Let $\alpha$ be a (strictly) positive real number. Consider the following tridiagonal Toeplitz matrix $$ A=\alpha\begin{bmatrix} 0 & 1 & 0 &\cdots & 0\\ 1 & 0 & 1 &\ddots & \vdots \\ 0 & 1 & 0 & \ddots & 0\\ \vdots & \ddots & \ddots & \ddots & 1 \\ 0 & \cdots & 0 & 1 & 0 \end{bmatrix}. $$

My question. Does there exist a closed-form expression for $\exp(A)$?

I played around a little bit with the truncated series $\sum_{k=0}^N \frac{A^k}{k!}$ but I didn't manage to provide an answer to my question. Pointers to the literature are also welcome!

  • $\begingroup$ Does it matter that, with the $\alpha,$ it can be called a "Toeplitz matrix," or is an answer just based on the definition of $A$ OK for your purposes? $\endgroup$ – coffeemath Jan 30 '17 at 15:29
  • 1
    $\begingroup$ @coffeemath: $A$ is a particular Toeplitz matrix. Since the class of Toeplitz matrices has been extensively studied, I think it could be helpful to stress that $A$ falls into this class. $\endgroup$ – Ludwig Jan 30 '17 at 15:38
  • $\begingroup$ It is as important to consider it as a tridiagonal matrix. Here is a pointer to approximate solutions:(math.stackexchange.com/q/54915) $\endgroup$ – Jean Marie Jan 30 '17 at 16:53
  • 1
    $\begingroup$ Silvia Noschese, Lionello Pasquini, and Lothar Reichel, Tridiagonal Toeplitz Matrices: Properties and Novel Applications, 2006. $\endgroup$ – Rodrigo de Azevedo Jan 30 '17 at 21:06
  • $\begingroup$ I assume you used Sylvester's shift matrix $\Sigma_1=$ and its N-1 powers in your expansion? Indices are distinct mod N , so $A=\alpha(\Sigma_1 + \Sigma_{N-1})$. The expansion of the exponential telescopes to just N/2 terms and the series are tractable. $\endgroup$ – Cosmas Zachos Nov 7 '17 at 15:59

Hint (too long for a comment): tridiagonal Toeplitz matrices are known to have distinct eigenvalues, which can be explicitly calculated (see e.g. here and here). For the matrix in question, for example, the eigenvalues are $\lambda_k=2 \alpha \cos\left(\cfrac{k \pi}{n+1}\right)\,$, $k=1,2,\cdots,n$.

The matrix is therefore diagonalizable, and since the eigenvectors can also be explicitly calculated, it is possible to determine the invertible matrix $P$ and diagonal matrix $D$ such that $A=P\,D\,P^{-1}$.

Given that $\,A^n=P\,D^n\,P^{-1}\,$ it follows that $\,e^A=P\,e^D\,P^{-1}\,$ where $e^D$ is the diagonal matrix with $e^{\lambda_k}$ on the diagonal.

  • $\begingroup$ [+1} Good answer. I thought I had a shorter exact answer. This was not the case. $\endgroup$ – Jean Marie Jan 30 '17 at 20:03
  • $\begingroup$ @JeanMarie Thanks. Your idea could still work out for small matrices, since there would be only a few $N^i \cdot {N^\intercal}^j$ to pre-calculate, but it becomes more tedious for large $n$. $\endgroup$ – dxiv Jan 30 '17 at 20:21
  • $\begingroup$ @dxiv In fact, I had a bigger problem: $N$ and $N^T$ are not commuting matrices, $C:=NN^T-NN^T$ is almost a null matrix but for $C_{11}=1$ and $C_{nn}=-1...$ $\endgroup$ – Jean Marie Jan 30 '17 at 20:27
  • $\begingroup$ @JeanMarie You are right. Too bad, I'd have hoped there could be a simpler way. $\endgroup$ – dxiv Jan 30 '17 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.