Proof. We begin by showing that $\textbf {u}_k$ is the left eigenvector corresponding to the eigenvalue $\lambda _k$ . It suffices to prove that, for each pair $(k,j)$ ,

$$ \begin{align*}\lambda_ku_k(j) &=\sum_{i=0}^m\sigma_{i,j}(x,y)u_k(i)\\ &=\sigma_{j,j}(x,y)u_k(j) +\sigma_{j-1,j}(x,y)u_k(j-1) +\sigma_{j+1,j}(x,y)u_k(j+1), \end{align*} $$

which can be, equivalently, restated as

(2.1) $$ \begin{align}(\lambda_k-x-2jy)u_k(j) =ju_k(j-1)+(m-j)u_k(j+1).\end{align} $$

Observing the functional relations

$$ \begin{align*} \rho=y+\sqrt{1+y^2} \quad\rightleftharpoons\quad y=\frac{\rho^2-1}{2\rho}, \end{align*} $$

we can manipulate the expression

$$ \begin{align*}\lambda_k-x-2jy&=(2n-2j+\delta)y+(2k-\delta)\sqrt{1+y^2}\\ &=(2n-2j-2k+2\delta)y+(2k-\delta)\rho, \end{align*} $$

which leads to the useful relation

(2.2) $$ \begin{align}\lambda_k-x-2jy =(n-j+k)\rho-(n-j-k+\delta)\rho^{-1}.\end{align} $$

According to the definition, the left-hand side of (2.1) can be written as

$$ \begin{align*} (\lambda_k-x-2jy)u_k(j)=\mathrm{P}-\mathrm{Q}, \end{align*} $$

where

$$ \begin{align*}\mathrm{P}&=(n-j+k)\sum_{\ell=0}^j(-1)^{j+\ell} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j+1},\\ \mathrm{Q}&=(n-j-k+\delta)\sum_{\ell=0}^j(-1)^{j+\ell} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j-1}. \end{align*} $$

Consequently, the equality (2.1) can be restated as

(2.3) $$ \begin{align}\mathrm{P}-ju_k(j-1)= \mathrm{Q}+(\delta+2n-j)u_k(j+1).\end{align} $$

The two sums on the left can be combined as

$$ \begin{align*} \mathrm{P} -ju_k(j-1) &=\sum_{\ell=0}^j(-1)^{j+\ell} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j+1} \\ &\quad \times\bigg\{(n-j+k)+\frac{(j-\ell)(1+\delta+2n-j)}{1+\delta+n-j-k+\ell}\bigg\}\\ &=\sum_{\ell=0}^j(-1)^{j+\ell} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j+1} \frac{(n+k-\ell)(1+\delta+n-k)}{1+\delta+n-j-k+\ell}\\ &=(1+\delta+n-k)\sum_{\ell=0}^j(-1)^{j+\ell} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell-1}\rho^{2\ell-j+1}, \end{align*} $$

while, similarly, for the two sums on the right,

$$ \begin{align*}\mathrm{Q} +(\delta+2n-j)u_k(j+1) &=\sum_{\ell=0}^{j+1}(-1)^{j+\ell} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j-1}\\ &\quad \times\bigg\{(n-j-k+\delta)-\frac{(j+1)(\delta+n-j-k+\ell)}{1+j-\ell}\bigg\}\\ &=\sum_{\ell=0}^{j+1}(-1)^{j+\ell-1} \binom{j}{\ell}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j-1} \frac{\ell(1+\delta+n-k)}{1+j-\ell}\\ &=(1+\delta+n-k)\sum_{\ell=0}^{j+1}(-1)^{j+\ell-1} \binom{j}{\ell-1}\binom{\delta+2n-j}{n+k-\ell}\rho^{2\ell-j-1}. \end{align*} $$

Shifting forward the summation index $\ell \to 1+\ell $ for the last sum, we see that the above two expressions for both sides of equation (2.3) coincide. This confirms item (a).

Likewise, (b) will be confirmed if we can show that $\textbf {v}_k$ is the right eigenvector corresponding to $\lambda _k$ . This can be done by showing, for each pair $(k,i)$ , that

$$ \begin{align*}\lambda_kv_k(i) &=\sum_{j=0}^m\sigma_{i,j}(x,y)v_k(j)\\ &=\sigma_{i,i-1}(x,y)v_k(i-1) +\sigma_{i,i}(x,y)v_k(i) +\sigma_{i,i+1}(x,y)v_k(i+1), \end{align*} $$

which can be, equivalently, restated as

(2.4) $$ \begin{align} (\lambda_k-x-2iy)v_k(i) =(m-i+1)v_k(i-1) +(i+1)v_k(i+1).\end{align} $$

Keeping in mind (2.2), we can express the left-hand side of (2.4) as

$$ \begin{align*} (\lambda_k-x-2iy)v_k(i)=\mathcal{P}-\mathcal{Q}, \end{align*} $$

where

$$ \begin{align*}\mathcal{P}&=(n-i+k)\sum_{\ell=0}^i(-1)^{i+\ell} \binom{n+k}{\ell}\binom{\delta+n-k}{i-\ell}\rho^{2\ell-i+1},\\ \mathcal{Q}&=(n-i-k+\delta)\sum_{\ell=0}^i(-1)^{i+\ell} \binom{n+k}{\ell}\binom{\delta+n-k}{i-\ell}\rho^{2\ell-i-1}. \end{align*} $$

Therefore, (2.4) can be reformulated as

$$ \begin{align*} \mathcal{P}-(\delta+2n-i+1)v_k(i-1)=\mathcal{Q}+(i+1)v_k(i+1). \end{align*} $$

According to the definitions, we first simplify the left-hand side of this equation,

$$ \begin{align*}\mathcal{P}&-(\delta+2n-i+1)v_k(i-1) \\ &=\sum_{\ell=0}^i(-1)^{i+\ell} \binom{n+k}{\ell}\binom{\delta+n-k}{i-\ell}\rho^{2\ell-i+1} \times\bigg\{(n-i+k)-\frac{(\delta+2n-i+1)(i-\ell)}{\delta+n-k-i+\ell+1}\bigg\}\\ &=\sum_{\ell=0}^i(-1)^{i+\ell} \binom{n+k}{\ell}\binom{\delta+n-k}{i-\ell}\rho^{2\ell-i+1} \frac{(n+k-\ell)(1+\delta+n-k)}{\delta+n-k-i+\ell+1}\\ &=(n+k)\sum_{\ell=0}^i(-1)^{i+\ell} \binom{n+k-1}{\ell}\binom{1+\delta+n-k}{i-\ell}\rho^{2\ell-i+1}, \end{align*} $$

and then the right-hand side of the same equation,

$$ \begin{align*}\mathcal{Q} & +(i+1)v_k(i+1) \\ &=\sum_{\ell=0}^{i+1}(-1)^{i+\ell} \binom{n+k}{\ell}\binom{\delta+n-k}{i-\ell}\rho^{2\ell-i-1} \times\bigg\{(n-i-k+\delta)-\frac{(i+1)(\delta+n-i-k+\ell)}{1+i-\ell}\bigg\}\\ &=\sum_{\ell=0}^{i+1}(-1)^{i+\ell-1} \binom{n+k}{\ell}\binom{\delta+n-k}{i-\ell}\rho^{2\ell-i-1} \frac{\ell(1+\delta+n-k)}{1+i-\ell}\\ &=(n+k)\sum_{\ell=0}^{i+1}(-1)^{i+\ell-1} \binom{n+k-1}{\ell-1}\binom{1+\delta+n-k}{1+i-\ell}\rho^{2\ell-i-1}. \end{align*} $$

The last two expressions become identical when the replacement $\ell \to 1+\ell $ is made in the latter one. This shows that $\textbf {v}_k$ is indeed the right eigenvector of the matrix $M_{\delta +2n}(x,y)$ corresponding to $\lambda _k$ .

Proof. To prove these orthogonality relations, write the scalar product as the triple sum

(3.1) $$ \begin{align} \langle{\textbf{u}_i,\textbf{v}_j}\rangle &=\sum_{k=0}^{m}u_i(k)v_j(k) \nonumber\\ &=\sum_{k=0}^{\delta+2n}\sum_{\imath=0}^k(-1)^{k+\imath} \binom{k}{\imath}\binom{\delta+2n-k}{n+i-\imath}\rho^{2\imath-k} \sum_{\jmath=0}^k(-1)^{k+\jmath} \binom{n+j}{\jmath}\binom{\delta+n-j}{k-\jmath}\rho^{2\jmath-k}.\ \end{align} $$

Observing that

$$ \begin{align*}\textbf{u}_i\Omega_m(x,y)\textbf{v}_j &=\langle{\textbf{u}_i\Omega_m(x,y),\textbf{v}_j}\rangle=\lambda_i\langle{\textbf{u}_i,\textbf{v}_j}\rangle\\ &=\langle{\textbf{u}_i,\Omega_m(x,y)\textbf{v}_j}\rangle=\lambda_j\langle{\textbf{u}_i,\textbf{v}_j}\rangle, \end{align*} $$

immediately, we have

$$ \begin{align*} \langle{\textbf{u}_i,\textbf{v}_j}\rangle=0 \quad\text{when } i\not=j \quad\text{for } \lambda_i\not=\lambda_j. \end{align*} $$

When $i=j=\ell $ , the scalar product (3.1) can be reformulated, by first making the replacement $\jmath \to k-\jmath $ and then interchanging the order of the triple sum, as

$$ \begin{align*}\langle{\textbf{u}_\ell,\textbf{v}_\ell}\rangle &=\sum_{k=0}^{\delta+2n}\sum_{\imath=0}^k(-1)^{k+\imath} \binom{k}{\imath}\binom{\delta+2n-k}{n+\ell-\imath}\rho^{2\imath-k} \sum_{\jmath=0}^k(-1)^{\jmath} \binom{n+\ell}{k-\jmath}\binom{\delta+n-\ell}{\jmath}\rho^{k-2\jmath}\\ &=\sum_{\imath,\jmath=0}^{\delta+2n}(-1)^{\imath+\jmath} \binom{\delta+n-\ell}{\jmath}\rho^{2\imath-2\jmath} \sum_{k=\max\{\imath,\jmath\}}^{\delta+2n}(-1)^k \binom{k}{\imath}\binom{\delta+2n-k}{n+\ell-\imath} \binom{n+\ell}{k-\jmath}. \end{align*} $$

The inner sum with respect to k can be reformulated under $k\to \kappa +\jmath $ and then evaluated in closed form as

$$ \begin{align*} \sum_{\kappa=0}^{\delta+2n-\jmath}(-1)^{\kappa+\jmath} \binom{n+\ell}{\jmath} \binom{\kappa+\jmath}{\imath} \binom{\delta+2n-\kappa-\jmath}{n+\ell-\imath} =(-1)^{\jmath} \binom{n+\ell}{\imath}, \end{align*} $$

since the last sum results substantially, apart from an alternating sign, in the finite differences of order $n+\ell $ for a polynomial of the same degree $n+\ell $ .

Therefore, the triple sum is reduced to a double one and can be evaluated further by the binomial theorem: that is,

$$ \begin{align*}\langle{\textbf{u}_\ell,\textbf{v}_\ell}\rangle &=\sum_{\imath=0}^{\delta+2n}\binom{n+\ell}{\imath}\rho^{2\imath} \sum_{\jmath=0}^{\delta+2n}\binom{\delta+n-\ell}{\jmath}\rho^{-2\jmath} =(1+\rho^2)^{n+\ell}(1+\rho^{-2})^{\delta+n-\ell}. \end{align*} $$

This is equivalent to the expression

$$ \begin{align*} \langle{\textbf{u}_\ell,\textbf{v}_\ell}\rangle =\frac{(1+\rho^2)^m}{\rho^{2\delta+2n-2\ell}} = \frac{(2+2y\rho)^m}{\rho^{2m-2n-2\ell}}.\\[-42.5pt] \end{align*} $$

Proof. We first evaluate the determinant $\det U_m$ . By making use of (2.2), we rewrite the three-term relation (2.1) as

(4.1) $$ \begin{align} \frac{(2k-m)(\rho+\rho^{-1})}{2(m-j)}u_{k-n}(j) &=u_{k-n}(j+1)+\frac{j}{m-j}u_{k-n}(j-1) \nonumber\\ &\quad +\frac{(2j-m)(\rho-\rho^{-1})}{2(m-j)}u_{k-n}(j). \end{align} $$

For the determinant of the matrix

$$ \begin{align*} \det U_m=\det_{0\le k\le m}[\textbf{u}_{k-n}] =\det_{0\le j, k\le m}[u_{k-n}(j)], \end{align*} $$

perform the following three operations:

• • add ${j}/{(m-j)}$ times the $(j-1)$ th row to the $(j+1)$ th row;

• • add ${(2j-m)(\rho -\rho ^{-1})}/{(2m-2j)}$ times the jth row to the $(j+1)$ th row; and

• • after the above two operations, the entry at position $(j+1,k)$ in the $(j+1)$ th row becomes $({(2k-m)(\rho +\rho ^{-1})}/{(2m-2j)})u_{k-n}(j)$ , in view of (4.1).

Repeating this operation upwards for all the rows except the first one and then pulling out the common factors in rows, we get the expression (with k indicating the kth column)

$$ \begin{align*} \det U_n=\prod_{j=0}^{m-1}\frac{(\rho+\rho^{-1})}{2(m-j)} \times\det_{0\le k\le m} \left[ \begin{array}{cl}&u_{k-n}(0)\\ (2k-m)&u_{k-n}(0)\\ (2k-m)&u_{k-n}(1)\\ (2k-m)&u_{k-n}(2)\\ \cdots&\cdots~\cdots\\ (2k-m)&u_{k-n}(m-1)\end{array}\right]. \end{align*} $$

By carrying out the same operations in the above matrix (except for the first two rows), we can further reduce the determinant: that is,

$$ \begin{align*} \det U_m=\prod_{i=1}^2\prod_{j=0}^{m-i}\frac{(\rho+\rho^{-1})}{2(m-j)} \times\det_{0\le k\le m} \left[ \begin{array}{cl}&u_{k-n}(0)\\ (2k-m)~&u_{k-n}(0)\\ (2k-m)^2&u_{k-n}(0)\\ (2k-m)^2&u_{k-n}(1)\\ \cdots&\cdots~\cdots\\ (2k-m)^2&u_{k-n}(m-2)\end{array}\right]. \end{align*} $$

Iterating the same procedure m times and extracting the common factors $ u_{k-n}(0)$ in the columns of the resulting matrix, we derive the simplified expression

$$ \begin{align*} \det U_n=\prod_{i=1}^m\prod_{j=0}^{m-i}\frac{(\rho+\rho^{-1})}{2(m-j)} \times\det_{0\le j,k \le m}[(2k-m)^j] \prod_{k=0}^mu_{k-n}(0). \end{align*} $$

Keeping in mind that $u_{k-n}(0)=\binom {m}{k}$ and the determinant in the middle is of Vandermonde type, we can evaluate separately

$$ \begin{align*} \prod_{i=1}^m\prod_{j=0}^{m-i} \frac{(\rho+\rho^{-1})}{2(m-j)} & =\prod_{i=1}^m\frac{(i-1)!}{m!} \bigg(\frac{\rho+\rho^{-1}}{2}\bigg)^{m-i+1} =\bigg(\frac{\rho+\rho^{-1}}{2}\bigg)^{\binom{m+1}2} \prod_{i=1}^m\frac{(i-1)!}{m!},\\[3pt] \det_{0\le j,k \le m}[(2k-m)^k] & =\prod_{0\le j<k\le m}(2k-2j) =2^{\binom{m+1}2}\prod_{k=1}^m(k!),\\[3pt] \prod_{k=0}^mu_{k-n}(0) & =\prod_{k=0}^m\binom{m}{k}=\prod_{k=1}^m\frac{m!}{k!(k-1)!}. \end{align*} $$

Multiplying these together, we find the closed formula

$$ \begin{align*} \det U_m =(\rho+\rho^{-1})^{\binom{m+1}2} =(4+4y^2)^{m(m+1)/4}. \end{align*} $$

Finally, to evaluate the other determinant $\det V_m$ , rewrite, analogously, the three-term relation (2.4) as

(4.2) $$ \begin{align} \frac{(2k-m)(\rho+\rho^{-1})}{2(i+1)}v_{k-n}(i) &=v_{k-n}(i+1)+\frac{m-i+1}{i+1}v_{k-n}(i-1)\nonumber\\ &\quad +\frac{(2i-m)(\rho-\rho^{-1})}{2(i+1)}v_{k-n}(i). \end{align} $$

Similarly, for the determinant

$$ \begin{align*} \det V_n=\det_{0\le k\le n}[\textbf{v}_k] =\det_{0\le i, k\le n}[v_k(i)], \end{align*} $$

make the following three operations:

• • add ${(m-i+1)}{/(i+1)}$ times the $(i-1)$ th row to the $(i+1)$ th row;

• • add ${(2i-m)(\rho -\rho ^{-1})}/{(2i+2)}$ times the ith row to the $(i+1)$ th row; and

• • after the above two operations, the entry at position $(i+1,k)$ in the $(i+1)$ th row becomes $({(2k-m)(\rho +\rho ^{-1})}/{(2i+2)})v_{k-n}(i)$ , taking into account (4.2).

Repeating this operation upwards for all the rows except the first one and then pulling out the common factors in rows, we get the expression (with k indicating the kth column)

$$ \begin{align*} \det V_m=\prod_{i=0}^{m-1} \frac{\rho+\rho^{-1}}{2(i+1)} \times\det_{0\le k\le n} \left[ \begin{array}{cl}&v_{k-n}(0)\\ (2k-m)&v_{k-n}(0)\\ (2k-m)&v_{k-n}(1)\\ (2k-m)&v_{k-n}(2)\\ \cdots&\cdots~\cdots\\ (2k-m)&v_{k-n}(n-1)\end{array}\right]. \end{align*} $$

By carrying out the same operations in the above matrix (except for the first two rows), we can reduce this further: that is,

$$ \begin{align*} \det V_m=\prod_{j=1}^2\prod_{i=0}^{m-j} \frac{\rho+\rho^{-1}}{2(i+1)} \times\det_{0\le k\le n} \left[ \begin{array}{cl}&v_k(0)\\ (2k-m)~&v_{k-n}(0)\\ (2k-m)^2&v_{k-n}(0)\\ (2k-m)^2&v_{k-n}(1)\\ \cdots&\cdots~\cdots\\ (2k-m)^2&v_{k-n}(m-2)\end{array}\right]. \end{align*} $$

Iterating the same procedure m times and extracting the common factors $ v_k(0)$ in the columns of the resulting matrix, we find the following simplified expression.

$$ \begin{align*} \det V_n=\prod_{j=1}^m\prod_{i=0}^{m-j} \frac{\rho+\rho^{-1}}{2(i+1)} \times\det_{0\le i,k\le n} [(2k-m)^i] \prod_{k=0}^mv_{k-n}(0). \end{align*} $$

Keeping in mind that $v_{k-n}(0)\equiv 1$ and then evaluating the product

$$ \begin{align*}\prod_{j=1}^m\prod_{i=0}^{m-j} \frac{\rho+\rho^{-1}}{2(i+1)} =\prod_{j=1}^m\frac{1}{(m-j+1)!} \bigg(\frac{\rho+\rho^{-1}}{2}\bigg)^{m-j+1} =\bigg(\frac{\rho+\rho^{-1}}{2}\bigg)^{\binom{m+1}{2}} \prod_{j=1}^m\frac{1}{j!}, \end{align*} $$

we derive, after substitution, the second determinantal evaluation

$$ \begin{align*} \det V_m =(\rho+\rho^{-1})^{\binom{m+1}{2}} =(4+4y^2)^{{m(m+1)}/{4}}.\\[-36.1pt] \end{align*} $$