Umbral calculus - eigenfunctions of operator I'm very new to umbral caluclus and I have come across a paper that makes use of some results in this area, which I do not quite understand.
The problem I have is the following. 
Consider the following operator
\begin{equation}
\mathcal{S} = a^{-1}(bI\Delta_{-1} + c\Delta_{1}), \quad I \in \mathbb{Z^+}
\end{equation}
where $\Delta_h[f(I)] =f(I+h)-f(I), h \in \mathbb{Z}$
It is claimed by the paper that given that $\Delta_{1}(I)_m=m(I)_{m-1}$ and $I\Delta_{-1}(I)_m = -m(I)_m$ [where with $(I)_m$ we denote the falling factorial] the eigenfunctions $\psi(I)$ of the operator are computable as:
\begin{equation}
\psi_n(I) = \sum_{m=0}^{n} {n \choose m} \left(-\frac{c}{b}\right)^m(I)_{n-m}
\end{equation}
and the eigenvalues 
\begin{equation}
\lambda_n=-n\frac{b}{a}
\end{equation}
Note that $\mathcal{S}(I)_m = -m\frac{b}{a}[(I)_m - \frac{c}{b}(I)_{m-1}]$ 
How is this caluculation performed? I have looked at Rota's book but I see no reference to the computation of the eigenfunctions. Can anybody point me out in the right direction?
Thank you all in advance
 A: Begin with the study of polynomials in $X$ and
some linear operators. Define the shift  linear operator
$$ E_h[f(X)] \!:=\! f(X\!+\!h), \tag{1} $$
and the difference linear operator
$$ \Delta_h[f(X)] := f(X\!+\!h)\!-\!f(X), \tag{2} $$
and the derivative linear operator
$$ D[f(X)] := \frac{d}{dX} f(X), \tag{3} $$
where $\,f()\,$ is any polynomial.
Define the falling factorial linear operator
for monomials
$$ L[X^n] := X(X-1)\dots(X-n+1). \tag{4} $$
Without loss of generality, define the $\,S\,$
linear operator by
$$ S[f(X)] := (X\Delta_{-1}+c\,\Delta_{1})[f(X)]. \tag{5}$$
Applying this to falling factorials gives
$$ S[(X)_n] = -n ( (X)_n - c\, (X)_{n-1}). \tag{6} $$
Rewriting this using the $\,L\,$ operator gives
$$ S[L[X^n]] = -n L[ X^{n-1} (X-c)]. \tag{7} $$
Rewrite this using the derivative operator gives
$$ S[L[f(X)]] = -L[ D[f(X)](X-c)]. \tag{8} $$
Apply this to the case $\,f(X)=E_h[X^n]\,$ to get
$$ S[L[E_h[X^n]]] = -L[D[E_h[X^n]](X-c)]. \tag{9} $$
Apply this to the case $\,h=-c\,$ to get
$$ S[L[(X-c)^n]] = -L[D[(X-c)^n](X-c)]. \tag{10} $$
But we know the derivative of $\,(X-c)^n\,$ and so get
$$ S[L[(X-c)^n]] = -n\,L[(X-c)^n]. \tag{11} $$
Define the eigenvector function
$$ \psi_n(X) \!:=\! L[(X\!-\!c)^n] \!=\! \sum_{m=0}^n
   {n \choose m} (-c)^m(X)_{n-m}. \tag{12} $$
with eigenvalue $\,-n\,$ and is expanded into
a finite sum using the binomial theorem.
A: In the article mentionned in comment and not in the question,
the author give details about the eigenvalues of $S$.
$$S_{x,y} = -y \frac{b}{a} \delta_{x,y} + y I \frac{b}{a} \delta_{x,y-1}$$
Now the author is implicitely using the fact that $S$ is triangular, therefore its eigenvalues are the diagonal components. I can provide a proof if needed but it is part of the properties you can find easily on any article about triangular matrices.
Now I did not perform the calculation of the eigenfunction, but I believe it is the general method.
