What are the Steps to Solve a Skew-Symmetric Matrix Problem?

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To demonstrate that the trace of the product of a symmetric matrix A and a skew-symmetric matrix B is zero, one approach involves using the properties of the trace. The discussion highlights that while the proof attempts to show this through summation notation, it lacks sufficient justification for certain steps. Participants suggest that leveraging algebraic properties of the trace could simplify the proof. Key points of contention include the need for clearer explanations of specific equalities used in the proof. Ultimately, the discussion emphasizes the importance of clarity and justification in mathematical proofs.
Kolahal Bhattacharya
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Homework Statement



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:


Homework Equations





The Attempt at a Solution



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
 
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A~ means the transpose of A?


I think your proof is right, although you skipped some steps and did not provide justification for what was skipped, so I can't be sure.

I will suggest, though, that you don't need to bother with summations at all: you can just use the algebraic properties of the trace instead.
 
Yes, ~ means transpose.
What are the algebraic properties of trace you are referring to?
also I do not uderstand which steps have I jumped?
Thank you.
 
Kolahal Bhattacharya said:
What are the algebraic properties of trace you are referring to?
The first few equations here


also I do not uderstand which steps have I jumped?
These are the two equalities I take issue with

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

-[#(j)] [#(i)] b_ji*a_ij=-[#(j)](AB)_jj

They are certainly true, but I don't think they're adequately explained.
 

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