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I have a problem finding a simple form for $\exp(M)$ (or $\exp(tM)$), where $$M = \begin{pmatrix} 1 & a & a^2 & \dots & a^{n-1} \\ 0 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & a^2 \\ \vdots & & \ddots & \ddots & a \\ 0 & \dots & \dots & 0 & 1 \end{pmatrix}$$ As $M$ has only one eigenvalue, I can write $$\exp(M) = \exp(I_n)\sum_{k=0}^{n-1} \frac{1}{k!}(M-I_n)^n$$ with $M-I_n$ nilpotent of order $n$. But I can't find a closed form for the sum.

With clear notations, I can write $M=\sum_{k=0}^{n-1} a^kJ^k$ ($J$ being the matrix with ones just above the diagonal, zeroes everywhere), but it doesn't seem to get me anywhere.

Some experimentation with MuPAD tells me $\exp(t(M-I_n))$ should be some polynomial function in $a$, $J$ and $t$ (which I can prove quite easily), but there is nothing really clear about the coefficients.

Does anyone have an idea where I could go from here ?

Thanks.

Nicolas FRANCOIS
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2 Answers2

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I use your notations.

  1. First idea: if $\ \ \forall x \in ]-1,1[ \ , \ \exp\left( \dfrac{x}{1-x}\right) = \displaystyle \sum_{n=0}^{+\infty} \alpha_k x^k \ \ $ , then: $$ \exp(M)= e\sum_{k=0}^{n-1}\alpha_k a^k J^k $$
  2. Second idea: Let $\ \ f(x)= \exp\left( \dfrac{x}{1-x}\right)\ \ $. Then $(1-x)^2 f'(x) =f(x)$ and we find that: $$\alpha_0=\alpha_1=1\ \text{ and } \ \forall k \in \mathbb N^{\star} \ , \ (k-1)\alpha_{k-1} + (k+1)\alpha_{k+1} = (2k+1)\alpha_k $$
  3. Third idea: Let $\beta_k= k! \alpha_k$. Then $$\beta_0=\beta_1=1\ \text{ and } \ \forall k \in \mathbb N^{\star} \ , \ \beta_{k+1} = (2k+1)\beta_{k} - k(k-1)\beta_{k-1} $$ We find that $\ (\beta_2,\beta_3,\beta_4,\beta_5,\beta_6)=(3,13,73,501,4051)$ $
  4. Fourth idea: Go to the OEIS, you will find A000262
perroquet
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The first thing that strikes me is that if $N$ is the standard nilpotent Jordan block of size$~n$ (so that $N^n=0$), then you have $I_n+aN+(aN)^2+\cdots(aN)^{n-1}$ which (because of the nilpotency) is $\sum_{i\in\Bbb N}(aN)^i=\frac1{1-aN}$ (in the ring of formal power series evaluated in $N$, which is in fact the $n$-dimensional algebra $\Bbb C[N]/(N^n)$). So it comes down to computing the power series $\exp(1/(1-aX))$ in $X$ up to degree $n$ and then substituting $N$ for $X$. Since the argument of $\exp$ has a constant term $1$ there is a multiplicative factor $\exp(1)=e$ that enters in the result. The remaining factor can be computed as a truncated power series in $T=aX$, namely $\exp(\frac1{1-aX}-1)=\exp\frac T{1-T}$, which for $n=5$ for instance gives $1+\frac X{1!}+3\frac{X^2}{2!}+13\frac{X^3}{3!}+73\frac{X^4}{4!}$. I computed this by hand (just expanding out powers and adding), but there are combinatorial interpretations to the coefficients of such an exponential generating series; the comments in the OEIS series A000262 (linked from the other answer) give such interpretations.

By the way, the first terms given above might suggest that the coefficient of $X^n$ in the series could be written as $n-1+\frac1{n!}=\frac{(n-1)n!+1}{n!}$ (except for $n=0$). However that fails from $n=5$ (for which the correct term is $501\frac{X^5}{5!}$ rather than $481\frac{X^5}{5!}$) onwards.

Marc van Leeuwen
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