1
$\begingroup$

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!

  • 0
    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? – 2017-01-30
  • 1
    @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. – 2017-01-30
  • 0
    It is as important to consider it as a tridiagonal matrix. Here is a pointer to approximate solutions:(http://math.stackexchange.com/q/54915) – 2017-01-30
  • 1
    Silvia Noschese, Lionello Pasquini, and Lothar Reichel, [Tridiagonal Toeplitz Matrices: Properties and Novel Applications](http://www.math.kent.edu/~reichel/publications/toep3.pdf), 2006. – 2017-01-30
  • 0
    I assume you used [Sylvester's shift matrix](https://en.wikipedia.org/wiki/Generalizations_of_Pauli_matrices#Construction:_The_clock_and_shift_matrices) $\Sigma_1=$ 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. – 2017-11-07

1 Answers 1

3

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.

  • 0
    [+1} Good answer. I thought I had a shorter exact answer. This was not the case. – 2017-01-30
  • 0
    @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 $n$. – 2017-01-30
  • 0
    @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 $C_{nn}=-1...$ – 2017-01-30
  • 0
    @JeanMarie You are right. Too bad, I'd have hoped there could be a simpler way. – 2017-01-30