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Matrix exponential

  1. Sep 6, 2011 #1

    syj

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    1. The problem statement, all variables and given/known data
    Given the matrix A=
    [0 1]
    [1 0]
    Find eA
    2. Relevant equations



    3. The attempt at a solution
    I have no idea how to approach this.
    I know the expansion for an exponential matrix, I was able to apply this for a nilpotent matrix.
    Could someone please explain to me how to find the matrix exponential of ANY matrix, with OUT using the expansion ( the expansion isn't really helpful unless my matrix is nilpotent.)

    Thanks a mil
     
  2. jcsd
  3. Sep 6, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't know where you got the idea that "the expansion isn't really helpful unless my matrix is nilpotent". Aren't infinite series useful?

    The Taylor's series expansion is very useful for many kinds of matrices. It is almost trivial to do, for example for diagonal matrices.

    If, for example
    [tex]A= \begin{bmatrix}\lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{bmatrix}[/tex]
    for any n,
    [tex]A^n= \begin{bmatrix}\lambda_1^n & 0 & 0 \\ 0 & \lambda_2^n & 0 \\ 0 & 0 & \lambda_3^n\end{bmatrix}[/tex]
    and so
    [tex]e^A= \sum \frac{1}{n!} A^n= \begin{bmatrix}\sum \frac{1}{n!}\lambda_1^n & 0 & 0 \\ 0 & \sum \frac{1}{n!}\lambda_2^n & 0 \\ 0 & 0 & \sum\frac{1}{n!}\end{bmatrix}= \begin{bmatrix}e^{\lambda_1} & 0 & 0 \\ 0 & e^{\lambda_2} & 0 \\ 0 & 0 & e^{\lambda_3}\end{bmatrix}[/tex]

    Many matrices are diagonalizable. If [itex]A= P^{-1}DP[/itex], where D is diagonal and P an invertible matrix (i.e. is A is "similar" to a diagona matrix), then it is easy to show that
    [tex]A^n= P^{-1}D^nP[/tex]
    and so
    [tex]e^A= P^{-1}e^DP[/tex]
    and [itex]e^D[/itex] is, as above, simple.

    If a matrix is not diagonalizable it has a "Jordan Normal form" for which the exponential expansion is more complcated but still doable.

    In any case, for the matrix you give
    [tex]A= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}[/tex]
    is symmetric and so diagonalizable.

    For this particular matrix the expansion is very simple. Did you at least calculate [itex]A^2[/itex]? It turns out to be easy to write [itex]A^n[/itex] for any n. But I would recommend diagonaizing A.
     
    Last edited: Sep 6, 2011
  4. Sep 6, 2011 #3

    syj

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    Im sorry if my question has offended u.
    What I meant was that it was not useful to just subsitute straight into the expansion.
    I did not know how to approach the question.
    Some further digging on the internet has lead me in the direction of eigenvalues and diagonalization.
    Thank you for your time, and guidance.
    I have the answer for the question, I just want to know how to get to the answer.
    I shall look over your explanation and rack your brains if I get lost.
    Thanks again
     
  5. Sep 6, 2011 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It did not offend me, I was just point out that you were mistaken!
     
  6. Sep 6, 2011 #5

    syj

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    Ok,
    I have diagonalised A
    The eigenvalues are 1 and -1
    My problem now is that I don't know the expansions.
    that is not major though. Some sifting through the internet shall yield the results.
    Im just happy I know HOW to answer the question.
    Thanks for all your help.
    :)
     
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