I am to show that the trace of the product of a symmetric and a skew-symmetric matrix is zero.Please check what I did is corect:
Let me assume:A~=A and B~=-B

(I will use # sign to denote the sum process)

trace(AB)=[#(i)](AB)_ii=[#(i)] [#(j)] a_ij*b_ji

trace(AB)=-[#(j)] [#(i)] b_ji*a_ij using conditions on A and B
=-[#(j)](AB)_jj
Since i and j are equivalent,
what we have is 2trace(AB)=0
hence,conclusion

Let me assume:A~=A and B~=-B


What does this mean?

Also, could you please type out your proof using LaTeX? The tutorial stuck to the top of this forum should get you through it in a minimal amout of time.

I don't think your approach is correct; where is the transpose introduced?

I would first notice that (1) tr(A^T) = tr(A) and (2) tr(AB) = tr(BA).

Then, tr(AB) = tr((AB)^T) = tr(B^T A^T) = -tr(BA) = -tr(AB) which implies that tr(AB) = 0.

Maybe you were trying to prove property (2) in your approach? I can't tell what your notation means.

quadraphonics, I see your method.However,I did not understand where I went wrong.I used ~ sign to mean transpose.Since A is symmetric and B is skew-symmetric,we must have: a_ij=a_ji and b_ji=-b_ij.
I do not think yhere s an error in the summation process.So,I will request you to recheck and tell me where I went wrong.

Currently,I am going to appear exams for next one week.So,let me be free from Latex upto that time.

It's the summation notation that I don't understand. The ~ as transpose was fairly clear, if non-standard. The derivation you have in mind may well be correct; it's just that I can't understand what you wrote.

And it's never too early to learn Latex.