1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Cool thanks very much, this looks promising. However, I'm not sure exactly what you mean by "circular permutation along the diagonals". Do you mean just "circular permutation" or (equivalently) do you mean that the diagonals, when wrapped, will be constant? When I try this I don't get $PP^{-1}=1$ $\endgroup$ – Ian Hincks May 8 '14 at 20:28
  • $\begingroup$ Also, can you point me in the direction of how you arrived at this conclusion? $\endgroup$ – Ian Hincks May 8 '14 at 20:29
  • $\begingroup$ Ah, okay, I figured out that "circular permutation along the diagonal" means that, for example, the second row of $P$ will be $(0,1,-(a_2-a_3)^{-1},((a_2-a_4)(a_3-a_4))^{-1},...)$. $\endgroup$ – Ian Hincks May 8 '14 at 21:12
  • $\begingroup$ Yes Ian, the second row is as you write it. Let $A_n$ be the matrix of dimension $n$. Then $e^{A_n}$ is the submatrix of $e^{A_{n+1}}$ that is constituted by its first $n$ rows and its first $n$ columns. $\endgroup$ – loup blanc May 9 '14 at 10:36
1
$\begingroup$

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.

$\endgroup$

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.