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

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Homework Help Overview

The discussion revolves around demonstrating that the trace of the product of a symmetric matrix and a skew-symmetric matrix is zero. The original poster presents an attempt at a proof involving matrix properties and trace operations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use summation notation and properties of the trace to show the result. Some participants question the clarity of the steps taken and seek further explanation on the algebraic properties of the trace.

Discussion Status

The discussion is ongoing, with participants providing feedback on the proof attempt. Some guidance has been offered regarding the use of algebraic properties instead of summation notation, but there is no explicit consensus on the correctness of the proof or the steps involved.

Contextual Notes

Participants are discussing the definitions and properties of symmetric and skew-symmetric matrices, as well as the implications of the trace operation in this context. There is a noted concern about the adequacy of explanations for certain steps in the proof.

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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