**Proposition A.15** (Solutions for $j = 2$ and $\alpha = 0$). The univariate Laguerre production matrix (1.41)/(A.36) with $\alpha = 0$ can be written as the production matrix $P^{(2;2)}$ where the $\alpha$ are polynomials in $x$ with nonnegative coefficients, in the following way (and only the following way):

$$ \alpha_2 = 0 \qquad (A.90a) $$

$$ \alpha_{3n} = x \qquad (A.90b) $$

$$ \alpha_{3n+1} = n \qquad (A.90c) $$

$$ \alpha_{3n+2} = n \qquad (A.90d) $$

**PROOF. Necessity.** In the proof of Proposition A.13 we saw that either $\alpha_4\alpha_6 = d$ or $\alpha_4\alpha_6 = dx$, for some real constant $d > 0$; and we saw that $\alpha_4\alpha_6 = d$ leads to $\alpha = 1$ whenever $\alpha \neq -1$. So, for $\alpha = 0$ we must have $\alpha_4\alpha_6 = dx$. Furthermore, we saw that $\alpha_3 + \alpha_4 = c_1 + c_2x$, and whenever $\alpha \neq -1$ and $\alpha_4\alpha_6 = dx$, to obtain $\alpha = 0$, we need $c_2 > 0$. In this case, we saw that we must have $d = 1$ with $\alpha_6 = x$ and thus, $\alpha_4 = 1$. We also obtained $c_2 = 1$ and $c_1 = 1$, and thus, $\alpha_3 = x$ and from $p_{0,0}$, $\alpha_2 = 0$.

From here on the proof is identical to that of Proposition A.12, with each $\alpha_i$ replaced by $\alpha_{i+1}$.

**Sufficiency.** This is easily checked. $\square$

**Proposition A.16** (Solutions for $j = 2$ and $\alpha = 1$). The univariate Laguerre production matrix (1.41)/(A.36) with $\alpha = 1$ can be written as the production matrix $P^{(2;2)}$ where the $\alpha$ are polynomials in $x$ with nonnegative coefficients, in the following ways (and only the following ways):

$$ \alpha_{3n-1} = x \qquad (A.91a) $$

$$ \alpha_{3n} = c_n n \qquad (A.91b) $$

$$ \alpha_{3n+1} = (2 - c_n)n \qquad (A.91c) $$

where $c_n$ is given by (A.39), with $\kappa \in [0, 1]$.

**PROOF. Necessity.** In the proof of Proposition A.13 we saw that either $\alpha_4\alpha_6 = d$ or $\alpha_4\alpha_6 = dx$, for some real constant $d > 0$; and we saw that $\alpha_4\alpha_6 = dx$ leads to $\alpha = 0$ whenever $\alpha \neq -1$. So, for $\alpha = 1$ we must have $\alpha_4\alpha_6 = d$. Furthermore, in this case we found that $c_2 = 0$ and $c_1 = 2$, hence $\alpha_3 + \alpha_4 = 2$. We also found that $d = (5 + \alpha)/3 = 2$, hence $\alpha_4\alpha_6 = 2$.

For $j = 2$ and $\alpha = 1$ we have

$$ p_{n,n} = 2n + 2 + x = \alpha_{3n+2} + \alpha_{3n+3} + \alpha_{3n+4} \qquad (A.92a) $$

$$ p_{n,n-1} = n(n+1) + 2nx = (\alpha_{3n} + \alpha_{3n+1})\alpha_{3n+2} + \alpha_{3n+1}\alpha_{3n+3} \qquad (A.92b) $$

$$ p_{n,n-2} = n(n-1)x = \alpha_{3n-2}\alpha_{3n}\alpha_{3n+2} \qquad (A.92c) $$

From $p_{0,0}$ and $\alpha_3 + \alpha_4 = 2$, we see that $\alpha_2 = x$. We will now prove by induction that for all $n \ge 1$, we have

$$ \alpha_{3n-1} = x \qquad (A.93a) $$

$$ \alpha_{3n} + \alpha_{3n+1} = 2n \qquad (A.93b) $$

$$ \alpha_{3n+1}\alpha_{3n+3} = n(n+1) \qquad (A.93c) $$
The base case $n = 1$ has just been proven. Using the inductive hypotheses $\alpha_{3n} + \alpha_{3n+1} = 2n$ and $\alpha_{3n+1}\alpha_{3n+3} = n(n+1)$ and the formula for $p_{n,n-1}$, we deduce $\alpha_{3n+2} = x$. Then using this together with the formula for $p_{n,n}$, we deduce $\alpha_{3n+3} + \alpha_{3n+4} = 2n+2$. Using $\alpha_{3n+1}\alpha_{3n+3} = n(n+1)$ and the formula for $p_{n+1,n-1}$, we deduce that $\alpha_{3n+5} = x$. And using this together with $\alpha_{3n+3} + \alpha_{3n+4} = 2n+2$ and the formula for $p_{n+1,n}$, we deduce that $\alpha_{3n+4}\alpha_{3n+6} = (n+1)(n+2)$. This completes the induction.

From here on the proof is identical to that of Proposition A.8.

**Sufficiency.** It is easy to verify that, for any $\kappa \in [0, 1]$, the parameters $\alpha$ defined by (A.91)/(A.39) satisfy (A.93) and thence (A.92). $\square$

## B When is $B_{\xi}^{-1}PB_{\xi}$ $(r, 1)$-banded?

We say that a matrix $A = (a_{ij})_{i,j \ge 0}$ is $(r, s)$-banded if $a_{ij} = 0$ whenever $j < i - r$ or $j > i + s$. Otherwise put, the nonzero elements $a_{ij}$ can occur only on the diagonal, in the first $r$ bands below the diagonal, and in the first $s$ bands above the diagonal. We say that a matrix is **lower-Hessenberg** if it is $(\infty, 1)$-banded.

Let $P = (p_{ij})_{i,j \ge 0}$ be a lower-Hessenberg matrix with entries in a commutative ring $R$. We are interested in the matrix $B_{\xi}^{-1}PB_{\xi}$, where $B_{\xi}$ is the $\xi$-binomial matrix (1.9) and $\xi$ is an indeterminate, understood as a matrix with entries in the polynomial ring $R[\xi]$. Obviously $B_{\xi}^{-1}PB_{\xi}$ is lower-Hessenberg; we want to know under what conditions it is $(r, 1)$-banded for some integer $r \ge 0$. An obvious necessary condition is that $P$ must itself be $(r, 1)$-banded, since $P$ can be recovered from $B_{\xi}^{-1}PB_{\xi}$ by specializing $\xi$ to zero. The full necessary and sufficient condition can be written in terms of the superdiagonal sequence $(p_{n,n+1})_{n \ge 0}$ and the $m$th subdiagonal sequences $(p_{n,n-m})_{n \ge m}$ for $0 \le m \le r$, and goes as follows:

**Proposition B.1.** Fix an integer $r \ge 0$, and let $P = (p_{ij})_{i,j \ge 0}$ be an $(r, 1)$-banded matrix with entries in a commutative ring $R$ containing the rationals. Then the following are equivalent:

(a) $B_{\xi}^{-1}PB_{\xi}$ is $(r, 1)$-banded. [That is, the $m$th subdiagonal of $B_{\xi}^{-1}PB_{\xi}$ vanishes for all $m \ge r + 1$.]

(b) The $(r+1)$st subdiagonal of $B_{\xi}^{-1}PB_{\xi}$ vanishes.

(c) The superdiagonal satisfies

$$p_{n,n+1} = f_{-1}(n) \qquad (B.1)$$

where $f_{-1}(\cdot)$ is a polynomial of degree at most $r$ with coefficients in $R$; and the $m$th subdiagonals for $0 \le m \le r$ satisfy

$$p_{n,n-m} = n(n-1)\cdots(n-m+1) f_m(n), \qquad (B.2)$$

where in each case $f_m(\cdot)$ is a polynomial of degree at most $r-m$ with coefficients in $R$.
PROOF. (a) $\implies$ (b) is trivial; we will prove (b) $\implies$ (c) and (c) $\implies$ (a).

We begin by computing the matrix element $(B_{\xi}^{-1}PB_{\xi})_{k+t,k}$ for $k, t \ge 0$:

$$ (B_{\xi}^{-1}PB_{\xi})_{k+t,k} = (B_{-\xi}PB_{\xi})_{k+t,k} = \sum_{m=-1}^{r} \sum_{j=k}^{k+t-m} \binom{k+t}{j+m} (-\xi)^{(k+t)-(j+m)} p_{j+m,j} \binom{j}{k} \xi^{j-k} \quad (B.3a) \\ = \sum_{m=-1}^{r} \xi^{t-m} \sum_{j=k}^{k+t-m} (-1)^{(t-m)-(j-k)} \binom{k+t}{j+m} \binom{j}{k} p_{j+m,j}. \quad (B.3b) $$

(Here we let the binomial coefficient $\binom{N}{-1} = 0$ and $p_{-1,0} = 0$. This happens in (B.3) when $j=k=0$ and $m=-1$.)

With $l = j - k$, we can rewrite equation (B.3) as

$$ \begin{aligned} [\xi^{t-m}] (B_{\xi}^{-1}PB_{\xi})_{k+t,k} &= \sum_{l=0}^{t-m} (-1)^{(t-m)-l} \binom{k+t}{k+l+m} \binom{k+l}{k} p_{k+l+m,k+l} \quad (B.4a) \\ &= \frac{(k+t)!}{k!(t-m)!} \sum_{l=0}^{t-m} (-1)^{(t-m)-l} \binom{t-m}{l} \frac{p_{k+l+m,k+l}}{(k+l+m)!(k+l)!}, \quad (B.4b) \end{aligned} $$

valid when $k, t \ge 0$ and $-1 \le m \le t$ (remembering that $p_{-1,0} = 0$).

So, for $m \ge -1$, let $g_m$ be the function defined on the domain $\mathbb{N}$ (with values in $\mathbb{R}$) by

$$ g_m(k) \stackrel{\text{def}}{=} \frac{p_{k+m,k}}{(k+m)!/k!} \quad (B.5) $$

with the convention that $g_{-1}(0) = 0$. Then we have

$$ \begin{aligned} [\xi^{t-m}] (B_{\xi}^{-1}PB_{\xi})_{k+t,k} &= \frac{(k+t)!}{k!(t-m)!} \sum_{l=0}^{t-m} (-1)^{(t-m)-l} \binom{t-m}{l} g_m(k+l) \quad (B.6a) \\ &= \frac{(k+t)!}{k!(t-m)!} (\Delta^{t-m} g_m)(k) \quad (B.6b) \end{aligned} $$

where $\Delta$ denotes the forward difference operator $(\Delta f)(n) = f(n+1) - f(n)$; this is valid when $k, t \ge 0$ and $-1 \le m \le t$. It follows that $[\xi^{t-m}] (B_{\xi}^{-1}PB_{\xi})_{k+t,k}$ vanishes for all $k \ge 0$ if and only if $g_m$ is a polynomial of degree $< t - m$.

Applying this now under hypothesis (b) for $t = r+1$ and $m \in [-1, r]$, we conclude that $g_m$ is a polynomial of degree $\le r - m$. When $m \ge 0$, we set $n = k + m$, which gives us (B.2). When $m = -1$, equation (B.5) gives us

$$ p_{k-1,k} = \frac{g_{-1}(k)}{k} \quad \text{for } k \ge 1 \quad (B.7) $$

where $g_{-1}$ is a polynomial of degree $\le r+1$ that satisfies $g_{-1}(0) = 0$. Setting $n = k-1$ gives us (B.1). We have therefore proven that (b) implies (c).

On the other hand, if $g_m$ is a polynomial of degree $\le r - m$ for all $m \in [0, r]$, and $p_{k-1,k} = f_{-1}(k)$ where $f_{-1}$ is a polynomial of degree at most $r$, it follows from (B.6b) that $(B_{\xi}^{-1}PB_{\xi})_{k+t,k} = 0$ for all $t \ge r+1$. So (c) implies (a). $\square$
C Total positivity for a variant class of quadridiagonal matrices

In this appendix we prove total positivity for a class of quadridiagonal matrices. The result is very similar to Theorem 6.1, but for a slightly different class of matrices; and the proof is also very similar, though slightly more difficult. We present this variant result because we think that it may be useful in future work. Indeed, in one current project of ours [20, 21], the total positivity of the production matrix can be proven using a special case of Theorem C.1, but not, as far as we can tell, using Theorem 6.1.

Consider the quadridiagonal lower-Hessenberg matrix $P$, defined by

$$P \stackrel{\text{def}}{=} L_1 L_2 U + L_1 D_1 + L_2 D_2 \quad (C.1a)$$

$$= L_1(L_2U + D_1) + L_2D_2 \quad (C.1b)$$

$$= L_2(L_1U + D_2) + L_1D_1 \quad (C.1c)$$

where

$$L_1 = \alpha I + xL \quad (C.2a)$$

$$L_2 = \beta I + yL \quad (C.2b)$$

and

* $L$ is the lower-bidiagonal matrix with the sequence $a_0, a_1, \dots$ on the diagonal, the sequence $b_1, b_2, \dots$ on the subdiagonal, and zeroes elsewhere;

* $U$ is the upper-bidiagonal matrix with the sequence $c_1, c_2, \dots$ on the superdiagonal, the sequence $d_0, d_1, \dots$ on the diagonal, and zeroes elsewhere;

* $D_1$ is the diagonal matrix with entries $e_0, e_1, \dots$;

* $D_2$ is the diagonal matrix with entries $f_0, f_1, \dots$;

and $\alpha, \beta, x, y, \mathbf{a} = (a_n)_{n \ge 0}$, $\mathbf{b} = (b_n)_{n \ge 1}$, $\mathbf{c} = (c_n)_{n \ge 1}$, $\mathbf{d} = (d_n)_{n \ge 0}$, $\mathbf{e} = (e_n)_{n \ge 0}$, $\mathbf{f} = (f_n)_{n \ge 0}$ are all indeterminates. Note that $(C.1b) = (C.1c)$ because by construction $L_1$ and $L_2$ commute:

$$L_1 L_2 = L_2 L_1. \quad (C.3)$$

This commutation is what makes the present situation more restrictive than that of Theorem 6.1; in compensation, the diagonal matrices $D_1$ and $D_2$ can both act on the same side (here the right).

The entries in the $k$th column of $P = (p_{n,k})_{n \ge 0}$ are given by

$$p_{k-1,k} = (\alpha + x a_{k-1})(\beta + y a_{k-1})c_k \quad (C.4a)$$

$$\begin{aligned} p_{k,k} = & (\alpha + x a_k)(\beta + y a_k)d_k + (\alpha + x a_k)y b_k c_k + x b_k(\beta + y a_{k-1})c_k \\ & + (\alpha + x a_k)e_k + (\beta + y a_k)f_k \end{aligned} \quad (C.4b)$$

$$\begin{aligned} p_{k+1,k} = & (\alpha + x a_{k+1}) y b_{k+1} d_k + x b_{k+1} (\beta + y a_k) d_k + x b_{k+1} y b_k c_k \\ & + x b_{k+1} e_k + y b_{k+1} f_k \end{aligned} \quad (C.4c)$$

$$p_{k+2,k} = xy b_{k+2} b_{k+1} d_k \quad (C.4d)$$

$$p_{n,k} = 0 \quad \text{if } n < k - 1 \text{ or } n > k + 2 \quad (C.4e)$$
where by definition $b_0 = c_0 = 0$ and $a_n = b_n = c_n = d_n = e_n = f_n = 0$ whenever $n < 0$. Our main result is:

**Theorem C.1 (Total positivity of the generalized production matrix).** The matrix $P$ defined by $(C.1)/(C.2)/(C.4)$ is totally positive, coefficientwise in the indeterminates $\alpha, \beta, x, y, \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}, \mathbf{f}$.

The proof of Theorem C.1 follows the same pattern as that of Theorem 6.1, but works with columns rather than rows (because the matrices $D_1$ and $D_2$ act on the right). More precisely, Theorem C.1 will be proven as follows: Define the matrix $Q = (q_{n,k})_{n,k \ge 0} = (\mathbf{q}_0, \mathbf{q}_1, \dots)$ by

$$Q \stackrel{\text{def}}{=} P|_{\mathbf{f}=0} = L_1(L_2U + D_1), \qquad (C.5)$$

so that

$$q_{k-1,k} = (\alpha + xa_{k-1})(\beta + ya_{k-1})c_k \qquad (C.6a)$$

$$\begin{aligned} q_{k,k} = & (\alpha + xa_k)(\beta + ya_k)d_k + (\alpha + xa_k)yb_kc_k + xb_k(\beta + ya_{k-1})c_k \\ & + (\alpha + xa_k)e_k \end{aligned} \qquad (C.6b)$$

$$\begin{aligned} q_{k+1,k} = & (\alpha + xa_{k+1})yb_{k+1}d_k + xb_{k+1}(\beta + ya_k)d_k + xb_{k+1}yb_kc_k \\ & +xb_{k+1}e_k \end{aligned} \qquad (C.6c)$$

$$q_{k+2,k} = xyb_{k+2}b_{k+1}d_k \qquad (C.6d)$$

$$q_{n,k} = 0 \quad \text{if } n < k - 1 \text{ or } n > k + 2 \qquad (C.6e)$$

Then

$$P = Q + L_2D_2. \qquad (C.7)$$

We will begin by proving (Lemma C.2) that $Q$ is coefficientwise totally positive; this proof uses the factorization $Q = L_1(L_2U + D_1)$ together with the tridiagonal comparison theorem. We omit the proof as it is almost identical to that of Lemma 6.6. It follows that for every integer $m \ge 0$, the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_m)$ is totally positive.

The rest of the proof shows how to restore the terms in $P$ involving $\mathbf{f}$. In terms of the column vectors $(\mathbf{p}_k)_{k \ge 0}$, $(\mathbf{q}_k)_{k \ge 0}$, $(\mathbf{l}_k)_{k \ge 0}$ associated to the matrices $P, Q, L_2$, equation (C.7) can be rewritten as

$$\mathbf{p}_k = \mathbf{q}_k + f_k \mathbf{l}_k, \qquad (C.8)$$

where

$$\mathbf{l}_k = \begin{bmatrix} \mathbf{0}_{k \times 1} \\ \beta + ya_k \\ yb_{k+1} \\ \mathbf{0}_{\infty \times 1} \end{bmatrix} \quad \text{for } k \ge 0. \qquad (C.9)$$

We will show (Lemma C.6) that for every pair of integers $0 \le k \le m+1$, the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{p}_k, \dots, \mathbf{p}_m)$ is totally positive; and we will do this, for each fixed $m \ge 0$, by induction on $k = m+1, m, m-1, \dots, 0$. The base case $k = m+1$ of this induction is thus Lemma C.2, and the final case $k = 0$ is Theorem C.1. Similar to the proof of Lemma 6.10, the proof of Lemma C.6 will involve the following steps:
**Lemma C.3:** The matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \ell_k)$ is totally positive.

**Lemma C.4:** If the matrix $(\mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ is totally positive, then so is $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \ell_k, \mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$.

**The induction step (Lemma C.5):** If the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_k, \mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ is totally positive, then so is $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{p}_k, \dots, \mathbf{p}_m)$.

Putting this all together will prove Lemma C.6 and hence Theorem C.1. Here Lemmas C.3 and C.4 are closely analogous to Lemmas 6.7 and 6.8, but their proofs are somewhat more difficult.

We now begin the proof of Theorem C.1.

**Lemma C.2 (Total positivity of Q).** The matrix $Q$ defined by (C.5)/(C.6) is totally positive, coefficientwise in the indeterminates $\alpha, \beta, x, y, \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}$.

In particular, for every integer $m \ge 0$, the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_m)$ is coefficientwise totally positive.

Lemma C.2 follows from the factorization $Q = L_1(L_2U + D_1)$ by the same argument as in Lemma 6.6.

**Lemma C.3.** For each integer $k \ge 0$, the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \ell_k)$ is totally positive.

**PROOF.** We prove this by induction on $k$. The base case $k = 0$ is trivial. So we need to show the inductive step: if $(\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \ell_{k-1})$ is totally positive, then so is $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \ell_k)$.

Define $\tilde{\mathbf{q}}_{k-1} \stackrel{\text{def}}{=} \mathbf{q}_{k-1}|_{d_{k-1}=0}$, so that $\tilde{\mathbf{q}}_{k-1} = (\tilde{q}_{n,k-1})_{n \ge 0}^\top$ where

$$ \tilde{q}_{k-2,k-1} = (\alpha + xa_{k-2})(\beta + ya_{k-2})c_{k-1} \quad (\text{C.10a}) $$

$$ \tilde{q}_{k-1,k-1} = (\alpha + xa_{k-1})yb_{k-1}c_{k-1} + xb_{k-1}(\beta + ya_{k-2})c_{k-1} + (\alpha + xa_{k-1})e_{k-1} \quad (\text{C.10b}) $$

$$ \tilde{q}_{k,k-1} = xb_kyb_{k-1}c_{k-1} + xb_k e_{k-1} \quad (\text{C.10c}) $$

$$ \tilde{q}_{n,k-1} = 0 \quad \text{if } n < k-2 \text{ or } n > k. \quad (\text{C.10d}) $$

Then

$$ \mathbf{q}_{k-1} = \tilde{\mathbf{q}}_{k-1} + d_{k-1}(\alpha + xa_{k-1})\ell_{k-1} + d_{k-1}xb_k\ell_k. \quad (\text{C.11}) $$

We will successively handle the second and third terms on the right-hand side of this formula.

By Lemma C.2 the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1})$ is totally positive; and specializing this matrix to $d_{k-1} = 0$ yields $M_1 \stackrel{\text{def}}{=} (\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \tilde{\mathbf{q}}_{k-1})$, which is therefore also totally positive. And by the induction hypothesis, the matrix $M_2 \stackrel{\text{def}}{=} (\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \ell_{k-1})$ is totally positive. Applying Lemma 6.3 (or rather its transpose) to the matrices $M_1$ and $M_2$, we conclude that the matrix $M_3 \stackrel{\text{def}}{=} (\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \tilde{\mathbf{q}}_{k-1} + d_{k-1}(\alpha + xa_{k-1})\ell_{k-1})$ is totally positive. Note that $M_3$ has $k+1$ nonzero rows (i.e. $0 \le n \le k$); all
subsequent rows are zero. On the other hand, the column vector $\ell_k$ begins with $k$ zeroes. Therefore, Lemma 6.5 implies that

$$ M_4 \stackrel{\text{def}}{=} (M_3 | \ell_k) = (\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \tilde{\mathbf{q}}_{k-1} + d_{k-1}(\alpha + x a_{k-1}) \ell_{k-1}, \ell_k) \quad (\text{C.12}) $$

is totally positive. Right-multiplying $M_4$ by the lower-bidiagonal matrix that has 1 on the diagonal, $d_{k-1}xb_k$ in position $(k, k-1)$ and zeroes elsewhere — in other words, adding $d_{k-1}xb_k$ times the last column of $M_4$ to its next-to-last column — we obtain the matrix

$$ (\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \tilde{\mathbf{q}}_{k-1} + d_{k-1}(\alpha + x a_{k-1}) \ell_{k-1} + d_{k-1} x b_k \ell_k, \ell_k) = (\mathbf{q}_0, \dots, \mathbf{q}_{k-2}, \mathbf{q}_{k-1}, \ell_k) \quad (\text{C.13}) $$

and prove its total positivity, completing the inductive step. $\square$

**Lemma C.4.** Fix integers $0 \le k \le m$. If the matrix $(\mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ is totally positive, then so is the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \ell_k, \mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$.

**PROOF.** The case $k = m$ is Lemma C.3; so assume that $k < m$. Let $\mathbf{t}_{k+1}$ be obtained from $\mathbf{p}_{k+1}$ by first specializing $b_{k+1} = 0$, and followed by the substitution $f_{k+1} \to f_{k+1} + x b_{k+1} c_{k+1}$: that is,

$$ \mathbf{t}_{k+1} = \left( \mathbf{p}_{k+1} \middle|_{b_{k+1}=0} \right) \bigg|_{f_{k+1} \to f_{k+1} + x b_{k+1} c_{k+1}} \quad (\text{C.14}) $$

The entries in the column vector $\mathbf{t}_{k+1}$ are $\mathbf{t}_{k+1} = (t_{n,k+1})_{n \ge 0}^\top$ where

$$ t_{k,k+1} = (\alpha + x a_k)(\beta + y a_k)c_{k+1} \quad (\text{C.15a}) $$

$$ t_{k+1,k+1} = (\alpha + x a_{k+1})(\beta + y a_{k+1})d_{k+1} + x y a_{k+1} b_{k+1} c_{k+1} + \beta x b_{k+1} c_{k+1} \\ + (\alpha + x a_{k+1})e_{k+1} + (\beta + y a_{k+1})f_{k+1} \quad (\text{C.15b}) $$

$$ t_{k+2,k+1} = (\alpha + x a_{k+2}) y b_{k+2} d_{k+1} + x b_{k+2} (\beta + y a_{k+1}) d_{k+1} + x b_{k+2} y b_{k+1} c_{k+1} \\ + x b_{k+2} e_{k+1} + y b_{k+2} f_{k+1} \quad (\text{C.15c}) $$

$$ t_{k+3,k+1} = x y b_{k+3} b_{k+2} d_{k+1} \quad (\text{C.15d}) $$

$$ t_{n,k+1} = 0 \quad \text{if } n < k \text{ or } n > k + 3 \quad (\text{C.15e}) $$

Note that these substitutions would not affect $\mathbf{p}_\ell$ for $\ell > k + 1$. Next, let $\tilde{\mathbf{p}}_{k+1}$ be identical to $\mathbf{t}_{k+1}$ except that the entry $t_{k,k+1}$ is now made equal to 0, i.e. $\tilde{\mathbf{p}}_{k+1} = (\tilde{\mathbf{p}}_{n,k+1})_{n \ge 0}^\top$ where

$$ \tilde{\mathbf{p}}_{k+1,k+1} = (\alpha + x a_{k+1})(\beta + y a_{k+1})d_{k+1} + x y a_{k+1} b_{k+1} c_{k+1} + \beta x b_{k+1} c_{k+1} \\ + (\alpha + x a_{k+1})e_{k+1} + (\beta + y a_{k+1})f_{k+1} \quad (\text{C.16a}) $$

$$ \tilde{\mathbf{p}}_{k+2,k+1} = (\alpha + x a_{k+2}) y b_{k+2} d_{k+1} + x b_{k+2} (\beta + y a_{k+1}) d_{k+1} + x b_{k+2} y b_{k+1} c_{k+1} \\ + x b_{k+2} e_{k+1} + y b_{k+2} f_{k+1} \quad (\text{C.16b}) $$

$$ \tilde{\mathbf{p}}_{k+3,k+1} = x y b_{k+3} b_{k+2} d_{k+1} \quad (\text{C.16c}) $$

$$ \tilde{\mathbf{p}}_{n,k+1} = 0 \quad \text{if } n < k + 1 \text{ or } n > k + 3 \quad (\text{C.16d}) $$

Notice that

$$ \mathbf{p}_{k+1} = \tilde{\mathbf{p}}_{k+1} + (\alpha + x a_k) c_{k+1} \ell_k . \quad (\text{C.17}) $$
By hypothesis $M_1 \stackrel{\text{def}}{=} (\mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ is totally positive; this implies, by substitution, that $M_2 \stackrel{\text{def}}{=} (\mathbf{t}_{k+1}, \mathbf{p}_{k+2}, \dots, \mathbf{p}_m)$ is totally positive; and finally, this implies that $M_3 \stackrel{\text{def}}{=} (\tilde{\mathbf{p}}_{k+1}, \mathbf{p}_{k+2}, \dots, \mathbf{p}_m)$ is totally positive, because the nonzero rows of $M_3$ form a submatrix of $M_2$.

Now observe that the matrix $S \stackrel{\text{def}}{=} (\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{\ell}_k, \tilde{\mathbf{p}}_{k+1}, \mathbf{p}_{k+2}, \dots, \mathbf{p}_m)$ consists of two blocks overlapping in a single row:

$$ S = (\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{\ell}_k | \tilde{\mathbf{p}}_{k+1}, \mathbf{p}_{k+2}, \dots, \mathbf{p}_m) = \left[ \begin{array}{c|c} * & \mathbf{0}_{k \times 1} \\ * & \beta + ya_k \\ \hline * & yb_{k+1} \\ \hline \mathbf{0}_{\infty \times k} & \mathbf{0}_{\infty \times 1} \end{array} \middle| \begin{array}{c} \mathbf{0}_{k \times (m-k)} \\ \mathbf{0}_{1 \times (m-k)} \\ \hline * \\ * \\ \hline * \end{array} \right] \quad (C.18) $$

where the asterisks stand for blocks of unspecified entries (which may be zero or nonzero). By Lemma C.3, the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{\ell}_k)$ is totally positive; and we have just shown that the matrix $(\tilde{\mathbf{p}}_{k+1}, \mathbf{p}_{k+2}, \dots, \mathbf{p}_m)$ is totally positive. So Lemma 6.5 implies that the matrix $S$ is totally positive.

On the other hand, using equation (C.17) we can conclude that the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{\ell}_k, \mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ can be obtained from $S$ by right-multiplying it by the upper-bidiagonal matrix that has 1 on the diagonal, $(\alpha+x a_k) c_{k+1}$ in position $(k, k+1)$ and zeroes elsewhere. This proves that the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{\ell}_k, \mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ is totally positive. $\square$

The next two lemmas are the following:

**Lemma C.5.** Fix integers $0 \le k \le m$. If the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_k, \mathbf{p}_{k+1}, \dots, \mathbf{p}_m)$ is totally positive, then so is $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{p}_k, \dots, \mathbf{p}_m)$.

**Lemma C.6.** For every pair of integers $0 \le k \le m+1$, the matrix $(\mathbf{q}_0, \dots, \mathbf{q}_{k-1}, \mathbf{p}_k, \dots, \mathbf{p}_m)$ is totally positive.

We omit the proofs of Lemmas C.5 and C.6 as they are analogous to the proofs of Lemmas 6.9 and 6.10, respectively, but using columns instead of rows.

This completes the proof of Theorem C.1.

We conclude by posing the following open problem:

**Problem C.7.** Find a combinatorial interpretation for the output matrix $A = O(P)$ generated by the production matrix (C.1)/(C.4), or by interesting specializations thereof.

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