\documentclass[11pt]{article} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \usepackage{amsthm,amssymb} \usepackage{psfig,epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\rk}{{\rm rk}} \newcommand{\bbox}{\nopagebreak[4] \hfill\rule{2mm}{2mm}} \newcounter{resno} \newenvironment{Proof}{{\sc Proof.\ }}{\hfill\bbox\bigskip} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo112.eps} \vskip 1cm {\LARGE \bf Hankel Matrices and Lattice Paths} \vskip 1.5cm {\large Wen-jin Woan} \smallskip \\ Department of Mathematics \\ Howard University \\ Washington, D.C. 20059, USA \medskip \\ Email address: \href{mailto:wwoan@howard.edu}{wwoan@howard.edu} \\ \vskip2.5cm {\bf Abstract} \end{center} {\em Let $H$ be the Hankel matrix formed from a sequence of real numbers $% S=\{a_0=1,a_1,a_2,a_3,...\},$ and let $L$ denote the lower triangular matrix obtained from the Gaussian column reduction of $H.$ This paper gives a matrix-theoretic proof that the associated Stieltjes matrix $S_L$ is a tri-diagonal matrix. It is also shown that for any sequence (of nonzero real numbers) $% T=\{d_0=1,d_1,d_2\ ,d_3,...\}$ there are infinitely many sequences such that the determinant sequence of the Hankel matrix formed from those sequences is $T$.} {\bf 1. Introduction}. In this paper we give a matrix-theoretic proof (Theorem 2.1) of one of the main theorems in \cite{PW}. In Section 2 we discuss the connection between the decomposition of a Hankel matrix and Stieltjes matrices, and in Section 3 we discuss the connection between certain lattice paths and Hankel matrices. Section 4 presents an explicit formula for the decomposition of a Hankel matrix. {\bf Definition 1.1.} Let $S=\{a_0=1,a_1,a_2,a_3,...\}$ be a sequence of real numbers. The Hankel matrix generated by $S$ is the infinite matrix \begin{center} \[ H=\left[ \begin{array}{cccccc} 1 & a_1\ & a_2\ & a_3\ & a_4\ & . \\ a_1\ & a_2 & a_3\ & a_4\ & a_5\ & . \\ a_2 & a_3 & a_4\ & a_5\ & a_6\ & . \\ a_3 & a_4\ & a_5\ & a_6\ & a_7\ & . \\ a_4\ & a_5\ & a_6\ & a_7\ & a_8 & . \\ . & . & . & . & . & . \end{array} \right] . \] \ \end{center} {\bf Definition 1.2}. A lower triangular matrix \[ L=\left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & . \\ l_{10} & 1 & 0 & 0 & 0 & . \\ l_{20} & l_{21} & 1 & 0 & 0 & . \\ l_{30} & l_{31} & l_{32} & 1 & 0 & . \\ l_{40} & l_{41} & l_{42} & l_{43} & 1 & . \\ . & . & . & . & . & . \end{array} \right] . \] \noindent is said to be a Riordan matrix if there exist Taylor series $g(x)=1+a_1x+a_2x^2+...+a_nx^n+...$ and $f(x)=x+b_2x^2+b_3x^3+...+b_nx^n+....$ such that for every $k \geq 0$ the $k$-th column has ordinary generating function $g(x)(f(x))^k$. {\bf Definition 1.3}. The Stieltjes matrix $\ $of a lower triangular matrix $% L$ is the matrix $S_L$ which satisfies $LS_L=L^r\;$where $L^r\;$is the matrix obtained from $L$ by deleting the first row of $L.$ Thus \[ \left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & . \\ l_{10} & 1 & 0 & 0 & 0 & . \\ l_{20} & l_{21} & 1 & 0 & 0 & . \\ l_{30} & l_{31} & l_{32} & 1 & 0 & . \\ l_{40} & l_{41} & l_{42} & l_{43} & 1 & . \\ . & . & . & . & . & . \end{array} \right] S_L=\left[ \begin{array}{cccccc} l_{10} & 1 & 0 & 0 & 0 & . \\ l_{20} & l_{21} & 1 & 0 & 0 & . \\ l_{30} & l_{31} & l_{32} & 1 & 0 & . \\ l_{40} & l_{41} & l_{42} & l_{43} & 1 & . \\ . & . & . & . & . & . \end{array} \right] \] and so \[ S_L=L^{-1}L^r=\ \left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & . \\ -l_{10} & 1 & 0 & 0 & 0 & . \\ \times & -l_{21} & 1 & 0 & 0 & . \\ \times & \times & -l_{32} & 1 & 0 & . \\ \times & \times & \times & -l_{43} & 1 & . \\ . & . & . & . & . & . \end{array} \right] \left[ \begin{array}{cccccc} l_{10} & 1 & 0 & 0 & 0 & . \\ l_{20} & l_{21} & 1 & 0 & 0 & . \\ l_{30} & l_{31} & l_{32} & 1 & 0 & . \\ l_{40} & l_{41} & l_{42} & l_{43} & 1 & . \\ . & . & . & . & . & . \end{array} \right] \] \[ =\left[ \begin{array}{cccccc} b_0 & 1 & 0 & 0 & 0 & . \\ c_0 & b_1 & 1 & 0 & 0 & . \\ \times & c_1 & b_2 & 1 & 0 & . \\ \times & \times & c_2 & b_3 & 1 & . \\ \times & \times & \times & c_3 & b_4 & . \\ . & . & . & . & . & . \end{array} \right] \] \ \noindent where \[ b_0=l_{10,}\ b_k=l_{k+1,k}-l_{k,k-1},\;k>0, \] \[ c_0=l_{2,0}-l_{1,0}^2,% \;c_k=(l_{k,k-1}l_{k+1,k}-l_{k+1,k-1})-l_{k+1,k}^2+l_{k+2,k},\ k>0. \] {\bf Definition 1.4}. Let $L$ and $S_L$ be as in Definition 1.3. We define \[ D_L=\left[ \begin{array}{cccccc} d_0 & 0 & 0 & 0 & 0 & . \\ 0 & d_1 & 0 & 0 & 0 & . \\ 0 & 0 & d_2 & 0 & 0 & . \\ 0 & 0 & 0 & d_3 & 0 & . \\ 0 & 0 & 0 & 0 & d_4 & . \\ . & . & . & . & . & . \end{array} \right] \] \noindent to be the diagonal matrix with diagonal entries given by $d_0=1,$ $% d_{k+1}=d_kc_k$ for $k>0$.\bigskip\ {\bf 2. Stieltjes and Hankel Matrices.} The following two theorems are proved in \cite{PW}. {\bf Theorem 2.1}. Let $L$ be a lower triangular matrix and let $D=D_L$ be the diagonal matrix with nonzero diagonal entries $\{d_i\}\;$as in Definition 1.4. Then $LDL^t$ is a Hankel matrix if and only if $S_L$ is a tri-diagonal matrix, i.e. if and only if \[ S_L=\left[ \begin{array}{cccccc} b_0 & 1 & 0 & 0 & 0 & . \\ c_0 & b_1 & 1 & 0 & 0 & . \\ 0 & c_1 & b_2 & 1 & 0 & . \\ 0 & 0 & c_2 & b_3 & 1 & . \\ 0 & 0 & 0 & c_3 & b_4 & . \\ . & . & . & . & . & . \end{array} \right] \] where $\ b_0=l_{1,0}\;,\quad c_0=d_1\;,\quad b_k=l_{k+1,k}-l_{k,k-1}\;,\quad c_k=\frac{d_{k+1}}{d_k}\;,\quad k\geq 1.$ \begin{Proof} Let $H=LDL^t$ be a Hankel matrix. Then $\ L=H(DL^t)^{-1},\ $ $L^r=(H(DL^t)^{-1})^r=H^r(DL^t)^{-1},\ $ $S_L=L^{-1}L^r=L^{-1}(H^r(DL^t)^{-1})=(L^{-1}H^r)(DL^t)^{-1}.$ $\ $Since $H$ is a Hankel matrix, deleting the first row has the same effect as deleting the first column. \[ L^{-1}H=DL^t=\left[ \begin{tabular}{llllll} $d_0$ & $d_0l_{10}$ & $d_0l_{20}$ & $d_0l_{3,0}$ & $d_0l_{4,0}$ & . \\ $0$ & $d_1\ $ & $d_1l_{21}$ & $d_1l_{31}$ & $d_1l_{41}$ & . \\ $0$ & $0$ & $d_2$ & $d_2l_{32}$ & $d_2l_{42}$ & . \\ $0$ & $0$ & $0$ & $d_3$ & $d_3l_{43}$ & . \\ $0$ & $0$ & $0$ & $0$ & $d_4$ & . \\ . & . & . & . & . & . \end{tabular} \right] , \] \ \[ L^{-1}H^r=L^{-1}H^c=(L^{-1}H)^c=\left[ \begin{tabular}{lllll} $d_0l_{10}$ & $d_0l_{20}$ & $d_0l_{30}$ & $d_0l_{4,0}$ & . \\ $d_1\ $ & $d_1l_{21}$ & $d_1l_{31}$ & $d_1l_{41}$ & . \\ $0$ & $d_2$ & $d_2l_{32}$ & $d_2l_{42}$ & . \\ $0$ & $0$ & $d_3$ & $d_3l_{43}$ & . \\ $0$ & $0$ & $0$ & $d_4$ & . \\ . & . & . & . & . \end{tabular} \right] , \] \[ S_L=(L^{-1}H)^c(DL^t)^{-1}=\left[ \begin{tabular}{lllll} $d_0l_{10}$ & $d_0l_{20}$ & $d_0l_{30}$ & $d_0l_{4,0}$ & . \\ $d_1\ $ & $d_1l_{21}$ & $d_1l_{31}$ & $d_1l_{41}$ & . \\ $0$ & $d_2$ & $d_2l_{32}$ & $d_2l_{42}$ & . \\ $0$ & $0$ & $d_3$ & $d_3l_{43}$ & . \\ $0$ & $0$ & $0$ & $d_4$ & . \\ . & . & . & . & . \end{tabular} \right] \left[ \begin{tabular}{llllll} $\frac 1{d_0}$ & $\times $ & $\times $ & $\times $ & $\times $ & . \\ $0$ & $\frac 1{d_1}\ $ & $\times $ & $\times $ & $\times $ & . \\ $0$ & $0$ & $\frac 1{d_2}$ & $\times $ & $\times $ & . \\ $0$ & $0$ & $0$ & $\frac 1{d_3}$ & $\times $ & . \\ $0$ & $0$ & $0$ & $0$ & $\frac 1{d_4}$ & . \\ . & . & . & . & . & . \end{tabular} \right] \] \ \[ =\left[ \begin{array}{cccccc} b_0 & 1 & 0 & 0 & 0 & . \\ c_0 & b_1 & 1 & 0 & 0 & . \\ 0 & c_1 & b_2 & 1 & 0 & . \\ 0 & 0 & c_2 & b_3 & 1 & . \\ 0 & 0 & 0 & c_3 & b_4 & . \\ . & . & . & . & . & . \end{array} \right] \] \ \noindent where \[ b_0=l_{1,0}\;,\quad c_0=\frac{d_1}{d_0}=d_1\;,\quad b_k=l_{k+1,k}-l_{k,k-1}\;,\quad c_k=\frac{d_{k+1}}{d_k}\;,\quad k\geq 1. \] $\;\;\;\;\;\;$Conversely$,$ let $S_L\;$be a tri-diagonal matrix and let $% H=LDL^t.$ Then $L^{-1}H^r=L^{-1}(LDL^t)^r=L^{-1}(L^rDL^t)=(L^{-1}L^r)DL^t=S_LDL^t$ \[ =\left[ \begin{array}{cccccc} b_0 & 1 & 0 & 0 & 0 & . \\ c_0 & b_1 & 1 & 0 & 0 & . \\ 0 & c_1 & b_2 & 1 & 0 & . \\ 0 & 0 & c_2 & b_3 & 1 & . \\ 0 & 0 & 0 & c_3 & b_4 & . \\ . & . & . & . & . & . \end{array} \right] \left[ \begin{tabular}{llllll} $d_0$ & $d_0l_{10}$ & $d_0l_{20}$ & $d_0l_{3,0}$ & $d_0l_{4,0}$ & . \\ $0$ & $d_1\ $ & $d_1l_{21}$ & $d_1l_{31}$ & $d_1l_{41}$ & . \\ $0$ & $0$ & $d_2$ & $d_2l_{32}$ & $d_2l_{42}$ & . \\ $0$ & $0$ & $0$ & $d_3$ & $d_3l_{43}$ & . \\ $0$ & $0$ & $0$ & $0$ & $d_4$ & . \\ . & . & . & . & . & . \end{tabular} \right] . \] Therefore\ $(L^{-1}H^r)_{n,k}=c_{n-1}d_{n-1}l_{k,n-1}+b_nd_nl_{k,n}+d_{n+1}l_{k,n+1}$ $=\frac{d_n}{d_{n-1}}d_{n-1}l_{k,n-1}+b_nd_nl_{k,n}+c_nd_nl_{k,n+1}$ $=d_n(l_{k,n-1}+b_nl_{k,n}+c_nl_{k,n+1})$ $% =d_nl_{k+1,n}=(DL^t)_{n,k+1}=(DL^t)_{n,k}^c=(L^{-1}H)_{n,k}^c=(L^{-1}H^c)_{n,k}. $ We have shown that $L^{-1}H^r=L^{-1}H^c,$ and so $H^r=H^c.$\ Hence $H$ is a Hankel matrix. \end{Proof} {\bf Theorem 2.2}. $L$ is a Riordan matrix (i.e. $b_k=b_1=b$ and $c_k=c_1=c$ for $k\geq 1)$ if and only if $f=x(1+bf+cf^2)\;$and \[ g=\frac 1{1-xb_0-xc_0f} ~, \] where $f,g$ are as in Definition 1.2. See \cite{PW} for the proof. {\bf Corollary 2.3}. Let $T=\{d_0=1,d_1,d_2\ ,d_3,...\}$ be any sequence of (nonzero) real numbers. Then there exists a sequence $S=\{a_0=1,a_1,a_2,a_3,...\}$ such that $T$ is equal to the sequence of diagonal entries of $D$ in the decomposition $H=LDL^t$ of the Hankel matrix generated by $S$ . \begin{Proof} As in Theorem 2.1, let $\ c_0=d_1\;,c_k=\frac{d_{k+1}}{d_k}\;, k\geq 1,\;$and form the Stieltjes matrix \[ S_L=\left[ \begin{array}{cccccc} b_0 & 1 & 0 & 0 & 0 & . \\ c_0 & b_1 & 1 & 0 & 0 & . \\ 0 & c_1 & b_2 & 1 & 0 & . \\ 0 & 0 & c_2 & b_3 & 1 & . \\ 0 & 0 & 0 & c_3 & b_4 & . \\ . & . & . & . & . & . \end{array} \right] \] \noindent where the $b_i$s are arbitrary. By Definition 1.3 there is a lower triangular matrix $L$ such that $LS_L=L^r$. Let $S$ be the sequence formed by the first column of $L$ and let $H$ denote the Hankel matrix generated by $S$. By Theorem 2.1 the diagonal entries of $D$ in the decomposition $H=LDL^t$ form the sequence $T$. \end{Proof} {\bf Example 2.4.} Let $T=\{1,1,2,5,14,42,132,...\}$ be the Catalan sequence (\htmladdnormallink{A000108}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000108} in \cite{OEIS}) and let \[ \ \ S_L=\left[ \begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & . \\ 1 & 0 & 1 & 0 & 0 & . \\ 0 & 2 & 0 & 1 & 0 & . \\ 0 & 0 & \frac 52 & 0 & 1 & . \\ 0 & 0 & 0 & \frac{14}5 & 0 & . \\ . & . & . & . & . & . \end{array} \right] . \] $\ $Then \[ L=\left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & . \\ 0 & 1 & 0 & 0 & 0 & . \\ 1 & 0 & 1 & 0 & 0 & . \\ 0 & 3 & 0 & 1 & 0 & . \\ 3 & 0 & \frac{11}2 & 0 & 1 & . \\ . & . & . & . & . & . \end{array} \right] , \] \[ LDL^t=\left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & . \\ 0 & 1 & 0 & 0 & 0 & . \\ 1 & 0 & 1 & 0 & 0 & . \\ 0 & 3 & 0 & 1 & 0 & . \\ 3 & 0 & \frac{11}2 & 0 & 1 & . \\ . & . & . & . & . & . \end{array} \right] \left[ \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & . \\ 0 & 1 & 0 & 0 & 0 & . \\ 0 & 0 & 2 & 0 & 0 & . \\ 0 & 0 & 0 & 5 & 0 & . \\ 0 & 0 & 0 & 0 & 14 & . \\ . & . & . & . & . & . \end{array} \right] \left[ \begin{array}{cccccc} 1 & 0 & 1 & 0 & 3 & . \\ 0 & 1 & 0 & 3 & 0 & . \\ 0 & 0 & 1 & 0 & \frac{11}2 & . \\ 0 & 0 & 0 & 1 & 0 & . \\ 0 & 0 & 0 & 0 & 1 & . \\ . & . & . & . & . & . \end{array} \right] \] \ \[ =\left[ \begin{array}{cccccc} 1 & 0 & 1 & 0 & 3 & . \\ 0 & 1 & 0 & 3 & 0 & . \\ 1 & 0 & 3 & 0 & 14 & . \\ 0 & 3 & 0 & 14 & 0 & . \\ 3 & 0 & 14 & 0 & \frac{167}2 & . \\ . & . & . & . & . & . \end{array} \right] =H.\ \] {\bf 3. Lattice Paths and Hankel Matrices} We consider those lattice paths in the Cartesian plane running from $(0,0)$ that use steps from $S=\{u=(1,1),\;h=(1,0),\;d=(1,-1)\}$ with assigned weights $1$ for $u$, $w_1$ for $h$ and $w_2$ for $d$. Let $L(n,k)$ be the set of paths that never go below the $x$-axis and end at $(n,k)$. The weight of a path is the product of the weights of its steps. Let $l_{n,k}\;$be\ the sum of the weights of all the paths in $L(n,k).$ See also \cite{Su}, \cite{We}. {\bf Theorem 3.1}. Let $L=(l_{n,k})_{n,k\geq 0}.$ Then $L$ is a lower triangular matrix, the Stieltjes matrix of $L$ is \[ S_L=\left[ \begin{array}{cccccc} w_1 & 1 & 0 & 0 & 0 & . \\ w_2 & w_1 & 1 & 0 & 0 & . \\ 0 & w_2 & w_1 & 1 & 0 & . \\ 0 & 0 & w_2 & w_1 & 1 & . \\ 0 & 0 & 0 & w_2 & w_1 & . \\ . & . & . & . & . & . \end{array} \right] \] \ \noindent and $H=LDL^t\;$is the Hankel matrix generated by the first column of $L$ and $d_k=w_2^k$ for $k>0.$ \begin{Proof} From Theorem 2.1. \end{Proof} {\bf Example 3.2.} For $w_1=0,\;w_2=1,$ $L$ is the Catalan matrix. For $% w_1=t,\;w_2=1$, $L$ is the $t$-Motzkin matrix. In both cases $D$ is the identity matrix. For example, when $t=1$, \[ L\ =\left[ \begin{tabular}{llllll} $1$ & $0$ & $0$ & $0$ & $0$ & . \\ $1$ & $1$ & $0$ & $0$ & $0$ & . \\ $2$ & $2$ & $1$ & $0$ & $0$ & . \\ $4$ & $5$ & $3$ & $1$ & $0$ & . \\ $9$ & $12$ & $9$ & $4$ & $1$ & . \\ . & . & . & . & . & . \end{tabular} \right] \ , \] \ \[ \ \ LDL^t=\left[ \begin{tabular}{llllll} $1$ & $1$ & $2$ & $4$ & $9$ & . \\ $1$ & $2$ & $4$ & $9$ & $21$ & . \\ $2$ & $4$ & $9$ & $21$ & $51$ & . \\ $4$ & $9$ & $21$ & $51$ & $127$ & . \\ $9$ & $21$ & $51$ & $127$ & $323$ & . \\ . & . & . & . & . & . \end{tabular} \right] =H \] \ \noindent where $S=\{1,1,2,4,9,21,51,...\}$ is the Motzkin sequence \htmladdnormallink{A001006}{http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A001006}. {\bf Theorem 3.3.} If $w_1,$ $w_2\;$depend on the height $k$, i.e. $% w_1(k)=b_k$ and $w_2(k+1)=c_k,$ then \[ S_L=\left[ \begin{array}{cccccc} b_0 & 1 & 0 & 0 & 0 & . \\ c_0 & b_1 & 1 & 0 & 0 & . \\ 0 & c_1 & b_2 & 1 & 0 & . \\ 0 & 0 & c_2 & b_3 & 1 & . \\ 0 & 0 & 0 & c_3 & b_4 & . \\ . & . & . & . & . & . \end{array} \right] \] \noindent and $H=LDL^t$ is the Hankel matrix generated by the first column of $L$ and $d_k=\Pi _{i\leq k}c_i.$ \begin{Proof} From Theorem 2.1. \end{Proof} \noindent See Example 2.4 for an illustration. {\bf 4. Gaussian Column Reduction } \ \ Let $S=\{a_0=1,a_1,a_2,a_3,...\}$ be a sequence of real numbers and let $% H$ denote the Hankel matrix generated by $S$. All the results in this section are well-known in matrix theory. We shall express the entries of $L$ in term of $S$. We assume that $H$ is positive definite. {\bf Lemma 4.1}. The decomposition of a positive definite Hankel matrix $% H=LDU$ is unique and $U=L^t$, where $L$ is a lower triangular matrix with diagonal entries 1, $D$ is a diagonal matrix and $U$ is an upper triangular matrix with diagonal entries 1. \begin{Proof} Let $LDU=H=L_1D_1U_1.$ Then $DUU_1^{-1}=L^{-1}L_1D_1$ is both an upper and lower triangular matrix, hence $UU_1^{-1}=L^{-1}L_1=I$ is the infinite identity matrix. \end{Proof} Let $H_n$ be the truncated submatrix of $H$ with $n\geq 0$ . For example, \begin{center} \[ H_3=\left[ \begin{array}{cccc} 1 & a_1\ & a_2\ & a_3 \\ a_1\ & a_2\ & a_3 & a_4\ \\ a_2\ & a_3 & a_4\ & a_5\ \\ a_3 & a_4\ & a_5\ & a_6\ \end{array} \right] ,\;\;\;\;\;H_4=\left[ \begin{array}{ccccc} 1 & a_1\ & a_2\ & a_3 & a_4\ \\ a_1\ & a_2\ & a_3 & a_4\ & a_5\ \\ a_2\ & a_3 & a_4\ & a_5\ & a_6\ \\ a_3 & a_4\ & a_5\ & a_6\ & a_7\ \\ a_4\ & a_5\ & a_6\ & a_7\ & a_8 \end{array} \right] . \] \ \end{center} Let $H_n(k)$ be the matrix obtained from $H_n$ by replacing the last column of $H_n$ by $a_k,a_{k+1},a_{k+2},...,a_{k+n}.$ For example, \[ H_3(1)=\left[ \begin{array}{cccc} 1 & a_1\ & a_2\ & a_1 \\ a_1\ & a_2\ & a_3 & a_2\ \\ a_2\ & a_3 & a_4\ & a_3\ \\ a_3 & a_4\ & a_5\ & a_4\ \end{array} \right] ,\;\;\;\;\;\;H_3(5)=\left[ \begin{array}{cccc} 1 & a_1\ & a_2\ & a_5 \\ a_1\ & a_2\ & a_3 & a_6\ \\ a_2\ & a_3 & a_4\ & a_7\ \\ a_3 & a_4\ & a_5\ & a_8 \end{array} \right] . \] .\ Let $h_i=\det H_i\;$and define an infinite upper triangular matrix $% R=(r_{n,k})$ in term of $(n,k)$-cofactor of $H_k$ by $r_{n,k}=0\;$for $k