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Relationship between determinant and trace

  1. Apr 30, 2010 #1
    Hi...

    We have all seen the equation det(M)=exp(tr(lnM)). I was taught the proof using diagonalisation. I was wondering if there was a proof for non-diagonalisable matrices also.
     
  2. jcsd
  3. Apr 30, 2010 #2

    Fredrik

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    Theorem 2.11, page 36. (And I don't think that logarithm is supposed to be there).
     
  4. Apr 30, 2010 #3
    Thanks...:smile: .....the way I have written it, the logarithm is supposed to be there...
     
  5. Apr 30, 2010 #4

    Fredrik

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    Ah, I see it now. The left-hand side in the book is det(exp(M)), not det(M).
     
  6. Apr 30, 2010 #5
    The book by Hall (linked above) uses the decomposition into diagonalisable + nilpotent which is very important in Lie group theory. As slightly more direct approach is to use http://en.wikipedia.org/wiki/Jordan_normal_form" [Broken].

    Schur decomposition: an arbitrary matrix M decomposes as QUQ-1 where U is upper-triangular and (therefore) has the eigenvalues of M on its diagonal.

    det(exp(M)) = det(exp(QUQ-1)) = det(Q exp(U) Q-1) = det(exp(U)) = ∏i exp(λi) = exp(∑λi)

    exp(tr(M)) = exp(tr(QMQ-1)) = exp(tr(MQ-1Q)) = exp(tr(M)) = exp(∑λi)

    btw, in general it is best to not use the logarithm form - because not all matrices will possess a logarithm.
     
    Last edited by a moderator: May 4, 2017
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