Matrix exponential of a simple bidiagonal matrix

Question

I am interested in finding an expression (closed form or recursive) for the matrix exponential of this banded matrix:

$$ \begin{pmatrix} 0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & a_1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & a_2 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 0 & a_3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots &\ddots&\vdots&\vdots \\ 0 & 0 & 0 & 0 & \cdots & a_{n-1} & 1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & a_n \end{pmatrix} $$

For simplicity, assume that $a_k>0~\forall k$. I am quite certain one must exist having played around with it for a while, and looking at the solution for small values of $n$.

Has anyone seen this structure before? Does it have a name? Do you know if there is a solution published somewhere?

If you go ahead and compute the answer for small values of $n$, you get:

$$ \exp\left( \begin{array}{cc} 0 & 1 \\ 0 & a_1 \\ \end{array} \right) = \left( \begin{array}{cc} 1 & \frac{-1+e^{a_1}}{a_1} \\ 0 & e^{a_1} \\ \end{array} \right) $$

$$ \exp \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & a_1 & 1 \\ 0 & 0 & a_2 \\ \end{array} \right) = \left( \begin{array}{ccc} 1 & \frac{-1+e^{a_1}}{a_1} & \frac{-e^{a_2} a_1+a_1+e^{a_1} a_2-a_2}{a_1 \left(a_1-a_2\right) a_2} \\ 0 & e^{a_1} & \frac{e^{a_1}-e^{a_2}}{a_1-a_2} \\ 0 & 0 & e^{a_2} \\ \end{array} \right) $$

Unfortunately, $n=3$ is to large to print here, but a pattern remains.

Answer 1

It suffices to diagonalize $A$ (if the $(a_i)$ are pairwise distinct). I change slightly the notation $A=D+J_n$ where $D=diag(a_1,\cdots,a_n)$ and $J_n$ is the nilpotent Jordan block of dimension $n$. $A=PDP^{-1}$ where the first row of $P$ is

$1,-\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_3)(a_2-a_3)},-\dfrac{1}{(a_1-a_4)(a_2-a_4)(a_3-a_4)},\cdots$

We obtain the other rows of $P$ by circular permutation along the diagonals. Finally $e^A=Pdiag(e^{a_1},\cdots,e^{a_n})P^{-1}$ where the first row of $P^{-1}$ is

$1,\dfrac{1}{a_1-a_2},\dfrac{1}{(a_1-a_2)(a_1-a_3)},\dfrac{1}{(a_1-a_2)(a_1-a_3)(a_1-a_4)},\cdots$

We obtain the other rows of $P^{-1}$ by circular permutation along the diagonals.

Answer 2

This response is very late but perhaps still useful to others who come across your question whilst browsing (as I did!).

Your banded (or upper bidiagonal) matrix is what Optiz termed a $\textit{Steigungsmatrix }$ (maybe ... "gradient matrix"). See McCurdy et al (1984), Mathematics of Computation, 43, 501-528. For any $n>0$, all you need is to apply $\textit{Opitz's formula}$. You will need a little knowledge of functions of matrices and divided differences to apply it but it will be worth the effort.

If we denote your matrix by A and its elements by $a_{ij}$ then, using Opitz's theorem, the matrix F defined as F = $g$(A) for a function $g$(.), has entries $f_{ij}$ as follows: $f_{ij}=0$ for $i>j$, $f_{ij} = g(a_{ii})$ for $i=j$, and $f_{ij} = g[a_{ii},a_{i+1,i+1},..,a_{jj}]$ for $i<j$ .

The notation $g[x_0,x_1,...,x_n]$ is the nth divided difference defined recursively as \begin{equation} g[x_0,x_1,...,x_{(n-1)},x_n] = \frac{g[x_1,...,x_n]-g[x_0,...,x_{(n-1)}]}{x_n-x_0}, \end{equation} $g[x_0,x_1] = (g[x_1]-g[x_0])/(x_1-x_0)$ and $g[x] = g(x)$.

Setting $g$(.) to be the exponential function, I get your answers for n=1 and n=2.

The above assumes distinct diagonal entries. If there are repeats in the diagonal terms then you should use the appropriate "confluent" form of the divided difference.