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Complex Eigenvalues - Proof

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data

    This is the second part of a multi-part question. Part (a) shows that:

    x'' = Ax = [tex]\left(\stackrel{-2}{4/3}\stackrel{3/2}{-3}\right)[/tex]x

    Part (b): Assume x = [tex]\epsilon[/tex]e[tex]^{rt}[/tex] and show that (A - r[tex]^{2}[/tex]I)[tex]\epsilon[/tex] = 0

    x is the solution to the second order differential equation above, and [tex]\epsilon[/tex] is an eigenvector corresponding to the eigenvalue r[tex]^{2}[/tex] of A.

    3. The attempt at a solution

    Part (b): Alright, so given the above, I stated that [tex]\epsilon[/tex] = e[tex]^{-rt}[/tex]x

    I then substituted everything into the left side of the equation I'm trying to prove to obtain:

    (A - r[tex]^{2}[/tex]I)[tex]\epsilon[/tex] = [tex]\left(\stackrel{-2-r^{2}}{4/3}\stackrel{3/2}{-3-r^{2}}\right)[/tex][tex]\left(\stackrel{x_{1}}{x_{2}}\right)[/tex]e[tex]^{-rt}[/tex]

    from here, I can see that I am not going in the right direction... any suggestions to get me moving along?


    EDIT: for clarification, the above are 2x2 matrices... latex put the entries fairly close together. The matrix A has entries (-2 3/2) on the top and (4/3 -3) on the bottom.
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2


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    What is [itex]\epsilon[/itex]? A vector? But since you also don't tell us what x is, what see no way to understand x= [itex]\epsilon e^{rt}[/itex].
  4. Mar 10, 2008 #3
    Sorry, thanks, I'll edit my above post to be more complete.

    Note: for some reason my [tex]\epsilon[/tex] appear to be superscripts... nowhere should this be the case.
    Last edited: Mar 10, 2008
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