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Math Help - "Eigenpair" Fundamental Matrix

  1. #1
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    "Eigenpair" Fundamental Matrix

    I came across this problem in my class notes. I can't solve this one and I was wondering how to solve it.

    Use the "eigenpair" method to find a fundamental matrix for the differential equation system. Use your answer to find e^At

    x'=Ax [0 1]
    [-2 -3] x
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  2. #2
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    Quote Originally Posted by meks08999 View Post
    I came across this problem in my class notes. I can't solve this one and I was wondering how to solve it.

    Use the "eigenpair" method to find a fundamental matrix for the differential equation system. Use your answer to find e^At

    x'=Ax [0 1]
    [-2 -3] x
    You notation is confusing. I know you mean \begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix} but what do you mean by " A x\begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix}x"? I suspect you have combined two notations and really mean that "Ax" is \begin{bmatrix}0 & 1 \\ -2 & -3\end{bmatrix}x.

    If that is the case, then you need to find all eigenvalues and a corresponding eigenvector for each eigenvalue (the "eigenpairs").

    Once you have that you can write A= PDP^{-1} where "P" is the matrix having eigevectors as columns and D is a diagonal matrix with the eigenvalues on the diagonal.

    It is easy to find powers of a diagonal matrix: D^n is just the diagonal matrix having the nth power of the eigenvalues on the diagonal. Further, A^2= (PDP^{-1})^2= (PDP^{-1})(PDP^{-1})= (PD)(P^{-1}P)Dp^{-1}= PD^2P^{-1}, A^3= A^2A= (PD^2P^{-1})(PDP^{-1})= PD^3P^{-1}, etc.

    e^{At}= I+ At+ \frac{1}{2!}A^2t^2+ \frac{1}{3!}A^3t^3+ \cdot\cdot\cdot= PP^{-1}+ PDtP^{-1}+ \frac{1}{2!}PD^2t^2P^{-1}+ \frac{1}{3!}PD^3t^3+ \cdot\cdot\cdot = P(I+ Dt+ \frac{1}{2}(Dt)^2+ \frac{1}{3!}(Dt)^3+ \cdot\cdot\cdot)P^{-1}= Pe^{Dt}P^{-1} where e^{Dt} is just the diagonal matrix having e^{\lambda t} on the diagonal with \lambda being the eigenvalues.
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  3. #3
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    Ya. Sorry about the notation. It is Ax= the matrix. So thanks for the help.
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