# Lie algebras of differential operators for Matrix valued Laguerre type
polynomials

Andrea L. Gallo, Pablo Román

###### Abstract.

We study algebras of differential and difference operators acting on matrix
valued orthogonal polynomials (MVOPs) with respect to a weight matrix of the
form $W^{(\nu)}_{\phi}(x)=x^{\nu}e^{-\phi(x)}W^{(\nu)}_{pol}(x)$, where
$\nu>0$, $W^{(\nu)}_{pol}(x)$ is certain matrix valued polynomial and $\phi$
an entire function. We introduce a pair differential operators $\mathcal{D}$,
$\mathcal{D}^{\dagger}$ which are mutually adjoint with respect to the matrix
inner product induced by $W^{(\nu)}_{\phi}(x)$. We prove that the Lie algebra
generated by $\mathcal{D}$ and $\mathcal{D}^{\dagger}$ is finite dimensional
if and only if $\phi$ is a polynomial, giving a partial answer to a problem by
M. Ismail. In the case $\phi$ polynomial, we describe the structure of this
Lie algebra. The case $\phi(x)=x$, is discussed in detail. We derive
difference and differential relations for the MVOPs. We give explicit
expressions for the entries of the MVOPs in terms of classical Laguerre and
Dual Hahn polynomials.

###### Key words and phrases:

orthogonal polynomials, ladder relations, Laguerre polynomials

2020 Mathematics Subject Classification. Primary 33C45.

Partially supported by CONICET, FONCyT and SECyT-UNC

##  1\. Introduction

The theory of matrix valued orthogonal polynomials (MVOPs) was initiated by
Krein 1940s, and it has since been used in various areas of mathematics and
mathematical physics. These areas include spectral theory, scattering theory,
tiling problems, integrable systems, and stochastic processes. For further
details and insights on these subjects, refer to [1], [4], [6], [10], [12],
[13], and the references therein.

Significant progress has been made in the past two decades towards
understanding how the differential and algebraic properties of classical
scalar orthogonal polynomials can be extended to the matrix valued setting. A
fundamental role has been played by the connection between harmonic analysis
of matrix valued functions on compact symmetric pairs and matrix valued
orthogonal polynomials. In [7], A. Durán poses the problem of determining
families of MVOPs which are eigenfunctions of a suitable second order
differential operator. In the scalar case, the answer to this problem is a
classical result due to Bochner [2]. The only families with this property are
those of Hermite, Laguerre and Jacobi. The matrix valued setting turns out to
be much more involved. The first explicit examples appeared in connection with
spherical functions of the compact symmetric pair
$(\mathrm{SU}(3),\mathrm{U}(2))$. Following [24], a direct approach was taken
in [22], [23] for the case of
$(\mathrm{SU}(2)\times\mathrm{SU}(2),\mathrm{diag})$, leading to a general
set-up in the context of multiplicity free pairs [15]. In this context,
certain properties of the orthogonal polynomials, such as orthogonality,
recurrence relations, and differential equations, are understood in terms of
the representation theory of the corresponding symmetric spaces. Recently,
Casper and Yakimov developed a framework in [3] to solve the matrix Bochner
problem. This involves the classification of all $N\times N$ weight matrix
$W(x)$ whose associated MVOPs are eigenfunctions of a second-order
differential operator.

Given $N\in\mathbb{N}$ we consider a matrix valued function
$W:[a,b]\rightarrow M_{N}(\mathbb{C})$ such that $W(x)$ is positive definite
for all $x\in[a,b]$ and $W$ has finite moments of all order. In such a case,
we say that $W$ is a weight function, which induces matrix valued inner
product

$\langle P,Q\rangle=\int_{a}^{b}P(x)W(x)Q(x)^{*}dx\in M_{N}(\mathbb{C}),$
(1.1)

such that for all $P,Q,R\in M_{N}(\mathbb{C})[x]$, $T\in M_{N}(C)$ and
$a,b\in\mathbb{C}$ the following properties are satisfied

$\langle aP+bQ,R\rangle=a\langle P,R\rangle+b\langle Q,R\rangle,\qquad\langle
TP,Q\rangle=T\langle P,Q\rangle,\langle P,Q\rangle^{*}=\langle Q,P\rangle.$

Moreover $\langle P,P\rangle=0$ if and only if $P=0$. Using standard
arguments, it can be shown that there exists a unique sequence $(P(x,n))_{n}$
monic MVOPs with respect to $W$ in the following sense:

$\langle P(x,n),P(x,m)\rangle=\mathcal{H}(n)\delta_{n,m}.$ (1.2)

where the squared norm $\mathcal{H}(n)$ is a positive definite matrix.

By orthogonality, the polynomials $P(x,n)$’s satisfy the following three-term
recurrence relation:

$xP(x,n)=P(x,n+1)+B(n)P(x,n)+C(n)P(x,n-1)$ (1.3)

where $B(n),C(n)\in M_{N}(\mathbb{C})$ and $n\geq 1$. Notice that $B(n)$ and
$C(n)$ satisfy

$B(n)=X(n)-X(n+1),\quad C(n)=\mathcal{H}(n)\mathcal{H}(n-1)^{-1},$ (1.4)

where $X(n)$ is the one-but-leading coefficient of $P(x,n)$ and
$\mathcal{H}(n)$ as in (1.2). Moreover, for $n\geq 2$, let $Y(n)$ denotes the
second-but-leading coefficient of $P(x,n)$. Then

$Y(n)=Y(n+1)+B(n)X(n)+C(n).$ (1.5)

In [5], the authors studied difference–differential relations for a specific
class of MVOPs associated with the weight $W(x)=e^{-v(x)}e^{xA}e^{xA^{\ast}}$,
where $x\in\mathbb{R}$, $v(x)$ is a scalar polynomial of even degree, and $A$
is a constant matrix. There is a way of obtaining information about the matrix
orthogonal polynomials by investigating two mutually adjoint operators
$\mathcal{D}$ and $\mathcal{D}^{\dagger}$. If $v(x)$ is a polynomial of degree
two, in addition to $\mathcal{D}$ and $\mathcal{D}^{\dagger}$, there exists a
second order differential operator $D$ having the MVOPs as eigenfunctions. It
turns out that $\mathcal{D},\mathcal{D}^{\dagger}$ and $D$ generate a finite
dimensional Lie algebra which is isomorphic to the Lie algebra of the
oscillator group. The Casimir operator for this algebra is given explicitly
and used to obtain information of the MVOPs. In this work, we solve the
analogous problem for Laguerre-type weights. This case is more involved than
the previous one due to the structure of the associated Lie algebra and the
non-diagonality of certain formulas that involve $W$.

In the scalar case, this problem is closely related to [16, Problem 24.5.2].
Here Ismail proposed to study the finite dimensionality of certain Lie algebra
generated by a pair of differential operators which are mutually adjoint
respect to a Laguerre-type weight. More precisely, given the scalar weight
$w_{1}(x)=x^{\alpha}e^{-\phi(x)}$ with $x>0,\,\alpha>1$ and differential
operators

$\mathcal{D}_{1,n}=x\partial_{x}+xB_{n}(x),\qquad\mathcal{D}_{2,n}=-x\partial_{x}+xB_{n}(x)+x\phi^{\prime}(x),$

where $\\{B_{n}\\}$ is a sequence of scalar polynomials, the problem asks to
prove that “ _The Lie algebra generated by $\mathcal{D}_{1,n}$ and
$\mathcal{D}_{2,n}$ is finite dimensional if and only if $\phi$ is a
polynomial_”.

In this paper we provide a partial answer to this problem in the context of
matrix valued orthogonal polynomials. We give an explicit matrix valued weight
$W^{(\nu)}_{\phi}(x)=x^{\nu}e^{-\phi(x)}W^{(\nu)}_{pol}(x)$, where
$W^{(\nu)}_{pol}(x)$ is a matrix polynomial depending on $\nu$, and
differential operators

$\mathcal{D}=\partial_{x}x+x(A-1),\quad\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J)+x\phi^{\prime}(x)-x.$

In this case, we prove that the Lie algebra generated by is finite dimensional
if and only if $\phi$ is a polynomial. As a consequence, this solves [16,
Problem 24.5.2] when $B_{n}(x)=-1$ for all $n\in\mathbb{N}$.

### Outline and main results

In Section $2$ we recall some preliminaries. In particular, we introduce the
left and right Fourier algebras related to the sequence of monic MVOPs.

In Section $3$ for a given analytic function $\phi$ on a neighborhood of the
interval $[0,\infty)$, we introduce a Laguerre type weight $W^{(\nu)}_{\phi}$
and the operators $\mathcal{D}$ , $\mathcal{D}^{\dagger}$ and prove that they
are mutually adjoint with respect to $W^{(\nu)}_{\phi}(x)$. For the MVOPs
$\\{P_{n}\\}$ respect to $W^{(\nu)}_{\phi}$, we find discrete operators
$M,\,M^{\dagger}$ associated to $\mathcal{D},\,\mathcal{D}^{\dagger}$
respectively, given by the relations $M\cdot P_{n}=P_{n}\cdot\mathcal{D}$ and
$M^{\dagger}\cdot P_{n}=P_{n}\cdot\mathcal{D}^{\dagger}$.

In Section $4$ we study the Lie algebra $\mathfrak{g}_{\phi}$ generated by the
differential operators $\mathcal{D}$, $\mathcal{D}^{\dagger}$. We prove that
$\mathfrak{g}_{\phi}$ is finite dimensional if and only if $\phi$ is a
polynomial. Also, for this family of Lie algebras $\\{\mathfrak{g}_{\phi}\\}$
we obtain that $\mathfrak{g}_{\phi}=\mathbb{C}^{2}\oplus\mathfrak{h}$ and
$\mathfrak{h}$ is a solvable Lie algebra with nilradical of codimension one.
Moreover, we obtain a classification of this family of Lie algebra up to
isomorphisms.

In Section $5$ we give an explicitly expression for $\mathcal{D}$,
$\mathcal{D}^{\dagger}$, $M$ and $M^{\dagger}$ in the case $\phi(x)=x$. In
this case, we also find a symmetric second-order differential operator $D$
which have $\\{P_{n}\\}$ as eigenfunctions. We describe the Lie algebra
$\mathcal{A}$ generated by $\mathcal{D}$, $\mathcal{D}^{\dagger}$ and $D$,
$\mathcal{A}=\mathcal{Z}_{\mathcal{A}}\oplus[\mathcal{A},\mathcal{A}]$ where
$\dim\mathcal{Z}_{\mathcal{A}}=2$ and $[\mathcal{A},\mathcal{A}]$ is
isomorphic to $\mathrm{SL}(2,\mathbb{C})$. Also, we obtain some relations
between $\mathcal{H}_{n}$, $B_{n}$ and $C_{n}$.

In Section 6, we consider the polynomials $R(x,n)=K_{n}P(x,n)e^{xA}$ where
$P(x,n)$ are the MVOPs associated with the weight $W^{(\nu)}$ and $K_{n}$
certain lower triangular matrices. Using the operator $D$, we show that the
matrix entries of $R_{n}$ can be put in terms of generalized Laguerre
polynomials and a family of constants $\xi(n,i,j)$’s. Finally, we give two-
terms recursions for the constants $\xi(n,i,j)$’s and for the squared norm
$\mathcal{H}_{n}$.

Finally, in Section $7$, in the case $A=-\sum_{k=1}^{N-1}E_{k+1,k}$, and
$\delta^{k}>0$ satisfying two non-linear conditions (related to Pearson’s
equations), we show that the constants $\xi(n,i,j)$’s are written in terms of
dual Hanh polynomials.

##  2\. Preliminaries

This section presents the left and right Fourier algebras associated with the
sequence of monic MVOPs, as developed by Casper and Yakimov in [3]. The
results discussed in this section have been previously covered in a more
comprehensive context in [3].

Let $Q(x,n)$ be a function $Q:\mathbb{C}\times\mathbb{N}_{0}\to
M_{N}(\mathbb{C})$ such that $Q(x,n)$ is a rational function of $x$ for fixed
$n$. A differential operator of the form

$\mathcal{D}=\sum_{j=0}^{n}\partial_{x}^{j}F_{j}(x),\qquad\partial_{x}^{j}:=\tfrac{d^{j}}{dx^{j}},$
(2.1)

where $F_{j}:\mathbb{C}\to M_{N}(\mathbb{C})$ is a rational function of $x$,
acts on $Q$ from the right by

$(Q\cdot\mathcal{D})(x,n)=\sum_{j=0}^{n}(\partial_{x}^{j}Q)(x,n)\,F_{j}(x).$

The algebra of all differential operators of the form (2.1) will be denoted by
$\mathcal{M}_{N}$. In addition to the right action by differential operators,
we also consider a left action on $Q$ by difference operators on the variable
$n$. For $j\in\mathbb{Z}$, let $\delta^{j}$ be the discrete operator which
acts on a sequence $A:\mathbb{N}_{0}\to M_{N}(\mathbb{C})$ by

$(\delta^{j}\cdot A)(n)=A(n+j).$

Here we assume that the value of a sequence at a negative integer is equal to
zero. For given sequences $A_{-\ell},\ldots,A_{k}$, a discrete operator of the
form

$M=\sum_{j=-\ell}^{k}A_{j}(n)\delta^{j},$ (2.2)

acts on $Q$ from the left by

$\displaystyle(M\cdot Q)(x,n)$
$\displaystyle=\sum_{j=-\ell}^{k}A_{j}(n)\,(\delta^{j}\cdot
Q)(x,n)=\sum_{j=-\ell}^{k}A_{j}(n)\,Q(x,n+j).$

We shall denote the algebra of difference operators (2.2) by
$\mathcal{N}_{N}$. As in [3, Definition 2.20] we define:

######  Definition 2.1.

The left and right Fourier algebras are given by:

$\begin{split}\mathcal{F}_{L}(P)&=\\{M\in\mathcal{N}_{N}\colon\exists\,\mathcal{D}\in\mathcal{M}_{N},\,M\cdot
P=P\cdot\mathcal{D}\\}\subset\mathcal{N}_{N},\\\
\mathcal{F}_{R}(P)&=\\{\mathcal{D}\in\mathcal{M}_{N}\colon\exists\,M\in\mathcal{N}_{N},\,M\cdot
P=P\cdot\mathcal{D}\\}\subset\mathcal{M}_{N}.\end{split}$ (2.3)

The definition of the Fourier algebras directly implies a connection between
the elements of $\mathcal{F}_{L}(P)$ and $\mathcal{F}_{R}(P)$. Moreover, the
map

$\varphi\colon\mathcal{F}_{L}(P)\to\mathcal{F}_{R}(P),\qquad\text{ defined by
}\quad M\cdot P=P\cdot\varphi(M),$

is an algebra isomorphism. In [3] this map is called the generalized Fourier
map. More precisely, $M_{1}M_{2}\cdot P=P\cdot\varphi(M_{1})\varphi(M_{2})$
for all $M_{1},M_{2}\in\mathcal{F}_{L}(P)$. On the other hand, by the
definition of $\varphi$, we have that $M_{1}M_{2}\cdot
P=P\cdot\varphi(M_{1}M_{2})$.

######  Remark 2.2.

In this context, the three term recurrence relation (1.3) can be written as

$xP=P\cdot x=L\cdot P,\qquad\text{where }\quad L=\delta+B(n)+C(n)\delta^{-1}.$

Therefore $x\in\mathcal{F}_{R}$, $L\in\mathcal{F}_{L}$ and $\varphi(L)=x$. For
every polynomial $v\in\mathbb{C}[x]$, we have

$P\cdot v(x)=P\cdot v(\varphi(L))=v(L)\cdot P.$

On of the crucial results from [3] is the existence of an adjoint operation
$\dagger$ in the Fourier algebras $\mathcal{F}_{L}(P)$ and
$\mathcal{F}_{R}(P)$ as described in [3, §3.1]. To define the adjoint
operation in $\mathcal{F}_{L}(P)$, we initially observe that the algebra of
difference operators $\mathcal{N}_{N}$ has a $\ast$-operation defined as
follows:

$\left(\sum_{j=-\ell}^{k}A_{j}(n)\,\delta^{j}\right)^{\ast}=\sum_{j=-\ell}^{k}A_{j}(n-j)^{\ast}\,\delta^{-j},$
(2.4)

where $A_{j}(n-j)^{\ast}$ is the conjugate transpose of $A_{j}(n-j)$. Now, the
adjoint of $M\in\mathcal{N}_{N}$ is given by

$M^{\dagger}=\mathcal{H}(n)M^{\ast}\mathcal{H}(n)^{-1},$ (2.5)

where $\mathcal{H}(n)$ is the squared norm which we view as an difference
operator of order zero. The following holds:

$\langle(M\cdot P)(x,n),P(x,m)\rangle=\langle P(x,n),(M^{\dagger}\cdot
P)(x,m)\rangle.$

In [3, Corollary 3.8] the authors show that every differential operator
$D\in\mathcal{F}_{R}(P)$ has a unique adjoint
$\mathcal{D}^{\dagger}\in\mathcal{F}_{R}(P)$ with the property

$\langle P\cdot\mathcal{D},Q\rangle=\langle
P,Q\cdot\mathcal{D}^{\dagger}\rangle,$

for all $P,Q\in M_{N}(\mathbb{C})[x]$. Moreover,
$\varphi(M^{\dagger})=\varphi(M)^{\dagger}$ for all $M\in\mathcal{F}_{L}(P)$.

##  3\. Semi-classical Laguerre type solutions

In the sequel, we consider the following two matrices $A,J\in
M_{N}(\mathbb{R})$ which satisfy

$J=\sum_{k=1}^{N}kE_{k,k}\qquad A=\sum_{k=1}^{N-1}a_{k}E_{k+1,k}.$ (3.1)

Notice that, it is straightforward to show that

$[J,A]=A\quad\text{and}\quad e^{xA}Je^{-xA}=J-Ax.$ (3.2)

Let us consider the following weight matrix supported on the interval
$[0,\infty)$:

$W^{(\nu)}_{\phi}(x)=e^{Ax}T^{(\nu)}_{\phi}(x)e^{A^{\ast}x},\qquad
T^{(\nu)}_{\phi}(x)=e^{-\phi(x)}\sum_{k=1}^{N}\delta^{(\nu)}_{k}x^{\nu+k}E_{k,k},$
(3.3)

where $\delta^{(\nu)}_{k}$ is a constant real number for $1\leq k\leq N$, and
$\phi$ be an analytic function on a neighborhood of the interval $[0,\infty)$.
In the sequel, we assume that $W(x)P(x)=0$ has vanishing limits at the
endpoints of support for any matrix polynomial $P$.

######  Proposition 3.1.

Let $A,J\in M_{N}(\mathbb{C})$ as in (3.1). Then, the first order differential
operators

$\mathcal{D}=\partial_{x}x+x(A-1),\quad\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J)+x\phi^{\prime}(x)-x,$

are mutually adjoint.

###### Proof.

Let $P,Q\in M_{N}(\mathbb{C}[x])$. In order to simplify notation, in the rest
of the proof, we denote by $W(x):=W^{(\nu)}_{\phi}(x)$ and
$T(x)=W^{(\nu)}_{\phi}(x)$

$\displaystyle\langle P\cdot\mathcal{D},Q\rangle$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}(P\cdot\mathcal{D})W(x)Q^{\ast}(x)dx$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\left(xP^{\prime}(x)+xP(x)(A-1)\right)W(x)Q^{\ast}(x)dx.$

Notice that, since $W(x)P(x)$ has vanishing limits at the endpoints $x=0$, and
$x=\infty$, integration by parts implies that

$\int_{0}^{\infty}xP^{\prime}(x)W(x)Q^{\ast}(x)dx=-\int_{0}^{\infty}P(x)\left(xW(x)Q^{\ast}(x)\right)^{\prime}dx.$

On the other hand, we have that

$\int_{0}^{\infty}xP(x)(A-1)W(x)Q^{\ast}(x)dx=\int_{0}^{\infty}P(x)W(x)\left(W^{-1}(x)x(A-1)W(x)\right)Q^{\ast}(x)dx,$

by linearity we obtain that

$\displaystyle\langle P\cdot\mathcal{D},Q\rangle$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)\left(xW(x)Q^{\ast}(x)\right)^{\prime}dx+\int_{0}^{\infty}P(x)W(x)\left(W^{-1}(x)x(A-1)W(x)\right)Q^{\ast}(x)dx$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)\left(W(x)Q^{\ast}(x)+xW^{\prime}(x)Q^{\ast}+xW(x)(Q^{\ast}(x))^{\prime}\right)dx$
$\displaystyle+\int_{0}^{\infty}P(x)W(x)\left(W^{-1}(x)x(A-1)W(x)\right)Q^{\ast}(x)dx.$

Notice that since
$\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)=Q^{\ast}(x)+x(Q^{\ast}(x))^{\prime}$,
we can put

$\int_{0}^{\infty}P(x)(W(x)Q^{\ast}(x)+W(x)x(Q^{\ast}(x))^{\prime})dx=\int_{0}^{\infty}P(x)W(x)\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)dx.$

On the other hand,

$\int_{0}^{\infty}P(x)xW^{\prime}(x)Q^{\ast}(x)dx=\int_{0}^{\infty}P(x)W(x)xW^{-1}(x)W^{\prime}(x)Q^{\ast}(x)dx.$

Hence, we obtain that

$\displaystyle\langle P\cdot\mathcal{D},Q\rangle$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)dx-\int_{0}^{\infty}P(x)W(x)xW^{-1}(x)W^{\prime}(x)Q^{\ast}(x)dx$
$\displaystyle+\int_{0}^{\infty}P(x)W(x)\left(W^{-1}(x)x(A-1)W(x)\right)Q^{\ast}(x)dx.$

By (3.3) we have that

$\displaystyle W^{-1}(x)W^{\prime}(x)$
$\displaystyle=e^{-A^{\ast}x}\left({T}^{-1}(x)\,A\,T(x)+T^{-1}(x)T^{\prime}(x)+A^{\ast}\right)e^{A^{\ast}x},$
$\displaystyle W^{-1}(x)x(A-1)W(x)$
$\displaystyle=e^{-A^{\ast}x}\left(xT^{-1}(x)\,A\,T(x)-x\right)e^{A^{\ast}x}.$

Thus, we obtain

$\displaystyle\langle P\cdot\mathcal{D},Q\rangle$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)dx$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)e^{-A^{\ast}x}\left(xT^{-1}(x)\,A\,T(x)+xT^{-1}(x)T^{\prime}(x)+xA^{\ast}\right)e^{A^{\ast}x}Q^{\ast}(x)dx$
$\displaystyle+\int_{0}^{\infty}P(x)W(x)e^{-A^{\ast}x}\left(xT^{-1}(x)\,A\,T(x)-x\right)e^{A^{\ast}x}Q^{\ast}(x)dx$
$\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)dx$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)e^{-A^{\ast}x}\left(xT^{-1}(x)T^{\prime}(x)+xA^{\ast}+x\right)e^{A^{\ast}x}Q^{\ast}(x)dx.$

By taking into account that $xT^{\prime}(x)=T(x)(-x\phi^{\prime}(x)+\nu+J)$,
we obtain that

$\displaystyle\langle P\cdot\mathcal{D},Q\rangle$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)dx$
$\displaystyle-$
$\displaystyle\int_{0}^{\infty}P(x)W(x)e^{-A^{\ast}x}(-x\phi^{\prime}(x)+\nu+J+xA^{\ast}+x)e^{A^{\ast}x}Q^{\ast}(x)dx.$

Notice that the second expresion of the right hand of the above equality is

$e^{-A^{\ast}x}(-x\phi^{\prime}(x)+\nu+J+xA^{\ast}+x)e^{A^{\ast}x}=-x\phi^{\prime}(x)+\nu+e^{-A^{\ast}x}Je^{A^{\ast}x}+xA^{\ast}+x.$

On the other hand, the equation $e^{xA}Je^{-xA}=J-Ax$ implies that
$e^{-A^{\ast}x}Je^{A^{\ast}x}=J-A^{\ast}x$. Hence, we obtain that

$\displaystyle\langle P\cdot\mathcal{D},Q\rangle$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)\left(Q\cdot(1+x\partial_{x})\right)^{\ast}(x)dx$
$\displaystyle-\int_{0}^{\infty}P(x)W(x)(x-x\phi^{\prime}(x)+(\nu+J-A^{\ast}x+xA^{\ast})Q^{\ast}(x)dx$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}P(x)W(x)\left(Q\cdot-(x\partial_{x}+x-x\phi^{\prime}(x)+(\nu+J+1)\right)^{\ast}(x)dx$
$\displaystyle=$ $\displaystyle\langle P,Q\cdot\mathcal{D}^{\dagger}\rangle.$

Therefore, the operators $\mathcal{D}$ and $\mathcal{D}^{\dagger}$ are
mutually adjoint, as asserted. ∎

By the above theorem, since $\mathcal{D}=\partial_{x}x+x(A-1)$ and
$\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J)+x\phi^{\prime}(x)-x$, then we
obtain that

$\mathcal{D}^{\dagger}=-\mathcal{D}+(Ax-J)-(1+\nu)+x\phi^{\prime}(x)-2x.$
(3.4)

######  Lemma 3.2.

Let $A,J\in M_{N}(\mathbb{C})$ be matrices as in (3.1). Then,
$\mathcal{C}=Ax-J$ is a symmetric operator respect to the weight
$W:=W^{(\nu)}_{\phi}$ as in (3.3). Moreover, if $P(x,n)$’s are monic MVOPs
associated with the weight $W$, such that

$P\cdot\mathcal{C}=M_{\mathcal{C}}\cdot P,\quad\text{then}\quad
M_{\mathcal{C}}=\sum_{i=-1}^{1}U_{j}(n)\delta^{j},$

with

$U_{1}(n):=A,\qquad U_{0}(n):=X(n)A-AX(n+1)-J,$
$U_{-1}(n):=Y(n)A-AY(n+1)+[J,X(n)]+(AX(n+1)-X(n)A)X(n),$

where $X(n)$ and $Y(n)$ are the coefficients of the $(n-1)$-term and
$(n-2)$-term of $P(x,n)$ respectively.

###### Proof.

In the same manner as Proposition 3.1, it can be shown that $\mathcal{C}$ is
an symmetric operator respect with $W$.

If we put $M_{\mathcal{C}}=\sum U_{j}(n)\delta^{j}$ such that
$P\cdot\mathcal{C}=M_{\mathcal{C}}\cdot P$. By taking into account that
$\mathcal{C}$ increases the degree of any polynomial in $1$, we obtain that
$U_{j}(n)=0$ for $j>1$. On the other hand, since $\mathcal{C}$ is an symmetric
operator respect to $W$, we have that
$M_{\mathcal{C}}=M^{\dagger}_{\mathcal{C}}$ and so $U_{j}(n)=0$ for $j<-1$.

The formulas for $U_{-1}(n),U_{0}(n)$ and $U_{1}(n)$ can be obtained by direct
computation from $P\cdot\mathcal{C}=M_{\mathcal{C}}\cdot P$. ∎

######  Remark 3.3.

Since $\mathcal{C}=Ax-J=(Ax-J)^{\dagger}$, then
$M_{\mathcal{C}}=M_{\mathcal{C}}^{\dagger}$. Thus, from equations (2.5) we
have that

$\displaystyle U_{1}(n)$
$\displaystyle=\mathcal{H}(n)U_{-1}(n+1)^{\ast}\mathcal{H}(n+1)^{-1},$
$\displaystyle U_{0}(n)$
$\displaystyle=\mathcal{H}(n)U_{0}(n)^{\ast}\mathcal{H}(n)^{-1},$

and so we obtain that

$A=\mathcal{H}(n)\Big{(}Y(n+1)A-AY(n+2)+[J,X(n+1)]+(AX(n+2)-X(n+1)A)X(n+1)\Big{)}^{\ast}\mathcal{H}(n+1)^{-1},$
$X(n)A-AX(n+1)-J=\mathcal{H}(n)\Big{(}X(n)A-AX(n+1)-J\Big{)}^{\ast}\mathcal{H}(n)^{-1}.$

######  Theorem 3.4.

Let $W:=W^{(\nu)}_{\phi}$ be a matrix weight as in (3.3), with monic MVOPs
$P(x,n)$ such that

$\mathcal{D}=\partial_{x}x+(A-1)x,\qquad\mathcal{D}^{\dagger}=-\mathcal{D}+\mathcal{C}+v^{\prime}(x)$

for some polynomial $v(x)$ of degree $k$ and $\mathcal{C}=Ax-J$. If $X(n)$ and
$Y(n)$ are the coefficients of the $(n-1)$-term and $(n-2)$-term of $P(x,n)$
respectively. Then, the monic polynomials $P(x,n)$ satisfy the following
relation

$P\cdot\mathcal{D}=M\cdot P,\qquad M=\sum_{j=-k+1}^{1}A_{j}(n)\delta^{j}$
(3.5)

with

$A_{1}(n)=A-1,\quad A_{0}(n)=n+X(n)A-AX(n+1)-B(n),$
$A_{-1}(n)=(n-1)X(n)+Y(n)(A-1)-(A-1)Y(n+1)-A_{0}(n)X(n),$
$A_{j}(n)=(v^{\prime}(L))_{j}(n),\quad-k+1<j<-1$

where $B(n)$ is given by (1.4).

###### Proof.

Clearly, the formulas for $A_{j}(n)$ with $j=-1,0,1$ can be derived from the
equalities in $\eqref{PDM}$ by using the definition of $\mathcal{D}$.

For $j<-1$, we have that

$\displaystyle A_{j}(n)$ $\displaystyle=$ $\displaystyle\langle
P\cdot\mathcal{D},\delta^{j}\cdot P\rangle\mathcal{H}(n-j)^{-1}=\langle
P,\delta^{j}\cdot P\cdot\mathcal{D}^{\dagger}\rangle\mathcal{H}(n-j)^{-1}$
$\displaystyle=$ $\displaystyle\langle P,\delta^{j}\cdot P\cdot
v^{\prime}(x)\rangle\mathcal{H}(n-j)^{-1}=\langle P\cdot
v^{\prime}(x),\delta^{j}\cdot P\rangle\mathcal{H}(n-j)^{-1}$ $\displaystyle=$
$\displaystyle\langle v^{\prime}(L)\cdot P,\delta^{j}\cdot
P\rangle\mathcal{H}(n-j)^{-1},$

where we have used that $\langle P,\delta^{j}\cdot P\cdot\mathcal{D}\rangle$
and $\langle P,\delta^{j}\cdot P\cdot\mathcal{C}\rangle$ are both zero for
$j<-1$ in the third equality, and the fact that $v^{\prime}(x)$ is a scalar
funtion in the fourth one. Then, we have that

$A_{j}(n)=(v^{\prime}(L))_{j}(n)\quad\text{for $j<-1$}.$

To complete the proof, notice that $(v^{\prime}(L))_{j}(n)=0$ for $j\leq-k$. ∎

As a direct consequence, we obtain the following corollary.

######  Corollary 3.5.

In the same hypothesis as in Theorem 3.4. If the polynomial $v$ has degree
$1$, then the discrete operator $M$ associated with $\mathcal{D}$ satisfies

$M=A_{0}(n)+(A-1)\delta$

with $A_{0}(n)$ as in Theorem 3.4. Moreover, in this case we have that

$(n-1)X(n)+Y(n)(A-1)-(A-1)Y(n+1)-\Big{(}n+X(n)A-AX(n+1)-B(n)\Big{)}X(n)=0.$
(3.6)

###### Proof.

By Theorem 3.4, since $\deg(v)<2$ we obtain that $A_{j}(n)=0$ for all $j<-1$.
On the other hand, notice that since $v$ has degree $1$, then $v^{\prime}(x)$
is a constant function and so the operator $\mathcal{D}^{\dagger}$ does not
increase degrees. This implies that $A_{1}^{\dagger}=0$ and therefore

$A_{-1}(n)=(n-1)X(n)+Y(n)(A-1)-(A-1)Y(n+1)-A_{0}(n)X(n)=0,$

the formula for $A_{1}(n)$ and $A_{0}(n)$ are consequence of Theorem 3.4. ∎

##  4\. Lie algebras associated to orthogonal polynomials

In this section, we solve a particular case of the problem proposed by Ismail
in [17, Problem 24.5.2], as described in the introduction. For this purpose,
we study the structure of a Lie algebra related with the operators
$\mathcal{D}$ and $\mathcal{D}^{\dagger}$.

Recall that if $\mathfrak{g}$ is a finite dimensional Lie algebra, and if
$\mathfrak{g}^{j}$ and $\mathfrak{g}_{j}$ denote the following recursions

$\mathfrak{g}^{0}=\mathfrak{g}_{0}=\mathfrak{g},\quad\mathfrak{g}^{j+1}=[\mathfrak{g}^{j},\mathfrak{g}^{j}]\quad\text{and
}\quad\mathfrak{g}_{j+1}=[\mathfrak{g},\mathfrak{g}_{j}],$

then $\mathfrak{g}$ is called solvable (nilpotent) if $\mathfrak{g}^{j}=0$ for
some $j$ (if $\mathfrak{g}_{j}=0$ for some $j$). Clearly, any nilpotent Lie
algebra is solvable. The radical (nilradical) of $\mathfrak{g}$ is its maximal
solvable ideal (maximal nilpotent ideal) of $\mathfrak{g}$. We will denote by
$\mathrm{Rad}(\mathfrak{g})$ and $\mathrm{Nil}(\mathfrak{g})$ to the radical
and nilradical of $\mathfrak{g}$, respectively.

### Lie algebra generated by $\mathcal{D}$ and $\mathcal{D}^{\dagger}$

######  Lemma 4.1.

Let $A,J\in M_{N}(\mathbb{C})$ as in (3.1) and let $\phi$ an entire function
over $\mathbb{C}$, let us consider the operators

$\mathcal{D}=\partial_{x}x+x(A-1),\quad\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J)+x\phi^{\prime}(x)-x.$

If $x$ and $x^{j}\phi^{(j)}(x)$ act over matrix valued polynomials by right
constant multiplication, then we have that

$[\mathcal{D},x]=-x,\qquad[\mathcal{D}^{\dagger},x]=x,\qquad[\mathcal{D},\mathcal{D}^{\dagger}]=-x^{2}\phi^{(2)}(x)+(2-\phi^{\prime}(x))x,$
$[\mathcal{D},\phi^{(j)}(x)x^{j}]=-(jx^{j}\phi^{(j)}(x)+x^{j+1}\phi^{(j+1)}(x))=-[\mathcal{D}^{\dagger},\phi^{(j)}(x)x]\quad\text{for
all $j\geq 1$}.$

In the sequel, given $\phi$ an entire function over $\mathbb{C}$, we denote by

$\mathfrak{g}_{\phi}=\langle
1,\mathcal{D},\mathcal{D}^{\dagger},x,x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots\rangle$
(4.1)

with bracket as above. We are interested in the case that this Lie algebra is
finite dimensional. The following proposition states that this happens if and
only if $\phi$ is a polynomial. We will need the following notation, given
$\phi$ a polynomial over $\mathbb{C}$ with $\ell$ non-zero coefficients

$k=\begin{cases}\ell+2&\text{if }\phi^{\prime}(0)=\phi(0)=0,\\\
\ell+1&\text{if }\phi^{\prime}(0)=0,\phi(0)\neq 0,\\\ \ell+1&\text{if
}\phi(0)=0,\phi^{\prime}(0)\neq 0,\\\ \ell&\text{if }\phi(0)\neq
0,\phi^{\prime}(0)\neq 0,\end{cases}$ (4.2)

######  Proposition 4.2.

Let $\phi$ an analitic function over $\mathbb{C}$ and let
$\mathfrak{g}:=\mathfrak{g}_{\phi}$ its associated Lie algebra as in (4.1).
Then, we have that $\dim(\mathfrak{g})$ is finite if and only if $\phi$ is a
polynomial. In such case, if $k$ is as in (4.2) then

$\dim(\mathfrak{g})=k+2.$

###### Proof.

Clearly, if $\phi$ is a polynomial, then the dimension of $\mathfrak{g}$ is
finite, since if $n$ is the degree of $\phi$ then $\phi^{(m)}(x)=0$ for all
$m>n$.

Conversely, assume now that $\phi$ is not a polynomial. Since $\phi$ is
analytic, we can express $\phi$ as follow

$\phi(x)=\sum_{i=0}^{\infty}a_{i}x^{i}.$

This implies that

$x^{j}\phi^{(j)}(x)=\sum_{i=0}^{\infty}b_{i,j}x^{i},\quad\text{with}\quad
b_{i,j}=\begin{cases}\tbinom{i}{j}j!\,a_{i}&\text{ if $j\leq i$},\\\
0&\text{if $i<j$}.\end{cases}$ (4.3)

In particular, if $a_{i}=0$ then $b_{i,j}=0$ for all $j\in\mathbb{N}$.

Let $\\{i_{t}\\}_{t\in\mathbb{N}}$ be the sequence of non-zero coefficients
indices of $\phi$, that is $i_{t}<i_{t+1}$ for all $t\in\mathbb{N}$, such that
$a_{i}\neq 0$ if and only if $i=i_{t}$ for some $t\in\mathbb{N}$.

Claim: The vector space $\langle
x^{i_{1}}\phi^{(i_{1})}(x),\ldots,x^{i_{\ell}}\phi^{(i_{\ell})}(x)\rangle$ has
dimension $\ell$.

Let $c_{1},\ldots,c_{\ell}\in\mathbb{C}$ such that

$c_{1}x^{i_{1}}\phi^{(i_{1})}(x)+\cdots+c_{\ell}x^{i_{\ell}}\phi^{(i_{\ell})}(x)=0,$

this induces the following system of equations

$\sum_{t=1}^{h}c_{t}\tbinom{i_{h}}{i_{t}}i_{h}!a_{i_{t}}=0\quad\text{for
$h=1,\ldots,\ell$}.$

By taking into account that $a_{i_{1}}\neq 0$, the equation for $h=1$ implies
that $c_{1}=0$. In the same way, since $c_{1}=0$, the equation for $h=2$
implies that $c_{2}a_{i_{2}}i_{2}!=0$ and so $c_{2}=0$ since $a_{i_{2}}\neq
0$. Inductively, if $c_{1}=c_{2}=\ldots=c_{\ell-1}=0$, then the equation for
$h={\ell}$ implies that $c_{\ell}a_{i_{\ell}}i_{\ell}!=0$ and so $c_{\ell}=0$
since $a_{i_{\ell}}\neq 0$, hence we obtain that $c_{h}=0$ for all
$h\in\\{1,\ldots,\ell\\}$.

By the claim, the space $\mathfrak{g}$ has subspaces of all of the possible
dimensions and so is non-finite dimensional, as asserted.

Now, assume that $\phi$ is a polynomial of degree $n$, in the same notation as
above, by (4.3) we have that

$\langle
x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots,x^{n}\phi^{(n)}(x)\rangle\subseteq\langle
x^{i_{1}},\ldots,x^{i_{\ell}}\rangle.$

The claim and the above statement imply that $\langle
x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots,x^{n}\phi^{(n)}(x)\rangle$ has
dimension $\ell$.

Finally, the last assertion follows from the fact that
$\mathcal{D},\mathcal{D}^{\dagger}$ are linearly independent respect to

$\langle
1,x,x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots,x^{n}\phi^{(n)}(x)\rangle$

and this vector space has dimension $k$, with $k$ as in the statement. ∎

######  Remark 4.3.

The above proposition solves the problem proposed by Ismail in [17, Problem
24.5.2] for the case $B_{n}=-1$ for all $n$ natural number. In the notation of
[17], the differential operators $\mathcal{D}$ and $\mathcal{D}^{\dagger}$
correspond to $\mathcal{D}=xL_{1,n}$ and
$\mathcal{D}^{\dagger}=xL_{2,n}+(1+\nu)$. Then, the algebra generated by
$\\{\mathcal{D},\mathcal{D}^{\dagger},1\\}$ is isomorphic to the algebra
generated by $\\{xL_{1,n},xL_{2,n},1\\}$.

######  Remark 4.4.

By the proof of the above theorem, if $\phi(x)=a_{0}+a_{1}x+\ldots+a_{n}x^{n}$
is a polynomial of degree $n$ with $\ell$ non-zero coefficients. If
$\\{i_{1},\ldots,i_{\ell}\\}\subseteq\\{0,\ldots,n\\}$ is the set of indices
such that $a_{i_{j}}\neq 0$. then we have that

$\langle
x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots,x^{n}\phi^{(n)}(x)\rangle=\langle
x^{i_{1}},\ldots,x^{i_{\ell}}\rangle.$ (4.4)

######  Example 4.5.

Let $\phi_{1}(x)=x^{3}$ and $\phi_{2}(x)=x^{3}+x^{2}$, by the above theorem
the associated Lie algebras $\mathfrak{g}_{\phi_{1}}$ and
$\mathfrak{g}_{\phi_{2}}$ have the dimensions $5$ and $6$, respectively. Then,
the algebras $\mathfrak{g}_{\phi_{1}}$ and $\mathfrak{g}_{\phi_{2}}$ are non-
isomorphic.

######  Lemma 4.6.

The element $z=\mathcal{D}+\mathcal{D^{\dagger}}+2x-x\phi^{\prime}(x)$ is a
symmetric differential operator which belongs to the center of the Lie algebra
$\mathfrak{g}$.

###### Proof.

It is follows immediately from the definition of the bracket of
$\mathfrak{g}_{\phi}$. ∎

######  Remark 4.7.

The central element that we found in the above lemma, it is related with the
symmetric operator $\mathcal{C}$ that was considered in Lemma 3.2, we will see
this in the following section.

In the sequel, given a polynomial $\phi(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ of
degree $n\geq 2$ with $\ell$ non-zero coefficients and $k$ as in (4.2), let us
consider the following notations.

$I_{\phi}=\\{i\in\\{2,\ldots,n\\}:a_{i}\neq
0\\}=\\{j_{1},\ldots,j_{k-2}\\}=\begin{cases}\\{i_{3},\ldots,i_{\ell}\\}&\text{if
$a_{0}\neq 0$, $a_{1}\neq 0$,}\\\ \\{i_{2},\ldots,i_{\ell}\\}&\text{if
$a_{0}=0$ and $a_{1}\neq 0$,}\\\ \\{i_{2},\ldots,i_{\ell}\\}&\text{if
$a_{0}\neq 0$ and $a_{1}=0$,}\\\ \\{i_{1},\ldots,i_{\ell}\\}&\text{if
$a_{0}=0$ and $a_{1}=0$,}\\\ \end{cases}$ (4.5)

with $j_{t}<j_{t+1}$ and $i_{t}<i_{t+1}$.

######  Theorem 4.8.

Let $\phi(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ be a polynomial of degree $n\geq
2$ with $\ell$ non-zero coefficients and $k$ as in (4.2). If
$\mathfrak{g}:=\mathfrak{g}_{\phi}$ is the associated Lie algebra of $\phi$ as
in (4.1), then we have that

$\mathfrak{g}\cong\mathbb{C}^{2}\oplus\mathfrak{h}$

where $\mathfrak{h}$ is a solvable Lie algebra of dimension $k$, with an
abelian nilradical of dimension $k-1$. More precisely, if $I_{\phi}$ is as in
(4.5) then

$\mathfrak{h}\cong\langle E\rangle\ltimes\langle E_{1},\ldots E_{k-1}\rangle$

where $\langle E_{1},\ldots E_{k-1}\rangle$ is abelian and the rest of the
brackets satisfy

$[E,E_{1}]=E_{1}\quad\text{and}\quad[E,E_{t}]=j_{t-1}E_{t}\quad\text{for
$t=2,\ldots,k-1$}.$ (4.6)

###### Proof.

By Lemma 4.6, the element
$z=\mathcal{D}+\mathcal{D^{\dagger}}+2x-x\phi^{\prime}(x)$ belongs to the
center of $\mathfrak{g}$, and so we obtain an element in the center which does
not belong to $\langle 1\rangle$, thus if

$\mathfrak{h}=\langle\mathcal{D},x,x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots,x^{n}\phi^{(n)}(x)\rangle$

then we obtain that

$\mathfrak{g}\cong\mathbb{C}^{2}\oplus\mathfrak{h},$

with $\mathfrak{h}$ a Lie algebra of dimension $k$.

Thus, it is enough to show that $\mathfrak{h}$ is solvable with nilradical of
dimension $k-1$. Let us consider

$\mathfrak{k}=\langle
x,x\phi^{\prime}(x),x^{2}\phi^{(2)}(x),\ldots,x^{n}\phi^{(n)}(x)\rangle,$

by definition of the bracket and by taking into account that

$[x,x^{l}\phi^{(l)}(x)]=[x^{i}\phi^{(i)}(x),x^{j}\phi^{(j)}(x)]=0\qquad\text{for
all $i\neq j$,}$ (4.7)

we obtain that $[\mathfrak{h},\mathfrak{h}]\subseteq\mathfrak{k}$. Hence, by
(4.7) we obtain that
$\Big{[}[\mathfrak{h},\mathfrak{h}],[\mathfrak{h},\mathfrak{h}]\Big{]}=0,$ and
so $\mathfrak{h}$ is solvable. Finally, notice that $\mathfrak{k}$ is an
abelian ideal of $\mathfrak{h}$ of dimension

$\dim(\mathfrak{k})=\dim(\mathfrak{h})-1=k-1,$

this implies that $\mathfrak{k}$ is the nilradical of $\mathfrak{h}$, as
desired.

In the same manner as in Remark 4.4 we have that

$\mathfrak{h}=\langle\mathcal{D},x,x^{j_{1}},\ldots,x^{j_{k-2}}\rangle,$

in this case $\langle x,x^{j_{1}},\ldots,x^{j_{k-2}}\rangle$ is an abelian
subalgebra of dimension $k-1$. It is enough to see the brackets
$[\mathcal{D},x^{j_{t}}]$ and $[\mathcal{D},x]$, in this case we obtain that

$[\mathcal{D},x]=-x\qquad\text{and}\qquad[\mathcal{D},x^{j_{t}}]=-j_{t}x^{j_{t}},$

so we obtain that $\langle x,x^{j_{1}},\ldots,x^{j_{k-2}}\rangle$ is an
abelian ideal. Finally, if we consider the following correspondence

$\mathcal{D}\longmapsto-E,\quad x\longmapsto E_{1},\quad x^{j_{i}}\longmapsto
E_{i+1}\quad\text{for $i=1,\ldots,k-2$},$

is an Lie Algebra isomorphism between $\mathfrak{h}$ and $\langle
E\rangle\ltimes\langle E_{1},\ldots E_{k-1}\rangle$ with brackets given as in
(4.6), as asserted. ∎

In the sequel we are going to study the structure of the solvable Lie algebra
$\mathfrak{h}_{\phi}$. In general, a Lie algebra $\mathfrak{h^{\prime}}$ with
an abelian ideal of codimension $1$ is called almost abelian. This kind of
algebra was studied by V.V. Gorbatsevich in [11]. The author asserted that in
general this kind of algebra it decomposes as

$\mathbb{C}\ltimes_{\psi}\mathbb{C}^{k-1},$

this semidirect product gives a linear transformation
$\psi:\mathbb{C}\rightarrow gl_{k-1}(\mathbb{C})$, moreover he asserted that
the structure of this kind of algebras it determines by the matrix
$\Psi=\psi(1)$. More precisely,

$\mathbb{C}\ltimes_{\psi}\mathbb{C}^{k-1}\cong\mathbb{C}\ltimes_{\psi^{\prime}}\mathbb{C}^{k-1}\Longleftrightarrow\quad\text{$\Psi$
and $\Psi^{\prime}$ are conformally similar},$ (4.8)

recall that $\Psi$ and $\Psi^{\prime}$ are conformally similar if and only if
there exist a matrix $P\in Gl_{k-1}(\mathbb{C})$ and a non-zero complex number
$\lambda\in\mathbb{C}\smallsetminus\\{0\\}$ such that $\Psi=\lambda
P\Psi^{\prime}P^{-1}$.

######  Lemma 4.9.

Let $k$ be an integer greater than $1$ and let $1<j_{1}<\ldots<j_{k-2}$ and
$1<j^{\prime}_{1}<\ldots<j^{\prime}_{k-2}$ be two sequences of positive
integers. Let us consider

$\Psi=\mathrm{diag}(1,j_{1},\ldots,j_{k-2})\quad\text{and}\quad\Psi^{\prime}=\mathrm{diag}(1,j^{\prime}_{1},\ldots,j^{\prime}_{k-2}).$

Then, $\Psi$ and $\Psi^{\prime}$ are conformally similar if and only if
$\Psi=\Psi^{\prime}$.

###### Proof.

Clearly, if $\Psi=\Psi^{\prime}$ then they are conformally similar trivially.

Now, assume that $\Psi$ and $\Psi^{\prime}$ are conformally similar, so there
exist $\lambda\in\mathbb{C}^{*}$ and $P\in GL_{k-1}(\mathbb{C})$ such that

$\Psi^{\prime}=\lambda P\Psi P^{-1},$

by similarity we obtain the following spectral relationship

$\mathrm{Spec}(\Psi^{\prime})=\lambda\cdot\mathrm{Spec}(\Psi),$

i.e. all of the eigenvalues of $\Psi^{\prime}$ can be obtained from the
eigenvalues of $\Psi$ by multiplication by $\lambda$. Since $\Psi$ and
$\Psi^{\prime}$ are both diagonal, we have that

$\mathrm{Spec}(\Psi)=\\{1,j_{1},\ldots,j_{k-2}\\}\quad\text{
and}\quad\mathrm{Spec}(\Psi^{\prime})=\\{1,j^{\prime}_{1},\ldots,j^{\prime}_{k-2}\\},$

since $1$ belong to both spectra and the rest of the eigenvalues of $\Psi$ and
$\Psi^{\prime}$ are greater than $1$, we obtain that $\lambda=1$ necessarily.
Hence, we obtain that

$\mathrm{Spec}(\Psi^{\prime})=\mathrm{Spec}(\Psi).$

Therefore, $\Psi=\Psi^{\prime}$ as asserted. ∎

We are in position to give the following theorem, which says when the Lie
algebra associated to two different polynomials (as in (4.1)) are isomorphic.

######  Theorem 4.10.

Let $\phi_{1}(x),\phi_{2}(x)$ be polynomials over $\mathbb{C}$ of degree
greater or equal than $2$. Let $\mathfrak{h}_{\phi_{1}}$ and
$\mathfrak{h}_{\phi_{2}}$ be its associated solvable Lie algebras given as in
Theorem 4.8. Then,

$\mathfrak{h}_{\phi_{1}}\cong\mathfrak{h}_{\phi_{2}}\Longleftrightarrow
I_{\phi_{1}}=I_{\phi_{2}},$

where $I_{\phi_{1}},I_{\phi_{2}}$ are as in (4.5). Moreover, we have that

$\mathfrak{g}_{\phi_{1}}\cong\mathfrak{g}_{\phi_{2}}\Longleftrightarrow
I_{\phi_{1}}=I_{\phi_{2}},$

where $\mathfrak{g}_{\phi_{1}}$ and $\mathfrak{g}_{\phi_{2}}$ are the
associated Lie algebras of $\phi_{1}$ and $\phi_{2}$, respectively.

###### Proof.

From Theorem 4.8, we have that
$\mathfrak{g}_{\phi_{1}}\cong\mathfrak{g}_{\phi_{2}}$ if and only if
$\mathfrak{h}_{\phi_{1}}\cong\mathfrak{h}_{\phi_{2}}$. So, it is enough to see
that

$\mathfrak{h}_{\phi_{1}}\cong\mathfrak{h}_{\phi_{2}}\Longleftrightarrow
I_{\phi_{1}}=I_{\phi_{2}}.$

Now by (4.8), it enough to see that the associated matrices $\Phi_{1}$ and
$\Phi_{2}$ are conformally similar. By Theorem 4.8 and (4.8) we have that
$\Phi_{1}$ and $\Phi_{2}$ are conformally similar to

$\Psi=\mathrm{diag}(1,j_{1},\ldots,j_{k-2})\quad\text{and}\quad\Psi^{\prime}=\mathrm{diag}(1,j^{\prime}_{1},\ldots,j^{\prime}_{k-2})\quad\text{respectively,}$

where $\\{j_{1},\ldots,j_{k-2}\\}=I_{\phi_{1}}$ and
$\\{j^{\prime}_{1},\ldots,j^{\prime}_{k-2}\\}=I_{\phi_{2}}$. Finally, by Lemma
4.9, we obtain that $\Psi$ and $\Psi^{\prime}$ are conformally similar if and
only if $\Psi=\Psi^{\prime}$ which is equivalent to say that
$I_{\phi_{1}}=I_{\phi_{2}}$. Therefore
$\mathfrak{h}_{\phi_{1}}\cong\mathfrak{h}_{\phi_{2}}$ if and only if
$I_{\phi_{1}}=I_{\phi_{2}}$ as asserted. ∎

As a direct consequence, we obtain the following.

######  Corollary 4.11.

Let $\phi_{1}(x),\phi_{2}(x)$ be polynomials over $\mathbb{C}$ with degree
greater than $2$. Then, we have the following cases:

  1. (1)

If $\deg\phi_{1}=\deg\phi_{2}=2$, then
$\mathfrak{g}_{\phi_{1}}\cong\mathfrak{g}_{\phi_{2}}$.

  2. (2)

If $\deg\phi_{1}\neq\deg\phi_{2}$, then
$\mathfrak{g}_{\phi_{1}}\not\cong\mathfrak{g}_{\phi_{2}}$.

######  Remark 4.12.

Notice that if we consider $\phi_{m}(x)=x^{m}+ax$ with $m\geq 2$, then
$\dim(\mathfrak{h}_{\phi_{m}})=3$. The structure of solvable Lie algebras of
dimension $3$ was studied by J. Patera and H. Zassenhaus in [25]. In page 4,
the authors define the Lie algebra $L_{3,6}=\langle a_{1},a_{2},a_{3}\rangle$
with brackets

$[a_{1},a_{2}]=a_{3}\qquad[a_{1},a_{3}]=a_{3}-\alpha\cdot a_{2}$

with parameter $\alpha$ satisfying $\alpha\neq 0$ and $1-4\alpha\neq 0$, this
parameter $\alpha$ is in one-to-one correspondence with isomorphism classes of
this kind of algebras. Its associated matrix $\psi_{\alpha}(1)$ is

$\psi_{\alpha}(1)=\begin{pmatrix}0&-\alpha\\\ 1&1\end{pmatrix}$

The eigenvalues of $\psi_{\alpha}(1)$ are

$\lambda_{0}=\tfrac{1-\sqrt{1-4\alpha}}{2}\quad\text{and}\quad\lambda_{1}=\tfrac{1+\sqrt{1-4\alpha}}{2}.$

On the other hand, if we consider $\phi_{m}(x)=x^{m}+ax$ with $m\geq 2$, then
$\mathfrak{h}_{\phi_{m}}$ has dimension $3$ and its associated matrix
$\Psi_{m}$ is conformally similar to

$\begin{pmatrix}1&0\\\ 0&m\end{pmatrix}$

thus, $\Psi_{m}$ is conformally similar to $\psi_{\alpha}(1)$ if and only if

$\lambda_{0}=r,\qquad\lambda_{1}=rm\qquad\text{for some complex number $r$},$

since diagonalizable matrices are similar if and only if its spectrum are
equal. This system of equation has a solution $r=\tfrac{1}{m+1}$ and
$\alpha=\tfrac{m}{(m+1)^{2}}$. Therefore $\mathfrak{h}_{\phi_{m}}\cong
L_{3,6}^{\alpha}$ with $\alpha=\tfrac{m}{(m+1)^{2}}$.

##  5\. Laguerre type solutions

Let $A\in M_{N}(\mathbb{C})$ be a constant matrix and let $\nu\in\mathbb{R}$
such that $\nu>0$. In this section and the sequel, we are going to consider
the weight matrix $W$ given by

$W^{(\nu)}(x)=e^{Ax}T^{(\nu)}(x)e^{A^{\ast}x},\qquad
T^{(\nu)}(x)=e^{-x}\sum_{k=1}^{N}\delta^{(\nu)}_{k}x^{\nu+k}E_{k,k}.$ (5.1)

If we denote $L(x)=e^{Ax}$, then

$W^{(\nu)}(x)=L(x)T^{(\nu)}(x)L^{*}(x).$ (5.2)

Recall that if $A,J$ are as in (3.1), the equation (3.2) in terms of $L$ says
that

$L(x)JL^{-1}(x)=J-Ax.$ (5.3)

In this case, we obtain the same kind of weight that was consider in section
3, with $\phi(x)=x$.

######  Proposition 5.1.

The first order differential operators

$\mathcal{D}=\partial_{x}x+x(A-1),\quad\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J),$

are mutually adjoint and satisfy

$\displaystyle M=\psi^{-1}(\mathcal{D})=$
$\displaystyle(A-1)\delta-(n+1+\nu)-\mathcal{H}(n)J\mathcal{H}^{-1}(n),$
$\displaystyle M^{\dagger}=\psi^{-1}(\mathcal{D}^{\dagger})=$
$\displaystyle-(n+\nu+J+1)+\mathcal{H}(n)(A-1)^{\ast}\mathcal{H}^{-1}(n-1)\delta^{-1}.$

###### Proof.

The operators $\mathcal{D}$ and $\mathcal{D^{\dagger}}$ are mutually adjoint
by taking $\phi(x)=x$ in Proposition 3.1.

On the other hand, since $v(x)$ has degree $1$ in this case, by Corollary 3.5
we obtain that $A_{j}(n)=0$ for $j\leq-1$ and $A_{1}(n)=A-1$.

Finally, the formula for $A^{\dagger}_{0}(n)$ can be obtained directly from
the relation $P\cdot\mathcal{D}^{\dagger}=M_{\mathcal{D}^{\dagger}}\cdot P$
and the term $A_{0}(n)$ can be obtained from equation (2.5). ∎

As a direct consequence of the above proposition and Corollary 3.5 we obtain
the following result.

######  Corollary 5.2.

Let $A,J\in M_{N}(\mathbb{C})$ as in (3.1) and let $W:=W^{(\nu)}_{\phi}$ be a
matrix weight as in (3.3) with $\phi(x)=x$ and MVOPs $P(x,n)$. If $X(n)$ and
$Y(n)$ are the coefficients of the $(n-1)$-term and $(n-2)$-term of $P(x,n)$
respectively, then

$n+X(n)A-AX(n+1)-B(n)=-(n+1+\nu)-\mathcal{H}(n)J\mathcal{H}^{-1}(n),$ (5.4)
$X(n)+[J,X(n)]=\mathcal{H}(n)(A^{\ast}-1)\mathcal{H}^{-1}(n-1),$ (5.5)

where $\mathcal{H}(n)$ and $B(n)$are as in (1.2) and (1.4).

###### Proof.

It follows from Proposition 5.1 and Corollary 3.5, by taking into account the
term $A_{0}(n)$ of $\mathcal{D}$.

On the other hand, from the equality
$P\cdot\mathcal{D}^{\dagger}=M^{\dagger}\cdot P$ with

$\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J)\quad\text{and}\quad
M^{\dagger}=-(n+\nu+J+1)+A^{\dagger}_{-1}(n)\delta^{-1},$

we can obtained the equation (5.5) from the above theorem. ∎

### Existence of the operator $D$

Families of matrix valued orthogonal polynomials which are eigenfunctions of a
second order differential operator are of great importance, see e.g. [8], [9],
[14], [19]. Using the approach of [19], we get a symmetric second-order
differential operator which preserves polynomials and its degree. For this we
establish a conjugation with a diagonal matrix differential operator.

Let us consider matrix valued polynomials $F_{2}$, $F_{1}$, $F_{0}$ of degrees
two, one and zero respectively and let us assume that we have a matrix valued
second-order differential operator $D$ such that

$Q\cdot
D=\left(\tfrac{d^{2}Q}{dx^{2}}\right)(x)\,F_{2}(x)+\left(\tfrac{dQ}{dx}\right)(x)\,F_{1}(x)+Q(x)F_{0}(x).$
(5.6)

for a matrix valued polynomial $Q$. It follows from the definition of the
matrix valued inner product (1.1) that a differential operator $D$ is
symmetric with respect to $W$ if for all matrix valued polynomials $G,H$ we
have

$\int_{0}^{\infty}(GD)(x)W(x)(H(x))^{\ast}\,dx=\int_{0}^{\infty}G(x)W(x)((HD)(x))^{\ast}dx.$

By [9, Thm 3.1], this symmetry condition is equivalent to the following
equations

$\displaystyle F_{2}(x)W(x)=W(x)\bigl{(}F_{2}(x)\bigr{)}^{\ast},\qquad
2\tfrac{d(F_{2}W)}{dx}(x)-F_{1}(x)W(x)=W(x)\bigl{(}F_{1}(x)\bigr{)}^{\ast},$
(5.7)
$\displaystyle\tfrac{d^{2}(F_{2}W)}{dx^{2}}(x)-\tfrac{d(F_{1}W)}{dx}(x)+F_{0}(x)W(x)=W(x)\bigl{(}F_{0}(x)\bigr{)}^{\ast},$
(5.8)

and the boundary conditions

$\displaystyle\lim_{x\to 0}F_{2}(x)W(x)=0=\lim_{x\to\infty}F_{2}(x)W(x),$
(5.9) $\displaystyle\lim_{x\to
0}F_{1}(x)W(x)-\tfrac{d(F_{2}W)}{dx}(x)=0=\lim_{x\to\infty}F_{1}(x)W(x)-\tfrac{d(F_{2}W)}{dx}(x).$
(5.10)

We have the following lemma.

######  Lemma 5.3.

The second order differential operator

$D_{Q}=\partial_{x}^{2}x+\partial_{x}(1+\nu-x+J)-J$

is symmetric respect to the weight $T^{(\nu)}(x)$.

###### Proof.

By taking into account that $F_{2}(x)=x$, $F_{1}(x)=1+\nu-x+J$ and
$F_{0}(x)=-J$, it is a straightforward computation see that (5.9), (5.10),
(5.7) and (5.8) are satisfied for $D_{Q}$ and $T^{(\nu)}(x)$. ∎

######  Proposition 5.4.

The second order differential operator

$D=\partial_{x}^{2}x+\partial_{x}((A-1)x+1+\nu+J)+A\nu+JA-J$

is symmetric respect to the weight $W(x)$. Moreover

$P\cdot D=\Gamma\cdot P\quad\text{where}\quad\Gamma(n)=A(n+\nu+1+J)-n-J.$

###### Proof.

It follows from Remark 4.1 in [21]. ∎

### The Lie Algebra associated to $W$

Recall that a Lie algebra $\mathfrak{g}$ is called reductive if its radical is
equal to its center.

######  Lemma 5.5.

Let $A,J\in M_{N}(\mathbb{C})$ as in (3.1) and let us consider the operators

$\mathcal{D}=\partial_{x}x+x(A-1),\,\mathcal{D}^{\dagger}=-\partial_{x}x-(1+\nu+J),\,D=\partial_{x}^{2}x+\partial_{x}((A-1)x+1+\nu+J)+A\nu+JA-J.$

If $x$ acts over matrix valued polynomials by right constant multiplication,
then we have that

$[\mathcal{D},x]=-x,\qquad[\mathcal{D}^{\dagger},x]=x,\qquad[\mathcal{D},\mathcal{D}^{\dagger}]=x,\qquad[D,x]=-\mathcal{D}+\mathcal{D}^{\dagger},$
$[\mathcal{D},D]=-\mathcal{D}+D-(1+\nu),\quad[\mathcal{D}^{\dagger},D]=\mathcal{D}^{\dagger}-D+(1+\nu).$

The subjacent Lie algebra generated by
$\\{\mathcal{D},\mathcal{D}^{\dagger},D,x,I\\}$ is isomorphic to the Lie
algebra $\mathcal{A}=\langle x_{1},x_{2},x_{3},x_{4},x_{5}\rangle$ with
brackets

$[x_{1},x_{2}]=-x_{4},\quad[x_{1},x_{4}]=-x_{4},\quad[x_{2},x_{4}]=x_{4},\quad[x_{3},x_{4}]=-x_{1}+x_{2},$
$[x_{1},x_{3}]=-x_{2}+x_{3}+x_{4}-(1+\nu)x_{5}\quad\text{and}\quad[x_{2},x_{3}]=x_{1}-x_{3}-x_{4}+(1+\nu)x_{5},$

with correspondence:

$x_{1}\longmapsto\mathcal{D}+x,\quad
x_{2}\longmapsto\mathcal{D}^{\dagger}+x,\quad x_{3}\longmapsto D,\quad
x_{4}\longmapsto x,\quad x_{5}\longmapsto 1.$

###### Proof.

Let $\mathcal{D}^{\prime}=\mathcal{D}+x$ and
$\mathcal{D}^{\prime\dagger}=\mathcal{D}^{\dagger}+x$. It can be shown by
direct computation that

$[\mathcal{D}^{\prime},x]=-x,\quad[\mathcal{D}^{\prime\dagger},x]=x,\quad[\mathcal{D}^{\prime},\mathcal{D}^{\prime\dagger}]=-x,\quad[D,x]=-\mathcal{D}^{\prime}+\mathcal{D}^{\prime\dagger},$
$[\mathcal{D}^{\prime},D]=-\mathcal{D}^{\prime\dagger}+D+x-(1+\nu),\quad[\mathcal{D}^{\prime\dagger},D]=\mathcal{D}^{\prime}-D-x+(1+\nu).$

Clearly, the subjacent Lie algebra associated to this representation, is the
Lie algebra $\mathcal{A}=\langle x_{1},x_{2},x_{3},x_{4},x_{5}\rangle$ with
brackets

$[x_{1},x_{2}]=-x_{4},\quad[x_{1},x_{4}]=-x_{4},\quad[x_{2},x_{4}]=x_{4},\quad[x_{3},x_{4}]=-x_{1}+x_{2},$
$[x_{1},x_{3}]=-x_{2}+x_{3}+x_{4}-(1+\nu)x_{5}\quad\text{and}\quad[x_{2},x_{3}]=x_{1}-x_{3}-x_{4}+(1+\nu)x_{5},$

with correspondence

$x_{1}\longmapsto\mathcal{D}+x,\quad
x_{2}\longmapsto\mathcal{D}^{\dagger}+x,\quad x_{3}\longmapsto D,\quad
x_{4}\longmapsto x,\quad x_{5}\longmapsto 1,$

as desired. ∎

We have the following structure result.

######  Proposition 5.6.

The Lie algebra $\mathcal{A}$ defined as above is a 5-dimensional reductive
algebra with center of dimension two, given by
$\mathcal{Z}_{\mathcal{A}}=\langle x_{1}+x_{2}-x_{4},x_{5}\rangle$. Moreover,
$\mathcal{A}=[\mathcal{A},\mathcal{A}]\oplus\mathcal{Z}_{\mathcal{A}}$ with
$[\mathcal{A},\mathcal{A}]$ isomorphic to $\mathrm{SL}(2,\mathbb{C})$. In
particular,

$\mathcal{C}_{1}=-4x_{4}(x_{1}-x_{3}-x_{4}+(1+\nu)x_{5})+(x_{4}-x_{1}+x_{2})^{2},\quad\mathcal{C}_{2}=x_{1}+x_{2}-x_{4}\quad\text{and}\quad\mathcal{C}_{3}=x_{5}$

are Casimir elements of $\mathcal{A}$.

###### Proof.

From Lie algebra theory (see [18]), the radical of $\mathcal{A}$ can be
computed from its Killing form, in this case we have that

$\mathrm{Rad}(\mathcal{A})=\mathcal{Z}_{\mathcal{A}}=\langle
x_{1}+x_{2}-x_{4},x_{5}\rangle,$

and so, the Lie algebra $\mathcal{A}$ is reductive. Now, from general theory,
since $\mathcal{A}$ is reductive, we obtain

$\mathcal{A}=[\mathcal{A},\mathcal{A}]\oplus\mathcal{Z}_{\mathcal{A}}.$

In this case, we obtain that

$[\mathcal{A},\mathcal{A}]=\langle
x_{4},\,x_{1}-x_{2},\,x_{1}-x_{3}-x_{4}+(1+\nu)x_{5}\rangle.$

By taking $a_{1}=x_{4}$, $a_{2}=x_{1}-x_{2}$ and
$a_{3}=x_{1}-x_{3}-x_{4}+(1+\nu)x_{5}$ we obtain that

$[a_{1},a_{2}]=-2a_{1},\quad[a_{1},a_{3}]=a_{1}-a_{2},\quad[a_{2},a_{3}]=-a_{3}$

and so if we take $\hat{a}_{2}=a_{1}-a_{2}$ and $\hat{a}_{3}=-a_{3}$ we have

$[a_{1},\hat{a}_{2}]=2a_{1},\quad[a_{1},\hat{a}_{3}]=-\hat{a}_{2},\quad[\hat{a}_{2},\hat{a}_{3}]=2\hat{a}_{3},$

and so $[\mathcal{A},\mathcal{A}]$ is isomorphic to
$\mathrm{SL}(2,\mathbb{C})$ by consider the map $a_{1}\mapsto e_{1}$
$\hat{a}_{2}\mapsto e_{2}$ and $\hat{a}_{3}\mapsto e_{3}$. In particular, the
Casimir element of $\mathrm{SL}(2,\mathbb{C})$ given by
$4e_{1}e_{3}+e_{2}^{2}$ induces a Casimir element of
$[\mathcal{A},\mathcal{A}]$

$\mathcal{C}_{[\mathcal{A},\mathcal{A}]}=-4x_{4}(x_{1}-x_{3}-x_{4}+(1+\nu)x_{5})+(x_{4}-x_{1}+x_{2})^{2}.$

By taking into account that $\mathcal{C}_{[\mathcal{A},\mathcal{A}]}$ commutes
with the central elements of $\mathcal{A}$, we obtain that
$\mathcal{C}_{[\mathcal{A},\mathcal{A}]}$ commutes with all of the elements of
$\mathcal{A}$ and so is a Casimir element of $\mathcal{A}$.

The last assertion is clear, since the central elements always are Casimir
elements of a given Lie algebra. ∎

######  Remark 5.7.

Notice that

$\mathcal{C}=\mathcal{C}_{2}+(1+\nu)\mathcal{C}_{3}=\mathcal{D}+\mathcal{D}^{\dagger}+x+(1+\nu)$

it is also a Casimir invariant of $\mathcal{A}$. Hence, under the
representation given by $\mathcal{D},\mathcal{D}^{\dagger},D,x,1$, the image
of this Casimir satisfies

$\mathcal{C}=Ax-J.$

Thus, in terms of $M,M^{\dagger}$ and $L$, we obtain that

$\varphi^{-1}(\mathcal{C})=M+M^{\dagger}+L+(1+\nu).$ (5.11)

On the other hand, it can be check that

$\mathcal{C}_{[\mathcal{A},\mathcal{A}]}=\dfrac{1}{8}(A^{2}x^{2}+\nu^{2}+J^{2}+Ax-2\nu
Ax-2xJA+2\nu J-1).$

Thus, $\mathcal{C}^{\prime}=A^{2}x^{2}-2xJA+J+J^{2}$ it is also a Casimir
since

$\mathcal{C}^{\prime}=8\mathcal{C}_{[\mathcal{A},\mathcal{A}]}-(1-2\nu)\mathcal{C}-(\nu^{2}-1)\mathcal{C}_{3},$

moreover, notice that the relation $[J,A]=A$ implies that
$\mathcal{C}^{\prime}=\mathcal{C}^{2}-\mathcal{C}$, and so we obtain that

$\mathcal{C}_{[\mathcal{A},\mathcal{A}]}=\tfrac{1}{8}\Big{(}\mathcal{C}^{2}-2\nu\mathcal{C}+(\nu^{2}-1)\mathcal{C}_{3}\Big{)}.$

Therefore, in this representation the Casimir element corresponding to
$\mathcal{C}_{[\mathcal{A},\mathcal{A}]}$ does not give more information than
$\mathcal{C}$. Hence, in the rest of the paper we will only consider the
Casimir element $\mathcal{C}$.

We can also consider the Lie subalgebra generated by
$\\{x_{1},x_{2},x_{4},x_{5}\\}$. Notice that a representation of this kind of
algebra was consider in the above section by taking $\phi(x)=x$. In this case,
we have the following structure result.

######  Proposition 5.8.

The Lie subalgebra $\mathcal{A}^{\prime}=\langle
x_{1},x_{2},x_{4},x_{5}\rangle$ of $\mathcal{A}$ is isomorphic to
$\mathfrak{g}_{2}\oplus\mathbb{C}^{2}$ where $\mathfrak{g}_{2}$ is the
2-dimensional solvable Lie algebra with bracket $[e_{1},e_{2}]=e_{2}$. In
particular, $\mathcal{A}^{\prime}$ has no non-central Casimir invariants.

###### Proof.

By taking the map

$x_{2}\mapsto e_{1},\quad x_{4}\mapsto e_{2},\quad x_{1}+x_{2}-x_{4}\mapsto
e_{3}\quad\text{and }\quad x_{5}\mapsto e_{4}.$

we obtain an isomorphic algebra of $\mathcal{A}^{\prime}$, in this case the
only non-vanishing bracket of $\langle e_{1},e_{2},e_{3},e_{4}\rangle$ is the
bracket $[e_{1},e_{2}]=e_{2}$ and so $\mathcal{A}^{\prime}$ is isomorphic to
$\mathfrak{g}_{2}\oplus\mathbb{C}^{2}$, as asserted.

The last assertion is a consequence of $\mathfrak{g}_{2}$ has not central
elements. Hence, the only Casimir invariants of $\mathcal{A}^{\prime}$ are the
central elements. ∎

In the sequel, in order to simplify the notation, we will consider

$B_{n}:=B(n),\quad
C_{n}:=C(n),\quad\mathcal{H}_{n}:=\mathcal{H}(n),\quad\Gamma_{n}=\Gamma(n).$
(5.12)

The following proposition is a consequence, of the relations between the
brackets of $M,M^{\dagger},L,\Gamma$.

######  Proposition 5.9.

Let $A,J\in M_{N}(\mathbb{C})$ be matrices as in (3.1) and let
$B_{n},C_{n},\mathcal{H}_{n}$ and $\Gamma_{n}$ be as in (5.12). Then,

$B_{n}(A-1)-(A-1)B_{n+1}=2+\mathcal{H}_{n+1}J\mathcal{H}_{n+1}^{-1}-\mathcal{H}_{n}J\mathcal{H}_{n}^{-1},$
(5.13)
$B_{n}=[B_{n},J]+\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}-\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1},$
(5.14)
$2\mathcal{H}_{n}\mathcal{H}_{n-1}^{-1}=[\mathcal{H}_{n}\mathcal{H}_{n-1}^{-1},J]-B_{n}\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}-\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}B_{n-1},$
(5.15)
$B_{n}=-[J,\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}]-\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}(A-1)+(A-1)\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1},$
(5.16)
$[\Gamma,C_{n}\delta^{-1}]=\mathcal{H}_{n}(A-1)^{*}\mathcal{H}_{n-1}^{-1},$
(5.17)
$[\Gamma_{n},\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}]=n+\Gamma_{n}+\mathcal{H}_{n}J\mathcal{H}_{n}^{-1},$
(5.18)
$[\Gamma,\mathcal{H}_{n}(A-1)^{*}\mathcal{H}_{n-1}^{-1}\delta^{-1}]=-\mathcal{H}_{n}(A-1)^{*}\mathcal{H}_{n-1}^{-1}.$
(5.19)

###### Proof.

The equation (5.13) is consequence of seeing the coefficient of $\delta^{1}$
in the bracket relation $[M,L]=L$.

In the same way, the equations (5.14), (5.15) are consequence of seeing the
coefficients of $\delta^{0}$ and $\delta^{-1}$ in the bracket relation
$[M^{\dagger},L]=L$.

On the other hand, the equation (5.16) it follows from the bracket relation
$[M,M^{\dagger}]=L$. The equation (5.17) is a consequence of the coefficient
of $\delta^{-1}$ in the bracket $[\Gamma,L]=-M+M^{\dagger}$. The equation
(5.18) is obtained from the coefficient of $\delta^{0}$ in the bracket
$[\Gamma,M]=M-\Gamma+(1+\nu)$. Finally, the equation (5.19) can be obtained
from the coefficient of $\delta^{-1}$ in the bracket relation
$[\Gamma,M^{\dagger}]=-M^{\dagger}+\Gamma-(1+\nu)$. ∎

######  Proposition 5.10.

Let $A,J\in M_{N}(\mathbb{C})$ be matrices as in (3.1). Then,

$\displaystyle[B_{n},J]$
$\displaystyle=B_{n}-\overline{\Gamma_{n}}\overline{C_{n}}+\overline{C_{n}\Gamma_{n-1}}+\overline{\Gamma_{n+1}C_{n+1}}-\overline{C_{n+1}\Gamma_{n}},$
$\displaystyle[C_{n},J]$
$\displaystyle=2C_{n}+B_{n}(\overline{\Gamma_{n}}\overline{C_{n}}-\overline{C_{n}\Gamma_{n-1}})+(\overline{\Gamma_{n}}\overline{C_{n}}-\overline{C_{n}\Gamma_{n-1}})B_{n-1},$

where $B_{n},C_{n},\mathcal{H}_{n}$ and $\Gamma_{n}$ be as in (5.12)

###### Proof.

By (5.14) we have that

$[B_{n},J]=B_{n}-\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}+\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1}.$

On the other hand, by (5.17) and taking into account that $\mathcal{H}_{n}\in
M_{N}(\mathbb{R})$, we obtain that

$\displaystyle\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}=\overline{[\Gamma,C_{n}\delta^{-1}]}$
$\displaystyle=\overline{\Gamma_{n}}\overline{C_{n}}-\overline{C_{n}\Gamma_{n-1}},$
$\displaystyle\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1}=\overline{[\Gamma,C_{n}\delta^{-1}]}$
$\displaystyle=\overline{\Gamma_{n+1}C_{n+1}}-\overline{C_{n+1}\Gamma_{n}},$

and so we have that

$[B_{n},J]=B_{n}-\overline{\Gamma_{n}}\overline{C_{n}}+\overline{C_{n}\Gamma_{n-1}}+\overline{\Gamma_{n+1}C_{n+1}}-\overline{C_{n+1}\Gamma_{n}},$

as asserted.

For the second equality, notice that since
$C_{n}=\mathcal{H}_{n}\mathcal{H}_{n-1}^{-1}$, thus the equation (5.15)
implies that

$[C_{n},J]=2C_{n}+B_{n}\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}+\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}B_{n-1}.$

In the same way as above, we obtain that

$[C_{n},J]=2C_{n}+B_{n}(\overline{\Gamma_{n}}\overline{C_{n}}-\overline{C_{n}\Gamma_{n-1}})+(\overline{\Gamma_{n}}\overline{C_{n}}-\overline{C_{n}\Gamma_{n-1}})B_{n-1},$

as desired. ∎

##  6\. Matrix entries of $P(x,n)$ as classical Laguerre polynomials

In this section we will give explicit expressions of the entries of the
Laguerre-type MVOPs in terms of the classical scalar Laguerre polynomials. For
this we use the approach of [19], [17], [20], which consists in observing that
a symmetric second order differential operator can be diagonalized via
conjugation with an appropriate matrix valued function. The present situation
is more involved than the previous cases because, although the differential
operator can be diagonalized, the eigenvalue remains non-diagonal.

In the rest of the section, the weight matrix $W(x)$ is as in the previous
section.

### Step I: Diagonalizing the differential operator:

Let $A,J\in M_{N}(\mathbb{C})$ be matrices as in (3.1). We can define the
following auxiliary matrix valued polynomials

$Q(x,n):=P(x,n)L(x),\qquad\text{where $L(x)=e^{xA}$.}$ (6.1)

The polynomials $Q_{n}$ satisfy the following relations,

$Q_{n}\cdot\mathcal{D}_{Q}=M\cdot Q_{n},\qquad Q_{n}\cdot
D_{Q}=\Lambda_{n}\cdot Q_{n},\qquad
Q_{n}\cdot\mathcal{C}_{Q}=M_{\mathcal{C}}\cdot Q_{n},$

where

$\mathcal{D}_{Q}=L(x)^{-1}\mathcal{D}L(x),\qquad
D_{Q}=L(x)^{-1}DL(x),\qquad\mathcal{C}_{Q}=L(x)^{-1}\mathcal{C}L(x).$ (6.2)

Here $\mathcal{D}$, $D$ are as in Proposition 5.1, Proposition 5.4
respectively and $\mathcal{C}=Ax-J$ . In the following lemma, we show that
$\mathcal{D}_{Q}$, $D_{Q}$ and $\mathcal{C}_{Q}$ are simple diagonal
operators.

######  Lemma 6.1.

The operators $\mathcal{D}_{Q},D_{Q}$ and $\mathcal{C}_{Q}$ as in (6.2) are
given explicitly as follows:

$\mathcal{D}_{Q}=\partial_{x}x-x,\qquad
D_{Q}=\partial_{x}^{2}x+\partial_{x}(1+\nu-x+J)-J,\qquad\mathcal{C}_{Q}=-J.$

###### Proof.

All of the equalities are follow directly from definition and the equation
(3.2). ∎

In the following proposition, we will use the expressions of
$\mathcal{C}_{Q}$, $\mathcal{C}$ and $\mathcal{D}$ in order to obtain an
equation which relates $Q(x,n)$ and the recurrent matrix
$H_{n}(A^{\ast}-1)H_{n-1}^{-1}$.

######  Proposition 6.2.

The auxiliary functions $Q(x,n)$’s defined in (6.1), satisfy the following
equation which depend on the squared norms:

$-Q(x,n)J=xQ^{\prime}(x,n)-(n+J)Q(x,n)+\mathcal{H}(n)(A-1)^{\ast}\mathcal{H}^{-1}(n-1)Q(x,n-1)\quad\text{for
$n\geq 1$}$

and for $n=0$

$-Q(x,0)J=-J-xQ^{\prime}(x,0).$

###### Proof.

By Lemmas 3.2 and 6.1 we have that

$-Q(x,n)J=Q(x,n)\cdot\mathcal{C}_{Q}=(M_{\mathcal{C}}\cdot P(x,n))e^{xA}$

where $M_{\mathcal{C}}$ is the discrete operator

$M_{\mathcal{C}}=A\delta+X(n)A-AX(n+1)-J+\Big{(}Y(n)A-AY(n+1)+[J,X(n)]+(AX(n+1)-X(n)A)X(n)\Big{)}\delta^{-1}.$

By equation (3.6), we have

$(n-1)X(n)+Y(n)(A-1)-(A-1)Y(n+1)-\Big{(}n+X(n)A-AX(n+1)-B(n)\Big{)}X(n)=0,$

and so

$Y(n)A-AY(n+1)+(AX(n+1)-X(n)A)X(n)=X(n)+Y(n)-Y(n+1)-B(n)X(n).$

By taking into account the relation $P\cdot x=L\cdot P$ with
$L=\delta+B(n)+C(n)\delta^{-1}$, thus we have that

$Y(n)=Y(n+1)+B(n)X(n)+C(n).$

Hence, we obtain that

$(M_{\mathcal{C}}\cdot
P(x,n))e^{xA}=AQ(x,n+1)+(X(n)A-AX(n+1)-J)Q(x,n)+(C(n)+[J,X(n)]+X(n))Q(x,n-1).$

On the other hand, by Corollary 3.5 we have that

$\displaystyle(M_{\mathcal{D}}\cdot P(x,n))e^{xA}$ $\displaystyle=$
$\displaystyle(A-1)Q(x,n+1)+\Big{(}n+X(n)A-AX(n+1)-B(n)\Big{)}Q(x,n),$
$\displaystyle(L\cdot P(x,n))e^{xA}$ $\displaystyle=$ $\displaystyle
Q(x,n+1)+B(n)Q(x,n)+C(n)Q(x,n-1),$

where $B(n)$ and $C(n)$ are as in equation (1.4). Hence, we have that

$((M_{\mathcal{D}}+L)\cdot
P(x,n))e^{xA}=AQ(x,n+1)+(n+X(n)A-AX(n+1))Q(x,n)+C(n)Q(x,n-1)$

and thus

$(M_{\mathcal{C}}\cdot P(x,n))e^{xA}=((M_{\mathcal{D}}+L)\cdot
P(x,n))e^{xA}-(n+J)Q(x,n)+(X(n)+[J,X(n)])Q(x,n-1).$

Notice that if we denote $L(x)=e^{xA}$, the Lemma 6.1 implies that

$((M_{\mathcal{D}}+L)\cdot
P(x,n))L(x)=(P(x,n)\cdot(\mathcal{D}+x))L(x)=Q(x,n)\cdot
L^{-1}(x)(\mathcal{D}+x))L(x)=Q(x,n)\cdot\partial_{x}x,$

and by (5.5), $X(n)+[J,X(n)]=\mathcal{H}(n)(A-1)^{\ast}\mathcal{H}^{-1}(n-1)$.
Thus, we obtain that

$-Q(x,n)J=xQ^{\prime}(x,n)-(n+J)Q(x,n)+\mathcal{H}(n)(A-1)^{\ast}\mathcal{H}^{-1}(n-1),$

as asserted. Finally, the equation for $n=0$, it follows from equation (5.3).
∎

### Step II: Diagonalizing the eigenvalue $\Lambda_{n}$:

Although the differential operator $D_{Q}$ is a diagonal operator, the system
of equations given by $Q_{n}\cdot D_{Q}=\Lambda_{n}Q_{n}$ is not a decoupled
system since $\Lambda_{n}$ is a lower triangular matrix. However,
$\Lambda_{n}$ can be diagonalized in a somewhat simple way:

$\Gamma_{n}=A(n+\nu+J+1)-(n+J)=K_{n}\Lambda_{n}K_{n}^{-1},$ (6.3)

where $\Lambda_{n}$ is the diagonal matrix $\Lambda_{n}=-(n+J)$.

######  Lemma 6.3.

We can choose $K_{n}$ such that is a lower triangular matrix with diagonal
elements $(K_{n})_{i,i}=1$ for all $i=1,\ldots,N$. Moreover, in this case
$K_{n}$ is the following matrix:

$(K_{n})_{i,j}=\begin{cases}(-1)^{i-j}\tfrac{(n+\nu+j+1)_{i-1}}{(i-j)!}\prod_{k=j}^{i-1}a_{k}&i>j,\\\
1&i=j,\\\ 0&i<j.\end{cases}$ (6.4)

###### Proof.

Since $A(n+\nu+J+1)-(n+J)$ is lower triangular, the its characteristic
polynomial satisfy

$\det\Big{(}\lambda I-\Big{(}A(n+\nu+J+1)-(n+J)\Big{)}\Big{)}=\det(\lambda
I+nI+J)=\prod_{r=1}^{N}(\lambda+n+r).$

Then, the eingevalues of $A(n+\nu+J+1)-(n+J)$ are

$\lambda_{1}=-(n+1),\ldots,\lambda_{N}=-(n+N)\quad\text{with multiplicity
$1$}.$

Since $K_{n}$ diagonalizes $A(n+\nu+J+1)-(n+J)$, then its $r$-th column can be
obtained from the eigenspace correspond to $\lambda_{r}=-(n+r)$, that is

$Nu(\lambda_{r}I-A(n+\nu+J+1)+n+J)=Nu(-A(n+\nu+J+1)+J-rI),\quad\text{ for
$r=1,\ldots,N$}.$

We can obtain (6.4) by a straightforward computation of these eigenspaces. ∎

With the matrix $K_{n}$ as in (6.4), we will consider the following matrix
polynomial

$R(x,n):=K_{n}^{-1}Q(x,n).$ (6.5)

The following result that shows a relationship between the non-zero matrix
entries of $R(x,n)$ and generalized Laguerre polynomials.

######  Theorem 6.4.

Let $n\in\mathbb{N}$ and let $\nu>0$. The matrix elements of $R(x,n)$ are
multiples of scalar Laguerre functions

$R(x,n)_{i,j}=\begin{cases}L^{(\nu+j)}_{n+i-j}(x)\xi(n,i,j)&\text{if
$n+i-j\geq 0$}\\\ 0&\text{if $n+i-j<0$}.\end{cases}$ (6.6)

###### Proof.

Notice that the polynomials $R(x,n)$’s are eigenfunctions of
$D_{Q}=\partial_{x}^{2}x+\partial_{x}(1+\nu-x+J)-J$ with associated
eigenvalues $\Lambda_{n}=-(n+J)$ where $J$ is the diagonal matrix
$\mathrm{diag}(1,2,\ldots,N)$. If we look the $(i,j)$-entry of

$xR^{\prime\prime}(x,n)+R^{\prime}(x,n)(1+\nu-x+J)+(nR(x,n)+JR(x,n)-R(x,n)J)=0,$

we obtain that the following expression

$xR^{\prime\prime}(x,n)_{i,j}+\sum_{k=1}^{N}R^{\prime}(x,n)_{i,k}(1+\nu-x+J)_{k,j}+nR(x,n)_{i,j}+\sum_{k=1}^{N}(J_{i,k}R(x,n)_{k,j}-R(x,n)_{i,k}J_{k,j})=0.$

Since $J=\mathrm{diag}(1,2,\ldots,N)$, the above equality is equivalent to

$xR^{\prime\prime}(x,n)_{i,j}+R^{\prime}(x,n)_{i,j}(1+\nu-x+j)+(n+i-j)R(x,n)_{i,j}=0.$

Hence, we obtain that

$R(x,n)_{i,j}=L^{(\nu+j)}_{n+i-j}(x)\xi(n,i,j),$

as asserted.

On the other hand, it is well-known that if $M<0$ the only solution for the
differential equation

$xP^{\prime\prime}(x)+P^{\prime}(x)(1+\nu-x)+MP(x)=0,$

is $P(x)=0$ and since

$xR^{\prime\prime}(x,n)_{i,j}+R^{\prime}(x,n)_{i,j}(1+\nu-x+j)+(n+i-j)R(x,n)_{i,j}=0,$

we obtain that $R(x,n)_{i,j}=0$ if $n+i-j<0$, as asserted. ∎

######  Remark 6.5.

Notice that by the above theorem the are only defined for $n+i-j\geq 0$, so we
extend its definition as follows

$\xi(n,i,j)=0\qquad\text{if $n+i-j<0$}.$ (6.7)

It is well-known that the generalized Laguerre polynomial satisifes

$L^{(\alpha)}_{N}(0)=\frac{(\alpha+1)_{N}}{N!},$

where $(a)_{N}$ is the Pochhammer symbol defined by

$(a)_{N}=\begin{cases}1&\text{if $N=0$},\\\ a(a+1)\cdots(a+N-1)&\text{if
$N>0$}.\end{cases}$ (6.8)

As a direct consequence of the above theorem we obtain the following
corollary.

######  Corollary 6.6.

Let $n\in\mathbb{N}$ and let $\nu>0$. Then, the coefficients of $R(0,n)$
satisfy

$R(0,n)_{i,j}=\begin{cases}\frac{(\nu+j+1)_{n+i-j}}{(n+i-j)!}\xi(n,i,j)&\text{if
$n+i-j\geq 0$,}\\\ 0&\text{if $n+i-j<0$}.\end{cases}$ (6.9)

In particular, the $(i,n+i)$-th and $(i,n+i-1)$-th coordinates of $R(0,n)$
satisfy

$R(0,n)_{i,n+i}=\xi(n,i,n+i)\quad\text{and}\quad
R(0,n)_{i,n+i-1}=(n+\nu+i)\xi(n,i,n+i-1).$

Now we need to identify the coefficients $\xi(n,i,j)$. For this, we exploit
the relation in Proposition 6.2. In the following lemma, we observe that the
factor $\mathcal{H}(n)(A-1)^{\ast}\mathcal{H}^{-1}(n-1)$ in this relation is
turned into a diagonal matrix via multiplication by appropriate matrices
$K_{n}$. This will allow us to obtain a simple recursion for $\xi(n,i,j)$. For
this purpose, we define the following

$G(n):=K_{n}^{-1}\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1}K_{n-1}\qquad
I(n):=K_{n}^{-1}\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}K_{n},$ (6.10)

where $\mathcal{H}_{n}$ and $K_{n}$ are as in (5.12) and (6.4), respectively.

######  Lemma 6.7.

For $n\geq 1$ and let $G(n)$, $I(n)$ be the matrices as in (6.10). Then,
$G(n)$ is diagonal and $I(n)$ satisfies

$(I(n))_{i,i}=i,\qquad(I(n))_{i,j}=0\ \text{ for }i\neq j\text{ and }i\neq
j-1.$

Moreover

$(G(n))_{i,i}=(\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1})_{i,i}\quad\text{and}\quad(I(n))_{i,i+1}=(\mathcal{H}_{n}J\mathcal{H}_{n}^{-1})_{i,i+1}.$
(6.11)

###### Proof.

By equation (5.19),

$[\Gamma_{n},\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1}\delta^{-1}]=-\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1}.$

Then,

$K_{n}^{-1}\Gamma_{n}(\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1})K_{n-1}-K_{n}^{-1}(\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1})\Gamma_{n-1}K_{n-1}=-G(n).$

Then, by $K_{n}^{-1}\Gamma_{n}K_{n}=\Lambda_{n}=-(n+J)$ we obtain that

$\Lambda_{n}G(n)-G(n)\Lambda_{n-1}=-G(n).$

The assertion it follows by observing the $(i,j)$-entry for $i\neq j$. On the
other hand, from definition we have that

$K_{n}G(n)K_{n-1}^{-1}=\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1}.$

Since $K_{n}$ and $K_{n-1}^{-1}$ are both lower triangular matrices with $1$’s
in its diagonal, then the left term in the above equation is a lower
triangular matrix with $(i,i)$-th coordinate equal to $(G(n))_{i,i}$ and so

$(\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1})_{i,i}=(K_{n}G(n)K_{n-1}^{-1})_{i,i}=(G(n))_{i,i},$

as asserted.

Now, for the second assertion, we can proceed as in the same way, in this case
by equation (5.18), we have that

$[\Gamma_{n},\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}]=n+\Gamma_{n}+\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}.$

Then, we obtain that

$K_{n}^{-1}\Gamma_{n}\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}K_{n}-K_{n}^{-1}\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}\Gamma_{n}K_{n}=K_{n}^{-1}(n+\Gamma_{n}+\mathcal{H}_{n}J\mathcal{H}_{n}^{-1})K_{n}.$

Hence, by $K_{n}^{-1}\Gamma_{n}K_{n}=\Lambda_{n}=-(n+J)$ we obtain that

$\Lambda_{n}I(n)-I(n)\Lambda_{n}=n+\Lambda_{n}+I(n)).$

The proof follows by observing the $(i,j)$-entry in the above matrix equality.
Finally, from definition we have

$K_{n}I(n)K_{n}^{-1}=\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}.$

So, in general we have that

$(K_{n}I(n)K_{n}^{-1})_{i,j}=\sum_{k=1}^{N}(K_{n})_{i,k}(I(n)K_{n}^{-1})_{k,j}\quad\text{and}\quad(I(n)K_{n}^{-1})_{k,j}=(I(n))_{k,k}(K_{n}^{-1})_{k,j}+(I(n))_{k,k+1}(K_{n}^{-1})_{k+1,j}.$

By taking into account that $(K_{n})_{i,j}=(K_{n}^{-1})_{i,j}=0$ for $j>i$, if
$j=i+1$ we obtain that

$(K_{n}I(n)K_{n}^{-1})_{i,i+1}=\sum_{k=1}^{N}(K_{n})_{i,k}(I(n)K_{n}^{-1})_{k,j}=\sum_{k=1}^{i}(K_{n})_{i,k}(I(n)K_{n}^{-1})_{k,i+1}=(I(n))_{i,i+1}(K_{n}^{-1})_{i+1,i+1},$

by taking into account that $(K_{n}^{-1})_{i+1,i+1}=1$ for any $i$. Therefore,
we have that $(K_{n}I(n)K_{n}^{-1})_{i,i+1}=(I(n))_{i,i+1}$, as desired. ∎

The following proposition is a consequence of the relation given by the
Casimir operator and Proposition 6.4

######  Proposition 6.8.

Let $\xi(n,i,j)$ as in Theorem 6.4. If $n+i-j>0$, then the constants
$\xi(n,i,j)$’s satisfy the following:

  1. ($a$)

$\xi(0,i,j)={(K_{0}^{-1})}_{i,j}\tfrac{(i-j)!}{(\nu+j+1)_{i-j}}$,

  2. ($b$)

If $i=1$ and $n>0$ then

$\xi(n,1,j)=\tfrac{(G(n))_{1,1}}{n+\nu+1}\xi(n-1,1,j),$ (6.12)

  3. ($c$)

If $i>1$ and $n>0$, then

$\xi(n,i,j)=a_{i-1}\xi(n,i-1,j)+\tfrac{(G(n))_{i,i}}{n+\nu+i}\xi(n-1,i,j),$
(6.13)

with $G(n)$ as in (6.10).

###### Proof.

For the first assertion, recall that

$R(0,0)=K_{0}^{-1}Q(0,0)=K_{0}^{-1}P(0,0)=K_{0}^{-1},$

thus we have that

${(K_{0}^{-1})}_{i,j}=\xi(0,i,j)L^{(\nu+j)}_{i-j}(0).$

By taking into account that $L^{(\nu)}_{N}(0)=\tfrac{(\nu+1)_{N}}{N!}$ for any
$N$, hence we obtain

${(K_{0}^{-1})}_{i,j}\tfrac{(i-j)!}{(\nu+j+1)_{i-j}}=\xi(0,i,j),$

as asserted.

For items ($b$) and ($c$), recall that Proposition 6.2 implies that

$-Q(0,n)J=-(n+J)Q(0,n)+\mathcal{H}(n)(A^{\ast}-1)\mathcal{H}(n-1)^{-1}Q(0,n-1).$

Then,

$-K_{n}R(0,n)J=-(n+J)K_{n}R(0,n)+\mathcal{H}(n)(A^{\ast}-1)\mathcal{H}(n-1)^{-1}K_{n-1}R(0,n-1),$

and so we obtain that

$-R(0,n)J=-K_{n}^{-1}(n+J)K_{n}R(0,n)+G(n)R(0,n-1)$

where $G(n)=K_{n}^{-1}\mathcal{H}(n)(A^{\ast}-1)\mathcal{H}(n-1)^{-1}K_{n-1}$.
By recalling that

$K_{n}^{-1}(n+J)K_{n}R(0,n)=(n+J)-A(n+\nu+J+1),$

we obtain that

$-R(0,n)J=-(n+J)R(0,n)+A(n+\nu+J+1)R(0,n)+G(n)R(0,n-1).$

Thus, in terms of coordinates we have that

$-jR(0,n)_{i,j}=-(n+i)R(0,n)_{i,j}+\sum_{k=1}^{N}(A(n+\nu+J+1))_{i,k}R(0,n)_{k,j}+\sum_{k=1}^{N}(G(n))_{i,k}R(0,n-1)_{k,j}.$
(6.14)

Since $A$ has only non-zero entries $a_{i}$’s in the place $(i,i+1)$ and
$n+\nu+J+1$ is diagonal, we obtain that

$\sum_{k=1}^{N}(A(n+\nu+J+1))_{i,k}R(0,n)_{k,j}=\begin{cases}(a_{i-1}(n+\nu+i))R(0,n)_{i-1,j}&\text{if
$i>1$},\\\ 0&\text{if $i=1$}.\end{cases}$ (6.15)

Thus, for $i>1$ the equation (6.14) takes the form

$-jR(0,n)_{i,j}=-(n+i)R(0,n)_{i,j}+a_{i-1}(n+\nu+i)R(0,n)_{i-1,j}+(G(n))_{i,i}R(0,n-1)_{i,j}$

where in the last term of the equality, we use Lemma 6.7. By Theorem 6.4, we
have that

$(n+i-j)L^{(\nu+j)}_{n+i-j}(0)\xi(n,i,j)=L^{(\nu+j)}_{n+i-1-j}(0)\Big{(}a_{i-1}(n+\nu+i)\xi(n,i-1,j)+(G(n))_{i,i}\xi(n-1,i,j)\Big{)}.$

Therefore, since $n+i-j>0$ we obtain a similar expression of (6.13) by
multiplication for $((n+i-j)L^{(\nu+j)}_{n+i-j}(0))^{-1}$, that is

$\xi(n,i,j)=C_{1}(n,i,j)\xi(n,i-1,j)+C_{2}(n,i,j)\xi(n-1,i,j),$

where

$C_{1}(n,i,j)=\tfrac{(a_{i-1}(n+\nu+i))L^{(\nu+j)}_{n+i-1-j}(0)}{(n+i-j)L^{(\nu+j)}_{n+i-j}(0)}\quad\text{and}\quad
C_{2}(n,i,j)=\tfrac{(G(n))_{i,i}L^{(\nu+j)}_{n-1+i-j}(0)}{(n+i-j)L^{(\nu+j)}_{n+i-j}(0)}.$

To finish the proof, recall that

$L^{(\alpha)}_{N}(0)=\frac{(\alpha+1)_{N}}{N!},$

where $(a)_{N}$ is the Pochhammer symbol defined by

$(a)_{N}=\begin{cases}1&\text{if $N=0$},\\\ a(a+1)\cdots(a+N-1)&\text{if
$N>0$}.\end{cases}$

Hence, we have that

$\frac{L^{(\nu+j)}_{n+i-1-j}(0)}{L^{(\nu+j)}_{n+i-j}(0)}=\frac{n+i-j}{n+\nu+i}.$

Therefore we obtain (6.13), as asserted.

Now, for $i=1$ the equation (6.14) takes the form

$-jR(0,n)_{1,j}=-(n+1)R(0,n)_{i,j}+(G(n))_{1,1}R(0,n-1)_{1,j}.$

So, the equation (6.12) it can be prooved in a similar way as in the case
$i>1$. ∎

######  Remark 6.9.

Notice that the item ($a$) in the above proposition still holds for $n+i-j=0$
since in this case we just use the definition of $\xi$’s.

Now, we are going to study the case $n+i-j=0$. In this case, we have the
following result.

######  Proposition 6.10.

Let $n\in\mathbb{N}_{0}$ and let $I(n),\,G(n)$ be as in (6.10). If $i\geq 1$,
then the constants $\xi$’s satisfy the following:

  1. ($a$)

$\xi(0,i,i)=1$ and $\xi(1,i,i+1)=I(0)_{i,i+1}$.

  2. ($b$)

If $i=1$ and $n>0$, then

$N_{1}(n,i)\xi(n,1,n+1)=N_{2}(n,i)\xi(n-1,2,n+1),\quad\text{ with }$
$N_{1}(n,i)=(\nu+2n+3)\tfrac{(G(n+1))_{1,1}}{n+\nu+2}+(n+2+\nu)+{I(n)}_{1,2}(\nu+2n+2)a_{1},\quad
N_{2}(n,i)={I(n)}_{1,2}(\nu+2n+2)\tfrac{(G(n))_{2,2}}{n+\nu+2}.$

  3. ($c$)

If $i>1$ and $n>0$, then

$M_{1}(n,i)\xi(n+1,i-1,n+i)=M_{2}(n,i)\xi(n,i,n+i)+M_{3}(n,i)\xi(n-1,i+1,n+i),$
(6.16) $\quad\text{ with }\quad
M_{1}(n,i)=a_{i-1}\Big{(}i(n+1+\nu+i)_{i-2}-(n+\nu+i)\Big{)},$
$M_{2}(n,i)=(G(n+1))_{i,i}+(n+\nu+i+1),\quad
M_{3}(n,i)=(I(n))_{i,i+1}G(n)_{i+1,i+1}.$

###### Proof.

The first assertion of item ($a$) it follows from item ($a$) of 6.8 (see
Remark 6.9). Indeed, since $K_{n}^{-1}$ has a diagonal of $1$’s, in this case
we have that

$\xi(0,i,i)=(K_{0}^{-1})_{i,i}\tfrac{0!}{(\nu+i+1)_{0}}=(K_{0}^{-1})_{i,i}=1.$

By Proposition 5.1, we have $\mathcal{D}=\partial_{x}x+x(A-1)$ and
$M=(A-1)\delta-(n+1+\nu)-\mathcal{H}(n)J\mathcal{H}^{-1}(n)$ satisfies

$P_{n}\cdot\mathcal{D}=M\cdot P_{n}.$

By recalling that $(P_{n}\cdot\mathcal{D})(0)=0$, we obtain that

$(A-1)P(0,n+1)-(n+1+\nu)P(0,n)-\mathcal{H}_{n}J\mathcal{H}^{-1}_{n}P(0,n)=0,$

since $P(0,m)=K_{m}R(0,m)$, we have that

$K_{n}^{-1}(A-1)K_{n+1}R(0,n+1)-(n+1+\nu)R(0,n)-I(n)R(0,n)=0$ (6.17)

where $I(n)=K_{n}^{-1}\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}K_{n}$.

If we consider the expression (6.17) with $n=0$, we arrive to

$0=K_{0}^{-1}(A-1)K_{1}R(0,1)-K_{0}^{-1}(1+\nu)K_{0}R(0,0)-I(0)R(0,0).$

By taking the $(i,i+1)$-th coordinate, we obtain that

$0=-R(0,1)_{i,i+1}-(1+\nu)R(0,0)_{i,i+1}-I(0)_{i,i}R(0,0)_{i,i+1}-I(0)_{i,i+1}R(0,0)_{i+1,i+1}.$

From Theorem 6.4 we have

$\xi(1,i,i+1)=I(0)_{i,i+1}\xi(0,i+1,i+1)=I(0)_{i,i+1}.$

as asserted.

Now, let us consider $n>0$ in the equation (6.17). By taking into account the
$(i,n+i)$-coordinate in (6.17),

$\sum_{r=1}^{N}{(K_{n}^{-1}(A-1)K_{n+1})}_{i,r}{R(0,n+1)}_{r,n+i}-(n+1+\nu){R(0,n)}_{i,n+i}-\sum_{r=1}^{N}{I(n)}_{i,r}{R(0,n)}_{r,n+i}=0.$

By Lemma (6.7) and since ${R(0,n+1)}_{r,n+i}=0$ for $r<i-1$ we have that

$\sum_{r=i-1}^{N}{(K_{n}^{-1}(A-1)K_{n+1})}_{i,r}{R(0,n+1)}_{r,n+i}-(n+1+\nu+i){R(0,n)}_{i,n+i}-{I(n)}_{i,i+1}{R(0,n)}_{i+1,n+i}=0.$

On the other hand, since the matrix $K_{n}^{-1}(A-1)K_{n+1}$ is lower
triangular we obtain that

$\sum_{r=i-1}^{N}(K_{n}^{-1}(A-1)K_{n+1})_{i,r}{R(0,n+1)}_{r,n+i}=\sum_{r=i-1}^{i}(K_{n}^{-1}(A-1)K_{n+1})_{i,r}{R(0,n+1)}_{r,n+i}.$

By taking into account that
$(K_{n}^{-1}(A-1)K_{n+1})_{i,i-1}=a_{i-1}-(K_{n}^{-1}K_{n+1})_{i,i-1}$ and
$(K_{n}^{-1}(A-1)K_{n+1})_{i,i}=-1$, we obtain that

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\big{(}a_{i-1}-(K_{n}^{-1}K_{n+1})_{i,i-1}\big{)}R(0,n+1)_{i-1,n+i}-R(0,n+1)_{i,n+i}$
$\displaystyle-$
$\displaystyle(n+1+\nu+i){R(0,n)}_{i,n+i}-{I(n)}_{i,i+1}{R(0,n)}_{i+1,n+i}.$

By Corollary 6.6 we have

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\big{(}a_{i-1}-(K_{n}^{-1}K_{n+1})_{i,i-1}\big{)}\xi(n+1,i-1,n+i)-(n+1+\nu+i)\xi(n+1,i,n+i)$
$\displaystyle-$
$\displaystyle(n+1+\nu+i)\xi(n,i,n+i)-{I(n)}_{i,i+1}(n+\nu+i+1)\xi(n,i+1,n+i).$

On the other hand, by Proposition 6.8 if $n+i-j>0$ we know that

$\xi(n,i,j)=a_{i-1}\xi(n,i-1,j)+\tfrac{(G(n))_{i,i}}{n+\nu+i}\xi(n-1,i,j).$

Thus, in particular we obtain

$\xi(n+1,i,n+i)=a_{i-1}\xi(n+1,i-1,n+i)+\tfrac{(G(n+1))_{i,i}}{n+1+\nu+i}\xi(n,i,n+i),$
$\xi(n,i+1,n+i)=a_{i}\xi(n,i,n+i)+\tfrac{(G(n))_{i+1,i+1})}{n+\nu+i+1})\xi(n-1,i+1,n+i)$

and so we have

$\displaystyle\big{(}a_{i-1}-{(K_{n}^{-1}K_{n+1})}_{i,i-1}\big{)}\xi(n+1,i-1,n+i)$
$\displaystyle-(\nu+n+i+1)\Big{(}a_{i-1}\xi(n+1,i-1,n+i)+\tfrac{(G(n+1))_{i,i}}{n+1+\nu+i}\xi(n,i,n+i)\Big{)}$
$\displaystyle=(n+1+\nu+i)\xi(n,i,n+i)$
$\displaystyle+{I(n)}_{i,i+1}(n+1+\nu+i)\Big{(}a_{i}\xi(n,i,n+i)+\tfrac{(G(n))_{i+1,i+1}}{n+\nu+i+1}\xi(n-1,i+1,n+i)\Big{)}.$

Hence

$M_{1}(n,i)\xi(n+1,i-1,n+i)=M_{2}(n,i)\xi(n,i,n+i)+M_{3}(n,i)\xi(n-1,i+1,n+i),$

with $M_{1}(n,i)=-{(K_{n}^{-1}K_{n+1})}_{i,i-1}-a_{i-1}(\nu+n+i)$,
$M_{2}(n,i)=(G(n+1))_{i,i}+(n+\nu+i+1),$ and
$M_{3}(n,i)=(I(n))_{i,i+1}G(n)_{i+1,i+1}.$ To finish the proof, notice that
since $K_{n}^{-1}$ and $K_{n+1}$ are both lower triangular matrix with $1$’s
in the diagonal, we have that

$(K_{n}^{-1}K_{n+1})_{i,i-1}=(K_{n}^{-1})_{i,i-1}+(K_{n+1})_{i,i-1}\quad\text{and}\quad(K_{n}^{-1})_{i,i-1}=-(K_{n})_{i,i-1},$

by definition of $K_{n}$’s we have that

$(K_{n}^{-1}K_{n+1})_{i,i-1}=a_{i-1}\Big{(}(n+\nu+i)_{i-1}-(n+1+\nu+i)_{i-1}\Big{)}=-a_{i-1}i(n+1+\nu+i)_{i-2}.$

Hence,

$M_{1}(n,i)=a_{i-1}\Big{(}i(n+1+\nu+i)_{i-2}-(n+\nu+i)\Big{)},$

as asserted.

For the case $i=1$, $j=n+1$ in the expression (6.17) we obtain

$0=-{R(0,n+1)}_{1,n+1}-(n+2+\nu){R(0,n)}_{1,n+1}-{I(n)}_{1,2}{R(0,n)}_{2,n+1},$

then, from Corollary 6.6 we have

$0=-(\nu+2n+3)\xi(n+1,1,n+1)-(n+2+\nu)\xi(n,1,n+1)-{I(n)}_{1,2}(\nu+2n+2)\xi(n,2,n+1).$
(6.18)

As a consequence of Proposition 6.8 we have that

$\xi(n,i,j)=a_{i-1}\xi(n,i-1,j)+\tfrac{(G(n))_{i,i}}{n+\nu+i}\xi(n-1,i,j)\quad\text{for
$j<n+i$},$

which implies that

$\xi(n,2,n+1)=a_{1}\xi(n,1,n+1)+\tfrac{(G(n))_{2,2}}{n+\nu+2}\xi(n-1,2,n+1),$

and

$\xi(n,1,j)=\tfrac{(G(n))_{1,1}}{n+\nu+1}\xi(n-1,1,j)\quad\text{for $j<n+1$}.$

In particular

$\xi(n+1,1,n+1)=\tfrac{(G(n+1))_{1,1}}{n+\nu+2}\xi(n,1,n+1).$

Thus, we obtain

$N_{1}(n,i)\xi(n,1,n+1)=N_{2}(n,i)\xi(n-1,2,n+1),$

with

$N_{1}(n,i)=(\nu+2n+3)\tfrac{(G(n+1))_{1,1}}{n+\nu+2}+(n+2+\nu)+{I(n)}_{1,2}(\nu+2n+2)a_{1},\quad
N_{2}(n,i)={I(n)}_{1,2}(\nu+2n+2)\tfrac{(G(n))_{2,2}}{n+\nu+2}$

as desired. ∎

As a consequence of Propositions 6.8 and 6.10 we obtain the following result.

######  Theorem 6.11.

Let $n\geq 0$ and let $G(n),I(n)$ as in (6.10). Then, all of the non-zero
entries of $R(x,n)$ can be found in terms of $G(\ell)$ and $I(\ell)$ and the
generalized Laguerre polynomials $L^{(\alpha)}_{\ell}$ for $\ell=0,\ldots,n$.

###### Proof.

The equality (6.6) implies that
$(R(x,n))_{i,j}=L^{(\nu+j)}_{n+i-j}(x)\xi(n,i,j)$ for $n+i-j\geq 0$. It is
enough to show that all of the non-zero constants $\xi(n,i,j)$ can be obtained
in terms of $G(\ell)$ and $I(\ell)$ for $\ell=0,\ldots,n+1$.

By items ($a$)’s of Propositions 6.8 and 6.10 we obtain the values of
$\xi(0,i,j)$. Thus, assume that $n>0$ and suppose that we want to determine
$\xi(n,i,j)$ with $n+i-j>0$. Notice that each time that we use items ($b$) and
($c$) of Proposition 6.8, the value of ”$n+i-j$” it reduces by $1$, so if we
use these items inductively we obtain that $\xi(n,i,j)$ can be determined by
the values of somes $\xi(n^{\prime},i^{\prime},j^{\prime})$’s with
$n^{\prime}+i^{\prime}-j^{\prime}=0$ and in each step also appear $G(\ell)$
with $\ell=0,\ldots,n$. So, it is enough to see that we can determine
$\xi(n,i,j)$ with $n+i-j=0$, in terms of $G(\ell)$ and $I(\ell)$. ∎

######  Remark 6.12.

Notice that by Lemma 6.7 and Propositions 6.8, 6.10, we can replace
$(G(n))_{i,i}$ and $(I(n))_{i,i+1}$ by the expressions
$(\mathcal{H}_{n}(A^{\ast}-1)\mathcal{H}_{n-1}^{-1}))_{i,i}$ and
$(\mathcal{H}_{n}J\mathcal{H}_{n}^{-1})_{i,i+1}$, respectively.

As a direct consequence of the above results and the equations obtained in
Proposition 5.9, we obtain a three-terms non-linear recursion for
$\mathcal{H}_{n}$. With this in mind, we need the following lemma.

######  Lemma 6.13.

Let $A,J\in M_{N}(\mathbb{C})$ be matrices as in (3.1). Then, the matrix
$\mathcal{H}_{1}$ can be obtained from $\mathcal{H}_{0}$.

###### Proof.

By (5.5), we have that

$\mathcal{H}_{1}=\big{(}X(1)+[J,X(1)]\big{)}\mathcal{H}_{0}(A^{*}-1)^{-1}.$

So, it is enough to show that we can obtain $X(1)$ in terms of
$\mathcal{H}_{0}$.

In order to show this, recall that $P(x,1)=x+X(1)$. Let us put
$P_{1}(x):=P(x,1)$, by applying the operator
$D=\partial_{x}^{2}x+\partial_{x}((A-1)x+1+\nu+J)+A\nu+JA-J$ to $P_{1}(x)$, we
have that

$(P_{1}\cdot D)(x)=(A-1)x+1+\nu+J+(x+X(1))(A\nu+JA-J),$

by taking into account that $P_{1}\cdot D=\Gamma_{1}\cdot P_{1}$, where
$\Gamma_{1}$ is the discrete constant operator $A(\nu+2+J)-1-J$. So we obtain
that

$(A-1)x+1+\nu+J+(x+X(1))(A\nu+JA-J)=\big{(}A(\nu+2+J)-1-J\big{)}(x+X(1)).$

After some computation, we obtain that

$X(1)(A\nu+JA-J)+1+\nu+J=\big{(}A(\nu+2+J)-1-J\big{)}X(1),$

by taking into account that $[J,A]=A$, we obtain that $JA=A+AJ$ and hence

$[X(1),A(1+\nu+J)]+X(1)+[J,X(1)]+1+\nu+J-AX(1)=0.$

Thus, by seeing the $(i,j)$-th coordinate in the above equality we obtain the
following recurrences

$a_{j}(\nu+j+1)X(1)_{1,j+1}+(2-j)X(1)_{1,j}+(1+\nu+J)_{1,j}=0,$ (6.19)
$a_{j}(\nu+j+1)X(1)_{i,j+1}-a_{i-1}(\nu+i+1)X(1)_{i-1,j}+(1+i-j)X(1)_{i,j}+(1+\nu+J)_{i,j}=0\quad\text{for
$i\geq 2$}.$ (6.20)

Recall that by Theorem 6.4 we have that $(R(0,1))_{i,j}=0$ if $i+1<j$.

Claim: If $i+1<j$ then $X(1)_{i,j}=0$.

We are going to prove this assertion by induction on $i$. Assume first that
$i=1$ and let us consider $j>2$, in this case

$0=(R(0,1))_{1,j}=(K_{1}P(0,1))_{1,j}=\sum_{r=1}^{N}(K_{1})_{1,r}X(1)_{r,j}=X(1)_{1,j}.$

Now, let $i>1$ and assume that the statement is true for $r<i$, that is
$X(1)_{r,j}=0$ when $r+1<j$. Thus, if $i+1<j$ then

$0=(R(0,1))_{i,j}=(K_{1}P(0,1))_{i,j}=\sum_{r=1}^{N}(K_{1})_{i,r}X(1)_{r,j}=(K_{1})_{i,i}X(1)_{i,j}=X(1)_{i,j}.$

By induction hypothesis and by taking into account that $K_{1}$ is a lower
triangular matrix with $1$’s in its diagonal, we have that

$\sum_{r=1}^{N}(K_{1})_{i,r}X(1)_{r,j}=\sum_{r=1}^{i}(K_{1})_{i,r}X(1)_{r,j}=(K_{1})_{i,i}X(1)_{i,j}=X(1)_{i,j}.$

Therefore, $X(1)_{i,j}=0$ when $i+1<j$ as claimed.

Hence, by equations (6.19) and (6.20), we can observe by a recursive argument,
that in order to compute $X(1)$, it is enough to know the value of
$X(1)_{i,i+1}$ for any $i\geq 1$. By taking into account that

$(R(0,1))_{i,i+1}=(K_{1}P(0,1))_{i,i+1}=\sum_{r=1}^{i}(K_{1})_{i,r}(X(1))_{r,i+1}=X(1)_{i,i+1},$

thus by Corollary 6.6, we obtain that

$X(1)_{i,i+1}=(R(0,1))_{i,i+1}=\xi(1,i,i+1).$

Thus, by item ($a$) of Proposition 6.10, we obtain that

$X(1)_{i,i+1}=\xi(1,i,i+1)=I(0)_{i,i+1}=(\mathcal{H}_{0}J\mathcal{H}_{0}^{-1})_{i,i+1}.$

∎

######  Proposition 6.14.

Let $A,J\in M_{N}(\mathbb{C})$ be matrices as in (3.1). Then,

$\displaystyle\mathcal{H}_{n+2}=(A-1)^{-2}\Big{(}(-[J,\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}]-\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}(A-1)+(A-1)\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1})(A-1)$
$\displaystyle-2-\mathcal{H}_{n+1}J\mathcal{H}_{n+1}^{-1}+\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}+(A-1)[J,\mathcal{H}_{n+1}J\mathcal{H}_{n+1}^{-1}]+(A-1)\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1}(A-1)\Big{)}\mathcal{H}_{n+1}(A^{t}-1)^{-1},$

where $B_{n},C_{n}$ and $\mathcal{H}_{n}$ be as in (5.12).

Moreover,

$(\mathcal{H}_{0})_{i,j}=\Gamma(\nu)\sum_{r=1}^{\min\\{i,j\\}}\frac{\delta_{r}^{(\nu)}}{(i-r)!(j-r)!}\Big{(}\prod_{k=r}^{i-1}a_{k}\Big{)}\Big{(}\prod_{s=r}^{j-1}a_{s}\Big{)}(\nu)_{i+j-r},$

where $\Gamma$ is the Gamma function.

###### Proof.

By (5.13), we have that

$B_{n+1}=(A-1)^{-1}\Big{(}B_{n}(A-1)-2-\mathcal{H}_{n+1}J\mathcal{H}_{n+1}^{-1}+\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}\Big{)}.$
(6.21)

On the other hand, the equation (5.16) implies

$B_{n}=-[J,\mathcal{H}_{n}J\mathcal{H}_{n}^{-1}]-\mathcal{H}_{n}(A^{t}-1)\mathcal{H}_{n-1}^{-1}(A-1)+(A-1)\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1}.$
$B_{n+1}=-[J,\mathcal{H}_{n+1}J\mathcal{H}_{n+1}^{-1}]-\mathcal{H}_{n+1}(A^{t}-1)\mathcal{H}_{n}^{-1}(A-1)+(A-1)\mathcal{H}_{n+2}(A^{t}-1)\mathcal{H}_{n+1}^{-1}$

The statement it follows by changing $B_{n+1}$, and $B_{n}$ in (6.21).

For the last assertion, by definition we have that

$\mathcal{H}_{0}=\int_{0}^{\infty}e^{xA}T^{(\nu)}(x)e^{xA^{\ast}}dx\quad\text{
with }\quad
T^{(\nu)}(x)=e^{-x}\sum_{k=1}^{N}\delta_{k}^{(\nu)}x^{\nu+k}E_{k,k}.$

Then,

$(\mathcal{H}_{0})_{i,j}=\sum_{r=1}^{N}\delta_{r}^{(\nu)}\int_{0}^{\infty}e^{-x}x^{(\nu+r)}(e^{xA})_{i,r}(e^{xA^{\ast}})_{r,j}.$

By taking into account that $A^{\ast}_{r,j}=A_{j,r}$, and
$(e^{xA})_{i,r}=\tfrac{1}{(i-r)!}\big{(}(xA)^{i-r}\big{)}_{i,r}$ if $r<i$ and
$0$ otherwise, we have

$(\mathcal{H}_{0})_{i,j}=\sum_{r=1}^{\min\\{i,j\\}}\delta_{r}^{(\nu)}\int_{0}^{\infty}e^{-x}x^{(\nu+r)}\tfrac{\big{(}(xA)^{i-r}\big{)}_{i,r}}{(i-r)!}\tfrac{\big{(}(xA)^{j-r}\big{)}_{j,r}}{(j-r)!}dx=\sum_{r=1}^{\min\\{i,j\\}}\delta_{r}^{(\nu)}\tfrac{(A^{i-r})_{i,r}}{(i-r)!}\tfrac{(A^{j-r})_{j,r}}{(j-r)!}\int_{0}^{\infty}e^{-x}x^{(\nu+i+j-r)}dx.$

Notice that $(A^{i-r})_{i,r}=\prod_{k=r}^{i-1}a_{k}$, and recall that
$\Gamma(z+1)=\int_{0}^{\infty}e^{-x}x^{z}dx,$ we can write

$(\mathcal{H}_{0})_{i,j}=\sum_{r=1}^{\min\\{i,j\\}}\frac{\delta_{r}^{(\nu)}}{(i-r)!(j-r)!}\Big{(}\prod_{k=r}^{i-1}a_{k}\Big{)}\Big{(}\prod_{s=r}^{j-1}a_{s}\Big{)}\Gamma(\nu+i+j-r+1).$

Taking into account that $\Gamma(z+1)=z\Gamma(z)$ we obtain that

$(\mathcal{H}_{0})_{i,j}=\Gamma(\nu)\sum_{r=1}^{\min\\{i,j\\}}\frac{\delta_{r}^{(\nu)}}{(i-r)!(j-r)!}\Big{(}\prod_{k=r}^{i-1}a_{k}\Big{)}\Big{(}\prod_{s=r}^{j-1}a_{s}\Big{)}(\nu)_{i+j-r},$

as asserted. ∎

######  Remark 6.15.

Notice that we also have the expression

$\mathcal{H}_{0}=(L^{(\nu)}_{\mu}(0))^{-1}\mathcal{H}^{(\nu,\nu)}_{0}(L^{(\nu)}_{\mu}(0))^{\ast},$

with $L^{(\nu)}_{\mu}(0)$ and $\mathcal{H}^{(\nu,\nu)}_{0}$ defined in [21],
where

$\mu=(\mu_{1},\ldots,\mu_{N})\quad\text{such that
}a_{i}=\tfrac{\mu_{i+1}}{\mu_{i}}\quad\text{for $i=1,\ldots,N-1$}.$

This assertion can be deduced from
$W^{\nu}(x)=(L^{(\nu)}_{\mu}(0))^{-1}W^{(\nu,\nu)}(x)(L^{(\nu)}_{\mu}(0))^{\ast}$,
where $W^{(\nu,\nu)}(x)$ are as in [21]. They also proved that
$\mathcal{H}^{(\nu,\nu)}_{0}$ is a diagonal matrix.

The following result gives a recursion for $X(n)$’s in terms of $I(n)$.

######  Proposition 6.16.

For $n\geq 1$, let $I(n)$ as in (6.10). If $\delta_{i,j}$ denotes the
kronecker delta function, then:

  1. ($a$)

$n\delta_{1,j}+X(n)_{1,j+1}a_{j}-X(n)_{1,j}=-(n+1+\nu)\delta_{1,j}-I(n)_{1,j}$.

  2. ($b$)

$n\delta_{i,j}+X(n)_{i,j+1}a_{j}-a_{i-1}X(n+1)_{i-1,j}-X(n)_{i,j}+X(n+1)_{i,j}=-(n+1+\nu)\delta_{i,j}-I(n)_{i,j}$
for $i\geq 2$.

###### Proof.

By Corollary 5.2, we have that

$n+X(n)A-AX(n+1)-B(n)=-(n+1+\nu)-\mathcal{H}(n)J\mathcal{H}^{-1}(n).$

The result is a direct consequense of taking $(i,j)$-th coordinate in the
above equation. ∎

######  Remark 6.17.

  1. ($a$)

By item $(a)$ of Corollary 5.2 we can obtain that $G(n)_{i,i}=X(n)_{i,i}$ for
any $i\geq 1$.

  2. ($b$)

In order to compute $P(x,n)$, for a given $n$, we can do the following:

    1. (a)

Compute $\mathcal{H}_{n}$ and the explicit form of
$\mathcal{H}_{0},\,\mathcal{H}_{1}$ using Lemma 6.13 and Proposition 6.14.

    2. (b)

Compute $G(n)$ and $I(n)$ using $\mathcal{H}_{n}$.

    3. (c)

Compute the $\xi(n,i,j)$ using the recursions of Propositions 6.8 and 6.10.

    4. (d)

Then we have $R(x,n)$ and so $P(x,n)=K_{n}R(x,n)e^{-xA}$.

##  7\. Matrix entries of $P(x,n)$ in terms of Laguerre and dual Hahn
polynomials.

In this section, we will show that under some hypothesis, the $\xi(n,i,j)$’s
can be expressed in terms of dual Hahn polynomials.

###  Some technical lemmas

Let $N\geq 1$ be a fixed integer and let $\mu=(\mu_{1},\ldots,\mu_{N})$ be a
sequence of non-zero coefficients and $\alpha>0$. Then $L_{\mu}^{(\alpha)}$ is
the $N\times N$ unipotent lower triangular matrix defined by

$L^{(\alpha)}_{\mu}(x)_{m,n}=\begin{cases}\frac{\mu_{m}}{\mu_{n}}L^{(\alpha+n)}_{m-n}(x),&m\geq
n,\\\ 0&n<m.\end{cases}$ (7.1)

For $\nu>0$ we consider the weight matrix

$W^{(\alpha,\nu)}_{\mu}(x)=L_{\mu}^{(\alpha)}(x)\,T^{(\nu)}(x)\,L_{\mu}^{(\alpha)}(x)^{\ast},\qquad
T^{(\nu)}(x)=e^{-x}\sum_{k=1}^{N}x^{\nu+k}\delta_{k}^{(\nu)}\,E_{k,k}.$ (7.2)

It can be showed that

$W_{\mu}^{(\alpha,\nu)}(x)=L^{(\alpha)}_{\mu}(0)e^{xA_{\mu}}T^{(\nu)}(x)e^{xA_{\mu}^{\ast}}L^{(\alpha)}_{\mu}(0)^{\ast}\quad\text{where}\quad
A_{\mu}=-\sum_{k=1}^{N-1}\tfrac{\mu_{k+1}}{\mu_{k}}E_{k+1,k}.$ (7.3)

We impose conditions on the sequence $\\{\mu_{i}\\}_{i=1}^{N}$ and the
coefficients $\delta_{k}^{(\nu)}$. First of all, we assume that the
coefficients $\mu_{i}$ are real and non-zero for all $i$ and
$\delta_{k}^{(\nu)}>0$, $1\leq k\leq N$, so that the weight matrix is positive
definite (see [21] for more information about the weight matrix
$W^{(\alpha,\nu)}_{\mu}$). On the other hand, we consider the diagonal matrix
$\Delta^{(\nu)}=\mathrm{diag}(\delta_{1}^{(\nu)},\ldots,\delta_{N}^{(\nu)})$,
so that $(T^{(\nu)})_{k,k}=e^{-x}x^{\nu+k}(\Delta^{(\nu)})_{k,k}$. We assume
that there exist coefficients $c^{(\nu)}$ and $d^{(\nu)}$ such that

$\Delta^{(\nu+1)}=(d^{(\nu)}J+c^{(\nu)})\,\Delta^{(\nu)}.$ (7.4)

We also assume that the coefficients $\mu_{k}$ and $\delta^{(\nu)}_{k}$
satisfy the relation

$\tfrac{\mu_{k+1}^{2}}{\mu_{k}^{2}}=d^{(\nu)}k(N-k)\tfrac{\delta_{k+1}^{(\nu)}}{\delta_{k}^{(\nu+1)}},\qquad
k=1,\ldots,N-1.$ (7.5)

Under the above conditions, Propositions 5.1 and 5.2 from [21] say that

$\Phi^{(\alpha,\nu)}(x)=(W_{\mu}^{(\alpha,\nu)}(x))^{-1}W_{\mu}^{(\alpha,\nu+1)}(x)\quad\text{and}\quad\Psi^{(\alpha,\nu)}(x)=(W_{\mu}^{(\alpha,\nu)}(x))^{-1}\tfrac{dW_{\mu}^{(\alpha,\nu+1)}}{dx}(x)$
(7.6)

are matrix polynomials of degree $2$ and $1$ respectively. Moreover, Corollary
6.3 from [21] asserts that the operator $D_{2}$ defined by

$D_{2}(x)=\dfrac{d^{2}}{dx^{2}}\Phi^{\ast}(x)+\dfrac{d}{dx}\Psi^{\ast}(x)$
(7.7)

is symmetric respect to $W_{\mu}^{(\alpha,\nu)}$.

We begin with the following technical lemma which relates the matrix
polynomials $R_{n}(x)$ with the constants $c^{(\nu)}$ and $d^{(\nu)}$. The
proof of Lemma 7.1 and Lemma 7.3 can be found in the appendix.

######  Lemma 7.1.

Let $\mu=(\mu_{1},\ldots,\mu_{N})$ and $\delta^{(\nu)}_{k}>0$ for $1\leq k\leq
N$, satisfying (7.4) and (7.5). Let $A:=A_{\mu}$ as in (7.3). If $R(x,n)$ are
the matrix polynomials defined in (6.5), then

$(R^{\prime}(0,n)-R(0,n)A)C^{(\nu)}=D^{(\nu)}R(0,n)$ (7.8)

where
$C^{(\nu)}=(d^{(\nu)}J+c^{(\nu)})(\nu+J+1)+((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}$
and $D^{(\nu)}=n(d^{(\nu)}(J-N-1)-c^{(\nu)})$ with $c^{(\nu)}$ and $d^{(\nu)}$
as in (7.4).

In the sequel, let $0\leq n$, $1\leq i\leq N$. We consider the sequence
$\epsilon_{j}:=\epsilon_{j}^{(n,i)}$ defined recursively by

$\epsilon_{0}=1,\quad\quad\epsilon_{j}=(n+i-j+1)d^{(\nu)}\Big{(}(j-1)+\tfrac{c^{(\nu)}}{d^{(\nu)}}\Big{)}\epsilon_{j-1}\quad\text{for
$n+i-j\geq 0$},$ (7.9)

with $c^{(\nu)},d^{(\nu)}$ defined as in (7.4). We have the following result.

######  Lemma 7.2.

Let $0\leq n$ and let $1\leq N$, $1\leq i\leq N$ integers and let
$\\{\epsilon_{l}\\}_{l=0}^{i+n}$ be the sequence defined as in (7.9). If
$\nu>0$, $\delta^{(\nu)}$ satisfies (7.4) and $0\leq j<n+i$, we have

$\tfrac{\epsilon_{j}}{\epsilon_{j+1}}M_{j}=1\qquad\text{where}\qquad
M_{j}=(n+i-j)(d^{(\nu)}j+c^{(\nu)})$

with $c^{(\nu)},d^{(\nu)}$ as in (7.4).

###### Proof.

It follows by a simple inductive argument from the definition of
$\epsilon_{j}$. ∎

Given $i=1,\ldots,N$, $n\geq 0$ and $j$ a positive integer such that
$n+i-j\geq 0$, in the sequel we consider the sequence $q_{j}:=q_{j}^{(n,i)}$
by the expression

$q_{j}^{(n,i)}:=\epsilon_{j}^{(n,i)}\xi(n,i,j),$ (7.10)

where $\epsilon_{j}$’s are as in (7.9).

######  Lemma 7.3.

Let $\mu=(\mu_{1},\ldots,\mu_{N})$ and $\delta^{(\nu)}_{k}>0$ for $1\leq k\leq
N$, satisfying (7.4) and (7.5). For $1\leq i\leq N$ and $0\leq n$, let
$\\{q_{l}\\}_{l=0}^{i+n}$ be the sequence as in (7.10). Then, the sequence
$\\{q_{l}\\}_{l=0}^{i+n}$ satisfies

$E_{j}q_{j}+F_{j}q_{j-1}+\tfrac{1}{d^{(\nu)}}q_{j+1}=0,\quad n+i-j>0,$

where

$\displaystyle E_{j}$
$\displaystyle=(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n((i-N-1)-\tfrac{c^{(\nu)}}{d^{(\nu)}}),$
$\displaystyle F_{j}$
$\displaystyle=(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}(n+i-j+1)\big{(}d^{(\nu)}(j-1)+c^{(\nu)}\big{)}$

with $c^{(\nu)},d^{(\nu)}$ as in (7.4).

### Dual Hahn polynomials

For $0\leq n$, $1\leq i\leq N$ let us consider the following sequence

$\widetilde{q_{j}}^{(n,i)}:={(d^{(\nu)})}^{-j}q_{j}^{(n,i)},\quad n+i-j>0,$
(7.11)

where $q_{j}$’s are as in (7.10) and $d^{(\nu)}$ as (7.4).

######  Lemma 7.4.

Let $\mu=(\mu_{1},\ldots,\mu_{N})$ and $0<\delta^{(\nu)}_{k}$, for $1\leq
k\leq N$, satisfying (7.4) and (7.5). Let $\widetilde{q_{l}}$ be as in (7.11).
If $\mu_{j}=1$ for all $j$ then the $\widetilde{q_{l}}$’s satisfy the
following equation

$\widetilde{E}_{j}\widetilde{q_{j}}+\widetilde{F}_{j}\widetilde{q}_{j-1}+\widetilde{q}_{j+1}=0,\quad
n+i-j>0,$

where

$\displaystyle\widetilde{E}_{j}$
$\displaystyle=(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n(i-N-1-\tfrac{c^{(\nu)}}{d^{(\nu)}}),$
$\displaystyle\widetilde{F}_{j}$
$\displaystyle=(j-1)(N-j+1)(n+i-j+1)\Big{(}j-1+\tfrac{c^{(\nu)}}{d^{(\nu)}}\Big{)}$

with $c^{(\nu)},d^{(\nu)}$ as in (7.4).

###### Proof.

By Lemma 7.3, we have that
$E_{j}q_{j}+F_{j}q_{j-1}+\tfrac{1}{d^{(\nu)}}q_{j+1}=0$ with

$\displaystyle E_{j}$
$\displaystyle=(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n((i-N-1)-\tfrac{c^{(\nu)}}{d^{(\nu)}}),$
$\displaystyle F_{j}$
$\displaystyle=(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}(n+i-j+1)\big{(}d^{(\nu)}(j-1)+c^{(\nu)}\big{)}.$

Now, since $\mu_{j}=1$ for all $j$ and
$\widetilde{q_{j}}(x):={(d^{(\nu)})}^{-j}q_{j}(x)$, we have that

$\displaystyle\Big{(}(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n((i-N-1)-\tfrac{c^{(\nu)}}{d^{(\nu)}})\Big{)}\widetilde{q_{j}}{(d^{(\nu)})}^{j}$
$\displaystyle+$
$\displaystyle(j-1)(N-j+1)(n+i-j+1)\big{(}d^{(\nu)}(j-1)+c^{(\nu)}\big{)}{(d^{(\nu)})}^{j-1}\widetilde{q}_{j-1}+\tfrac{1}{d^{(\nu)}}{(d^{(\nu)})}^{j+1}\widetilde{q}_{j+1}=0.$

By multiplication for ${(d^{(\nu)})}^{-j}$, we obtain that

$\widetilde{E}_{j}\widetilde{q_{j}}+\widetilde{F}_{j}\widetilde{q}_{j-1}+\widetilde{q}_{j+1}=0,$

with

$\displaystyle\widetilde{E}_{j}$
$\displaystyle=(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n(i-N-1-\tfrac{c^{(\nu)}}{d^{(\nu)}}),$
$\displaystyle\widetilde{F}_{j}$
$\displaystyle=(j-1)(N-j+1)(n+i-j+1)\Big{(}j-1+\tfrac{c^{(\nu)}}{d^{(\nu)}}\Big{)},$

as asserted. ∎

Recall that if the sequence of polynomials $\\{s_{k}\\}$ satisfies the
normalized recurrence relations

$xs_{k}(x)=s_{k+1}(x)-(u_{k}+v_{k})s_{k}(x)+u_{k-1}v_{k}s_{k-1}(x),$ (7.12)

with

$u_{k}=(k+\gamma+1)(k-M),\qquad v_{k}=k(k-\delta-M-1),$

then $s_{k}$ satisfies

$T_{k}(\lambda(x);\gamma,\delta,M)=\tfrac{1}{(\gamma+1)_{k}(-M)_{k}}s_{k}(\lambda(x))$

where $\lambda(x)=x(x+\gamma+\delta+1)$ and $\\{T_{k}\\}$ is the sequence of
dual Hahn polynomials defined by

$T_{k}(\lambda(x);\gamma,\delta,M)={}_{3}F_{2}\left(\begin{matrix}-k,&&-x,&&x+\gamma+\delta+1\\\\[1.00006pt]
&&\gamma+1,&&-M&\end{matrix};1\right)\qquad\text{for }k=0,1,\ldots,M.$

######  Proposition 7.5.

Let $\mu=(\mu_{1},\ldots,\mu_{N})$ and $\delta^{(\nu)}_{k}>0$ for $1\leq k\leq
N$, satisfying (7.4) and (7.5). Let $\widetilde{q_{j}}$ be as in (7.11). If
$\mu_{\ell}=1$ for all $\ell=1,\ldots,N$, then the $\widetilde{q_{j}}$’s
satisfy

$\widetilde{q}_{j}=(\gamma+1)_{j-1}(-(N-1))_{j-1}T_{j-1}(\lambda(x^{(n,i)});\gamma,\delta,N-1)\quad
n+i-j>0$ (7.13)

where $\gamma=\tfrac{c^{(\nu)}}{d^{(\nu)}}$, $\delta=n+i-N$ and
$x^{(n,i)}=(\gamma+1)(N+i-2)-n(N-i)$.

###### Proof.

By Lemma 7.4, we have that

$\widetilde{E}_{j}\widetilde{q_{j}}+\widetilde{F}_{j}\widetilde{q}_{j-1}+\widetilde{q}_{j+1}=0,\quad
n+i-j>0,$

where

$\displaystyle\widetilde{E}_{j}$
$\displaystyle=(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n(i-N-1-\tfrac{c^{(\nu)}}{d^{(\nu)}}),$
$\displaystyle\widetilde{F}_{j}$
$\displaystyle=(j-1)(N-j+1)(n+i-j+1)\Big{(}j-1+\tfrac{c^{(\nu)}}{d^{(\nu)}}\Big{)}$

with $c^{(\nu)},d^{(\nu)}$ as in (7.4). Now, if we consider

$u_{k}=(k+\gamma+1)(k-M)\text{ and }v_{k}=k(k-\delta-M-1),$

with $\gamma=\tfrac{c^{(\nu)}}{d^{(\nu)}}$, $\delta=n+i-N$ and $M=N-1$. It is
straightforward to check that

$\widetilde{F}_{j}=u_{j-2}v_{j-1}\quad\text{and}\quad\widetilde{E}_{j}=-(u_{j-1}+v_{j-1}+x^{(n,i)})$

where $x^{(n,i)}=(\gamma+1)(N+i-2)-n(N-i)$. Thus, since $\widetilde{q_{j}}$
can be seen as constants polynomial, we have that

$x^{(n,i)}\widetilde{q_{j}}(x^{(n,i)})=\widetilde{q}_{j+1}(x^{(n,i)})-(u_{j-1}+v_{j-1})\widetilde{q}_{j}(x^{(n,i)})+u_{j-2}v_{j-1}\widetilde{q}_{j-1}(x^{(n,i)}).$

Therefore, by definition of dual Hahn polynomials and by taking into account
that $\widetilde{q}_{j}$ is a constant polynomial, we obtain (7.13) as
desired. ∎

By recalling that
$\widetilde{q}_{j}=(d^{(\nu)})^{-j}q_{j}=\epsilon_{j}(d^{(\nu)})^{-j}\xi(n,i,j)$
for $n+i-j>0$, we obtain the following result

######  Theorem 7.6.

Let $\mu=(\mu_{1},\ldots,\mu_{N})$ and $\delta^{(\nu)}_{k}>0$, for $1\leq
k\leq N$, satisfying (7.4) and (7.5). Let $\epsilon_{j}$ as in (7.9). If
$\mu_{\ell}=1$ for all $\ell=1,\ldots,N$, then the constants $\xi(n,i,j)$’s
satisfy

$\xi(n,i,j)=\tfrac{(d^{(\nu)})^{j}(\gamma+1)_{j-1}(-(N-1))_{j-1}}{\epsilon_{j}}T_{j-1}(\lambda(x^{(n,i)});\gamma,\delta,N-1),\quad
n+i-j>0$ (7.14)

where $\gamma=\tfrac{c^{(\nu)}}{d^{(\nu)}}$, $\delta=n+i-N$ and
$x^{(n,i)}=(\gamma+1)(N+i-2)-n(N-i)$. Moreover, if $n+i=j$, the constants
$\xi(n,i,j)$’s satisfy the recursion

$\xi(0,1,1)=\tfrac{1}{\nu+2},\qquad\xi(n,i,j-1)=\Big{(}\tfrac{n(N+1-i)(i-1)}{(j-1)(N-j+1)}+1\Big{)}\xi(n,i,j),\quad
j>1,$ (7.15)

###### Proof.

If $n+i-j>0$, by Proposition 7.5 we have that

$\xi(n,i,j)=\tfrac{(d^{(\nu)})^{j}}{\epsilon_{j}}\widetilde{q}_{j}=\tfrac{(d^{(\nu)})^{j}(\gamma+1)_{j-1}(-(N-1))_{j-1}}{\epsilon_{j}}T_{j-1}(\lambda(x^{(n,i)});\gamma,\delta,N-1)$

with $\gamma=\tfrac{c^{(\nu)}}{d^{(\nu)}}$, $\delta=n+i-N$ and
$x^{(n,i)}=(\gamma+1)(N+i-2)-n(N-i)$ as asserted.

Now, if we take $j=n+i$ in the expression (7.8), we obtain

$\displaystyle
R_{n}^{\prime}(0)_{i,j}(d^{(\nu)}j+c^{(\nu)})(\nu+j+1)+{R_{n}^{\prime}(0)}_{i(j-1)}{((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}}_{(j-1)j}$
$\displaystyle-({R_{n}(0)}_{i,j}A_{j(j-1)}{((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}}_{(j-1),j}=n(d^{(\nu)}(J-N-1)-c^{(\nu)})R(n,0)_{i,j}.$

By taking into account that $R_{n}(x)_{i,j}=L^{(\nu+j)}_{n+i-j}(x)\xi(n,i,j)$,

$L^{\alpha}_{n}(0)=\tfrac{(\alpha+1)_{n}}{n!}\qquad\text{and}\qquad\tfrac{\partial}{\partial
x}L^{\alpha}_{n}(x)=(-1)L^{\alpha+1}_{n-1}(x),$

we have

$\displaystyle-\xi(n,i,j-1){((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}}_{(j-1)j}$
$\displaystyle-$
$\displaystyle(\xi(n,i,j)A_{j(j-1)}{((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}}_{(j-1),j}$
$\displaystyle=$ $\displaystyle n(d^{(\nu)}(i-N-1)-c^{(\nu)})\xi(n,i,j).$

Recall that by (7.5) we have that

${{({(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu+1)}\big{)}}^{\ast}}_{(j-1,j)}=d^{(\nu)}(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}},$

then, taking in account that $\mu_{k}=1$ for all $k$, we have

$\displaystyle-\xi(n,i,j-1)d^{(\nu)}(j-1)(N-j+1)$ $\displaystyle-$
$\displaystyle(\xi(n,i,j)A_{j(j-1)}d^{(\nu)}(j-1)(N-j+1)$ $\displaystyle=$
$\displaystyle n(d^{(\nu)}(i-N-1)-c^{(\nu)})\xi(n,i,j).$

Thus, using the conditions (7.4), (7.5), and taking in account that
$A_{s,s-1}=-1$, we obtain

$-d^{(\nu)}(j-1)(N-j+1)\xi(n,i,j-1)=\Big{(}-nd^{(\nu)}(N+1-i)(i-1)-d^{(\nu)}(j-1)(N-j+1)\Big{)}\xi(n,i,j),$

as desired.

The last assertion it follows from

$\xi(0,1,1)(\nu+2)=\xi(0,1,1)L^{1}_{1}(0)=R(0,0)_{1,1}=(K_{1}P(0,0))_{1,1}=1$

∎

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## Appendix A

In this apprendix we give the proofs of the Lemmas 7.1 and 7.3.

###### Proof of Lemma 7.1.

Let $P_{n}(x)=P(x,n)$ be the sequence of monic orthogonal polynomials respect
to the weight $W^{(\nu)}$ as in (5.1). Since
$W_{\mu}^{(\alpha,\nu)}(x)=L_{\mu}^{(\alpha)}(0)W^{(\nu)}(x)L_{\mu}^{(\alpha)}(0)^{*}$
we have that

$\int_{0}^{\infty}P_{n}(x)(L_{\mu}^{(\alpha)}(0)^{-1})W_{\mu}^{(\alpha,\nu)}(x)(P_{m}(x)L_{\mu}^{(\alpha)}(0)^{-1})^{*}dx=\int_{0}^{\infty}P_{n}(x)W^{(\nu)}(x)P_{m}(x)^{*}dx=\delta_{n,m}\mathcal{H}_{n}.$

Hence, if
$P_{n}^{(\alpha,\nu)}(x)=L_{\mu}^{(\alpha)}(0)P_{n}(x)L_{\mu}^{(\alpha)}(0)^{-1}$
then $P^{(\alpha,\nu)}_{n}(x)$ is the sequence of monic orthogonal polynomials
with respect to the weight $W_{\mu}^{(\alpha,\nu)}$. Now, if $D_{2}$ is the
operator defined in (7.7), the Corollary 6.3 of [21] implies that
$P^{(\alpha,\nu)}_{n}D_{2}=n{\hat{K}_{n}}^{(\alpha,\nu)}P^{(\alpha,\nu)}_{n}$
for certain matrix ${\hat{K}_{n}}^{(\alpha,\nu)}$ defined recursively in
section 6 from [21], and so

$P_{n}L_{\mu}^{(\alpha)}(0)^{-1}D_{2}L_{\mu}^{(\alpha)}(0)=nL_{\mu}^{(\alpha)}(0)^{-1}{\hat{K}_{n}}^{(\alpha,\nu)}L_{\mu}^{(\alpha)}(0)P_{n}.$
(A.1)

By Proposition 6.1 of [21], we obtain that

${(L_{\mu}^{(\alpha)}(0))}^{-1}{\hat{K}_{n}}^{(\alpha,\nu)}L_{\mu}^{(\alpha)}(0)=d^{(\nu)}(J-(J+\nu+n)A-N-1)-c^{(\nu)},$

and by taking into account that $JA-AJ=A$, we have that

$J-JA-(\nu+n)A-N-1=J-AJ-A-(\nu+n)A-N-1.$

On the other hand, since $\Gamma_{n}=A(n+\nu+J+1)-(n+J)=-J+(n+\nu)A+A+AJ-n$,
then

${(L_{\mu}^{(\alpha)}(0)})^{-1}{\hat{K}_{n}}^{(\alpha,\nu)}L_{\mu}^{(\alpha)}(0)=d^{(\nu)}(-\Gamma_{n}-n-N-1)-c^{(\nu)}.$

By (A.1), we have that

$K_{n}R_{n}e^{-xA}{(L_{\mu}^{(\alpha)}(0))}^{-1}D_{2}L_{\mu}^{(\alpha)}(0)=n{(L_{\mu}^{(\alpha)}(0))}^{-1}{\hat{K}_{n}}^{(\alpha,\nu)}L_{\mu}^{(\alpha)}(0)K_{n}R_{n}e^{-xA}$

and thus

$R_{n}e^{-xA}{(L_{\mu}^{(\alpha)}(0))}^{-1}D_{2}L_{\mu}^{(\alpha)}(0)e^{xA}=nK_{n}^{-1}(d^{(\nu)}(-\Gamma_{n}-n-N-1)-c^{(\nu)})K_{n}R_{n}.$

By recalling that $K_{n}^{-1}\Gamma_{n}K_{n}=-n-J$, we have that

$R_{n}e^{-xA}{(L_{\mu}^{(\alpha)}(0))}^{-1}D_{2}L_{\mu}^{(\alpha)}(0)e^{xA}=n(d^{(\nu)}(J-N-1)-c^{(\nu)})R_{n}.$
(A.2)

Now, by taking into account that
$D_{2}=\dfrac{d^{2}}{dx^{2}}\Phi^{\ast}(x)+\dfrac{d}{dx}\Psi^{\ast}(x)$ with
$\Phi,\Psi$ polynomials of degree $2$ and $1$. In general, if $U(x)$ is a
polynomial, we have that

$\displaystyle U(x)\cdot
e^{-xA}{(L_{\mu}^{(\alpha)}(0))}^{-1}D_{2}L_{\mu}^{(\alpha)}(0)e^{xA}$
$\displaystyle=$
$\displaystyle\dfrac{d^{2}}{dx^{2}}\Big{(}U(x)e^{-xA}\Big{)}{(L_{\mu}^{(\alpha)}(0))}^{-1}\Phi^{\ast}(x)L_{\mu}^{(\alpha)}(0)e^{xA}$
$\displaystyle+\dfrac{d}{dx}\Big{(}U(x)e^{-xA}\Big{)}{(L_{\mu}^{(\alpha)}(0))}^{-1}\Psi^{\ast}(x)L_{\mu}^{(\alpha)}(0)e^{xA}$
$\displaystyle=$
$\displaystyle\Big{(}U^{\prime\prime}(x)-2U^{\prime}(x)A+U(x)A^{2}\Big{)}e^{-xA}{(L_{\mu}^{(\alpha)}0))}^{-1}\Phi^{\ast}(x)L_{\mu}^{(\alpha)}(0)e^{xA}$
$\displaystyle+\Big{(}U^{\prime}(x)-U(x)A\Big{)}e^{-xA}{(L_{\mu}^{(\alpha)}(0))}^{-1}\Psi^{\ast}(x)L_{\mu}^{(\alpha)}(0)e^{xA}.$

Hence, we want to find some easy expression for

$e^{-xA}L(0)^{-1}\Phi^{\ast}(x)L(0)e^{xA}\quad\text{and}\quad
e^{-xA}L(0)^{-1}\Psi^{\ast}(x)L(0)e^{xA}.$

Some similar expressions was studied by Koelink and Roman (see [21]). By
Corollary 5.3 in [21], we have that

$\displaystyle
L_{\mu}^{(\alpha)}(0)^{\ast}\Phi(x)((L_{\mu}^{(\alpha)}(0))^{\ast})^{-1}$
$\displaystyle=$
$\displaystyle-d^{(\nu)}x^{2}A^{\ast}+x(d^{(\nu)}J+c^{(\nu)})$ $\displaystyle
L_{\mu}^{(\alpha)}(0)^{\ast}\Psi(x)((L_{\mu}^{(\alpha)}(0))^{\ast})^{-1}$
$\displaystyle=$ $\displaystyle
x\Big{(}d^{(\nu)}(J-A^{\ast}(J+\nu+1)-N-1)-c^{(\nu)})\Big{)}$
$\displaystyle+(\nu+J+1)(d^{(\nu)}J+c^{(\nu)})+(\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)}.$

Hence, we have that

$\displaystyle
e^{-xA}L_{\mu}^{(\alpha)}(0)^{-1}\Phi^{\ast}(x)L_{\mu}^{(\alpha)}(0)e^{xA}$
$\displaystyle=$
$\displaystyle-d^{(\nu)}x^{2}e^{-xA}Ae^{xA}+xd^{(\nu)}e^{-xA}Je^{xA}+xc^{(\nu)}$
$\displaystyle=$ $\displaystyle-d^{(\nu)}x^{2}A+xd^{(\nu)}(xA+J)+xc^{(\nu)}$
$\displaystyle=$ $\displaystyle x(d^{(\nu)}J+c^{(\nu)}).$ $\displaystyle
e^{-xA}L_{\mu}^{(\alpha)}(0)^{-1}\Psi^{\ast}(x)L_{\mu}^{(\alpha)}(0)e^{xA}$
$\displaystyle=$ $\displaystyle
x\Big{(}d^{(\nu)}(xA+J-(xA+J+\nu+1)A-N-1)-c^{(\nu)})\Big{)}$
$\displaystyle+(d^{(\nu)}(xA+J)+c^{(\nu)})(\nu+xA+J+1)+e^{-xA}((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}e^{xA}.$

By evaluation in $x=0$ in (A.2) and by taking into account the above
expressions, we obtain (7.8), as desired. ∎

###### Proof of Lemma 7.3.

By taking into account that $(R_{n}(x))_{i,j}$ satisfies the expression given
by Theorem 6.4, the $(i,j)$-coordinate of the matrix in the right hand of
(7.8) is

$n(d^{(\nu)}(i-N-1)-c^{(\nu)}){(R_{n}(0))}_{i,j}=n(d^{(\nu)}(i-N-1)-c^{(\nu)})L^{(\nu+j)}_{n+i-j}(0)\xi(n,i,j).$

On the other hand, for $n+i-j>0$ the $(i,j)$-coordinate of the matrix in the
left hand of (7.8) is

$\displaystyle
R_{n}^{\prime}(0)_{i,j}(d^{(\nu)}j+c^{(\nu)})(\nu+j+1)+{R_{n}^{\prime}(0)}_{i(j-1)}{((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}}_{(j-1)j}$
$\displaystyle-{R_{n}(0)}_{i(j+1)}(d^{(\nu)}j+c^{(\nu)})(\nu+j+1)A_{(j+1)j}-({R_{n}(0)}_{i,j}A_{j(j-1)}{((\Delta^{(\nu)})^{-1}A\Delta^{(\nu+1)})^{\ast}}_{(j-1),j}.$

By taking into account that $R_{n}(x)_{i,j}=L^{(\nu+j)}_{n+i-j}(x)\xi(n,i,j)$,

$L^{\alpha}_{n}(0)=\tfrac{(\alpha+1)_{n}}{n!}\qquad\text{and}\qquad\tfrac{\partial}{\partial
x}L^{\alpha}_{n}(x)=(-1)L^{\alpha+1}_{n-1}(x),$

we obtain that the above expression is equivalent to

$\displaystyle-n\Big{(}d^{(\nu)}(i-N-1)-c^{(\nu)}\Big{)}\tfrac{{(\nu+j+1)}_{n+i-j}}{(n+i-j)!}\xi(n,i,j)$
$\displaystyle=$
$\displaystyle\tfrac{{(\nu+j+2)}_{n+i-j-1}}{(n+i-j-1)!}(d^{(\nu)}j+c^{(\nu)})(\nu+j+1)\xi(n,i,j)$
$\displaystyle+\tfrac{{(\nu+j+1)}_{n+i-j}}{(n+i-j)!}{{\big{(}{(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu+1)}\big{)}}^{\ast}}_{(j-1,j)}\xi(n,i,j-1)$
$\displaystyle+\tfrac{{(\nu+j+2)}_{n+i-j-1}}{(n+i-j-1)!}(d^{(\nu)}j+c^{(\nu)})(\nu+j+1)\xi(n,i,j+1)$
$\displaystyle+\tfrac{{(\nu+j+1)}_{(n+i-j)}}{(n+i-j)!}A_{j(j-1)}{{\Big{(}{(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu)}\Big{)}}^{\ast}}_{(j-1)j}\xi(n,i,j).$

Thus, by taking into account that
$(\nu+j+2)_{n+i-j-1}(\nu+j+1)=(\nu+j+1)_{n+i-j-1}$, and multiplying both sides
by $\tfrac{(n+i-j-1)!}{(\nu+j+1)_{n+i-j-1}}$, we obtain

$\displaystyle-n\Big{(}d^{(\nu)}(i-N-1)-c^{(\nu)}\Big{)}\tfrac{(n+i-j-1)}{n+i-j}\xi(n,i,j)$
$\displaystyle=$ $\displaystyle(d^{(\nu)}j+c^{(\nu)})\xi(n,i,j)$
$\displaystyle+\tfrac{(n+i-j-1)}{n+i-j}{{\big{(}{(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu+1)}\big{)}}^{\ast}}_{(j-1,j)}\xi(n,i,j-1)$
$\displaystyle+(d^{(\nu)}j+c^{(\nu)})\xi(n,i,j+1)$
$\displaystyle+\tfrac{(n+i-j-1)}{n+i-j}A_{j(j-1)}{{\Big{(}{(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu)}\Big{)}}^{\ast}}_{(j-1)j}\xi(n,i,j).$

Finally, we obtain that

$\displaystyle\Big{(}(n+i-j)(d^{(\nu)}j+c^{(\nu)})+A_{j(j-1)}{{\big{(}{(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu)}\big{)}}^{\ast}}_{(j-1)j}+n(d^{(\nu)}(i-N-1)-c^{(\nu)})\Big{)}\xi(n,i,j)$
$\displaystyle+{{\big{(}{(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu+1)}\big{)}}^{\ast}}_{(j-1,j)}\xi(n,i,j-1)+(n+i-j)(d^{(\nu)}j+c^{(\nu)})\xi(n,i,j+1)=0.$

By (7.5) we have that
$\tfrac{{\delta^{(\nu)}_{k+1}}}{{\delta^{(\nu+1)}_{k}}}=\tfrac{\mu_{k+1}^{2}}{d^{(\nu)}k(N-k)\mu_{k}^{2}}$,
and then

$\displaystyle{{({(\Delta^{(\nu)})}^{-1}A\Delta^{(\nu+1)}\big{)}}^{\ast}}_{(j-1,j)}$
$\displaystyle=$
$\displaystyle\tfrac{{\delta^{(\nu+1)}_{j-1}}}{\delta^{(\nu)}_{j}}A_{j(j-1)}$
$\displaystyle=$ $\displaystyle
d^{(\nu)}(j-1)(N-j+1)\tfrac{{\mu_{j-1}}^{2}}{{\mu_{j}}^{2}}\tfrac{\mu_{j}}{\mu_{j-1}}$
$\displaystyle=$ $\displaystyle
d^{(\nu)}(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}.$

Thus, we can rewrite the above equation as follows

$\displaystyle\Big{(}(n+i-j)(d^{(\nu)}j+c^{(\nu)})+d^{(\nu)}(j-1)(N-j+1)+n(d^{(\nu)}(i-N-1)-c^{(\nu)})\Big{)}\xi(n,i,j)$
$\displaystyle+d^{(\nu)}(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}\xi(n,i,j-1)+(n+i-j)(d^{(\nu)}j+c^{(\nu)})\xi(n,i,j+1)=0.$

Now, if we consider $M_{j}=(n+i-j)(d^{(\nu)}j+c^{(\nu)})$. Since
$\epsilon^{(n,i)}_{j}\xi(n,i,j)=q_{j}^{(n,i)}$, we can rewrite the above
equation as follows

$\displaystyle\Big{(}(n+i-j)(d^{(\nu)}j+c^{(\nu)})+d^{(\nu)}(j-1)(N-j+1)+n(d^{(\nu)}(i-N-1)-c^{(\nu)})\Big{)}\tfrac{q_{j}}{\epsilon_{j}}$
$\displaystyle+d^{(\nu)}(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}\tfrac{q_{j-1}}{\epsilon_{j-1}}+M_{j}\tfrac{q_{j+1}}{\epsilon_{j+1}}=0.$

If we multiply by $\epsilon_{j}$, we have that

$\displaystyle\Big{(}(n+i-j)(d^{(\nu)}j+c^{(\nu)})+d^{(\nu)}(j-1)(N-j+1)+n(d^{(\nu)}(i-N-1)-c^{(\nu)})\Big{)}q_{j}$
$\displaystyle+d^{(\nu)}(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}\tfrac{\epsilon_{j}}{\epsilon_{j-1}}q_{j-1}+M_{j}\tfrac{\epsilon_{j}}{\epsilon_{j+1}}q_{j+1}=0.$

By Lemma 7.2, we have that $M_{j}\tfrac{\epsilon_{j}}{\epsilon_{j+1}}=1$ and
so we obtain

$\displaystyle
d^{(\nu)}\Big{(}(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n((i-N-1)-\tfrac{c^{(\nu)}}{d^{(\nu)}})\Big{)}q_{j}$
$\displaystyle+$ $\displaystyle
d^{(\nu)}(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}(n+i-j+1)\big{(}d^{(\nu)}(j-1)+c^{(\nu)}\big{)}q_{j-1}+q_{j+1}=0,$

which is equivalent to

$E_{j}q_{j}+F_{j}q_{j-1}+\tfrac{1}{d^{(\nu)}}q_{j+1}=0,$

with

$\displaystyle E_{j}$
$\displaystyle=(n+i-j)(j+\tfrac{c^{(\nu)}}{d^{(\nu)}})+(j-1)(N-j+1)+n((i-N-1)-\tfrac{c^{(\nu)}}{d^{(\nu)}}),$
$\displaystyle F_{j}$
$\displaystyle=(j-1)(N-j+1)\tfrac{\mu_{j-1}}{\mu_{j}}(n+i-j+1)\big{(}d^{(\nu)}(j-1)+c^{(\nu)}\big{)},$

as asserted. ∎