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Eigenvalues and Laplace

  1. Mar 9, 2008 #1
    hi, can some one give me any hints how to solve this problem? thank you


    i tried to type it here but it dint come up so i uploaded a picture with the problem.

    Thank you so much


    Recall that for an nxn matrix A with distinct eigenvalues [tex]\lambda[/tex] [tex]_{F}[/ktex], k=1,2,...,n


    e^{At} = \sum^{n}_{k=1} Z_{k}e^{\lambda_{k}t}

    By taking the Laplace Transform of both sides (or otherwise) show that
    \sum^{n}_{k=1}Z_{k}= I_{n}

    Where I_{n} is the nxn identity matrix
     
  2. jcsd
  3. Mar 9, 2008 #2
    Although stated in a very horrible way, what is asked here boils down to show that, if A is diagonalizable, one can compute [itex]e^{At}[/itex] by only computing the diagonal entries after diagonalizing. Keep in mind that [itex]\mathcal{L}(e^{At}) = (sI - A)^{-1}[/itex].

    So once you have the diagonal form employ the inverse laplace transformation.

    I am sure that you can do it, so think about it for a while...
     
  4. Mar 9, 2008 #3
    okay maybe i read it totally wrong but if you know that

    [tex] e^{At} = \sum^{n}_{k=1} Z_{k}e^{\lambda_{k}t} [/tex]

    for all t then use it fot t=0 and get

    [tex] I = e^{0} = e^{A0} = \sum^{n}_{k=1} Z_{k}e^{\lambda_{k}0} = \sum^{n}_{k=1} Z_{k}[/tex]

    and you are done?
     
  5. Mar 9, 2008 #4
    The sum works with a general form of the following
    [tex]
    \left[ {\begin{array}{*{20}c}
    1 & 0 \\
    0 & 0 \\
    \end{array}} \right] + \left[ {\begin{array}{*{20}c}
    0 & 0 \\
    0 & 1 \\
    \end{array}} \right] = I
    [/tex]

    So it is directly related with the diagonal form... Thus you can also prove for non-zero t
     
  6. Mar 10, 2008 #5
    ok i have proven this so far
    (SI-A)L(e^(At)) = Identity...but i do not how to prove that this is equal to [tex]\sum^{n}_{k=1}Z_{k}[/tex]

    Any hints? thank you so much
     
  7. Mar 10, 2008 #6
    OK, but that one tells you nothing, how about this? Let A is a diagonal matrix, then for 2x2 case,

    [tex]
    e^{At}= e^{\left[ {\begin{array}{*{20}c}
    \alpha & 0 \\
    0 & \beta \\
    \end{array}} \right]t} = \left[ {\begin{array}{*{20}c}
    e^{\alpha t} & 0 \\
    0 & e^{\beta t} \\
    \end{array}} \right] = \left[ {\begin{array}{*{20}c}
    1 & 0 \\
    0 & 0 \\
    \end{array}} \right]e^{\alpha t} + \left[ {\begin{array}{*{20}c}
    0 & 0 \\
    0 & 1 \\
    \end{array}} \right]e^{\beta t}
    [/tex]
    And, then from the previous post,
    [tex]

    \left[ {\begin{array}{*{20}c}
    1 & 0 \\
    0 & 0 \\
    \end{array}} \right] + \left[ {\begin{array}{*{20}c}
    0 & 0 \\
    0 & 1 \\
    \end{array}} \right] = I

    [/tex]

    Can you see the pattern now?
     
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