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Matrix exponentials

  1. Oct 3, 2008 #1
    [tex] A = \[ \left( \begin{array}{ccc}
    0 & 0 & 1 \\
    0 & 1 & 0 \\
    1 & 0 & 0 \end{array} \right)\] [/tex]

    eigenvalues are [tex] \lambda_{1} = -1, \ \lambda_{2} = \lambda_{3} = 1 [/tex]

    [tex](A-\lambda_{1}I)u^{(1)} = 0 \ => \ u^{(1)} = \[ \left( \begin{array}{c}
    -1 \\
    0 \\
    1 \end{array} \right)\] [/tex]

    [tex] (A-\lambda_{2}I)u^{(2)} = 0 \ => u^{(1)} = \[ \left( \begin{array}{c}
    1 \\
    0 \\
    1 \end{array} \right)\] [/tex]

    [tex] (A -\lambda_{3}I)u^{(3)} = u^{(2)} \ => \
    \left(
    \begin{array}{ccc|c}
    1&0&-1&-1\\
    0&0 &0&1\\
    0&0&0&0
    \end{array}
    \right)
    [/tex]

    since we cannot have 0 = 1, we can say that there is only one eigenvector for

    [tex]\lambda = 1 [/tex]

    which means that the Jordan form will be

    [tex] \[ \left( \begin{array}{ccc}
    1 & 1 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & -1 \end{array} \right)\] [/tex]

    Am I correct?

    Now I need to find exp(Jt) and I'm not sure how.

    If I only have 2 eigenvectors, how can I find the fund. matrix T such that

    [tex]e^{At} = Te^{Jt}T^{-1} [/tex]?
     
    Last edited: Oct 3, 2008
  2. jcsd
  3. Oct 3, 2008 #2

    HallsofIvy

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    In general, to find such a function of other than a "number", use the Taylor's series

    eA= I+ A+ (1/2)A2+ (1/3!)A3+ ...

    If A happens to be diagonal, that's easy: An is just the diagonal matrix with exponentials of the diagonal elements of A on its diagonal.

    For Jordan form, its a little more complicated. Try calculating J2, J3, ... for this particular J yourself and see if you spot a pattern.
     
  4. Oct 3, 2008 #3
    I noticed that A2k = I and A(2k+1)'s = A, but I don't know how to get eA from that.
     
  5. Oct 3, 2008 #4

    Hurkyl

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    How exactly are you having trouble summing the series?
     
  6. Oct 3, 2008 #5
    I'm just an idiot. I didn't realize that the Taylor expansions' limits came out to be cosh and sinh. I think I finished this problem. Thanks for the help so far.
     
  7. Oct 3, 2008 #6

    Hurkyl

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    For the record, there's another method that can be useful. Note that you can separate that Jordan block into

    [tex]
    \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]
    =
    \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]
    +
    \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]
    [/tex]

    ....
     
  8. Oct 3, 2008 #7
    Ah, I see, and the second matrix is nilpotent.

    But still, if I got that, how would I compute

    [tex]
    e^{At} = Te^{Jt}T^{-1}
    [/tex]

    Not sure what T would be in this case.
     
  9. Oct 3, 2008 #8

    Hurkyl

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    Right. And you'll find the results of doing arithmetic look a lot like derivatives -- a nilpotent matrix behaves a little bit like an infinitessimal.

    (Again, look to the series to get started)
     
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