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For an arbitrary tridiagonal matrix of the form

$$ A = \begin{pmatrix} b_1 & c_1 & 0 & 0 & ... \\ a_2 & b_2 & c_2 & 0 & ... \\ 0 & a_3 & b_3 & c_3 & ... \\ \vdots &&\ddots&\ddots&\ddots\end{pmatrix} $$

is there a formula to calculate $\exp(A)$? Or at least for some special tridiagonal matrices?

The special case I am most interested in is a $(2n+1)^2$ matrix with $b_k = i(k-n-1)$ and $c_k = (a_{2n+2-k})^*$, i.e.

$$\begin{pmatrix} -in & c_1 & 0 & \\ c_{2n}^* & -i(n-1) & c_2 & \\ 0 & c_{2n-1}^* & -i(n-2) & \ddots \\ &&\ddots&\ddots \end{pmatrix}$$

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    $\begingroup$ A closed form for that exponential would entail finding a closed form for the characteristic polynomial of the tridiagonal matrix, since the eigenvectors can be expressed in terms of derivatives of the characteristic polynomial evaluated at appropriate values... $\endgroup$ Aug 10, 2011 at 8:53
  • $\begingroup$ Did you ever find a solution to your problem? $\endgroup$ Jan 10, 2013 at 16:55
  • $\begingroup$ @JohnSalvatier I'm afraid not :-/ $\endgroup$ Jan 10, 2013 at 17:00
  • $\begingroup$ I'm looking for a way to compute exp(At)*x_0 cheaply when A's a symmetric tridiagonal matrix. I think I may just have to eigen-decompose A and do it that way. Luckily I only have to decompose A once, and then it's O(n**2), which I guess is okay. Since you should be able to compute Ax_0 in O(n) steps since its tridiagonal, I was hoping for something better, but maybe that's not possible. $\endgroup$ Jan 10, 2013 at 20:26

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I don't know about closed formulas, but there are several ways to find good approximations of the matrix see e.g. http://math.cityu.edu.hk/~mayylu/papers/matexp.pdf

Most of these aproaches are using Padè approximations of the exponential function of the matrix. These consist of a simple fracture, and don't need much terms to be near machine precision.

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  • $\begingroup$ Thanks for the interesting link; however link-only answers are strongly discouraged here since links can go stale (even to papers), so at least a short summary would be helpful... $\endgroup$ Nov 11, 2013 at 8:46

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