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Determinant of the exponential of a matrix

  1. Jun 19, 2006 #1
    Let [itex]B \in \mathbb{R}^{n\times n}. [/itex]Show that [itex]\det e^B = e^{tr B}[/itex] where tr B is the trace of of B.

    Clearly [itex]e^{tr B}[/itex] is the product of the diagonal entries of [itex]e^{B}[/itex].

    By the Jordan canonical for theorem, [itex]\exists P,J \in \mathbb{C}^{n \times n}[/itex] where P is invertible and J is a diagonal sum of Jordan blocks, such that

    [itex]BP = PJ[/itex]
    [itex]P^{-1}BP = J[/itex]
    [itex]P^{-1}e^{B}P = e^{J}[/itex]
    [itex]e^{B} = Pe^{J}P^{-1}[/itex]
    [itex]\det e^{B} = \det P \det e^{J} \det P^{-1}[/itex]
    [itex]\det e^{B} = \det e^{J}[/itex]

    This is where I get stuck. Since [itex]e^{J}[/itex] is upper triangular, its determinant is the product of its diagonal entries, but why should this be equal to the product of the diagonal entries of [itex]e^{B}[/itex]?

    Last edited: Jun 19, 2006
  2. jcsd
  3. Jun 19, 2006 #2


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    Can you compute [itex]\det e^J[/itex] explicitly (in terms of the eigenvalues of B)? If so, then you can use the fact that the traces of similar matrices are equal.
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