The discussion highlights the correct application of the transpose operation, emphasizing that Transpose(AB) = Transpose(B) * Transpose(A). By substituting Transpose(S) with -S, it is necessary to demonstrate that Transpose(B) * S * B satisfies the condition for skew-symmetry, specifically that Transpose(C) = -C. The conversation revolves around clarifying these matrix operations to reach the proof.
frostshoxx
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Homework Statement


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


Homework 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


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

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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