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Solving Exponential of a Matrix

  1. Oct 29, 2015 #1
    Please help me understand the following step

    upload_2015-10-29_18-48-12.png
     
  2. jcsd
  3. Oct 29, 2015 #2

    Geofleur

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    When ## a ## is just a real number, one can use a Taylor series to represent ## e^{at} ## as

    ## e^{at} = \sum_{n=0}^{\infty} \frac{(at)^n}{n!} = 1 + at + \frac{1}{2}(at)^2 + \ldots ##

    By analogy, one can define the exponential of ## \mathbf{A}t ##, where ## \mathbf{A} ## is now a matrix, as

    ## e^{\mathbf{A}t} = \sum_{n=0}^{\infty} \frac{(\mathbf{A}t)^n}{n!} ##.

    Because multiplying a matrix by itself is perfectly well defined, the above sum makes sense. Now if ## t ## is not too large, we can truncate the series and write

    ## e^{\mathbf{A}t} \approx (\mathbf{A}t)^0 + \mathbf{A}t \equiv \mathbf{I} + \mathbf{A}t ##,

    where ##\mathbf{I}## is the identity matrix. Using this truncated sum, your formula can be derived, except for the time dependent functions ##\alpha_1(t)## and ##\alpha_2(t)##. Perhaps someone else can shed some light on where those might be coming from. What is the context in which you are seeing this equation?
     
  4. Nov 2, 2015 #3

    jasonRF

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    Geofleur's infinite series is the place to start. The trick is to use the Cayley-Hamilton theorem, which tells you that a matrix satisfies it's own characteristic equation. Sine you have a 2x2 matrix this is a second order polynomial, so you can write ##\mathbf{A}^2=\alpha_0 \mathbf{I} + \alpha_1 \mathbf{A}## for some ##\alpha_0## and ##\alpha_1## . It follows that any power of your matrix can also be represented by a linear combination of ##\mathbf{I}## and ## \mathbf{A}##. Hopefully that shows you where the result comes from.

    This is a common approach used in electrical engineering.

    Jason
     
  5. Nov 2, 2015 #4
    Thank you so much for your help!
     
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