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Math Help - Trace problem

  1. #1
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    Trace problem

    Prove that if A and B are similar nxn matrices, then trace of (A) = trace of (B).

    Proof.

    Write A = Q^{-1}BQ

    Then we have QA = BQ, tr(QA) = tr(BQ), since trace is commutative, I have tr(AQ) = tr(BQ).

    Now, can I say that tr(A) = tr(B)?
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    Super Member Rebesques's Avatar
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    Remember that the determinant of a matrix is equal to the sum of its eigenvalues.

    So looking at

    A-sI=Q^{-1}BQ-sQ^{-1}Q=Q^{-1}(B-sI)Q

    it's evident that A and B have the same eigenvalues, hence equal trace.
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  3. #3
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    our class haven't got to engenvalue yet, is there another way to do it?

    Thanks!
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tttcomrader View Post
    Prove that if A and B are similar nxn matrices, then trace of (A) = trace of (B).

    Proof.

    Write A = Q^{-1}BQ

    Then we have QA = BQ, tr(QA) = tr(BQ), since trace is commutative, I have tr(AQ) = tr(BQ).

    Now, can I say that tr(A) = tr(B)?
    Remember that the trace is cyclic:
    tr(A) = tr(Q^{-1}BQ) = tr(QQ^{-1}B) = tr(IB) = tr(B)

    -Dan
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