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Computing a complex matrix exponential

  1. Oct 17, 2006 #1
    Dear Members,

    I have been recently introduced with the concept of matrix exponentials in class. I have been successful with computing such functions as [tex]e^{At}[/tex] where say, A is a 2x2 invertible matrix that has real eigenvalues.

    When I am presented with a problem such as: D =

    \D = \left(
    -1+j2 & 0\\
    0 & -1-j2

    And I wish to find [tex]e^{Dt}[/tex], I am having difficulty going about finding a pattern.

    The upper left term, -1+j2, ends up being -3-j4, 11-j2, -7+j24, and so on for [tex]D^{2}[/tex], [tex]D^{3}[/tex], [tex]D^{4}[/tex].

    With that in mind, I am trying to represent a pattern for the real part and complex part as taylor series in order to find out what the solution is, and repeat the process for each entry within the 2x2 matrix D. The problem I am having is that the numbers do not seem to follow any easy pattern, other than the fact that the real values are negative, negative, positive, negative, negative, positive... etc. And a similar pattern with the complex part.

    Matlab gives an answer to [tex]e^{Dt}[/tex] as:

    e^-t*cos(2t) + je^-t*sin(2t) in the top left entry and e^-t*cos(2t) - je^-t*sin(2t) in the bottom left entry, with zeros in the other two entries,

    \D = \left(
    e^{-t}cos(2t) + je^{-t}sin(2t) & 0\\
    0 & e^{-t}cos(2t) - je^{-t}sin(2t)

    So my assumption is that I need to find a pattern that involves the taylor series of e^-t combined with cosine or sine.

    Does anyone have a suggestion as to how I can obtain this? Thank you in advance for your response.
    Last edited: Oct 17, 2006
  2. jcsd
  3. Oct 17, 2006 #2


    User Avatar
    Homework Helper

    For any diagonal matrix you have:

    [tex] \left[ \begin{array}{cc} p & 0 \\ 0 & q \end{array} \right]^n = \left[ \begin{array}{cc} p^n & 0 \\ 0 & q^n \end{array} \right][/tex].

    Plug this into:

    [tex] exp \left( \left[ \begin{array}{cc} p & 0 \\ 0 & q \end{array} \right] \right)= \sum_{n=0}^\infty \frac{1}{n!} \left[ \begin{array}{cc} p & 0 \\ 0 & q \end{array} \right]^n [/tex]

    And you'll end up with:

    [tex]exp \left( \left[ \begin{array}{cc} p & 0 \\ 0 & q \end{array} \right] \right)=\left[ \begin{array}{cc} e^p & 0 \\ 0 & e^q \end{array} \right] [/tex].

    You don't need it here, but if you want to find a general formula for (a+bi)n, try writing it in the form rei θ.
    Last edited: Oct 17, 2006
  4. Oct 17, 2006 #3
    Excellent, thank you very much I understand the problem completely.
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