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Math Help - Matrix e^(tA)

  1. #1
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    Matrix e^(tA)

    I'm trying to figure out if I am doing this correctly.

    Given the matrix A:
    [1 -.5]
    [.5 2]
    write down e^(tA) and calculate e^(tA)X, X = (1,1)

    So far what I did was, I found the eigenvalues (1.5 repeated) and the eigenvectors ( [-1 1], [1 1]). (The eigenvectors are transposed)
    Then I set e^(tA) = T^(-1)AT, where T = [eigenvector1, eigenvector 2]

    Which leads to e^(tA):
    [1.75 .75]
    [0 1.5]


    I Feel that I do not have the correct equation for e^(tA), can anyone help me?
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  2. #2
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    For a 2 X 2 system, how can you have two eigenvectors for a repeated eigenvalue?
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  3. #3
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    The second eigenvector is the adjoint vector.
    I calculated it from (A -1.5I)W = V, where V is the eigenvector of the eigenvalue 1.5
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  4. #4
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    First, it is very easy for a 2 by 2 matrix, having only one eigenvalue, to have two independent eigenvectors- the matrix
    \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}
    has the single eigenvalue 3 and has <1, 0> and <0, 1> as independent eigenvectors.


    However, this matrix does not. It has only multiples of <1, -1> as eigenvectors. <1, 1> is a "generalized eigenvector".

    But " T^{-1}AT", even for a diagonalizable matrix, is NOT " e^{At}". Rather, T^{-1}AT= D where D is the diagonal matrix having the eigenvalues on the diagonal. Then e^{At}= Te^DT^{-1} where e^{D} is the diagonal matrix having e to the eigenvalue powers on the diagonal.

    That is, if D= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix} then e^D= \begin{bmatrix}e^{\lambda_1} & 0 \\ 0 & e^{\lambda_2} \end{bmatrix}.

    However, when we do not have a "complete set of eigenvectors" as here, we cannot diagonalize the matrix. What we can do is write it in "Jordan Normal Form" which, here, would be J= \begin{bmatrix}\frac{3}{2} & 1 \\ 0 & \frac{3}{2}\end{bmatrix}.

    And if B= \begin{bmatrix}\lambda_1 & 1 \\ 0 & \lambda_1\end{bmatrix} then
    e^{Bt}= \begin{bmatrix}e^{\lambda_1t} & te^{\lambda_1t} \\ 0 & e^{\lambda_1t}\end{bmatrix}
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  5. #5
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    You mentioned that e^D will not be the diagonal matrix with the eigenvalues as the diagonal elements.

    Am I supposed to solve e^(tA) = Te^(D)T^(-1)
    where D:
    (1.5 1)
    (0 1.5)

    So that e^D:
    (e^(1.5t) te^1.5t)
    (0 e^(1.5t))
    or
    e^D:
    (e^(1.5) e^(1.5))
    (0 e^(1.5))?

    I did it the first way and solved it out to e^(tA)=
    (-.5te^(1.5t)+e^(1.5t) -.5te^(1.5t))
    (.5te^(1.5t) .5te^(1.5t)+e^(1.5t))
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  6. #6
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    Quote Originally Posted by Borkborkmath View Post
    I did it the first way and solved it out to e^{tA}= \begin{bmatrix} -.5te^{1.5t}+e^{1.5t} & -.5te^{1.5t} \\ .5te^{1.5t} & .5te^{1.5t}+e^{1.5t} \end{bmatrix}
    Yes, that's the same answer that I got!
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