Matrix exponential of a subdiagonal matrix Given a subdiagonal matrix (every element is zero except the elements directly below the main diagonal), is there an analytic form to calculate the elements of the matrix exponential?
In particular, I want an analytic expression of the elements $m_{ij}$ of $M = e^{\alpha\hat{a}^\dagger}$ where $\alpha$ is an arbitrary complex number, and $\hat{a}^\dagger$ is the $N\times N$-matrix
$$
a^{\dagger} = \begin{pmatrix}           
0 & 0 & 0 & \dots & 0 \\
\sqrt{1} & 0 & 0 & \dots & 0\\
0 & \sqrt{2} & 0 & \dots & 0\\
\vdots & \vdots & \ddots & \ddots  & \vdots\\
0 & 0 & \dots & \sqrt{N-1} & 0 \\
\end{pmatrix},
$$
which is also known as the creation operator.
Alternatively, if there are analytic expressions for $P$ and $\lambda$ in the eigen-decomposition $\alpha\hat{a}^{\dagger} = P \lambda P^{-1}$, then that approach might suffice.
 A: An $n\times n$ subdiagonal matrix $A$ has all elements of $A^k$ zero except along the diagonal that is $k$ below the main diagonal. This means we have $A^n=0$, and the finite sum:
$$e^{\alpha A}=\sum_{k=0}^{n-1}\frac{\alpha^k}{k!}A^k$$
The entires along the $k$th subdiagonal of $A^k$ are given by the product of $k$ consecutive terms in $A$'s subdiagonal. That is, for your case,
$$A^k=\begin{bmatrix}0&\cdots&\cdots&\cdots&0\\\vdots&\ddots&\ddots&\ddots&\vdots\\0&\ddots&\ddots&\ddots&\vdots\\\sqrt{1\times\dots\times k}&0&\ddots&\ddots&\vdots\\0&\sqrt{2\times\dots\times(k+1)}&\ddots&\ddots&\vdots\\\vdots&0&\ddots&\ddots&\vdots\\\vdots&\ddots&\ddots&\ddots&0\\0&\cdots&\cdots&0&\sqrt{(n-k)\times\dots\times(n-1)}\end{bmatrix}$$
Which gives a direct closed form for your exponential (the $k$th subdiagonal of $e^{\alpha A}$ comes from $A^k$).
A: Let $D=\operatorname{diag}(\sqrt{0!},\sqrt{1!},\ldots,\sqrt{n!})$ and let $J$ be the lower triangular nilpotent Jordan block of size $n$ (i.e. the entries on the first subdiagonal of $J$ are ones and all other entries are zero). Then $a^\dagger=DJD^{-1}$. Therefore
$$
M:=e^{\alpha a^\dagger}=De^{\alpha J}D^{-1}=D\left(\sum_{k=0}^{n-1}\frac{\alpha^k}{k!}J^k\right)D^{-1}.
$$
Since $(J^k)_{ij}=1$ when $i-j=k$ and zero elsewhere, $M$ is a lower triangular matrix with
$$
m_{ij}=\frac{\alpha^{i-j}}{(i-j)!}\frac{d_{ii}}{d_{jj}}=\frac{\alpha^{i-j}}{(i-j)!}\sqrt{\frac{i!}{j!}}
$$
when $i\ge j$.
A: An example will help to understand the structure of the exponential of such a category of matrices (a particular case of nilpotent matrix : $(a^{\dagger})^n=0$ for some $n$, here $n$ because we are dealing with with $6\times 6$ matrices).
Let us make a colored parallel between 
1) the matrix exponential using definition :
$$\exp(a^{\dagger})= \color{cyan}{I}+ \color{blue}{a^{\dagger}}+ \color{magenta}{\tfrac12 (a^{\dagger})^2}+\color{green}{\tfrac16 (a^{\dagger})^3}+\color{red}{\tfrac{1}{24} (a^{\dagger})^4}+\tfrac{1}{120} (a^{\dagger})^5$$
(we stop here because $(a^{\dagger})^6=0$, giving  $(a^{\dagger})^n=0$ for any $n \ge 6$). 
2) and the different subdiagonals of the result :
$$\exp(a^{\dagger})=\begin{pmatrix}
           \color{cyan}{1}&           0&          0&       0&       0& 0 \\
            \color{blue}{1}&             \color{cyan}{1}&          0&       0&       0& 0 \\
    \color{magenta}{\sqrt{2}/2}&     \color{blue}{\sqrt{2}}&           \color{cyan}{1}&       0&       0& 0 \\
  \color{green}{\sqrt{6}/6}&    \color{magenta}{\sqrt{6}/2}&     \color{blue}{\sqrt{3}}&         \color{cyan}{1}&       0& 0 \\
   \color{red}{\sqrt{6}/12}&    \color{green}{\sqrt{6}/3}&    \color{magenta}{\sqrt{3}}&        \color{blue}{2}&         \color{cyan}{1}& 0 \\
  \sqrt{30}/60& \color{red}{\sqrt{30}/12}& \color{green}{\sqrt{15}/3}& \color{magenta}{\sqrt{5}}&  \color{blue}{\sqrt{5}}&   \color{cyan}{1} 
\end{pmatrix}$$
exhibiting the fact that the powers of $a^{\dagger}$ make progressively "receding" contributions, each one on a specific subdiagonal, till a certain rank beyond which there is no longer any contribution.

Edit : in fact, the preceding result can be cast into a more general one :
If $M$ has values $M_{k,k+1}=a_k$ and $0$ otherwise, its exponential is :
$$\begin{pmatrix}
           \color{cyan}{1}&           0&          0&       0&       0& 0 \\
            \color{blue}{a_1}&             \color{cyan}{1}&          0&       0&       0& 0 \\
    \color{magenta}{\dfrac12 a_1a_2}&     \color{blue}{a_2}&           \color{cyan}{1}&       0&       0& 0 \\
  \color{green}{\dfrac16 a_1a_2a_3}&    \color{magenta}{\dfrac12 a_2a_3}&     \color{blue}{a_3}&         \color{cyan}{1}&       0& 0 \\
  \color{red}{\dfrac{1}{24} a_1a_2a_3a_4}&    \color{green}{\dfrac16 a_2a_3a_4}&    \color{magenta}{\dfrac12 a_3a_4}&        \color{blue}{a_4}&         \color{cyan}{1}& 0 \\
\dfrac{1}{120} a_1a_2a_3a_4a_5  & \color{red}{\dfrac{1}{24} a_2a_3a_4a_5}& \color{green}{\dfrac16 a_3a_4a_5}& \color{magenta}{\dfrac12 a_4a_5}&  \color{blue}{a_5}&   \color{cyan}{1} 
\end{pmatrix}$$
giving at once the formula for the general entry of the exponential.
