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\begin{center}
\vskip 1cm{\LARGE\bf Note on Total Positivity for a Class of \\
\vskip .1in
Recursive Matrices}
\vskip 1cm
{\large Liang Zhao and Fengyao Yan\\
School of Mathematical Sciences \\
Dalian University of Technology \\
Dalian 116024 \\
P. R. China \\
\href{mailto:light0617@mail.dlut.edu.cn}{\tt light0617@mail.dlut.edu.cn}\\
\href{mailto:yanfengyao@mail.dlut.edu.cn}{\tt yanfengyao@mail.dlut.edu.cn}\\
}
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\begin{abstract}
In this note, we study the total positivity of a class of infinite
recursive matrices that depend on three infinite sets of independent
variables and on an integer parameter. We give a simple algebraic proof
and provide a few examples.
\end{abstract}

\section{Introduction}

A (finite or infinite) matrix $M$ is said to be {\it totally positive} ($TP$)
if its minors of all orders are nonnegative.
Total positivity is a powerful concept that arises frequently
in various branches of mathematics, statistics, probability, mechanics,
economics, and computer science \cite{Bre,5}. The totally positive matrices play an
important role in the theory of total positivity \cite{1,4,2,3,5}.
Recently, Chen et al.~\cite{wy1,wy2} presented some sufficient conditions
for the total positivity of Riordan arrays and recursive matrices.
Brenti \cite{Bre} introduced a class of recursive matrices that depend on
three infinite sets of independent variables and on an integer parameter.
More precisely, let the infinite matrix $M=[M_{n,k}]_{n,k\geq0}$ be defined by
\begin{equation}\label{equ1}
M_{0,0}=1,\qquad M_{n,k}=z_{n}M_{n-t,k-1}+y_{n}M_{n-1-t,k-1}+x_{n}M_{n-1,k}
\end{equation}
for $n+k\geq1$, where $t\in \mathbb{N}$, $M_{n,k}=0$ if either $n&lt;0$ or $k&lt;0$,
and
 $(x_{n})_{n\geq0}$, $(y_{n})_{n\geq0}$, and $(z_{n})_{n\geq0}$
are nonnegative sequences.
Brenti \cite{Bre} associated a planar network with the matrix
and used its planarity to prove the total positivity of $M$.

\begin{theorem}[{\cite[Theorem 4.3]{Bre}}]\label{main-thm}
 Let $M=[M_{n,k}]_{n,k\geq0}$ be defined by \eqref{equ1}. Then $M$ is $TP$.
\end{theorem}

In this note we give a simple algebraic proof of Theorem \ref{main-thm}.
With the method of our proof,
the total positivity of many recursive matrices can be easily proved,
and a few examples are provided in Section \ref{secRemark}.

\section{Proof of Theorem \ref{main-thm}}

We first review some basic facts about $TP$ matrices.

\begin{lemma}[{\cite[Proposition 1.6]{Pin}}]\label{le1}
Assume $A$ is a nonsingular totally positive matrix.
Then so is $JA^{-1}J$, where $J$ is the diagonal matrix with
diagonal entries alternately $1$ and $-1$.
\end{lemma}

An operation that preserves total positivity is the following form of iteration.
Let $A=[a_{ij}]_{i,j=1}^n$ be an $n\times n$ matrix.
Define the matrix $B=[b_{ij}]_{i,j=1}^n$
by $b_{1j}=a_{1j}, j=1,\ldots ,n$,
and for $i\geq2, b_{ij}=\sum_{k=1}^{n}b_{i-1,k}a_{kj}, j=1,\ldots ,n$.
Pinkus \cite{Pin} showed that if $A$ is $TP$,
then $B$ is also $TP$.
Actually, this result can be generalized to infinite matrices
when $A=[a_{i,j}]_{i,j\geq 0}$ is an upper triangular matrix.
It is clear that $B=[b_{i,j}]_{i,j\geq 0}$ is $TP$ if and only if
its leading principal submatrices $[b_{i,j}]_{0\leq i,j\leq n}$ are all $TP$ {\cite[Lemma 2.1]{wy1}},
and $[b_{i,j}]_{0\leq i,j\leq n}$ is obtained directly from $[a_{i,j}]_{0\leq i,j\leq n}$.
Then we have the following.

\begin{lemma}\label{le2}
Let $A^T = {[\alpha_1,\alpha_2, \ldots ]}$ be an infinite upper triangular matrix,
where for $j\geq1$, $\alpha _j$ denotes the infinite column vector.
Define the matrix $B^T = {[\beta _1,\beta _2, \ldots ]}$ by ${\beta _1} = {\alpha _1}$,
and for $i\geq2$, ${\beta _i}^T = {\beta _{i - 1}}^T A$. If $A$ is $TP$, then so is $B$.
\end{lemma}

To prove Theorem \ref{main-thm}, we construct two matrices
\[A = {[{a_{ij}}]_{i \ge 0,j \ge 0}} = \left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;{{x_1}}&amp;{{x_1}{x_2}}&amp;{{x_1}{x_2}{x_3}}&amp; \cdots \\
{}&amp;{{z_0}}&amp;{{y_1} + {x_1}{z_0}}&amp;{{x_2}({y_1} + {x_1}{z_0})}&amp;{{x_2}{x_3}({y_1} + {x_1}{z_0})}&amp;{}\\
{}&amp;{}&amp;{{z_1}}&amp;{{y_2} + {x_2}{z_1}}&amp;{{x_3}({y_2} + {x_2}{z_1})}&amp;{}\\
{}&amp;{}&amp;{}&amp;{{z_2}}&amp;{{y_3} + {x_3}{z_2}}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{{z_3}}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right]\]
and
\[{I_t} = \left[ {\begin{array}{*{20}{c}}
1&amp;O\\
O&amp;U
\end{array}} \right], \]
 where $U={{[{\delta _{i + t,j}}]}_{i,j \ge 0}}$ and $\delta _{i,j}$
is the Kronecker delta, i.e.,
\begin{displaymath}
\delta _{i,j} = \begin{cases}
	1, &amp; \text{if $i= j$;} \\
	0, &amp; \text{if $i \ne j$.}
	\end{cases}
\end{displaymath}
Clearly, $I_{0}=I$ and $I_{t}A$ is the matrix obtained from $A$ by removing rows
of $A$ from the second row to the $(t+1)$th row.
Applying Lemma~\ref{le2} to the matrix $I_{t}A$,
we get $B = [O,{\widetilde B}]$,
where $\widetilde B =[b_{i,j}]_{i\geq0,j\geq0}$.

\begin{proof}[Proof of Theorem \ref{main-thm}]
We first show that $M=\widetilde B^{T}$.
For $n=0$, it is readily verified
that $b_{0,k}=M_{k,0}$. For $n\geq1$, we have
$$
\begin{aligned}
{b_{n,k}} = &amp;\prod\limits_{i = t + 2}^{k}x_{i} {({y_{t + 1}} + {x_{t + 1}}{z_{t}})} {b_{n - 1,0}}
+ \prod\limits_{i = t + 3}^{k}x_{i} {({y_{t + 2}} + {x_{t + 2}}{z_{t + 1}})} {b_{n - 1,1}}\\
&amp;+\cdots  + ({y_{k}} + {x_{k}}{z_{k - 1}}){b_{n - 1,k - t - 1}} + {z_k}{b_{n - 1,k - t}}\\
 =&amp; {x_{k}}{b_{n,k - 1}} + {y_{k}}{b_{n - 1,k - t - 1}} + {z_k}{b_{n - 1,k - t}}.
\end{aligned}
$$
Thus $M=\widetilde B^{T}$.

By the definition of $TP$ and the fact that $I_{t}A$ is a submatrix of $A$,
it suffices to show that $A$ is $TP$.
Let
\[X = \left[ {\begin{array}{*{20}{c}}
1&amp;0&amp;{}&amp;{}&amp;{}&amp;{}\\
{}&amp;1&amp;{{x_1}}&amp;{}&amp;{}&amp;{}\\
{}&amp;{}&amp;1&amp;{{x_2}}&amp;{}&amp;{}\\
{}&amp;{}&amp;{}&amp;1&amp;{{x_3}}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;1&amp; \ddots \\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right],\ Y = \left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;{}&amp;{}&amp;{}&amp;{}\\
{}&amp;{{z_0}}&amp;{{y_1}}&amp;{}&amp;{}&amp;{}\\
{}&amp;{}&amp;{{z_1}}&amp;{{y_2}}&amp;{}&amp;{}\\
{}&amp;{}&amp;{}&amp;{{z_2}}&amp;{{y_3}}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{{z_3}}&amp; \ddots \\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right].\]
It is obvious that $X$ and $Y$ are $TP$.
Note that by column  operations, we can turn $A$ into $Y$.
Thus $A$ has a factorization of the form
 $$A=YJX^{-1}J,$$
  where $J$ is defined in Lemma~\ref{le1},
 which implies that $JX^{-1}J$ is $TP$.
It is known that the product of two $TP$ matrices is still $TP$
by the classic Cauchy-Binet formula.
Thus $A$ is $TP$. This completes the proof.
\end{proof}

\section{Applications}\label{secRemark}

In this section we consider some examples for different $t$.
A basic example for case $t=0$ and $t=1$ is
the Delannoy square and the Delannoy
triangle \cite[\seqnum{A008288}]{oeis}.
 Many properties and applications of Delannoy numbers have been discussed \cite{Ban,Bar,Sul,Yang}.
The Delannoy numbers $d_{n,k}$ are defined as the numbers of lattice paths
from $(0,0)$ to $(n,k)$ with steps $(1,0)$, $(0,1)$, and $(1,1)$.
It is well known that the Delannoy numbers satisfy the recursion $$d_{n,k}=d_{n,k-1}+d_{n-1,k-1}+d_{n-1,k}.$$
Brenti \cite{Bre} showed that the Delannoy square
\[D = {[{d_{n,k}}]_{n,k \ge 0}} = \left[ {\begin{array}{*{20}{c}}
1&amp;1&amp;1&amp;1&amp; \ldots \\
1&amp;3&amp;5&amp;7&amp;{}\\
1&amp;5&amp;{13}&amp;{25}&amp;{}\\
1&amp;7&amp;{25}&amp;{63}&amp;{}\\
 \vdots &amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right]\]
is $TP$.
Let $d(n,k) = {d_{n - k,k}}$
denote the corresponding Delannoy
triangle $\bar D = {[d(n,k)]_{n,k \ge 0}}$,
i.e.,
\[\bar D = \left[ {\begin{array}{*{20}{c}}
1&amp;{}&amp;{}&amp;{}&amp;{}\\
1&amp;1&amp;{}&amp;{}&amp;{}\\
1&amp;3&amp;1&amp;{}&amp;{}\\
1&amp;5&amp;5&amp;1&amp;{}\\
 \vdots &amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right].\]
Then the entries in $\bar D$ satisfy the recurrence relation
$$d(n+1,k+1)=d(n,k)+d(n-1,k)+d(n,k+1).$$
Wang and Yang \cite{wangyang} showed that $\bar D$ is $TP$.

With the method of our proof, applying Lemma~\ref{le2} to the matrices
\[\left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;1&amp;1&amp;1&amp; \ldots \\
{}&amp;1&amp;2&amp;2&amp;2&amp;{}\\
{}&amp;{}&amp;1&amp;2&amp;2&amp;{}\\
{}&amp;{}&amp;{}&amp;1&amp;2&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;1&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right]\text{and}\left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;1&amp;1&amp;1&amp; \ldots \\
{}&amp;0&amp;1&amp;2&amp;2&amp;{}\\
{}&amp;{}&amp;0&amp;1&amp;2&amp;{}\\
{}&amp;{}&amp;{}&amp;0&amp;1&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;0&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right],\]
we get
\[\left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;1&amp;1&amp;1&amp; \cdots \\
0&amp;1&amp;3&amp;5&amp;7&amp;{}\\
0&amp;1&amp;5&amp;{13}&amp;{25}&amp;{}\\
0&amp;1&amp;7&amp;{25}&amp;{63}&amp;{}\\
0&amp;1&amp;9&amp;{41}&amp;{129}&amp;{}\\
 \vdots &amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right]\text{and}\left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;1&amp;1&amp;1&amp; \cdots \\
{}&amp;0&amp;1&amp;3&amp;5&amp;{}\\
{}&amp;{}&amp;0&amp;1&amp;5&amp;{}\\
{}&amp;{}&amp;{}&amp;0&amp;1&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;0&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right].\]
Then the total positivity of the Delannoy square and the
Delannoy triangle immediately follows.

An example for case $t=2$ is the Harer-Zagier numbers $g(n,k)$ \cite[\seqnum{A035309}]{oeis}, which count the number of
ways to glue all edges of a $2n$-gon pairwise, so as to
produce a surface of given genus $k$.
The first three columns of $G=[g(n,k)]_{n,k\geq0}$ 
(for $k=0,1,2$) are respectively
the Catalan numbers \cite[\seqnum{A000108}]{oeis}, \seqnum{A002802}, and \seqnum{A006298}.
The entries of $G$ satisfy the recursion
\begin{equation}\label{equ2}
(n+2)g(n+1,k)=(4n+2)g(n,k)+(4n^{3}-n)g(n-1,k-1),
\end{equation}
where $g(n,k)=0$ if either $n&lt;0$ or $k&lt;0$, $g(0,0)=1$, and $g(0,k)=0$. So,
\[G = \left[ {\begin{array}{*{20}{c}}
1&amp;{}&amp;{}&amp;{}&amp;{}\\
1&amp;{}&amp;{}&amp;{}&amp;{}\\
2&amp;1&amp;{}&amp;{}&amp;{}\\
5&amp;{10}&amp;{}&amp;{}&amp;{}\\
{14}&amp;{70}&amp;{21}&amp;{}&amp;{}\\
{42}&amp;{420}&amp;{483}&amp;{}&amp;{}\\
{132}&amp;{2310}&amp;{6468}&amp;{1485}&amp;{}\\
 \vdots &amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right].\]


Using Lemma~\ref{le2} with
\[A = \left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;1&amp;2&amp;5&amp;{14}&amp;{42}&amp; \ldots \\
{}&amp;{}&amp;{}&amp;1&amp;{\frac{5}{2}}&amp;7&amp;{21}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{\frac{{15}}{2}}&amp;{21}&amp;{63}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{21}&amp;{63}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{42}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right],\]
we get
\[B = \left[ {\begin{array}{*{20}{c}}
0&amp;1&amp;1&amp;2&amp;5&amp;{14}&amp;{42}&amp; \cdots \\
{}&amp;{}&amp;{}&amp;1&amp;{10}&amp;{70}&amp;{420}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{21}&amp;{483}&amp;{}\\
{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{}&amp;{}&amp; \ddots
\end{array}} \right].\]
Then we have the following.

\begin{corollary}
Let $G=[g(n,k)]_{n,k\geq0}$ be defined by \eqref{equ2}. Then $G$ is $TP$.
\end{corollary}

\section{Acknowledgments}

The authors thank the anonymous referees for their careful reading and valuable suggestions.
\begin{thebibliography}{99}

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\bibitem{Ban} C. Banderier and S. Schwer, Why Delannoy numbers?, {\it J. Statist. Plann. Inference} {\bf 135} (2005), 40--54.

\bibitem{Bar} P. Barry, On integer-sequence-based constructions of generalized Pascal triangles,
{\it J. Integer Seq.} {\bf 9} (2006), \href{http://www.emis.de/journals/JIS/VOL9/Barry/barry91.html}{Article 06.2.4}.

\bibitem{Bre} F. Brenti, Combinatorics and total positivity, {\it J. Combin. Theory Ser. A} {\bf 71} (1995), 175--218.

\bibitem{4} F. Brenti, The applications of total positivity to combinatorics, and conversely,
in {\it Total Positivity and Its Applications}, in Math. Appl., Vol.~359,
Kluwer, 1996, pp.~451--473.

\bibitem{wy1} X. Chen, H. Liang, and Y. Wang, Total positivity of recursive matrices, {\it Linear Algebra Appl.} {\bf 471} (2015), 383--393.

\bibitem{wy2} X. Chen, H. Liang, and Y. Wang, Total positivity of Riordan arrays, {\it European J. Combin.} {\bf 46} (2015), 68--74.

\bibitem{2} S. M. Fallat and C. R. Johnson, {\it Totally Nonnegative
Matrices}, Princeton University Press, 2011.

\bibitem{3} M. Gasca and J. M. Pe\~{n}a,  Total positivity and Neville elimination, {\it Linear Algebra Appl.} {\bf 165} (1992), 25--44.

\bibitem{5} S. Karlin, {\it Total Positivity}, Vol.~1, Stanford
University Press, 1968.

\bibitem{Pin} A. Pinkus, {\it Totally Positive Matrices}, Cambridge
University Press, 2010.

\bibitem{oeis}N. J. A. Sloane, The On-Line Encyclopedia of Integer 
Sequences, \url{http://oeis.org}.

\bibitem{Sul}R. A. Sulanke, Objects counted by the central Delannoy
numbers, {\it J. Integer Seq.} {\bf 6} (2003),
\href{http://www.emis.de/journals/JIS/VOL6/Sulanke/delannoy.html}{Article
03.1.5}.

\bibitem{wangyang} Y. Wang and A. L. B. Yang, Total positivity of some
symmetric triangular and square arrays, in preparation.

\bibitem{Yang} S.-L. Yang, S.-N. Zheng, S.-P. Yuan, and T.-X. He,
Schr\"{o}der matrix as inverse of Delannoy matrix, {\it Linear Algebra
Appl.} {\bf 439} (2013), 3605--3614.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A20; Secondary 15A45, 15B36.

\noindent \emph{Keywords: }
totally positive matrix,
recursive matrix.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000108},
\seqnum{A002802},
\seqnum{A006298},
\seqnum{A008288}, and
\seqnum{A035309}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received August 17 2015;
revised versions received  April 11 2016; June 4 2016; July 2 2016.
Published in {\it Journal of Integer Sequences}, July 5 2016.

\bigskip
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\noindent
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