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Math Help - Exponential Matrix

  1. #1
    Newbie Chief65's Avatar
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    Exponential Matrix

    Find e^{At} of : \begin{pmatrix}0&2&1\\-1&-3&-1\\1&1&-1\end{pmatrix} . I have found \lambda=2, i, -i but I get stuck trying to find e-vectors and so on. Thanks.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Chief65 View Post
    Find e^{At} of : \begin{pmatrix}0&2&1\\-1&-3&-1\\1&1&-1\end{pmatrix} . I have found \lambda=2, i, -i but I get stuck trying to find e-vectors and so on. Thanks.
    I find the characteristic polynomial : -(\lambda+1)^2(\lambda+2)

    Maybe this would simplify your research of the eigenvectors
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    Quote Originally Posted by Moo View Post
    Hello,

    I find the characteristic polynomial : -(\lambda+1)^2(\lambda+2)

    Maybe this would simplify your research of the eigenvectors
    D'oh! Mixed up problems as I was writing them down. The actual matrix is \begin{pmatrix}2&0&1\\1&0&1\\1&-2&0\end{pmatrix} with \lambda=2, i, -i
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    Quote Originally Posted by Chief65 View Post
    D'oh! Mixed up problems as I was writing them down. The actual matrix is \begin{pmatrix}2&0&1\\1&0&1\\1&-2&0\end{pmatrix} with \lambda=2, i, -i
    Okay

    Then, you have to find the vectors \bold{x}_i=\begin{pmatrix} a \\ b \\ c \end{pmatrix}

    Such that A\bold{x}_i=\lambda_i \bold{x}_i

    \begin{pmatrix}2&0&1\\1&0&1\\1&-2&0\end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} 2a+c \\ a+c \\ a-2b \end{pmatrix}

    Let's do it for \lambda=i (for \lambda=2, it's common techniques !)

    \begin{pmatrix} 2a+c \\ a+c \\ a-2b \end{pmatrix}=i \begin{pmatrix} a \\ b  \\ c \end{pmatrix}=\begin{pmatrix} ai \\ bi \\ ci \end{pmatrix}

    That is :
    \begin{cases} 2a+c=ai \quad (1)\\ a+c=bi \quad (2)\\ a-2b=ci \quad (3)\end{cases}

    From (2), you get b=\frac{a+c}{i}=-i (a+c)
    Substitute in (3) :
    a+2i(a+c)=ci
    a(1+2i)+ci=0

    Now let for example a=1 (that's what's good with eigenvectors, you can choose arbitrarily one component)
    So 1+2i+ci=0 \Rightarrow c=-\frac{1+2i}{i}=i-2

    And now compute b=-i(a+c)

    that's all



    Similar one for \lambda=-i
    For \lambda=2, it should be easy, because all the coefficients are real.


    Once you get the 3 eigenvectors, place them in columns to form matrix P=[x_1,x_2,x_3] such that A=PDP^{-1}
    and where D=\begin{pmatrix} \lambda_1 &0&0 \\ 0&\lambda_2 &0 \\ 0&0&\lambda_3 \end{pmatrix}
    (the order of the eigenvalues has to be the same as the order of the eigenvectors in P)
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    Just want to confirm my thought process on backsolving for A given an exponential matrix e^{At}.

    A matrix of sums is also a sum of matrices, and vice versa. Right? For example,
    would it go like this?
    e^{At} = I + \sum_{n=1}^\infty \frac{A^nt^n}{n!} = I + \sum_{n=1}^\infty\begin{pmatrix}a_{11}&a_{12}&a_{1  3}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end  {pmatrix}^n\frac{t^n}{n!}


     <br />
= I + \begin{pmatrix}\sum_{n=1}^\infty \frac{(a_{11}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{12}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{13}t)^n}{n!}\\ \sum_{n=1}^\infty \frac{(a_{21}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{22}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{23}t)^n}{n!}\\\sum_{n=1}^\infty \frac{(a_{31}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{32}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{33}t)^n}{n!}\end{pmatrix}<br />

    Once I find my A^n, just plug in n=1 to get A. Right?
    Last edited by seadog; May 2nd 2009 at 09:30 PM.
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  6. #6
    Moo
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    Quote Originally Posted by seadog View Post
    Just want to confirm my thought process on backsolving for A given an exponential matrix e^{At}.

    A matrix of sums is also a sum of matrices, and vice versa. Right? For example,
    would it go like this?
    e^{At} = I + \sum_{n=1}^\infty \frac{A^nt^n}{n!} = I + \sum_{n=1}^\infty\begin{pmatrix}a_{11}&a_{12}&a_{1  3}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end  {pmatrix}^n\frac{t^n}{n!}


     <br />
= I + \begin{pmatrix}\sum_{n=1}^\infty \frac{(a_{11}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{12}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{13}t)^n}{n!}\\ \sum_{n=1}^\infty \frac{(a_{21}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{22}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{23}t)^n}{n!}\\\sum_{n=1}^\infty \frac{(a_{31}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{32}t)^n}{n!}&\sum_{n=1}^\infty \frac{(a_{33}t)^n}{n!}\end{pmatrix}<br />

    Once I find my A^n, just plug in n=1 to get A. Right?
    Nope.

    The n-th power of a matrix is not the matrix formed by the elements to the n-th power ~

    Try A=\begin{pmatrix} 2&3 \\ 3&4 \end{pmatrix}
    Do you have A^2=\begin{pmatrix} 2^2 & 3^2 \\ 3^2 & 4^2 \end{pmatrix} ?
    You will see that it's not the case.

    You don't need to find A, you need to find an expression for A^n
    And by finding its eigenvalues and the transition matrix, P, you'll have :
    A=PDP^{-1}
    And hence A^n=PDP^{-1}PDP^{-1}\dots PDP^{-1}PDP^{-1}=P\underbrace{D\dots D}_{n ~times} P^{-1}

    But as I said above, the n-th power of a diagonal matrix is a matrix formed by the n-th power of its diagonal elements.

    So finally, you have :
    e^{At}=I+\sum_{n \geq 1} \frac{A^n t^n}{n!}=I+P \sum_{n\geq 1} \frac{D^nt^n}{n!} P^{-1}=I+P\begin{pmatrix} \sum_{n\geq 1} \frac{\lambda_1^n t^n}{n!} & 0 \\ 0 & \sum_{n\geq 1} \frac{\lambda_2^n t^n}{n!} \end{pmatrix} P^{-1}

    Does it look clear ?
    Last edited by Moo; May 3rd 2009 at 12:26 AM.
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