Matrix exponential of a simple bidiagonal matrix 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.
 A: 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.
A: 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.
