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Skew-symmetric matrix problem

  1. Mar 10, 2007 #1
    1. The problem statement, all variables and given/known data

    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:

    2. Relevant equations

    3. 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
    Since i and j are equivalent,
    what we have is 2trace(AB)=0
  2. jcsd
  3. Mar 10, 2007 #2


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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.
  4. Mar 10, 2007 #3
    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.
  5. Mar 10, 2007 #4


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    The first few equations here

    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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