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The Matrix Exponent of the Identity Matrix, I

  1. May 23, 2015 #1
    So, essentially, all I wonder is: What is the The Matrix Exponent of the Identity Matrix, [itex]I[/itex]?

    Silly question perhaps, but here follows my problem. Per definition, the Matrix Exponent of the matrix [itex]A[/itex] is,

    e^{A} = I + A + \frac{A^2}{2} + \ldots = I + \sum_{k=1}^{\infty} \frac{A^k}{k!} = \sum_{k=0}^{\infty} \frac{A^k}{k!}

    as [itex]e^0 = I[/itex]. I suspected that, since [itex]I^k = I[/itex] for any integer [itex]k[/itex], we would get

    e^{I} = I + I + \frac{I}{2} + \ldots = I \cdot \left( \sum_{k=0}^{\infty} \frac{1}{k!} \right) = I \cdot e,\quad e\approx 2.72

    such that for an arbitrary constant [itex]a[/itex] we could write

    e^{aI} = I \left( \sum_{k=0}^{\infty} \frac{a^k}{k!} \right) = I e^{a}

    However, apparently this is not the case as a (suggested) solution to some (homework) problem I've been working on claims that

    e^{aI} = e^{-a} I

    With a sign change of a!! I think I'm just missing something trivial and fundamental, but I'd really appreciate some help to sort this one out. Might it also be a misprint in the solution?
  2. jcsd
  3. May 23, 2015 #2


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    This is correct.

    This is wrong.
  4. May 23, 2015 #3


    Staff: Mentor

    @mhsd91, when you post a question, please do not delete the three parts of the homework template. The template is required.
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