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How can e^{Diag Matrix} not be an infinite series?

  1. Nov 12, 2015 #1
    So, in a section on applying Eigenvectors to Differential Equations (what a jump in the learning curve), I've encountered

    [tex]e^{At} \vec{u}(0) = \vec{u}(t)[/tex]

    as a solution to certain differential equations, if we are considering the trial substitution [itex]y = e^{\lambda t}[/itex] and solving for constant coefficients. I understand the idea more or less, but the math gets quite involved, particularly with complex eigenvalues and eigenvectors.

    I can see how (or at least I can follow and accept the explanation for) [itex]e^{At}[/itex] being an infinite series.

    But how does [itex]e^{\Lambda t} = [/itex] a single matrix with [itex]e^{\lambda_i t}[/itex] on it's diagonal?
     
  2. jcsd
  3. Nov 12, 2015 #2

    Geofleur

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    If we write the series expansion of the exponential, we get

    ## e^{\Lambda t} = \mathbf{1} + \Lambda t + \frac{1}{2}t^2 \Lambda^2 + \cdots ##,

    where ## \mathbf{1} ## is the identity matrix. Now ## \Lambda ## is a diagonal matrix, so the result of multiplying it by itself is simple, e.g., ## \Lambda^2 = diag(\lambda_1^2, ..., \lambda_n^2)##. Therefore,

    ## e^{\Lambda t} = \mathbf{1} + diag(\lambda_1, ...,\lambda_n) t + \frac{1}{2}t^2 diag(\lambda_1^2,...,\lambda_n^2) + \cdots ##.

    If you write this out in matrix form, summing to form a single matrix, you will see that each entry in the resulting diagonal is an infinite series that itself equals ## e^{t \lambda_j } ## for the ## jj ## matrix entry.
     
  4. Nov 12, 2015 #3

    DrClaude

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    Staff: Mentor

    It works because for ##D = \mathrm{diag}(d_1, d_2, \dots d_N)##, ##D^n = \mathrm{diag}(d_1^n, d_2^n, \dots d_N^n)##.

    Looks like Geofleur beat me to it :smile:
     
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