$(I+A)^{-n-\lambda+1}L_{0}$ are polynomials in $n$, we conclude from (6.24)
that $C_{0}^{(\lambda)}=0$. ∎

###  6.4. Difference operators and Darboux transforms

Throughout this paper, we have proved that the matrix Charlier polynomials
$P_{n}^{(\lambda)}$ are solutions to four second order difference equations.
The difference equation in Corollary 4.5 involves multiplications by
nontrivial matrices on both left and right side and we will not discuss it
further in this section.

We are now left with three nontrivial difference operators
$\mathfrak{D}^{(\lambda)},\Delta
S^{(\lambda)},S^{(\lambda-1)}\Delta\in\mathcal{B}_{R}(P)$, given respectively
in (5.14) and Theorem 2.11. We note that the operators $\Delta S^{(\lambda)}$
and $S^{(\lambda-1)}\Delta$ are each others Darboux transforms. In the
following theorem we show that there exists a linear relation between these
operators.

######  Theorem 6.7.

The second order difference operators $\mathfrak{D}^{(\lambda)},\Delta
S^{(\lambda)},S^{(\lambda-1)}\Delta$ are related by

$2(\Delta
S^{(\lambda)}-S^{(\lambda-1)}\Delta)=\left(a-N-2\lambda\right)I-\mathfrak{D}^{(\lambda)}.$

###### Proof.

The operator $\mathfrak{D}^{(\lambda)}$ is given in (5.13) and the operators
$\Delta S^{(\lambda)}$, $S^{(\lambda-1)}\Delta$ are constructed using Theorem
2.8, yielding

$\displaystyle\Delta S^{(\lambda)}$
$\displaystyle=-\eta\Phi^{(\lambda)\ast}(x)+2\Phi^{(\lambda)\ast}(x)-\Psi^{(\lambda)\ast}(x)+\eta^{-1}\left(\Psi^{(\lambda)\ast}(x)-\Phi^{(\lambda)\ast}(x)\right),$
(6.25) $\displaystyle S^{(\lambda-1)}\Delta$
$\displaystyle=-\eta\Phi^{(\lambda-1)\ast}(x+1)+\Phi^{(\lambda-1)\ast}(x+1)+\Phi^{(\lambda-1)\ast}(x)-\Psi^{(\lambda-1)\ast}(x+1)$
$\displaystyle\hskip
227.62204pt+\eta^{-1}\left(\Psi^{(\lambda-1)\ast}(x)-\Phi^{(\lambda-1)\ast}(x)\right).$

With the help of (6.17), we obtain

$\displaystyle\Phi^{(\lambda)\ast}(x)$
$\displaystyle=\frac{a}{2}(I+A)^{x+\lambda+1}(J(I+A^{\ast})+\lambda)(I+A)^{-x-\lambda}$
(6.26) $\displaystyle\Phi^{(\lambda)\ast}(x)-\Psi^{(\lambda)\ast}(x)$
$\displaystyle=\frac{x}{2}(I+A)^{x+\lambda}(J+\lambda I)(I+A)^{-x-\lambda}.$

The coefficients of $\eta$ and $\eta^{-1}$ in $2(\Delta
S^{(\lambda)}-S^{(\lambda-1)}\Delta)$ are

$\displaystyle
2\left(\Phi^{(\lambda-1)\ast}(x+1)-\Phi^{(\lambda)\ast}(x)\right)$
$\displaystyle=-a(I+A),$ $\displaystyle
2\biggl{(}\Psi^{(\lambda)\ast}(x)-\Phi^{(\lambda)\ast}(x)-\left(\Psi^{(\lambda-1)\ast}(x)-\Phi^{(\lambda-1)\ast}(x)\right)\biggr{)}$
$\displaystyle=-x(I+A)^{-1},$

respectively, where we have used Lemma 5.5 in the last step. The above agrees
with the corresponding $\eta^{\pm}$ coefficients of
$-\mathfrak{D}^{(\lambda)}$. For the $\eta^{0}$ term in $2(\Delta
S^{(\lambda)}-S^{(\lambda-1)}\Delta)$, we need to consider

$4\Phi^{(\lambda)\ast}(x)-2\Psi^{(\lambda)\ast}(x)-2\left(\Phi^{(\lambda-1)\ast}(x+1)+\Phi^{(\lambda-1)\ast}(x)-\Psi^{(\lambda-1)\ast}(x+1)\right),$

which can be regrouped into quantities we have calculated above:

$\displaystyle 2\biggl{(}$
$\displaystyle\Phi^{(\lambda)\ast}(x)-\Psi^{(\lambda)\ast}(x)-\left(\Phi^{(\lambda-1)\ast}(x)-\Psi^{(\lambda-1)\ast}(x)\right)\biggr{)}$
$\displaystyle\hskip
85.35826pt+2\left(\Phi^{(\lambda)\ast}(x)-\Phi^{(\lambda-1)\ast}(x+1)\right)+2\left(\Psi^{(\lambda-1)\ast}(x+1)-\Psi^{(\lambda-1)\ast}(x)\right)$
$\displaystyle=x(I+A)^{-1}+a(I+A)+2\mathcal{J}_{1}^{(\lambda-1)\ast}$
$\displaystyle=\mathfrak{J}^{(\lambda)}+(a-N-2\lambda)I.$

In the last line we have used (5.12) and that the leading coefficient of
$\Psi^{(\lambda)}$ simplifies to

$2\mathcal{J}_{1}^{(\lambda)}=J-aA^{\ast}-(\lambda+1)A^{\ast}(I+A^{\ast})^{-1}-N-1-\lambda.$

This completes the proof. ∎

##  7\. Explicit expression for the entries of the matrix Charlier polynomials

In Section 5.3 we found that a difference equation for $P_{n}^{(\lambda)}$
could be significantly simplified by considering the following modification of
the polynomials

$R_{n}^{(\lambda)}(x)=L_{0}(I+A)^{-n-\lambda}P_{n}^{(\lambda)}(x)(I+A)^{\lambda+x}.$
(7.1)

Recall from Proposition 5.13 that the entries of $R_{n}^{(\lambda)}(x)$ are
given by

$R_{n}^{(\lambda)}(x)_{j,k}=\xi_{j,k,n}^{(\lambda)}c_{n+j-k}^{(a)}(x),$ (7.2)

with the $\xi_{j,k,n}^{(\lambda)}$ independent of $x$. The goal of this
section is to give explicit expressions for $\xi_{j,k,n}^{(\lambda)}$ which in
turn gives us the explicit entries of the monic orthogonal polynomials
$P_{n}^{(\lambda)}$ due to Corollary 5.14. The main ingredients in this
calculation are two difference equations. The first one is the difference
equation (4.18). The second one is $(P^{(\lambda)}\cdot\Delta
S^{(\lambda)})_{n}=nG_{n}^{(\lambda)}P_{n}^{(\lambda)}$, the left hand side of
which we have seen in (6.25).

######  Remark 7.1.

It is important to emphasize that the $\xi_{j,k,n}^{(\lambda)}$ have only been
defined for $n+j-k\geq 0$, c.f. (5.15). It will be convenient to define
$\xi_{j,k,n}^{(\lambda)}=0$ whenever $j<1$, $k<1$, $j>N$, $k>N$ or $n+j-k<0$.

###  7.1. A recurrence relation in $k$ for $\xi_{j,k,n}^{(\lambda)}$

Firstly, we will write $P_{n}^{(\lambda)}\cdot\Delta
S^{(\lambda)}=nG_{n}^{(\lambda)}P_{n}^{(\lambda)}$ in terms of
$R_{n}^{(\lambda)}$.

######  Proposition 7.2.

The difference equation $(P^{(\lambda)}\cdot\Delta
S^{(\lambda)})_{n}=nG_{n}^{(\lambda)}P_{n}^{(\lambda)}$ in terms of
$R_{n}^{(\lambda)}$ from (7.1) is given by

$\displaystyle n\left((N+\lambda+1)I-J\right)R_{n}^{(\lambda)}(x)$
$\displaystyle=-R_{n}^{(\lambda)}(x+1)a\left(J(I+A^{*})+\lambda I\right)$
$\displaystyle\hskip
56.9055pt+R_{n}^{(\lambda)}(x)\left(a(I+A)(J(I+A^{*})+\lambda I)+x(J+\lambda
I)\right)$ $\displaystyle\hskip 142.26378pt-
R_{n}^{(\lambda)}(x-1)x(I+A)(J+\lambda I).$

###### Proof.

It follows directly from (7.1), (6.21), (6.25) and Lemma 6.3. ∎

In the following proposition we show that the difference equation for
$R_{n}^{(\lambda)}$ in Proposition 7.2 induces a three term recurrence
relation for $\xi_{j,k,n}^{(\lambda)}$.

######  Proposition 7.3.

The coefficients $\phi_{k}\vcentcolon=\xi_{j,k,n}^{(\lambda)}$ satisfy the
following recursion in $k$

$0=(\sqrt{a}\left(k+\lambda\right)\sqrt{N-k})\,\phi_{k+1}\\\
\qquad+\left(j\lambda-n(N+1-j)-N-1+k(N+j-\lambda+n+2)-2k^{2}\right)\,\phi_{k}\\\
\qquad+\left(a^{-\frac{1}{2}}(n+j-k+1)(k-1)\sqrt{N-k+1}\right)\,\phi_{k-1},$

which holds for $1\leq k\leq K$ with $K\vcentcolon=\min(N,n+j)$.

###### Proof.

The main ingredient of the proof is the difference equation in Proposition 7.2
but with $x=0.$ This reduces to

$n\left((N+\lambda+1)I-J\right)R_{n}^{(\lambda)}(0)=-aR_{n}^{(\lambda)}(1)\left(J(I+A^{*})+\lambda
I\right)+R_{n}^{(\lambda)}(0)a(I+A)(J(I+A^{*})+\lambda I).$ (7.3)

From (7.2) we obtain $R_{n}^{(\lambda)}(0)_{j,k}=\xi_{j,k,n}^{(\lambda)}$ and
$R_{n}^{(\lambda)}(1)_{j,k}=\xi_{j,k,n}^{(\lambda)}c_{n+j-k}^{(a)}(1)=\xi_{j,k,n}^{(\lambda)}(1-\frac{n+j-k}{a})$.
For the other matrices involved in the above equation, recall that

$\displaystyle A_{k+1,k}$ $\displaystyle=\frac{1}{\sqrt{a}}\sqrt{N-k},\qquad$
$\displaystyle(AJ)_{k+1,k}=\frac{1}{\sqrt{a}}k\sqrt{N-k},$ $\displaystyle
1\leq k\leq N-1,$ $\displaystyle(JA^{\ast})_{k-1,k}$
$\displaystyle=\frac{1}{\sqrt{a}}(k-1)\sqrt{N-k+1},\qquad$
$\displaystyle(AJA^{\ast})_{k,k}=\frac{1}{a}(k-1)(N-k+1),$ $\displaystyle
2\leq k\leq N,$

where the expression for $AJA^{\ast}$ does also hold for $k=1$. Now we use
these to look at the $(j,k)$ entry of (7.3) to arrive at the desired result. ∎

###  7.2. A recurrence relation in $j$ for $\xi_{j,1,n}^{(\lambda)}$

The recurrence relation in Proposition 7.3 determines the coefficients
$\xi_{j,k,n}^{(\lambda)}$ up to the initial values $\xi_{j,1,n}^{(\lambda)}$.
In order to find a recurrence relation for these coefficients, we will write
(4.18) in terms of $R_{n}^{(\lambda)}$.

######  Proposition 7.4.

The difference equation (4.18) in terms of $R_{n}^{(\lambda)}$ is given by

$\displaystyle R_{n}^{(\lambda)}(x+1)$
$\displaystyle+\left(\frac{1}{a^{2}}\mathcal{D}_{n}^{(\lambda)}(I+A^{\ast})^{-1}\mathcal{D}_{n-1}^{(\lambda)-1}-\frac{x}{a}\mathcal{D}_{n}(I+A^{\ast})^{-1}\mathcal{D}_{n}^{-1}-(I+A)\right)R_{n}^{(\lambda)}(x)$
$\displaystyle\hskip
142.26378pt+\frac{x}{a}\mathcal{D}_{n}^{(\lambda)}(I+A^{\ast})^{-1}\mathcal{D}_{n}^{(\lambda)-1}(I+A)R_{n}^{(\lambda)}(x-1)=0.$

###### Proof.

Recall the LDU decomposition from Corollary 5.7 and further that
$\mathscr{A}=I+A$. The rest follows from direct computation. ∎

As in the previous subsection, in the following proposition we derive a three
term recurrence relation for $\xi_{j,1,n}^{(\lambda)}$ from Proposition 7.4.

######  Proposition 7.5.

The coefficients $\psi_{j}\vcentcolon=\xi_{j,1,n}^{(\lambda)}$ satisfy the
following recursion in $j$

$\displaystyle 0$
$\displaystyle=\psi_{j+1}\left(\frac{1}{\sqrt{a}}j(N+\lambda-j)(n+j)\sqrt{N-j}\right)$
$\displaystyle\qquad+\psi_{j}\biggl{(}(n+j-1)(N+n+\lambda-j)(N+n+\lambda-j+1)$
$\displaystyle\qquad\qquad-n(\lambda+n)(N+\lambda+n)+j(N-j)(N+\lambda-j)\biggr{)}$
$\displaystyle\qquad+\psi_{j-1}\left(\sqrt{a}(N+n+\lambda-j)(N+n+\lambda-j+1)\sqrt{N-(j-1)}\right),$

for $1\leq j\leq N$.

###### Proof.

The main ingredient of the proof is Proposition 7.4 but with $x=0$. This
reduces to

$a\mathcal{D}_{n}^{(\lambda)}(I+A^{\ast})\mathcal{D}_{n}^{(\lambda)-1}R_{n}^{(\lambda)}(1)=\left(a\mathcal{D}_{n}^{(\lambda)}(I+A^{\ast})\mathcal{D}_{n}^{(\lambda)-1}(I+A)-\frac{1}{a}\mathcal{D}_{n}^{(\lambda)}\mathcal{D}_{n-1}^{(\lambda)-1}\right)R_{n}^{(\lambda)}(0).$
(7.4)

As before (7.2) gives us $R_{n}^{(\lambda)}(0)_{jk}=\xi_{j,k,n}^{(\lambda)}$
and
$R_{n}^{(\lambda)}(1)_{jk}=\xi_{j,k,n}^{(\lambda)}c_{n+j-k}^{(a)}(1)=\xi_{j,k,n}^{(\lambda)}(1-\frac{n+j-k}{a})$.
For the other matrix entries involved in the above equation we look to (6.16)
for $A_{j+1,j}=\frac{1}{\sqrt{a}}\sqrt{N-j}$ and (6.18) for

$\displaystyle\left(\mathcal{D}_{n}^{(\lambda)}\right)_{jj}\left(\mathcal{D}_{n-1}^{(\lambda)-1}\right)_{jj}$
$\displaystyle=\frac{na(N+\lambda+n)(\lambda+n)}{(N+\lambda+n-j)(N+\lambda+n+1-j)},$
$\displaystyle\left(\mathcal{D}_{n}^{(\lambda)}\right)_{jj}\left(A^{\ast}\right)_{j,j+1}\left(\mathcal{D}_{n-1}^{(\lambda)-1}\right)_{j+1,j+1}$
$\displaystyle=\frac{1}{\sqrt{a}}\frac{j(N+\lambda-j)\sqrt{N-j}}{(N+\lambda+n-j)(N+\lambda+n+1-j)}.$

Now we use these expressions to look at the $(j,1)$ entries of (7.4) and
multiply everything by $(N+\lambda+n-j)(N+\lambda+n+1-j)$ to arrive at the
desired result. ∎

###  7.3. The coefficients $\xi_{j,k,n}^{(\lambda)}$ as dual Hahn polynomials

In the final step of the determination of the coefficients
$\xi_{j,k,n}^{(\lambda)}$, we solve the recurrence relations of the previous
subsections in terms of the dual Hahn polynomials.

The dual Hahn polynomials are given by

$\mathcal{R}_{k}\bigl{(}\ell(x);\,\gamma,\delta,\mathcal{N}\bigr{)}=\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{-k,-x,x+\gamma+\delta+1}{\gamma+1,-\mathcal{N}};1\right),\qquad
k=0,1,\ldots,\mathcal{N},$

see for instance [42].

######  Proposition 7.6.

Let $K=\min(N,n+j)$ and $\mathscr{K}=\max(N,n+j)$. Then for
$k\in\\{1,\dots,K\\}$ we have

$\xi_{j,k,n}^{(\lambda)}=\xi_{j,1,n}^{(\lambda)}a^{-(k-1)/2}(1-K)_{k-1}\sqrt{\frac{(N-k)!}{(N-1)!}}\mathcal{R}_{k-1}\left((K-j)(\mathscr{K}-j+\lambda+1);\lambda,\mathscr{K}-K,K-1\right),$

and $0$ for $k>K$.

###### Proof.

First observe that $\xi_{j,k,n,}^{(\lambda)}=0$ when $k>K$, because of Remark
7.1. Next we take Proposition 7.3 and substitute
$\zeta_{k}=(\lambda+1)_{k}a^{k/2}\sqrt{\frac{(N-k-1)!}{(N-1)!}}\frac{\xi_{j,k+1,n}^{(\lambda)}}{\xi_{j,1,n}^{(\lambda)}}$,
to arrive at a recursion that is more reminiscent of a three term recurrence
of monic orthogonal polynomials

$\displaystyle 0$
$\displaystyle=\zeta_{k+1}+\left(-2k^{2}+k(j-\lambda+n+N-2)+j(n+\lambda+1)-(nN+\lambda+1)\right)\zeta_{k}$
$\displaystyle\hskip 227.62204pt+k(k+\lambda)(k-(n+j))(k-N)\zeta_{k-1},$

for $0\leq k\leq K-1$. It is easy to check, comparing with [42], that this is
the three term recurrence for the monic dual Hahn with $\gamma=\lambda$,
$\delta=\mathscr{K}-K$, $\mathcal{N}=K-1$ where
$(K-j)(\mathscr{K}-j+\lambda+1)$ plays the role of the variable. Lastly the
monic polynomials are related to the standard ones by
$p_{k-1}=(\lambda+1)_{k-1}(1-K)_{k-1}\mathcal{R}_{k-1}$. ∎

######  Proposition 7.7.

For $j\in\\{1,\dots,N\\}$ we have

$\xi_{j,1,n}^{(\lambda)}=(-1)^{j-1}\xi_{1,1,n}^{(\lambda)}\sqrt{\frac{(N-1)!}{(N-j)!}}\frac{a^{(j-1)/2}(1-N-\lambda-n)_{j-1}}{(j-1)!(1-N-\lambda)_{j-1}}.$

###### Proof.

We take Proposition 7.5 and substitute
$\sigma_{j}=\frac{j!(n+1)_{j}(1-N-\lambda)_{j}}{a^{j/2}(1-N-\lambda-n)_{j}}\sqrt{\frac{(N-1)!}{(N-j-1)!}}\frac{\xi_{j+1,1,n}^{(\lambda)}}{\xi_{1,1,n}^{(\lambda)}}$.
Just like in the proof of Proposition 7.6, we arrive at a three term
recurrence for monic dual Hahn polynomials

$0=\sigma_{j+1}+\left(-2j^{2}+j(2N+\lambda-n-2)+n(N-1)+N-1\right)\sigma_{j}+j(j-(N+\lambda))(j-N)(j+n)\sigma_{j-1},$

but with simpler parameters. So now we have
$\sigma_{j}=p_{j}(0;n,\lambda,N-1)=(n+1)_{j}(1-N)_{j}$. Substituting back
gives us the desired result. ∎

Now all that we are left to obtain is the initial value.

######  Lemma 7.8.

The final and initial values are

$\xi_{N,1,n}^{(\lambda)}=(-1)^{n+N-1}\frac{a^{n+(N-1)/2}}{\sqrt{(N-1)!}},\qquad\qquad\xi_{1,1,n}^{(\lambda)}=(-a)^{n}\frac{(\lambda+1)_{N-1}}{(\lambda+n)_{N-1}}.$

###### Proof.

We can write

$(I+A)^{x+\lambda}=\exp((x+\lambda)B),\qquad
B\vcentcolon=\log(I+A)=\sum_{s=1}^{N-1}\frac{(-1)^{s+1}}{s}A^{s}.$

Then since $A^{N}=0$, we have $B^{N-1}=A^{N-1}$. This shows us that both are
polynomial in $x$ and in particular we have a simple leading term

$(I+A)^{x+\lambda}=x^{N-1}\frac{A^{N-1}}{(N-1)!}+\mathcal{O}(x^{N-2}).$

The monicity of $P_{n}^{(\lambda)}$ and (7.1) give

$R_{n}^{(\lambda)}(x)=x^{n+N-1}L_{0}(I+A)^{-n-\lambda}\frac{A^{N-1}}{(N-1)!}+\mathcal{O}(x^{n+N-2}).$

This leading term has only one non-zero entry because
$L_{0}(I+A)^{-n-\lambda}$ is lower triangular and $(A^{N-1})_{jk}\neq 0$ only
for $(j,k)=(N,1)$. So then entrywise we get

$(R_{n}^{(\lambda)}(x))_{N,1}=\xi_{N,1,n}^{(\lambda)}c_{n+N-1}^{(a)}(x)=x^{n+N-1}\frac{(A^{N-1})_{N,1}}{(N-1)!}+\mathcal{O}(x^{n+N-2})$

which gives us the first result by comparing leading terms

$\xi_{N,1,n}^{(\lambda)}=(-a)^{n+N-1}\frac{(A^{N-1})_{N,1}}{(N-1)!}=(-1)^{n+N-1}\frac{a^{n+(N-1)/2}}{\sqrt{(N-1)!}}.$

We can then directly obtain $\xi_{1,1,n}^{(\lambda)}$ from Proposition 7.7
with $j=N$. ∎

Finally, we summarize the content of Propositions 7.6, 7.7 and Lemma 7.8 in
the main Theorem of this section.

######  Theorem 7.9.

The coefficients $\xi_{j,k,n}^{(\lambda)}$ in (7.2) are given in terms of the
dual Hahn polynomials by

$\displaystyle\xi_{j,k,n}^{(\lambda)}$
$\displaystyle=\mathfrak{X}(j,k,n,\lambda)\times\begin{cases}(1-N)_{k-1}\,\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,j-N,n+\lambda+1}{\lambda+1,1-N};1\right)&n+j\geq
N\\\
(1-n-j)_{k-1}\,\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,-n,N-j+\lambda+1}{\lambda+1,1-n-j};1\right)&n+j<N,\end{cases}$
$\displaystyle=\mathfrak{X}(j,k,n,\lambda)\times\begin{cases}(1-N)_{k-1}\mathcal{R}_{k-1}\bigl{(}\ell(N-j);\,\lambda,n+j-N,N-1\bigr{)}&n+j\geq
N,\\\
(1-n-j)_{k-1}\mathcal{R}_{k-1}\bigl{(}\ell(n);\,\lambda,N-(n+j),n+j-1\bigr{)}&n+j<N,\end{cases}$

with

$\mathfrak{X}(j,k,n,\lambda)=\sqrt{\frac{(N-k)!}{(N-j)!}}\frac{(-1)^{n+j-1}a^{n+(j-k)/2}(\lambda+n)!(N+\lambda-j)!}{(j-1)!\lambda!(N+\lambda+n-j)!}.$

##  8\. Dual Charlier Polynomials

In this section we carry out the construction of three non–equivalent families
of matrix valued polynomials in the sense of (3.5) which are dual to the
matrix valued Charlier polynomials $P_{n}^{(\lambda)}$. It follows from
Theorem 3.9 that each family is related to a suitable operator in the set
$\mathcal{B}^{2}_{R}(P)$.

###  8.1. Explicit expression for $P_{n}(0)$

The goal of this subsection is to give an explicit expression $P_{n}(0)$ which
is a key ingredient of the dual families, see Theorem 3.9. Firstly, we recall
the matrix sequence $\mathcal{A}_{\lambda}$ introduced in (6.23). In the
following remark we establish some basic properties for
$\mathcal{A}_{\lambda}$.

######  Remark 8.1.

As we have seen in Lemma 5.5, the matrices $J$ and $A$ have simple commutation
relations. We can easily derive similar results for $\mathcal{A}_{\lambda}$

  1. (1)

$[J,\mathcal{A}_{\lambda}]=-\mathcal{A}_{\lambda},$

  2. (2)

$[J,(I+\mathcal{A}_{\lambda})^{k}]=-k\mathcal{A}_{\lambda}(I+\mathcal{A}_{\lambda})^{k-1},$

  3. (3)

$(I+\mathcal{A}_{\lambda})^{-k}J(I+\mathcal{A}_{\lambda})^{k}=J-k\mathcal{A}_{\lambda}(I+\mathcal{A}_{\lambda})^{k-1}.$

######  Proposition 8.2.

Let $R_{n}^{(\lambda)}$ be the function defined in (7.1). Then the following
relations hold:

  1. (1)

$R_{n}^{(\lambda)}(0)=-a(I+\mathcal{A}_{n+\lambda})R_{n-1}^{(\lambda+1)}(0)$.

  2. (2)

$R_{n}^{(\lambda)}(0)=(-a)^{n}(I+\mathcal{A}_{n+\lambda})^{n}L_{0}.$

###### Proof.

We relegate the proof of $(1)$ to Appendix A.2. Part $(2)$ follows by
iterating part $(1)$ and applying the fact that
$R_{0}^{(\lambda+n)}(0)=L_{0}.$ ∎

######  Corollary 8.3.

For the monic Charlier polynomials we get that

$P_{n}^{(\lambda)}(0)=(-a)^{n}L_{0}^{-1}(I+A)^{n+\lambda}(I+\mathcal{A}_{n+\lambda})^{n}(I+A)^{-\lambda}L_{0}.$

###### Proof.

The Corollary follows directly from the previous Proposition and equation
(7.1) which defines
$R_{n}^{(\lambda)}(x)=L_{0}(I+A)^{-n-\lambda}P_{n}^{(\lambda)}(x)(I+A)^{\lambda+x}$.
∎

######  Remark 8.4.

Note that since $A^{*}$ is _strictly upper triangular_ , so is
$\mathcal{A}_{n+\lambda}$. From the explicit expression of
$\mathcal{A}_{n+\lambda}$ in (6.23), we can verify that it has rational
entries in $n$. Then $(I+\mathcal{A}_{n+\lambda})$ is an upper triangular
matrix with diagonal entries equal to $1$. This allows us to conclude that
$(I+\mathcal{A}_{n+\lambda})^{n}$ is rational in $n$. In particular, this
implies that every entry of $(-a)^{-n}P_{n}^{(\lambda)}(0)$ is a rational
function of $n$.

###  8.2. Three families of dual polynomials

In Theorem 3.9 we have established a one to one correspondence between the
equivalence classes of families of dual polynomials and the subset of second
order difference operators $\mathcal{B}^{2}_{R}(P)$, see Definition 3.7. In
this section we give three families of dual polynomials, each one
corresponding to an specific operator of $\mathcal{B}^{2}_{R}(P)$.

Recall from (5.13) and (6.25) that

$\displaystyle\mathfrak{D}^{(\lambda)}$ $\displaystyle=\eta
a(I+A)-J-(x+\lambda)(I+A)^{-1}+\eta^{-1}x(I+A)^{-1},$ (8.1)
$\displaystyle\Delta S^{(\lambda)}$
$\displaystyle=-\eta\Phi^{(\lambda)}(x)^{\ast}+2\Phi^{(\lambda)}(x)^{\ast}-\Psi^{(\lambda)}(x)^{\ast}+\eta^{-1}\left(\Psi^{(\lambda)}(x)^{\ast}-\Phi^{(\lambda)}(x)^{\ast}\right),$
(8.2) $\displaystyle S^{(\lambda-1)}\Delta$
$\displaystyle=-\eta\Phi^{(\lambda-1)}(x+1)^{\ast}+\Phi^{(\lambda-1)}(x+1)^{\ast}+\Phi^{(\lambda-1)}(x)^{\ast}-\Psi^{(\lambda-1)}(x+1)^{\ast}$
(8.3) $\displaystyle\hskip
199.16928pt+\eta^{-1}\left(\Psi^{(\lambda-1)}(x)^{\ast}-\Phi^{(\lambda-1)}(x)^{\ast}\right),$

where $\Phi^{(\lambda)\ast}$ is a matrix polynomial of degree two given
explicitly in Theorem 6.2. However in this section we will have more use for
the decomposition given in (6.26),

$\Phi^{(\lambda)}(x)^{\ast}=\frac{a}{2}(I+A)^{x+\lambda+1}(J(I+A^{\ast})+\lambda
I)(I+A)^{-x-\lambda}.$

We label these as $D_{1}=\mathfrak{D}^{(\lambda)}$, $D_{2}=\Delta
S^{(\lambda)}$ and $D_{3}=S^{(\lambda-1)}\Delta$, to simplify notation and to
be able to discuss all three at the same time.

######  Lemma 8.5.

The difference operators $D_{1}=\mathfrak{D}^{(\lambda)},D_{2}=\Delta
S^{(\lambda)}$ and $D_{3}=S^{(\lambda-1)}\Delta$ are elements of
$\mathcal{B}^{2}_{R}(P)$. For $D_{1}$ and $D_{2}$ this holds for
$\lambda\in\mathbb{N}_{0}$ and for $D_{3}$ for $\lambda\in\mathbb{N}$.

###### Proof.

In (8.1) the raising coefficient $a(I+A)$ can immediately be seen to be
invertible. The same follows for the raising coefficients in (8.2) and (8.3)
by considering (6.26), although we must take $\lambda\geq 1$ for $D_{3}$. So
the invertibility condition is met for all three cases. Is easy to check that
in the three cases the lowering coefficient evaluated at $x=0$ is equal to
zero, then the condition over $P_{n}(0)^{-1}P_{n}(-1)F_{-1}(0)$ is
automatically satisfied. We just need to check that the
$\rho_{i}^{(\lambda)}(n)=P_{n}^{(\lambda)}(0)^{-1}\psi^{-1}(D_{i})P_{n}^{(\lambda)}(0)$
satisfy Condition 3.1 for all $i=1,2,3.$ Recall from (2.21)

$(P_{n}^{(\lambda)}\cdot
S^{(\lambda-1)}\Delta)(x)=(n+1)G_{n+1}^{(\lambda-1)}P_{n}(x),\qquad(P_{n}^{(\lambda)}\cdot\Delta
S^{(\lambda)})(x)=nG_{n}^{(\lambda)}P_{n}(x),$

where by Lemma 6.3 and equation (6.21)

$G_{n}^{(\lambda)}=\frac{1}{2}L_{0}^{-1}(I+A)^{n+\lambda}((N+\lambda+1)I-J)(I+A)^{-n-\lambda}L_{0}.$

Additionally Theorem 5.10 gives us

$P_{n}^{(\lambda)}\cdot\mathfrak{D}^{(\lambda)}=\Gamma_{n}^{(\lambda)}P_{n}^{(\lambda)},\qquad\Gamma_{n}^{(\lambda)}=a(I+A)-J-(n+\lambda)(I+A)^{-1}.$

Applying Corollary 8.3 and Remark 8.1 we get

$\displaystyle\rho_{1}^{(\lambda)}(n)$
$\displaystyle=L_{0}^{-1}(I+A)^{\lambda}\left((a-n-\lambda)I-J+n\mathcal{A}_{n+\lambda}(I+\mathcal{A}_{n+\lambda})^{n-1}\right)(I+A)^{-\lambda}L_{0},$
(8.4) $\displaystyle\rho_{2}^{(\lambda)}(n)$
$\displaystyle=\frac{n}{2}L_{0}^{-1}(I+A)^{\lambda}\left((N+\lambda+1)I-J+n\mathcal{A}_{n+\lambda}(I+\mathcal{A}_{n+\lambda})^{n-1}\right)(I+A)^{-\lambda}L_{0},$
(8.5) $\displaystyle\rho_{3}^{(\lambda)}(n)$
$\displaystyle=\frac{n+1}{2}L_{0}^{-1}(I+A)^{\lambda}\left((N+\lambda)I-J+n\mathcal{A}_{n+\lambda}(I+\mathcal{A}_{n+\lambda})^{n-1}\right)(I+A)^{-\lambda}L_{0}.$
(8.6)

From these expressions we can see that each $\rho_{i}^{(\lambda)}$ is an
$n$-dependent matrix of the form

$\rho_{i}^{(\lambda)}(n)=L_{0}^{-1}(I+A)^{\lambda}(n\mathcal{T}+\mathcal{T}_{0}+\mathrm{s.u.t.})(I+A)^{-\lambda}L_{0},$

where $\mathcal{T}$ and $\mathcal{T}_{0}$ are $n$-independent diagonal
matrices, and s.u.t. stands for strictly upper triangular. In addition, Remark
8.4 tells us that the s.u.t. part is rational in $n$.

That means that each $\rho_{i}^{(\lambda)}$ is of the form described in
Corollary A.2 in Appendix A.1. This then allows us to see that the
determinants of the block Vandermonde matrices in Condition 3.1 and Condition
3.20 are nonzero. ∎

As a consequence, by Theorem 3.9, we can find a dual family of polynomials for
each operator $S^{(\lambda-1)}\Delta,\Delta S^{(\lambda)},\text{ and
}\mathfrak{D}^{(\lambda)}.$

######  Remark 8.6.

Given $D_{1},D_{2},D_{3}\in\mathcal{B}_{R}^{2}(P)$, such that

$P_{n}\cdot D_{i}=\Lambda_{n}^{(i)}P_{n},\quad\text{for $i$ =1,2,3}$

for some matrices $\Lambda_{n}^{(i)}$. Suppose that there exist nonzero
complex numbers $\alpha,\beta,\gamma$ such that

$D_{3}=\alpha D_{1}+\beta D_{2}+\gamma
I,\qquad\Lambda_{n}^{(3)}=\alpha\Lambda_{n}^{(1)}+\beta\Lambda_{n}^{(2)}+\gamma
I.$

The corresponding $\rho_{3}(n)$ is then given by

$\rho_{3}(n)=P_{n}(0)^{-1}\Lambda_{n}^{(3)}P_{n}(0)=\alpha\rho_{1}(n)+\beta\rho_{2}(n)+\gamma
I.$

In Theorem 6.7 we have seen that this is indeed true in our case with
$\alpha=\frac{1}{2}$, $\beta=1$ and $\gamma=-\frac{1}{2}(a-N-2\lambda)$. This
is consistent with (8.6).

We explicitly give three families of dual polynomials associated with the
difference operators (8.1), (8.2) and (8.3).

####  8.2.1. Dual family associated to the operator $\mathfrak{D}^{(\lambda)}$

We start by finding the dual family which corresponds to the operator
$\mathfrak{D}^{(\lambda)}.$ Applying Theorem 3.9 we get that the dual
polynomials are

$Q_{x,1}^{(\lambda)}(\rho_{1}^{(\lambda)}(n))=P^{(\lambda)}_{n}(0)^{-1}P_{n}^{(\lambda)}(x)\Upsilon_{1}^{(\lambda)}(x)^{-1},$
(8.7)

where $P_{n}^{(\lambda)}(0)$ is given in Corollary 8.3, $\rho_{1}^{(\lambda)}$
is given in (8.4) and

$\Upsilon_{1}^{(\lambda)}(x)=a^{-x}(I+A)^{-x},$

which does not depend on $\lambda$ in this case.

####  8.2.2. Dual family associated to the operator $\Delta S^{(\lambda)}$

Similarly, using Theorem 3.9 we get

$\displaystyle Q_{x,2}^{(\lambda)}(\rho_{2}^{(\lambda)}(n))$
$\displaystyle=P^{(\lambda)}_{n}(0)^{-1}P_{n}^{(\lambda)}(x)\Upsilon^{(\lambda)}_{2}(x)^{-1},$
$\displaystyle\Upsilon^{(\lambda)}_{2}(x)$
$\displaystyle=\frac{(-2)^{x}}{a^{x}}(I+A)^{\lambda}(J(I+A^{*})+\lambda
I)^{-x}(I+A)^{-x-\lambda},$

where $\rho_{2}^{(\lambda)}$ is given in (8.5).

####  8.2.3. Dual family associated to the operator $S^{(\lambda-1)}\Delta$

As with the previous two families we use Theorem 3.9 to get the dual
polynomials and the related quantities, but in this case only for $\lambda>0$,

$\displaystyle Q_{x,3}^{(\lambda)}(\rho_{3}^{(\lambda)}(n))$
$\displaystyle=P^{(\lambda)}_{n}(0)^{-1}P_{n}^{(\lambda)}(x)\Upsilon^{(\lambda)}_{3}(x)^{-1},$
$\displaystyle\Upsilon^{(\lambda)}_{3}(x)$
$\displaystyle=\frac{(-2)^{x}}{a^{x}}(I+A)^{\lambda}(J(I+A^{*})+(\lambda-1)I)^{-x}(I+A)^{-x-\lambda},$

where $\rho_{3}^{(\lambda)}$ is given in (8.6).

###  8.3. Dual orthogonality relations

We are now ready to identify the orthogonality relations for the dual
polynomials $(Q^{(\lambda)}_{x,i})_{x}$.

######  Theorem 8.7.

The dual weight matrix for the Charlier polynomials is

$U^{(\lambda)}(n)=a^{2n}L_{0}^{\ast}(I+A^{\ast})^{-\lambda}(I+(\mathcal{A}_{n+\lambda})^{\ast})^{n}(\mathcal{D}_{n}^{(\lambda)})^{-1}(I+\mathcal{A}_{n+\lambda})^{n}(I+A)^{-\lambda}L_{0},$
(8.8)

and (for each $\rho_{i}$) it satisfies the hypothesis of Theorem 3.14.

######  Remark 8.8.

From Remark 8.4 and the explicit expression of $\mathcal{D}_{n}^{(\lambda)}$
(see Theorem 6.4), there exists a function $F^{(\lambda)}$ such that
$F^{(\lambda)}(n)_{i,j}$ is rational in $n$ for all $i,j$ and

$U^{(\lambda)}(n)=\frac{a^{n}}{n!}F^{(\lambda)}(n),$

and from Remark 8.4 and its explicit expression we get that
$\rho_{i}^{(\lambda)}(n)^{k}$ is rational in $n$ for all fixed
$k\in\mathbb{N}_{0}$.

###### Proof.

Equation (8.8) follows from Theorem 3.14, Corollary 8.3 and the LDU
decomposition of $\mathcal{H}_{n}^{(\lambda)}$ in Corollary 5.7. By Theorem
3.14 we just need to verify

$\sum_{n=0}^{\infty}\rho^{(\lambda)}_{i}(n)^{k}U^{(\lambda)}(n)<\infty,\qquad\forall
k\in\mathbb{N}_{0},$
$\lim_{n\to\infty}F_{1}(\rho_{i}^{(\lambda)}(n))P_{n}(0)\mathcal{H}_{n}^{-1}P_{n+1}(0)^{\ast}F_{2}(\rho_{i}^{(\lambda)}(n))=0,\qquad\text{for
all }F_{1},F_{2}\in M_{N}(\mathbb{C})[n].$ (8.9)

For a fixed $k\in\mathbb{N}_{0}$ and $s,t\in\\{1,\ldots,N\\}$, using Remark
8.4 we have

$\left(\sum_{n=0}^{\infty}\rho^{(\lambda)}_{i}(n)^{k}U^{(\lambda)}(n)\right)_{s,t}=\sum_{k=0}^{\infty}\frac{a^{n}}{n!}\left(\rho^{(\lambda)}_{i}(n)^{k}F^{(\lambda)}(n)\right)_{s,t}<\infty,$

since each entry of $\rho^{(\lambda)}_{i}(n)^{k}F^{(\lambda)}(n)$ is rational
in $n$. On the other hand, for $F_{1},F_{2}\in M_{N}(\mathbb{C})[n]$, we have
that $F_{1}(\rho_{i}^{(\lambda)}(n))$ and $F_{2}(\rho_{i}^{(\lambda)}(n))$ are
rational functions in $n$, and therefore using the explicit expressions of
$P_{n}(0)$ and $\mathcal{H}_{n}$ we have that

$F_{1}(\rho_{i}^{(\lambda)}(n))P_{n}(0)\mathcal{H}_{n}^{-1}P_{n+1}(0)^{\ast}F_{2}(\rho_{i}^{(\lambda)}(n))=\frac{a^{n}}{n!}R(n),\qquad\text{for
all }F_{1},F_{2}\in M_{N}(\mathbb{C})[n],$

where $R(n)$ is a rational function in $n$. Equation (8.9) then follows. ∎

So each dual family is orthogonal with respect to their own distinct dual
inner-product. For matrix polynomials $\mathscr{P},\mathscr{Q}$ we have

$\left\langle\mathscr{P},\mathscr{Q}\right\rangle_{i,d}^{(\lambda)}=\sum_{n=0}^{\infty}\mathscr{P}\left(\rho_{i}^{(\lambda)}(n)\right)^{\ast}U^{(\lambda)}(n)\mathscr{Q}\left(\rho_{i}^{(\lambda)}(n)\right).$

Note the weight $U^{(\lambda)}$ is the same for each family, but the inner-
product requires us to compose with $\rho_{i}^{(\lambda)}$ which is different
for each $i$.

###  8.4. Dual shift operators

In this subsection we show that the ladder relations for the polynomials
$P_{n}^{(\lambda)}$ associated to the Charlier weight induce shift operators
for the dual polynomials $Q_{x}^{(\lambda)}$. We recall the following
relations from Proposition 4.1:

$\displaystyle P^{(\lambda)}_{n}\cdot D=M\cdot P^{(\lambda)}_{n},\qquad$
$\displaystyle
M=(I+A)+\frac{1}{a}\mathcal{H}^{(\lambda)}_{n}(I+A^{*})^{-1}\mathcal{H}_{n-1}^{(\lambda)-1}\delta^{-1},$
(8.10) $\displaystyle P_{n}^{(\lambda)}\cdot D^{\dagger}=M^{\dagger}\cdot
P_{n}^{(\lambda)},\qquad$ $\displaystyle
M^{\dagger}=\frac{1}{a}(I+A)^{-1}\delta+\mathcal{H}_{n}^{(\lambda)}(I+A^{\ast})\mathcal{H}_{n}^{(\lambda)-1},$
(8.11)

where $D$ and $D^{\dagger}$ are

$D=\eta\,(I+A),\qquad D^{\dagger}=\eta^{-1}\frac{x}{a}(I+A)^{-1}.$

Since the operators $D$ and $D^{\dagger}$ are associated to the weak Pearson
equation 4.1, it follows from Remark 3.18 that
$D,D^{\dagger}\in\hat{\mathcal{F}}_{R}(P^{(\lambda)})$. Each dual family has
two relevant isomorphisms

$\displaystyle\tau\colon\hat{\mathcal{F}}_{L}(P^{(\lambda)})$
$\displaystyle\rightarrow{}$
$\displaystyle\hat{\mathcal{F}}^{d}_{L}(Q^{(\lambda)}),$
$\displaystyle\sigma_{i}\colon$
$\displaystyle\hat{\mathcal{F}}_{R}(P^{(\lambda)})$
$\displaystyle\rightarrow{}$
$\displaystyle\hat{\mathcal{F}}^{d}_{R}(Q^{(\lambda)}),$ $\displaystyle M$
$\displaystyle\mapsto{}$ $\displaystyle
P_{n}^{(\lambda)}(0)^{-1}MP_{n}^{(\lambda)}(0),$ $\displaystyle D$
$\displaystyle\mapsto{}$
$\displaystyle\Upsilon_{i}^{(\lambda)}(x)D\Upsilon_{i}^{(\lambda)}(x)^{-1},$

see (3.22) and (3.25). Note that by definition, $\tau$ is a conjugation by
$P_{n}^{(\lambda)}(0)$ and is therefore independent of the family of dual
polynomials. In this case we omit the subindex $i$. On the other hand, for
each $i=1,2,3$ we have the isomorphism $\sigma_{i}$ which involves conjugation
by $\Upsilon_{i}^{(\lambda)}$ and is different for each family of dual
polynomials.

######  Theorem 8.9.

Let $P_{n}^{(\lambda)}$ be the monic matrix Charlier polynomials and Let
$D,D^{\dagger},M,M^{\dagger}$ be given by (8.10) and (8.11). We fix a family
of dual polynomials $Q_{x}^{(\lambda)}=Q_{x,i}^{(\lambda)}$ for $i=1,2,3$.
Then the following relations hold true:

$\displaystyle\tau(M)\cdot Q_{x}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))$
$\displaystyle=Q_{x+1}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))\,\Upsilon^{(\lambda)}_{i}(1)(I+A),$
$\displaystyle\tau(M^{\dagger})\cdot
Q_{x}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))$
$\displaystyle=Q_{x-1}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))\,\frac{x}{a}(I+A)^{-1}\Upsilon^{(\lambda)}_{i}(1)^{-1}.$

###### Proof.

From Theorem 3.22 we get

$\displaystyle\tau(M)\cdot
Q_{x}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))=Q_{x}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))\cdot\sigma_{i}(D),\qquad\tau(M^{\dagger})\cdot
Q_{x}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))=Q_{x}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))\cdot\sigma_{i}(D^{\dagger}).$

Let us fix $i\in\\{1,2,3\\}$ and suppose that $D_{i}=\eta
F_{1}(x)+F_{0}(x)+\eta^{-1}F_{-1}(x)$, so that
$\Upsilon_{i}(x)=F_{1}(0)^{-1}\cdots F_{1}(x-1)^{-1}$ as in Theorem 3.9. Then
we have that

$\sigma_{i}(D)=\eta\Upsilon_{i}(x+1)(I+A)\Upsilon_{i}(x)^{-1},\qquad\sigma_{i}(D^{\dagger})=\eta^{-1}\frac{x}{a}\Upsilon_{i}(x-1)(I+A)^{-1}\Upsilon_{i}(x)^{-1}.$

Using that $F_{1}(x)=a(I+A)$ for $i=1$, $F_{1}(x)=-\Phi^{(\lambda)\ast}(x)$
for $i=2$ and $F_{1}(x)=-\Phi^{(\lambda-1)\ast}(x+1)$ for $i=3$ and using the
explicit expression in (6.26) we obtain

$\Upsilon_{i}(x+1)(I+A)\Upsilon_{i}(x)^{-1}=\Upsilon_{i}(1)(I+A),\qquad\Upsilon_{i}(x-1)(I+A)^{-1}\Upsilon_{i}(x)^{-1}=(I+A)^{-1}\Upsilon_{i}(1)^{-1}.$

This completes the proof of the theorem. ∎

######  Remark 8.10.

Applying the isomorphism $\tau,$ the LDU decomposition of the norm (see
Corollary 5.7), and properties of the matrix $J$ (see Lemma 5.5), we get

$\displaystyle\tau(M)$
$\displaystyle=L_{0}^{-1}(I+A)^{\lambda}(I+\mathcal{A}_{n+\lambda})^{-n}\left((I+A)-\frac{1}{a^{2}}\mathcal{D}_{n}^{(\lambda)}(\mathcal{D}_{n-1}^{(\lambda)})^{-1}\delta^{-1}\right)(I+\mathcal{A}_{n+\lambda})^{n}(I+A)^{-\lambda}L_{0},$
$\displaystyle\tau(M^{\dagger})$
$\displaystyle=L_{0}^{-1}(I+A)^{\lambda}(I+\mathcal{A}_{n+\lambda})^{-n}\left(\mathcal{D}_{n}^{(\lambda)}(I+A^{*})(\mathcal{D}_{n}^{(\lambda)})^{-1}-\delta\right)(I+\mathcal{A}_{n+\lambda})^{n}(I+A)^{-\lambda}L_{0}.$

These expressions are independent of the choice of the dual family.

######  Remark 8.11.

For the dual family that corresponds to $D_{1}$ we have the simple
expressions:

$\sigma_{1}(D)=\eta\frac{1}{a},\quad\sigma_{1}(D^{\dagger})=\eta^{-1}x,\quad\sigma_{1}(\mathfrak{J}^{(\lambda)})=J+xI+\lambda(I+A)^{-1}.$

######  Remark 8.12.

For the dual family corresponding to $D_{2}$ we have

$\displaystyle\sigma_{2}(D)$ $\displaystyle=-\eta
2a^{-1}(I+A)^{\lambda}(J(I+A^{*})+\lambda I)^{-1}(I+A)^{-\lambda},$
$\displaystyle\sigma_{2}(D^{\dagger})$
$\displaystyle=-\eta^{-1}\frac{x}{2}(I+A)^{\lambda}(J(I+A^{*})+\lambda
I)(I+A)^{-\lambda},$ $\displaystyle\sigma_{2}(\mathfrak{J}^{(\lambda)})$
$\displaystyle=(I+A)^{\lambda}(J(I+A^{*})+\lambda
I)^{-x}(J+(x+\lambda)I)(J(I+A^{*})+\lambda I)^{x}(I+A)^{-\lambda}.$

For the dual family corresponding to $D_{3}$ we have

$\displaystyle\sigma_{3}(D)$ $\displaystyle=-\eta
2a^{-1}(I+A)^{\lambda}(J(I+A^{*})+(\lambda-1)I)^{-1}(I+A)^{-\lambda},$
$\displaystyle\sigma_{3}(D^{\dagger})$
$\displaystyle=-\eta^{-1}\frac{x}{2}(I+A)^{\lambda}(J(I+A^{*})+(\lambda-1)I)(I+A)^{-\lambda},$
$\displaystyle\sigma_{3}(\mathfrak{J}^{(\lambda)})$
$\displaystyle=(I+A)^{\lambda}(J(I+A^{*})+(\lambda-1)I)^{-x}(J+(x+\lambda)I)(J(I+A^{*})+(\lambda-1)I)^{x}(I+A)^{-\lambda}.$

We note that expressions of
$\sigma_{i}(D),\sigma_{i}(D^{\dagger}),\sigma_{i}(\mathfrak{J}^{(\lambda)})$
for $i=1,2$ are very similar. This is due to the fact that the coefficients of
the operators $\Delta S^{(\lambda)}$ and $S^{(\lambda-1)}\Delta$ are related
by simple shifts in $\lambda$ and $x$, see (8.2) and (8.3).

######  Remark 8.13.

We finally include the expressions for the duals of the following relevant
operators which appeared in (2.8) and Remark 5.12.

$\displaystyle\tau(\mathcal{L})$
$\displaystyle=P_{n}^{(\lambda)}(0)^{-1}P_{n+1}^{(\lambda)}(0)\delta+P_{n}^{(\lambda)}(0)^{-1}B_{n}^{(\lambda)}P_{n}^{(\lambda)}(0)+P_{n}^{(\lambda)}(0)^{-1}C_{n}^{(\lambda)}P_{n-1}^{(\lambda)}(0)\delta^{-1},$
$\displaystyle\tau(\psi^{-1}(\mathfrak{J}^{(\lambda)}))$
$\displaystyle=a\tau(M)+a\tau(M^{\dagger})-\rho_{1}^{(\lambda)}(n).$

For the last expression we have used that
$\mathfrak{J}^{(\lambda)}=aD+aD^{\dagger}-\mathfrak{D}^{(\lambda)}$ which
follows from the definition of $\mathfrak{D}^{(\lambda)}$ in (5.13).

###  8.5. Dual square norm

In Theorem 3.16 we proved an expression for the square norms of the dual MVOP
assuming that certain functions are in the closure of the span of the MVOP
$(P_{n})_{n}$. Since we haven’t proven this assumption for our special case of
Charlier type MVOP, we will derive the square norms in a more direct fashion
that will ultimately agree with the earlier result.

It is more efficient to write the following proofs in terms of the

$R_{n}^{(\lambda)}(x)=L_{0}(I+A)^{-n-\lambda}P_{n}^{(\lambda)}(x)(I+A)^{x+\lambda},$

rather than the $Q_{x}^{(\lambda)}$. Using (7.1) and the fact that for
$i\in\\{1,2,3\\}$ we have

$Q_{x,i}^{(\lambda)}(\rho_{i}^{(\lambda)}(n))=P_{n}^{(\lambda)}(0)^{-1}P_{n}^{(\lambda)}(x)\Upsilon_{i}^{(\lambda)}(x)^{-1},$

we get that the square norm of the $Q_{x,i}^{(\lambda)}$ is given by

$\mathscr{W}_{x,i}^{(\lambda)}=\left((I+A^{\ast})^{x+\lambda}\Upsilon_{i}^{(\lambda)\ast}(x)\right)^{-1}\mathscr{W}_{R}^{(\lambda)}(x)\left(\Upsilon_{i}^{(\lambda)}(x)(I+A)^{x+\lambda}\right)^{-1},$
(8.12)

where

$\mathscr{W}_{R}^{(\lambda)}(x)=\sum_{n=0}^{\infty}R_{n}^{(\lambda)}(x)^{\ast}\mathcal{D}_{n}^{(\lambda)-1}R_{n}^{(\lambda)}(x).$

######  Lemma 8.14.

The 0-th moment of the dual weight $U^{(\lambda)}$ is given by

$\mathscr{W}_{0}^{(\lambda)}=W^{(\lambda)}(0)^{-1}.$

###### Proof.

The proof is given in Appendix A.3. ∎

######  Proposition 8.15.

The norm of the dual MVOP $Q_{x,i}^{(\lambda)}$ is given by

$\mathscr{W}_{x,i}^{(\lambda)}=(\Upsilon_{i}^{(\lambda)}(x)W^{(\lambda)}(x)\Upsilon_{i}^{(\lambda)}(x)^{\ast})^{-1}.$

###### Proof.

We can derive a recursion for $\mathscr{W}_{R}^{(\lambda)}$ by rewriting the
shift equations in (6.12) for $R_{n}^{(\lambda)}$. First
$P_{n}^{(\lambda)}\cdot\Delta=nP_{n-1}^{(\lambda+1)}$ is swiftly rewritten as

$R_{n}^{(\lambda)}(x+1)-R_{n}^{(\lambda)}(x)(I+A)=nR_{n-1}^{(\lambda+1)}(x),$

and for $P_{n-1}^{(\lambda+1)}\cdot
S^{(\lambda)}=G_{n}^{(\lambda)}P_{n}^{(\lambda)}$ we need

$L_{0}(I+A)^{-n-\lambda}G_{n}^{(\lambda)}(I+A)^{n+\lambda}L_{0}^{-1}=n\mathcal{D}_{n-1}^{(\lambda+1)}(\mathcal{D}_{n}^{(\lambda)})^{-1},$

and

$(I+A)^{-x-\lambda-1}\Phi^{(\lambda)\ast}(x)(I+A)^{x+\lambda}=T^{(\lambda+1)}(I+A^{\ast})(T^{(\lambda)})^{-1},$
$(I+A)^{-x-\lambda}(\Phi^{(\lambda)\ast}(x)-\Psi^{(\lambda)\ast}(x))(I+A)^{x+\lambda}=\frac{x}{a}T^{(\lambda+1)}(T^{(\lambda)})^{-1},$

to get to

$-R_{n-1}^{(\lambda+1)}(x)T^{(\lambda+1)}(I+A^{\ast})(T^{(\lambda)})^{-1}+R_{n-1}^{(\lambda+1)}(x-1)\frac{x}{a}T^{(\lambda+1)}(T^{(\lambda)})^{-1}=n\mathcal{D}_{n-1}^{(\lambda+1)}(\mathcal{D}_{n}^{(\lambda)})^{-1}R_{n}^{(\lambda)}(x).$

We now take $x\geq 1$, since the $x=0$ was covered in the previous result. We
can then use the first of the shifts to get

$\mathscr{W}_{R}^{(\lambda)}(x)=(I+A^{\ast})\sum_{n=0}^{\infty}R_{n}^{(\lambda)}(x-1)^{\ast}\mathcal{D}_{n}^{(\lambda)-1}R_{n}^{(\lambda)}(x)+\sum_{n=1}^{\infty}nR_{n-1}^{(\lambda+1)}(x-1)^{\ast}\mathcal{D}_{n}^{(\lambda)-1}R_{n}^{(\lambda)}(x).$

The first of the two sums is proportional to $\langle
Q_{x-1}^{(\lambda)},Q_{x}^{(\lambda)}\rangle_{d}^{(\lambda)}$ which is zero
due to the dual orthogonality. Inserting
$\mathcal{D}_{n-1}^{(\lambda+1)-1}\mathcal{D}_{n-1}^{(\lambda+1)}$ we get

$\mathscr{W}_{R}^{(\lambda)}(x)=\sum_{n=1}^{\infty}R_{n-1}^{(\lambda+1)}(x-1)^{\ast}\mathcal{D}_{n-1}^{(\lambda+1)-1}n\mathcal{D}_{n-1}^{(\lambda+1)}\mathcal{D}_{n}^{(\lambda)-1}R_{n}^{(\lambda)}(x),$

which allows us to apply the second shift

$\mathscr{W}_{R}^{(\lambda)}(x)=-\sum_{n=1}^{\infty}R_{n-1}^{(\lambda+1)}(x-1)^{\ast}\mathcal{D}_{n-1}^{(\lambda+1)-1}R_{n-1}^{(\lambda+1)}(x)T^{(\lambda+1)}(I+A^{\ast})(T^{(\lambda)})^{-1}$
$\hskip
142.26378pt+\sum_{n=1}^{\infty}R_{n-1}^{(\lambda+1)}(x-1)^{\ast}\mathcal{D}_{n-1}^{(\lambda+1)-1}R_{n-1}^{(\lambda+1)}(x-1)\frac{x}{a}T^{(\lambda+1)}(T^{(\lambda)})^{-1}.$

Here again the first sum vanishes due to the dual orthogonality, since it is
proportional to $\langle
Q_{x-1}^{(\lambda+1)},Q_{x}^{(\lambda+1)}\rangle_{d}^{(\lambda+1)}$. So we
then arrive at

$\mathscr{W}_{R}^{(\lambda)}(x)=\mathscr{W}_{R}^{(\lambda+1)}(x-1)\frac{x}{a}T^{(\lambda+1)}(T^{(\lambda)})^{-1},$

which after iteration gives
$\mathscr{W}_{R}^{(\lambda)}(x)=\mathscr{W}_{R}^{(\lambda+x)}(0)\frac{x!}{a^{x}}T^{(\lambda+x)}T^{(\lambda)-1}=\frac{x!}{a^{x}}T^{(\lambda)-1}$.
Putting this result back into (8.12), together with Lemma 8.14, gives the
desired result. ∎

###  8.6. Lie Algebras of difference operators associated to the Charlier
weight

As an aside, we describe Lie algebras generated by the operators of interest
in this paper. We denote by $\mathfrak{g}$ the linear span of
$\\{D,D^{\dagger},\mathfrak{J^{(\lambda)}},x,I\\}$, where $I$ is the identity
matrix.

######  Proposition 8.16.

The linear space $\mathfrak{g}$ is a Lie algebra that contains the operator
$\mathfrak{D}^{(\lambda)}$. More precisely, the following relations hold true

$[\mathfrak{J}^{(\lambda)},D]=D,\qquad[\mathfrak{J}^{(\lambda)},D^{\dagger}]=-D^{\dagger},\qquad[D,x]=-D,\qquad[D^{\dagger},x]=D^{\dagger},\qquad[D,D^{\dagger}]=-\frac{1}{a}I,$

where $[,]$ is the standard bracket $[E_{1},E_{2}]=E_{1}E_{2}-E_{2}E_{1}$.

###### Proof.

The commutations relations in the lemma follow directly from the definitions
of $\mathfrak{J}^{(\lambda)},$ in (5.12) and $D,D^{\dagger}$ in (5.11). This
shows that $\mathfrak{g}$ is closed under the bracket operation. Finally it
follows directly from (5.13) that $\mathfrak{D}^{(\lambda)}\in\mathfrak{g}$. ∎

We observe that since each of the generators of $\mathfrak{g}$ is related to a
weak Pearson equation, we have that
$\mathfrak{g}\subset\hat{\mathcal{F}}_{R}(P)$. Therefore, we can transport
this Lie algebra to
$\hat{\mathcal{F}}_{L}(P),\hat{\mathcal{F}}^{d}_{R}(Q),\hat{\mathcal{F}}^{d}_{L}(Q)$
via the isomorphisms $\psi,\sigma,\tau$ and we obtain the diagram:

${\psi^{-1}(\mathfrak{g})}$${\mathfrak{g}}$${\psi^{d}(\sigma(\mathfrak{g}))}$${\sigma(\mathfrak{g})}$$\scriptstyle{\psi}$$\scriptstyle{\tau}$$\scriptstyle{\sigma}$$\scriptstyle{\psi^{d}}$

The four dimensional Lie algebra $\mathfrak{g}_{1}$ generated by
$\\{D,D^{\dagger},\mathfrak{J^{(\lambda)}},I\\}$ is isomorphic to the Lie
algebra $\mathcal{G}(0,1)$ introduced by W. Miller in [41, chapter 2-5] via
the identifications:

$J^{+}\rightarrow D,\qquad J^{-}\rightarrow D^{\dagger},\qquad
J^{3}\rightarrow\mathfrak{J}^{(\lambda)},\qquad\mathcal{E}\rightarrow\frac{1}{a}I.$

The corresponding Casimir operator is given by

$aC_{0,1}=a\left(J^{+}J^{-}-\mathcal{E}J^{3}\right)\rightarrow
aDD^{\dagger}-\mathfrak{J}^{(\lambda)}=x-\mathfrak{J}^{(\lambda)},$

and commutes with every element in $\mathfrak{g}_{1}$. We observe that the
Casimir operator is indeed an element of the center
$\mathfrak{Z}(\mathfrak{g})$ of $\mathfrak{g}$, which is the two dimensional
abelian subalgebra generated by $\\{C_{0,1},I\\}$.

We observe that $C_{0,1}\in\hat{\mathcal{F}}_{R}(P)$ and therefore
$\psi^{-1}(C_{0,1})$ is a nontrivial second order operator in the bispectral
algebra $\mathcal{B}_{L}(P)$ which commutes with the ladder operators
$M,M^{\dagger}$ and the eigenvalue $\Gamma_{n}^{(\lambda)}$ of
$\mathfrak{D}^{(\lambda)}$.

######  Proposition 8.17.

Let $\mathfrak{k}$ be the four dimensional Lie algebra generated by
$\\{D,D^{\dagger},x,I\\}.$ Then $\mathfrak{k}$ is isomorphic with
$\mathcal{G}(0,1)$ via the identifications

$J^{+}\rightarrow D,\qquad J^{-}\rightarrow D^{\dagger},\qquad
J^{3}\rightarrow x,\qquad\mathcal{E}\rightarrow\frac{1}{a}I.$

Moreover, under this identification we have have

$\mathfrak{g}=\mathcal{G}(0,1)\oplus\mathbb{C}\,C_{0,1},$

###### Proof.

The isomorphism between $\mathfrak{k}$ and $\mathcal{G}(0,1)$ follows directly
from the relations in Proposition 8.16. Since $C_{0,1}\notin\mathfrak{k}$, we
immediately have that $\mathfrak{g}=\mathfrak{k}\oplus\mathbb{C}\,C_{0,1}$ as
vectors spaces. Now we only need to show that both summands are ideals in
$\mathfrak{g}$. Proposition 8.16 gives
$[\mathfrak{k},\mathfrak{g}]\subset\mathfrak{k}$ and $\mathbb{C}\,C_{0,1}$ is
an ideal since $C_{0,1}\in\mathfrak{z}(\mathfrak{g})$. This completes the
proof of the proposition. ∎

######  Remark 8.18.

Let $\mathfrak{h}$ be the three dimensional subspace generated by
$\\{D,D^{\dagger},I\\}$, then $\mathfrak{h}$ is isomorphic with the the
Heisenberg Lie algebra of dimension 3, see for instance [24, p. 57].

######  Remark 8.19.

The quotient Lie algebra $\mathfrak{g}/\mathfrak{z}(\mathfrak{g})$ is
generated by the elements:

$e_{1}=D^{\dagger}+\mathfrak{z}(\mathfrak{g}),\qquad
e_{2}=D+\mathfrak{z}(\mathfrak{g}),\qquad e_{3}=x+\mathfrak{z}(\mathfrak{g}),$

with commutation relations

$[e_{1},e_{2}]=0,\qquad[e_{1},e_{3}]=e_{1},\qquad[e_{2},e_{3}]=-e_{2}.$

Therefore $\mathfrak{g}/\mathfrak{z}(\mathfrak{g})$ is a three dimensional Lie
algebra which is isomorphic with the Lie algebra $\mathfrak{p}(1,1)$ of the
Poincaré group $P(1,1)$. This is the group of affine transformations of
$\mathbb{R}^{2}$ which preserve the Lorentz metric, see [29, §2.5.9].

##  9\. Dual–Dual polynomials

In this final section we will we investigate 4–tuples
$(\widetilde{P}_{n}^{(\lambda)},\widetilde{M}_{1},\widetilde{M}_{2},\widetilde{\nu}^{(\lambda)})$
that are dual to the dual 4–tuple
$(Q_{x,1}^{(\lambda)},P_{n}^{(\lambda)}(0),\Upsilon_{1}^{(\lambda)}(x),\rho_{1}^{(\lambda)})$
described in Section 8.2.1. These families will be called _dual–dual
polynomials_. The goal of this section is to indicate the steps in the
construction of dual–dual families, but we will not describe in detail some
sets that are analogues of those discussed in Section 3 (like
$\mathcal{B}_{L}^{d,2}(Q)$). More precisely, we will discuss two distinct
dual-dual families $\widetilde{P}_{n}^{(\lambda)}$ : one will be trivial,
giving us back $P_{n}^{(\lambda)}$, and the second will not be.

The dual–dual polynomials will be a sequence of matrix polynomials
$(\widetilde{P}_{n}^{(\lambda)})_{n}$ with a matrix argument $\mathcal{X}$ but
on the opposite side compared to the $Q_{x}^{(\lambda)}$

$\widetilde{P}_{n}^{(\lambda)}(\mathcal{X})=A_{n,n}\mathcal{X}^{n}+A_{n,n-1}\mathcal{X}^{n-1}+\ldots+A_{n,0},$

see (3.1) for the case corresponding to the $Q^{(\lambda)}_{x}$. To find such
a dual–dual family boils down to finding a sequence of polynomials
$(P_{n}^{(\lambda)})_{n}$ together with matrix functions
$\widetilde{\nu}^{(\lambda)}(x)$, $\widetilde{M}_{1}^{(\lambda)}(n)$,
$\widetilde{M}_{2}^{(\lambda)}(x)$ such that

$Q_{x}^{(\lambda)}(\rho^{(\lambda)}(n))=\widetilde{M}_{1}^{(\lambda)}(n)\widetilde{P}_{n}^{(\lambda)}(\widetilde{\nu}^{(\lambda)}(x))\widetilde{M}_{2}^{(\lambda)}(x),\qquad\forall
x,n\in\mathbb{N}_{0},$ (9.1)

and such that the polynomials $\widetilde{P}_{n}^{(\lambda)}$ satisfy the
monic three term recurrence

$\widetilde{P}_{n}^{(\lambda)}(\widetilde{\nu}^{(\lambda)}(x))\widetilde{\nu}^{(\lambda)}(x)=\widetilde{P}_{n+1}^{(\lambda)}(\widetilde{\nu}^{(\lambda)}(x))+\widetilde{B}_{n}^{(\lambda)}\widetilde{P}_{n}^{(\lambda)}(\widetilde{\nu}^{(\lambda)}(x))+\widetilde{C}_{n}^{(\lambda)}\widetilde{P}_{n-1}^{(\lambda)}(\widetilde{\nu}^{(\lambda)}(x)).$
(9.2)

As in Section 3, a dual–dual family is determined by a 4–tuple
$(\widetilde{P}^{(\lambda)}_{n},\widetilde{M}_{1}^{(\lambda)},\widetilde{M}_{2}^{(\lambda)},\widetilde{\nu}^{(\lambda)})$
and we will also restrict the superfluous degrees of freedom in a similar
fashion by setting $\widetilde{M}_{1}(0)=\widetilde{M}_{2}(0)=I$. This
immediately leads to
$\widetilde{M}_{2}(x)=Q_{x}^{(\lambda)}(\rho^{(\lambda)}(0))=\Upsilon^{(\lambda)}(x)^{-1}$
and
$\widetilde{M}_{1}(n)=(\widetilde{P}_{n}^{(\lambda)}(\widetilde{\nu}^{(\lambda)}(0)))^{-1}$.

###  9.1. First family of dual–dual polynomials: $P_{n}^{(\lambda)}$ as its
own dual–dual

The dual condition (9.1) implies trivially that

$(\widetilde{P}^{(\lambda)}_{n},\widetilde{M}_{1}^{(\lambda)},\widetilde{M}_{2}^{(\lambda)},\widetilde{\nu}^{(\lambda)})=(P^{(\lambda)}_{n},P_{n}^{(\lambda)}(0)^{-1},(I+A)^{x}a^{x},x)$

is a dual family for
$(Q^{(\lambda)}_{x},P_{n}^{(\lambda)}(0),(I+A)^{-x}a^{-x},\rho_{1}^{(\lambda)}(n))$,
i.e.

$Q_{x}^{(\lambda)}(\rho_{1}^{(\lambda)}(n))=P^{(\lambda)}_{n}(0)^{-1}P_{n}^{(\lambda)}(x)(I+A)^{x}a^{x},\qquad\forall
x,n\in\mathbb{N}_{0}.$

We observe that since $\widetilde{\nu}^{(\lambda)}=x$, the recurrence relation
for $P_{n}^{(\lambda)}$ in (9.2) is precisely the three term recurrence
relation (2.6).

As in Lemma 3.5, this dual–dual family is related to a difference operator,
namely $\tau(\mathcal{L})\in\hat{\mathcal{F}}_{L}^{d}(Q)$, where
$\mathcal{L}=\psi^{-1}(x)$ is defined in (2.8). Explicitly we have

$\tau(\mathcal{L})=P_{n}^{(\lambda)}(0)^{-1}P_{n+1}^{(\lambda)}(0)\delta+P_{n}^{(\lambda)}(0)^{-1}B_{n}^{(\lambda)}P_{n}^{(\lambda)}(0)+P_{n}^{(\lambda)}(0)^{-1}C_{n}^{(\lambda)}P_{n-1}^{(\lambda)}(0)\delta^{-1}.$

We note that Corollary 8.3 implies that the coefficient
$P_{n}^{(\lambda)}(0)^{-1}P_{n+1}^{(\lambda)}(0)$ of $\delta$ in
$\tau(\mathcal{L})$ is a rational function of $n$ which is invertible for all
$n\in\mathbb{N}_{0}$. On the other hand, the explicit expression in Theorem
6.5 implies that $C_{n}$ is a rational function of $n$ which vanishes at
$n=0$. Therefore, the coefficient of $\delta^{-1}$ in $\tau(\mathcal{L})$ is a
rational function of $n$ and vanishes at $n=0$. These conditions are analogous
to those in Definition 3.7. For this dual-dual family, the commutative diagram
in Theorem 3.22 is extended trivially in the following way:

${\hat{\mathcal{F}}_{L}(P_{n}^{(\lambda)})}$${\hat{\mathcal{F}}^{d}_{L}(Q_{n}^{(\lambda)})}$${\hat{\mathcal{F}}_{L}(P_{n}^{(\lambda)})}$${\hat{\mathcal{F}}_{R}(P)}$${\hat{\mathcal{F}}^{d}_{R}(Q_{n}^{(\lambda)})}$${\hat{\mathcal{F}}_{R}(P_{n}^{(\lambda)})}$$\scriptstyle{\tau}$$\scriptstyle{\psi}$$\scriptstyle{\tau^{-1}}$$\scriptstyle{\psi}$$\scriptstyle{\sigma_{1}}$$\scriptstyle{\psi^{d}}$$\scriptstyle{\sigma_{1}^{-1}}$

In conclusion $P_{n}^{(\lambda)}$ is its own dual–dual.

###  9.2. Second family of dual–dual polynomials

In order to construct a nontrivial family of dual–dual polynomials, we look
for second order difference operators in $\mathcal{B}_{L}^{d}(Q)$. There are
two natural candidates for these operators, namely
$\tau(\psi^{-1}(\mathfrak{J}^{(\lambda)}))$ and the Casimir operator
$\tau(\psi^{-1}(C_{0,1}))$. Next we follow the construction for the first
operator. We have already seen in Remark 5.12 that

$\psi^{-1}(\mathfrak{J}^{(\lambda)})\cdot
P_{n}^{(\lambda)}=P_{n}^{(\lambda)}\cdot\mathfrak{J}^{(\lambda)},$

so that $\mathfrak{J}^{(\lambda)}\in\hat{\mathcal{F}}_{R}(P)$ and
$\psi^{-1}(\mathfrak{J}^{(\lambda)})\in\hat{\mathcal{F}}_{L}(P)$. Let us write
out

$\psi^{-1}(\mathfrak{J}^{(\lambda)})=G_{1}(n)\delta+G_{0}(n)+G_{-1}(n)\delta^{-1},$

where the explicit expressions of the coefficients $G_{1},G_{0},G_{-1}$ are
given in Remark 5.12. On the other hand,
$\sigma(\mathfrak{J}^{(\lambda)})=\Upsilon^{(\lambda)}(x)\mathfrak{J}^{(\lambda)}\Upsilon^{(\lambda)}(x)^{-1}$
and

$\tau\bigl{(}\psi^{-1}(\mathfrak{J}^{(\lambda)})\bigr{)}=P_{n}^{(\lambda)}(0)^{-1}G_{1}(n)P_{n+1}^{(\lambda)}(0)\delta+P_{n}^{(\lambda)}(0)^{-1}G_{0}(n)P_{n}^{(\lambda)}(0)+P_{n}^{(\lambda)}(0)^{-1}G_{-1}(n)P_{n-1}^{(\lambda)}(0)\delta^{-1}.$

It follows directly from Remark 5.12 that the coefficient
$P_{n}^{(\lambda)}(0)^{-1}G_{1}(n)P_{n+1}^{(\lambda)}(0)$ is is a rational
function which is invertible for all $n\in\mathbb{N}_{0}$. Moreover, by
Corollary 5.7 we have that

$G_{-1}(n)=L_{0}^{-1}(I+A)^{n+\lambda}\mathcal{D}_{n}^{(\lambda)}(\mathcal{D}_{n-1}^{(\lambda)})^{-1}(I+A)^{-n-\lambda+1}L_{0}.$

Therefore (6.18) implies that $G_{-1}(n)$ is a rational function of $n$ which
vanishes at $n=0$.

######  Theorem 9.1.

Let $\widetilde{P}^{(\lambda)}_{n}$ be the sequence of monic polynomials
defined by the recurrence relation

$x\widetilde{P}^{(\lambda)}_{n}(x)=\widetilde{P}^{(\lambda)}_{n+1}(x)\\\
+(J+n+a+\lambda(I+A)^{-1}+aL_{0}^{-1}(I+A)^{\lambda}\mathcal{D}_{n}^{(\lambda)}A^{\ast}\mathcal{D}_{n}^{(\lambda)-1}(I+A)^{-\lambda}L_{0})\widetilde{P}^{(\lambda)}_{n}(x)\\\
+L_{0}^{-1}(I+A)^{\lambda}\mathcal{D}_{n}^{(\lambda)}\mathcal{D}_{n-1}^{(\lambda)-1}(I+A)^{-\lambda}L_{0}\widetilde{P}^{(\lambda)}_{n-1}(x).$

Then
$(\widetilde{P}^{(\lambda)}_{n},P^{(\lambda)}_{n}(0)^{-1}(I+A)^{n},\Upsilon^{(\lambda)}(x)^{-1},\mathfrak{J}^{(\lambda)}(x))$
is dual to $Q_{x}^{(\lambda)}$ and the following duality condition holds true

$Q^{(\lambda)}_{x}(\rho(n))=P^{(\lambda)}_{n}(0)^{-1}(I+A)^{n}\widetilde{P}^{(\lambda)}_{n}(\mathfrak{J}^{(\lambda)}(x)))\Upsilon^{(\lambda)}(x)^{-1},\qquad
n,x\in\mathbb{N}_{0}.$

###### Proof.

If we replace the dual condition in the relation
$\tau\bigl{(}\psi^{-1}(\mathfrak{J}^{(\lambda)})\bigr{)}\cdot
Q_{x}^{(\lambda)}=Q_{x}^{(\lambda)}\cdot\sigma(\mathfrak{J}^{(\lambda)})$, we
obtain the recurrence relation of theorem with $x$ replaced by
$\mathfrak{J}^{(\lambda)}(x)$. Since $\mathfrak{J}^{(\lambda)}$ satisfies the
block Vandermonde condition with $n$ replaced by $x$, an argument similar to
the one in (3.9) completes the proof of the theorem. ∎

######  Corollary 9.2.

The dual-dual family
$(\widetilde{P}^{(\lambda)}_{n},P^{(\lambda)}_{n}(0)^{-1}(I+A)^{n},\Upsilon^{(\lambda)}(x)^{-1},\mathfrak{J}^{(\lambda)}(x))$
is related to the original monic polynomials $P_{n}^{(\lambda)}$ in the
following way

$P_{n}^{(\lambda)}(x)=(I+A)^{n}\widetilde{P}_{n}^{(\lambda)}(\mathfrak{J}^{(\lambda)}(x)).$

###### Proof.

Using the dual relation for $Q_{x}^{(\lambda)}$ and
$\widetilde{P}_{n}^{(\lambda)}$ in Theorem 9.1 as well as the dual relation
for $P_{n}^{(\lambda)}$ and $Q_{x}^{(\lambda)}$ in (8.7) we get the result. ∎

##  Appendix A Miscellaneous Proofs

###  A.1. Block Vandermonde Determinant

This subsection will mostly involve upper triangular matrices and we will
focus our attention on their diagonal parts. So when two upper triangular
matrices $\mathcal{M}_{1},\mathcal{M}_{2}$ have the same diagonal, we will
write

$\mathcal{M}_{1}=\mathcal{M}_{2}+\textrm{s.u.t.}$

where s.u.t. stands for strictly upper triangular. All the matrices
$\rho_{i}^{(\lambda)}(n)$ in Section 8 satisfy the conditions in Corollary
A.2, but we will collect a preliminary result for a slightly simpler case
first.

######  Lemma A.1.

Consider the $n$-dependent upper triangular matrix of the following form

$\mathcal{M}(n)=n\mathcal{T}+\mathcal{T}_{0}+\mathrm{s.u.t.}$ (A.1)

where $\mathcal{T}$ and $\mathcal{T}_{0}$ are constant diagonal matrices, and
the s.u.t. part is allowed to depend on $n$. The determinant of its block
Vandermonde matrix does not depend on the s.u.t. part or $\mathcal{T}_{0}$,
and is given by

$\det\begin{pmatrix}I&\mathcal{M}(n_{0})&\dots&\mathcal{M}(n_{0})^{x}\\\
I&\mathcal{M}(n_{1})&\dots&\mathcal{M}(n_{1})^{x}\\\
\vdots&\vdots&\,&\vdots\\\
I&\mathcal{M}(n_{x})&\dots&\mathcal{M}(n_{x})^{x}\end{pmatrix}=\det(\mathcal{T})^{\frac{1}{2}x(x+1)}\left(\prod_{0\leq
s<t\leq x}(n_{t}-n_{s})\right)^{N},\qquad x\in\mathbb{N}_{>0},$

where $(n_{j})_{j=0}^{x}$ is a list of complex values for which
$\mathcal{M}(n_{j})$ is defined.

###### Proof.

This proof will be by induction on $x$. On occasion we will denote the block
Vandermonde as $\det\left(\mathcal{M}(n_{j})^{k}\right)_{j,k=0}^{x}$ even
though this a slight abuse of notation in case $\mathcal{M}(n_{j})$ is
singular. We apply a block analogue of an elementary row operation to get

$\displaystyle\det\begin{pmatrix}I&\mathcal{M}(n_{0})&\dots&\mathcal{M}(n_{0})^{x}\\\
I&\mathcal{M}(n_{1})&\dots&\mathcal{M}(n_{1})^{x}\\\
\vdots&\vdots&\,&\vdots\\\
I&\mathcal{M}(n_{x})&\dots&\mathcal{M}(n_{x})^{x}\end{pmatrix}$
$\displaystyle=\det\begin{pmatrix}I&\mathcal{M}(n_{0})&\dots&\mathcal{M}(n_{0})^{x}\\\
I&\mathcal{M}(n_{1})&\dots&\mathcal{M}(n_{1})^{x}\\\
\vdots&\vdots&\,&\vdots\\\
I&\mathcal{M}(n_{x})&\dots&\mathcal{M}(n_{x})^{x}\end{pmatrix}\det\begin{pmatrix}I&-\mathcal{M}(n_{0})&0&\dots&0\\\
0&I&-\mathcal{M}(n_{0})&\dots&0\\\ \vdots&\ddots&\ddots&\ddots&\vdots\\\
0&\dots&0&I&-\mathcal{M}(n_{0})\\\ 0&\dots&\dots&0&I\end{pmatrix}$
$\displaystyle=\det\begin{pmatrix}I&0&\dots&0\\\
I&(n_{1}-n_{0})\mathcal{T}+\textrm{s.u.t.}&\dots&\mathcal{M}(n_{1})^{x-1}(n_{1}-n_{0})\mathcal{T}+\textrm{s.u.t.}\\\
\vdots&\vdots&\,&\vdots\\\
I&(n_{x}-n_{0})\mathcal{T}+\textrm{s.u.t.}&\dots&\mathcal{M}(n_{x})^{x-1}(n_{x}-n_{0})\mathcal{T}+\textrm{s.u.t.}\end{pmatrix}.$

In the last step we have used that
$\mathcal{M}(n_{j})-\mathcal{M}(n_{0})=(n_{j}-n_{0})\mathcal{T}+\textrm{s.u.t}$.
To proceed we notice that
$\mathcal{M}(n_{j})^{k}-\mathcal{M}(n_{j})^{k-1}\mathcal{M}(n_{0})=(n_{j}-n_{0})\mathcal{T}\left(n_{j}\mathcal{T}+\mathcal{T}_{0}\right)^{k-1}+\textrm{s.u.t.}$
and reduce the size of the determinent using the Schur complement

$\displaystyle=\det\begin{pmatrix}(n_{1}-n_{0})\mathcal{T}+\textrm{s.u.t.}&\dots&(n_{1}-n_{0})\mathcal{T}(n_{1}\mathcal{T}+\mathcal{T}_{0})^{x-1}+\textrm{s.u.t.}\\\
\vdots&\,&\vdots\\\
(n_{x}-n_{0})\mathcal{T}+\textrm{s.u.t.}&\dots&(n_{x}-n_{0})\mathcal{T}(n_{x}\mathcal{T}+\mathcal{T}_{0})^{x-1}+\textrm{s.u.t.}\end{pmatrix}$
$\displaystyle=\det\left(\textrm{diag}((n_{j}-n_{0})\mathcal{T}+\textrm{s.u.t.})_{j=1}^{x}\right)\det\begin{pmatrix}I&n_{1}\mathcal{T}+\textrm{s.u.t.}&\dots&(n_{1}\mathcal{T}+\mathcal{T}_{0})^{x-1}+\textrm{s.u.t.}\\\
\vdots&\vdots&\,&\vdots\\\
I&n_{x}\mathcal{T}+\textrm{s.u.t.}&\dots&(n_{x}\mathcal{T}+\mathcal{T}_{0})^{x-1}+\textrm{s.u.t.}\end{pmatrix}$

The block diagonal matrix in the second line is meant to have exactly the
entries of the first column of the matrix in the previous line; in particular
it has the same s.u.t. parts. We need this so that the first column of the
second determinant is exactly just identity matrices without any s.u.t. part.

So we have a reduction of the block Vandermonde to one of a smaller size

$\det\left(\mathcal{M}(n_{j})^{k}\right)_{j,k=0}^{x}=\det(\mathcal{T})^{x}\left(\prod_{j=1}^{x}(n_{j}-n_{0})\right)^{N}\det\left(\mathcal{N}(n_{j+1})^{k}\right)_{j,k=0}^{x-1},$
(A.2)

with $\mathcal{N}(n)=n\mathcal{T}+\mathcal{T}_{0}+\textrm{s.u.t.}$ So
$\mathcal{N}(n)$ has the same diagonal entries as $\mathcal{M}(n)$. Then since
the factor in front of the block Vandermonde determinant with $\mathcal{N}(n)$
in (A.2), does not depend on the specifics of the s.u.t. we can continue
reducing the size of the block Vandermonde until we get the desired result. ∎

######  Corollary A.2.

The determinant formula in Lemma A.1 still holds if we conjugate
$\mathcal{M}(n)$ in (A.1) by an invertible matrix $\mathcal{Q}$ that does not
depend on $n$.

###### Proof.

Let $\mathcal{R}(n)=\mathcal{Q}\mathcal{M}(n)\mathcal{Q}^{-1}$. Then it is
easy to see that

$\det\begin{pmatrix}I&\mathcal{R}(n_{0})&\dots&\mathcal{R}(n_{0})^{x}\\\
I&\mathcal{R}(n_{1})&\dots&\mathcal{R}(n_{1})^{x}\\\
\vdots&\vdots&\,&\vdots\\\
I&\mathcal{R}(n_{x})&\dots&\mathcal{R}(n_{x})^{x}\end{pmatrix}=\frac{\det\left(\textrm{diag}(\mathcal{Q})_{j=0}^{x}\right)}{\det\left(\textrm{diag}(\mathcal{Q})_{j=0}^{x}\right)}\det\begin{pmatrix}I&\mathcal{M}(n_{0})&\dots&\mathcal{M}(n_{0})^{x}\\\
I&\mathcal{M}(n_{1})&\dots&\mathcal{M}(n_{1})^{x}\\\
\vdots&\vdots&\,&\vdots\\\
I&\mathcal{M}(n_{x})&\dots&\mathcal{M}(n_{x})^{x}\end{pmatrix}$

∎

###  A.2. Proof of Proposition 8.2

In this section we prove the first part of Proposition 8.2 which states that

$R_{n}^{(\lambda)}(0)=-a(I+\mathcal{A}_{n+\lambda})R_{n-1}^{(\lambda+1)}(0),$
(A.3)

using the matrix introduced in (6.23) to express
$\mathcal{A}_{n+\lambda}=(N+\lambda+n-J)^{-1}JA^{\ast}.$

###### Proof.

Recall that the entries $R_{n}^{(\lambda)}(0)_{jk}=\xi_{j,k,n}^{(\lambda)}$
are given explicitly in Theorem 7.9 and that
$A_{j+1,j}=\frac{\sqrt{N-j}}{\sqrt{a}}$. So we start off by writing (A.3) in
terms of its entries

$\xi_{j,k,n}^{(\lambda)}=-a\xi_{j,k,n-1}^{(\lambda+1)}-\frac{j\sqrt{a}\sqrt{N-j}}{(N+\lambda+n-j)}\xi_{j+1,k,n-1}^{(\lambda+1)}.$

Note that since the $\xi_{j,k,n}^{(\lambda)}$ are given by two different
expressions (for $n+j\geq N$ and for $n+j<N$), we should prove the desired
entrywise recursion for both cases. But it is not necessary to prove a mixed
case because the two cases actually coincide for $n+j=N$, so for any values of
the parameters the three terms can always be considered as belonging to the
same case.

Before specifying to either case, we compute the following expressions which
will come in handy later on in the proof

$\displaystyle\frac{\mathfrak{X}(j,k,n-1,\lambda+1)}{\mathfrak{X}(j,k,n,\lambda)}$
$\displaystyle=-\frac{N+\lambda+1-j}{a(\lambda+1)},\qquad\frac{\mathfrak{X}(j+1,k,n-1,\lambda+1)}{\mathfrak{X}(j,k,n,\lambda)}=\frac{\sqrt{N-j}}{\sqrt{a}}\frac{N+\lambda+n-j}{j(\lambda+1)}.$

To derive the $n+j>N$ case we will need a fairly standard hypergeometric
identity that is easy to check in general (for parameters with which the
series either converge or truncate)

$\mathfrak{d}\left(\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b},\mathfrak{c}}{\mathfrak{d},\mathfrak{e}};1\right)-\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b},\mathfrak{c}}{\mathfrak{d}+1,\mathfrak{e}};1\right)\right)=\mathfrak{b}\left(\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b}+1,\mathfrak{c}}{\mathfrak{d}+1,\mathfrak{e}};1\right)-\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b},\mathfrak{c}}{\mathfrak{d}+1,\mathfrak{e}};1\right)\right),$

It follows from both sides being equal to

$\sum_{s=1}\frac{(\mathfrak{a})_{s}(\mathfrak{b})_{s}(\mathfrak{c})_{s}s}{s!(\mathfrak{d}+1)_{s}(\mathfrak{e})_{s}}.$

When we fill in the values $\mathfrak{a}=1-k$, $\mathfrak{b}=j-N$,
$\mathfrak{c}=n+\lambda+1$, $\mathfrak{d}=\lambda+1$ and $\mathfrak{e}=1-N$
and collect some similar terms we get

$\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,j-N,n+\lambda+1}{\lambda+1,1-N};1\right)$
$\displaystyle=\frac{N+\lambda+1-j}{\lambda+1}\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,j-N,n+\lambda+1}{\lambda+2,1-N};1\right)$
$\displaystyle\qquad-\frac{N-j}{\lambda+1}\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,j-N+1,n+\lambda+1}{\lambda+2,1-N};1\right).$

After multiplying this equation by $\mathfrak{X}(j,k,n,\lambda)(1-N)_{k-1}$ we
get the desired result for $n+j>N$ by using the ratios of different
$\mathfrak{X}$ expressed earlier in this proof.

The $n+j\leq N$ case is slightly different. Note that now instead of the
factor of $(1-N)_{k-1}$ we have $(1-n-j)_{k-1}$ so it is no longer a common
factor in the three terms of our desired result. The necessary hypergeometric
identity is somewhat less standard,

$\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b},\mathfrak{c}}{\mathfrak{d},\mathfrak{e}};1\right)=\frac{\mathfrak{c}}{\mathfrak{d}}\frac{\mathfrak{e}-\mathfrak{a}}{\mathfrak{e}}\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b}+1,\mathfrak{c}+1}{\mathfrak{d}+1,\mathfrak{e}+1};1\right)-\frac{\mathfrak{c}-\mathfrak{d}}{\mathfrak{d}}\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{\mathfrak{a},\mathfrak{b}+1,\mathfrak{c}}{\mathfrak{d}+1,\mathfrak{e}};1\right),$

but it can be shown to hold using the methods described in [23] for the cases
where the series either converge or truncate. We then put in the values
$\mathfrak{a}=1-k$, $\mathfrak{b}=-n$, $\mathfrak{c}=N-j+\lambda+1$,
$\mathfrak{d}=\lambda+1$ and $\mathfrak{e}=1-n-j$, to get

$\,{}_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,-n,N-j+\lambda+1}{\lambda+1,1-n-j};1\right)$
$\displaystyle=\frac{N+\lambda+1-j}{\lambda+1}\frac{n+j-k}{n+j-1}\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,-n+1,N-j+\lambda+2}{\lambda+2,2-n-j};1\right)$
$\displaystyle\qquad-\frac{N-j}{\lambda+1}\,_{3}F_{2}\left(\genfrac{.}{.}{0.0pt}{}{1-k,-n+1,N-j+\lambda+1}{\lambda+2,1-n-j};1\right).$

After multiplying this equation by $\mathfrak{X}(j,k,n,\lambda)(1-n-j)_{k-1}$
we get the desired result for $n+j\leq N$ by using the ratios of different
$\mathfrak{X}$ expressed earlier in this proof.

∎

###  A.3. Proof of Lemma 8.14

We want to find $\mathscr{W}_{R}^{(\lambda)}(0)$ so that we will have the zero
moment courtesy of (8.12). For this we need
$(\mathcal{D}_{n}^{(\lambda)})_{jj}$ and
$R_{n}^{(\lambda)}(0)_{j,k}=\xi_{j,k,n}^{(\lambda)}$. The entrywise expression
is

$\left(\mathscr{W}_{R}^{(\lambda)}(0)\right)_{j,k}=\sum_{n=0}^{\infty}\sum_{r=1}^{N}\frac{\xi_{r,j,n}^{(\lambda)}\xi_{r,k,n}^{(\lambda)}}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{rr}}.$

Since this expression is clearly symmetric, we will take $j\geq k$ from now
on. It will prove useful to have the standard weight of the scalar dual Hahn
polynomials nearby c.f. [42],

$w_{dH}(x;\gamma,\delta,\mathcal{N})=\frac{(2x+\gamma+\delta+1)(\gamma+x)!(x+\gamma+\delta)!\delta!(\mathcal{N}!)^{2}}{x!\gamma!(\delta+x)!(\mathcal{N}-x)!(x+\gamma+\delta+\mathcal{N}+1)!}.$

Let us introduce the following shorthand to distinguish between two different
expressions for $\xi$ that hold for different parameter values,

$\xi_{r,k,n}^{(\lambda)}=\mathfrak{X}(r,k,n,\lambda)\times\begin{cases}\alpha_{r,k,n}^{(\lambda)}&n+r\geq
N,\\\ \beta_{r,k,n}^{(\lambda)}&k\leq n+r<N,\\\ 0&n+r<k.\end{cases}$

The explicit expressions for $\alpha$ and $\beta$ can be found in Theorem 7.9.
We then write out the double sum, which we note will not have any cross terms

$\displaystyle\left(\mathscr{W}_{R}^{(\lambda)}(0)\right)_{j,k}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{r=r_{0}}^{N}\frac{\mathfrak{X}(r,j,n,\lambda)\mathfrak{X}(r,k,n,\lambda)}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{rr}}\alpha_{r,j,n}^{(\lambda)}\alpha_{r,k,n}^{(\lambda)}$
$\displaystyle\hskip
113.81102pt+\sum_{n=0}^{N-1}\sum_{r=r_{1}}^{N-n-1}\frac{\mathfrak{X}(r,j,n,\lambda)\mathfrak{X}(r,k,n,\lambda)}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{rr}}\beta_{r,j,n}^{(\lambda)}\beta_{r,k,n}^{(\lambda)}.$

We denote $r_{0}=\max(1,N-n)$ and $r_{1}=\max(1,j-n)$ and whenever a sum is
over the empty set, we take it to be zero. It will come in handy to introduce
another summation variable $s=n+r$ and rearrange the summations

$\displaystyle\left(\mathscr{W}_{R}^{(\lambda)}(0)\right)_{j,k}$
$\displaystyle=\sum_{s=N}^{\infty}\sum_{r=1}^{N}\frac{\mathfrak{X}(r,j,s-r,\lambda)\mathfrak{X}(r,k,s-r,\lambda)}{\left(\mathcal{D}_{s-r}^{(\lambda)}\right)_{rr}}\alpha_{r,j,s-r}^{(\lambda)}\alpha_{r,k,s-r}^{(\lambda)}$
(A.4)
$\displaystyle\qquad+\sum_{s=j}^{N-1}\sum_{n=0}^{s-1}\frac{\mathfrak{X}(s-n,j,n,\lambda)\mathfrak{X}(s-n,k,n,\lambda)}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{s-n,s-n}}\beta_{s-n,j,n}^{(\lambda)}\beta_{s-n,k,n}^{(\lambda)}.$

The common part of the summands is (note that the rightmost factor is not a
factorial)

$\frac{\mathfrak{X}(r,j,n,\lambda)\mathfrak{X}(r,k,n,\lambda)}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{rr}}=\frac{2^{\lambda}\sqrt{(N-j)!(N-k)!}(\lambda+n)!}{e^{a}a^{\lambda+\frac{1}{2}(j+k)}n!(\lambda!)^{2}}\frac{a^{r+n}(N+\lambda-r)!(N+1-r+n+\lambda)}{(r-1)!(N-r)!(N+n+\lambda)!},$

where we have used the expression of the inverse of the diagonal part of the
square norm in terms of factorials

$\left(\mathcal{D}_{n}^{(\lambda)}\right)_{rr}^{-1}=e^{-a}\left(\frac{2}{a}\right)^{\lambda}\frac{(r-1)!}{n!a^{n}}\frac{(N-r+n+\lambda)!(N+1-r+n+\lambda)!}{(\lambda+n)!(N-r+\lambda)!(N+n+\lambda)!}.$

The distinct parts of the summands are

$\displaystyle\alpha_{r,j,s-r}^{(\lambda)}\alpha_{r,k,s-r}^{(\lambda)}$
$\displaystyle=(-1)^{j+k}\frac{(N-1)!^{2}}{(N-k)!(N-j)!}$
$\displaystyle\qquad\times\mathcal{R}_{j-1}\bigl{(}\ell(N-r);\,\lambda,s-N,N-1\bigr{)}\mathcal{R}_{k-1}\bigl{(}\ell(N-r);\,\lambda,s-N,N-1\bigr{)}$
$\displaystyle\beta_{s-n,j,n}^{(\lambda)}\beta_{s-n,k,n}^{(\lambda)}$
$\displaystyle=(-1)^{j+k}\frac{(s-1)!^{2}}{(s-k)!(s-j)!}$
$\displaystyle\qquad\times\mathcal{R}_{j-1}\bigl{(}\ell(n);\,\lambda,N-s,s-1\bigr{)}\mathcal{R}_{k-1}\bigl{(}\ell(n);\,\lambda,N-s,s-1\bigr{)}.$

Fortunately we can find a corresponding dual Hahn weight in the common parts
as follows

$\displaystyle\frac{\mathfrak{X}(r,j,s-r,\lambda)\mathfrak{X}(r,k,s-r,\lambda)}{\left(\mathcal{D}_{s-r}^{(\lambda)}\right)_{rr}}$
$\displaystyle=\frac{2^{\lambda}\sqrt{(N-j)!(N-k)!}}{e^{a}a^{\lambda-s+\frac{1}{2}(j+k)}}\frac{w_{dH}(N-r;\lambda,s-N,N-1)}{\lambda!(s-N)!((N-1)!)^{2}},$

and similarly in

$\displaystyle\frac{\mathfrak{X}(s-n,j,n,\lambda)\mathfrak{X}(s-n,k,n,\lambda)}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{s-n,s-n}}$
$\displaystyle=\frac{2^{\lambda}\sqrt{(N-j)!(N-k)!}}{e^{a}a^{\lambda-s+\frac{1}{2}(j+k)}}\frac{w_{dH}(n;\lambda,N-s,s-1)}{\lambda!(N-s)!((s-1)!)^{2}}.$

The first sum in (A.4) is then

$\displaystyle\sum_{s=N}^{\infty}\sum_{r=1}^{N}\frac{\mathfrak{X}(r,j,s-r,\lambda)\mathfrak{X}(r,k,s-r,\lambda)}{\left(\mathcal{D}_{s-r}^{(\lambda)}\right)_{rr}}\alpha_{r,j,s-r}^{(\lambda)}\alpha_{r,k,s-r}^{(\lambda)}$
$\displaystyle\qquad=\frac{e^{-a}2^{\lambda}(-1)^{j+k}a^{-\lambda-\frac{1}{2}(j+k)}}{\lambda!\sqrt{(N-j)!(N-k)!}}\sum_{s=N}^{\infty}\frac{a^{s}}{(s-N)!}\sum_{r=1}^{N}w_{dH}(N-r;\lambda,s-N,N-1)$
$\displaystyle\qquad\qquad\times\mathcal{R}_{j-1}\bigl{(}\ell(N-r);\,\lambda,s-N,N-1\bigr{)}\mathcal{R}_{k-1}\bigl{(}\ell(N-r);\,\lambda,s-N,N-1\bigr{)},$

which vanishes when $j\neq k$ due to the orthogonality of the dual Hahn
polynomials. When we do have $j=k$ we get

$\displaystyle\frac{e^{-a}2^{\lambda}a^{-\lambda-j}}{\lambda!(N-j)!}\sum_{s=N}^{\infty}\frac{a^{s}}{(s-N)!}\sum_{r=1}^{N}w_{dH}(N-r;\lambda,s-N,N-1)\mathcal{R}_{j-1}\bigl{(}\ell(N-r);\,\lambda,s-N,N-1\bigr{)}^{2}$
$\displaystyle\qquad=e^{-a}\left(\frac{2}{a}\right)^{\lambda}\frac{(j-1)!}{(\lambda+j-1)!}\sum_{s=N}^{\infty}\frac{a^{s-j}}{(s-j)!}.$

The second sum in (A.4) is then

$\displaystyle\sum_{s=j}^{N}\sum_{n=0}^{s-1}\frac{\mathfrak{X}(s-n,j,n,\lambda)\mathfrak{X}(s-n,k,n,\lambda)}{\left(\mathcal{D}_{n}^{(\lambda)}\right)_{s-n,s-n}}\beta_{s-n,j,n}^{(\lambda)}\beta_{s-n,k,n}^{(\lambda)}$
$\displaystyle\qquad=(-1)^{j+k}\sum_{s={j}}^{N}\frac{e^{-a}a^{s-\frac{1}{2}(j+k)}}{(s-k)!(s-j)!}\left(\frac{2}{a}\right)^{\lambda}\frac{\sqrt{(N-j)!(N-k)!}}{(N-s)!\lambda!}\sum_{n=0}^{s-1}w_{dH}(n;\lambda,N-s,s-1)$
$\displaystyle\qquad\qquad\times\mathcal{R}_{j-1}\bigl{(}\ell(n);\,\lambda,N-s,s-1\bigr{)}\mathcal{R}_{k-1}\bigl{(}\ell(n);\,\lambda,N-s,s-1\bigr{)}.$

which also vanishes for $j\neq k$ and when we do have $j=k$ we get

$\displaystyle\sum_{s={j}}^{N-1}\frac{e^{-a}a^{s-j}}{(s-j)!^{2}}\left(\frac{2}{a}\right)^{\lambda}\frac{(N-j)!}{(N-s)!\lambda!}\sum_{n=0}^{s-1}w_{dH}(n;\lambda,N-s,s-1)\mathcal{R}_{j-1}\bigl{(}\ell(n);\,\lambda,N-s,s-1\bigr{)}^{2}$
$\displaystyle\qquad\qquad=e^{-a}\left(\frac{2}{a}\right)^{\lambda}\frac{(j-1)!}{(\lambda+j-1)!}\sum_{s={j}}^{N-1}\frac{a^{s-j}}{(s-j)!}.$

The only thing that is left to do is combine the sums and perform the
summation over $s$

$\left(\mathscr{W}_{R}^{(\lambda)}(0)\right)_{jj}=\frac{e^{-a}(j-1)!}{(\lambda+j-1)!}\left(\frac{2}{a}\right)^{\lambda}\sum_{s=j}^{\infty}\frac{a^{s-j}}{(s-j)!}=\frac{(j-1)!}{(\lambda+j-1)!}\left(\frac{2}{a}\right)^{\lambda}=\left(T^{(\lambda)}\right)_{jj}^{-1}.$

The entries of $T^{(\lambda)}$ were given in (6.16). Now all that is left is
to use (8.12) to arrive at the zero moment

$\mathscr{W}_{0}^{(\lambda)}=(I+A^{\ast})^{-\lambda}T^{(\lambda)-1}(I+A)^{-\lambda}=\left(W^{(\lambda)}(0)\right)^{-1}.$

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