\documentclass[12pt, reqno]{amsart} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, plainpages]{hyperref} \textheight 22.5truecm \textwidth 14.5truecm \setlength{\oddsidemargin}{0.35in}\setlength{\evensidemargin}{0.35in} \setlength{\topmargin}{-.5cm} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \begin{document} \setcounter{page}{1} \noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}} \noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent Banach J. Math. Anal. 2 (2008), no. 2, 1--8\\ $\frac{\rule{4.55in}{0.05in}}{{}}$\\ {\footnotesize \textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach \textbf{\large{J}}ournal of \textbf{\large{M}}athematical \textbf{\large{A}}nalysis}}\\ ISSN: 1735-8787 (electronic)\\ \textcolor[rgb]{0.00,0.00,0.84}{\textbf{http://www.math-analysis.org }}\\ $\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]} \title{Positivity of operator-matrices of Hua-type} \author[T. Ando]{Tsuyoshi Ando} \address{Shiroishi-ku, Hongo-dori 9, Minami 4-10-805, Sapporo 003-0024, Japan.} \email{\textcolor[rgb]{0.00,0.00,0.84}{ando@es.hokudai.ac.jp}} \dedicatory{This paper is dedicated to Professor Josip E. Pe\v{c}ari\'{c}\\ \vspace{.5cm} {\rm Submitted by F. Kittaneh}} \subjclass[2000]{Primary 47B63; Secondary 47B15, 15A45.} \keywords{Positivity, Strict contraction, Operator-matrix, Hua theorem.} \date{Received: 1 March 2008; Accepted 25 March 2008.} \begin{abstract} Let $A_j\,\,(j = 1, 2,\ldots , n)$ be strict contractions on a Hilbert space. We study an $n \times n$ operator-matrix: \[\textbf{H}_n(A_1,A_2,\ldots ,A_n) = [(I - A^*_j A_i)^{-1}]^n_{i,j=1}.\] For the case $n = 2$, Hua [Inequalities involving determinants, Acta Math. Sinica, 5 (1955), 463--470 (in Chinese)] proved positivity, i.e., positive semi-definiteness of $\textbf{H}_2(A_1,A_2)$. This is, however, not always true for $n = 3$. First we generalize a known condition which guarantees positivity of $\textbf{H}_n$. Our main result is that positivity of $\textbf{H}_n$ is preserved under the operator M\"obius map of the open unit disc $\mathcal D$ of strict contractions. \end{abstract} \maketitle \section{Introduction and preliminaries} \noindent Let $A_j\,\,(j = 1, 2,\ldots , n)$ be \emph{strict contractions}, that is, $\|A_j\| < 1$, on a Hilbert space $\mathcal H$. Since all $I -A^*_j A_i$ and $I -A_iA^*_j$ are invertible, let us consider an $n\times n$ operator-matrix \[\textbf{H}_n(A_1,A_2,\ldots ,A_n) = [(I - A^*_j A_i)^{-1}]^n_{i,j=1},\] and its cousin \[\textbf{G}_n(A_1,A_2,\ldots ,A_n) = [(I - A_iA^*_j )^{-1}]^n_{i,j=1}.\] Here $\textbf{X} = [X_{i,j} ]^n_{i,j=1}$ means that $X_{i,j}$ is the $(i, j)$-operator entry of $\textbf{X}$. (Notice that Xu et al. \cite{X-X-Z} used $\textbf{H}_n(A_1,A_2, \ldots,A_n)$ for our $\textbf{G}_n(A^*_1 ,A^*_2,\ldots,A^*_n)$.) In this paper our interest is in \emph{positivity}, i.e., positive semi-definiteness, of the operator-matrix $\textbf{H}_n$ (and also that of $\textbf{G}_n$). We will use the notation $\textbf{X} \geq \textbf{Y}$ to mean that both $\textbf{X},\textbf{Y}$ are selfadjoint and $\textbf{X}-\textbf{Y}$ is positive. In particular $\textbf{X} \geq 0$ means that $\textbf{X}$ is positive. Here let us use $\textbf{X} > 0$ to denote its positive definiteness, that is, $\textbf{X}$ is positive and invertible. For an operator-matrix $\textbf{X} = [X_{i,j} ]^n_{i,j=1}$ with invertible $X_{n,n}$, the \emph{Schur complement} of the $(n, n)$-operator entry $X_{n,n}$ in $\textbf{X}$, denoted by $\textbf{X}/(n)$ in this paper, is the $(n - 1) \times (n - 1)$ operator-matrix defined by \begin{equation} \textbf{X}/(n) = [X_{i,j} - X_{i,n}X^{-1}_{n,n}X_{n,j} ]^{n-1}_{i,j=1}. \end{equation} In this case, $\textbf{X}$ is invertible if and only if $\textbf{X}/(n)$ is invertible. Further the following relation holds (see \cite[Section 7.7]{H-J}) \begin{equation}\label{1.2} (\textbf{X}/(n))^{-1} = \text{the top } (n - 1) \times (n - 1) \text{ operator-submatrix of } \textbf{X}^{-1}. \end{equation} For our purpose the following \emph{Schur criteria} are quite useful. For selfadjoint $\textbf{X}$ with invertible $X_{n,n}$ the positivity of $\textbf{X}$ is equivalent to that $X_{n,n} \geq 0$ and $\textbf{X}/(n) \geq 0$. Further $\textbf{X} > 0$ if and only if $X_{n,n} > 0$ and $\textbf{X}/(n) > 0$. Let us return to $\textbf{H}_n(A_1,A_2,\ldots ,A_n)$ and $\textbf{G}_n(A_1,A_2,\ldots,A_n)$. In the case $n = 2$, for simplicity, let us write $A = A_1$ and $A_2 = B$. Hua \cite{HUA} showed $\textbf{H}_2(A,B) \geq 0$. Since $(I - B^*B)^{-1} > 0$, by the Schur criteria the Hua's positivity result is equivalent to the following inequality: \begin{equation}\label{1.3} (I - A^*A)^{-1} - (I - B^*A)^{-1}(I - B^*B)(I - A^*B)^{-1} \geq 0. \end{equation} With help of the identity \eqref{1.2}, Xu et al. \cite{X-X-Z} gave a simple proof for the following identity due to Hua \cite{HUA} which guarantees the positivity \eqref{1.3}: \begin{eqnarray*} &&(I-A^*A)^{-1} - (I - B^*A)^{-1}(I - B^*B)(I - A^*B)^{-1}\cr &&= (I - B^*A)^{-1}(A - B)^*(I - AA^*)^{-1}(A - B)(I - A^*B)^{-1}. \end{eqnarray*} In \cite{AND} we proved also \begin{equation}\label{1.4} (I - AA^*)^{-1} - (I - AB^*)^{-1}(I - BB^*)(I - BA^*)^{-1} \geq 0, \end{equation} consequently $\textbf{G}_2(A,B) \geq 0$. In this connection, let us point out that the following relation exists behind the inequality \eqref{1.4}: \begin{eqnarray*} &&(I - AA^*)^{-1} - (I - AB^*)^{-1}(I - BB^*)(I - BA^*)^{-1}\\ &&= (I - AB^*)^{-1}\{A(A - B)^*(I - AA^*)^{-1}(A - B)A^*\\ &&+ (A - B)(A - B)^*\}(I - BA^*)^{-1}. \end{eqnarray*} What happens when $n \geq 3$ ? In \cite{AND} we showed that $\textbf{H}_3(A_1,A_2,A_3) \geq 0$ is not always true, while Xu et al. \cite{X-X-Z} has shown that the situation is the same for $\textbf{G}_3(A_1,A_2,A_3)$. Let us start with a relation between $\textbf{H}_n(A_1,A_2, \ldots ,A_n)$ and $\textbf{G}_n(A_1,A_2,\ldots ,A_n).$ \begin{eqnarray}\label{1.5} \textbf{G}_n(A_1,A_2, \ldots ,A_n) = [\overbrace{I, I,\ldots , I}^n]^*[\overbrace{ I, I,\ldots , I}^n] + \text{diag}(A_1,A_2,\ldots ,A_n)\cr\times \textbf{H}_n(A_1,A_2,\ldots ,A_n) \cdot \text{diag}(A_1,A_2,\ldots ,A_n)^*. \end{eqnarray} In fact, since $A(I - BA)^{-1} = (I - AB)^{-1}A$ for any strict contractions $A,B$, \[I + A_i(I - A^*_j A_i)^{-1}A^*_j = I + (I - A_iA^*_j )^{-1}A_iA^*_j = (I - A_iA^*_j )^{-1}.\] Since $[\overbrace{I,I,\ldots,I}^n]^*[\overbrace{I, I,\ldots , I}^n] \geq 0$, we can conclude from \eqref{1.5} the following. \begin{theorem}\label{th1.1} $\textbf{H}_n(A_1,A_2,\ldots,A_n) \geq 0$ implies $\textbf{G}_n(A_1,A_2,\ldots ,A_n) \geq 0$. \end{theorem} \begin{remark} The idea of the proof of Theorem \ref{th1.1} is implicit in Xu et al. \cite{X-X-Z}. \end{remark} However, $\textbf{G}_n(A_1,A_2,\ldots ,A_n) \geq 0$ does not imply $\textbf{H}_n(A_1,A_2, \ldots,A_n) \geq 0$. \begin{example}\label{ex1.3} When $\mathcal H$ is of 2-dimension, every operator is represented by a $2 \times 2$ matrix. Take $0 < \lambda < 1$ and let \[ A_1=\lambda \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right],\ A_2=\lambda \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right] \text{ and } A_3=0. \] Then $\textbf{G}_3(A_1,A_2,A_3) \geq 0$ but $\textbf{H}_3(A_1,A_2,A_3)\ngeq 0$. In fact, simple computation will show that, with $\alpha \equiv \lambda^2$, \[\textbf{G}_3(A_1,A_2,A_3)/(3)=\frac{\alpha}{1-\alpha} \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \geq 0. \] hence $\textbf{G}_3(A_1,A_2,A_3)\geq 0$ by the Schur criteria. On the other hand \[\textbf{H}_3(A_1,A_2,A_3)/(3)=\frac{\alpha}{1-\alpha} \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 1-\alpha & 0 \\ 0 & 1-\alpha & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right] \] is not positive semi-definite, because it has a $2 \times 2$ principal submatrix {\scriptsize $\left[ \begin{array}{cc} 0 & \alpha \\ \alpha & 0 \\ \end{array} \right]$}, which is not positive semi-definite. Therefore $\textbf{H}_3(A_1,A_2,A_3)\not\geq 0$ by the Schur criteria. \end{example} In \cite{AND} we showed that if $A_j\,\,(j = 1, 2, \ldots ,n)$ are commuting normal operators, then $\textbf{H}_n(A_1,A_2,\ldots ,A_n) \geq 0$ and also $\textbf{G}_n(A_1,A_2,\ldots ,A_n) \geq 0$. In the next section we give a generalization of this result. Our main result of this paper is that positivity of $\textbf{H}_n$ is preserved under an operator M\"obius map of the open unit disc $\mathcal D$ of strict contractions. \section{Main Results} \begin{theorem}\label{th2.1} Let $A_j\,\,(j = 1, 2,\ldots , n)$ be strict contractions. If the products $A^*_j A_i\,\,(i, j = 1, 2,\ldots, n)$ are commuting normal operators, $\textbf{H}_n(A_1,A_2,\ldots ,A_n) \geq 0$. \begin{proof} Our idea of the proof is parallel to that of Xu et al. \cite{X-X-Z}. The assumption means that there is a commutative unital *-subalgebra $\mathcal{C} \subset B(\mathcal{H})$ such that $A^*_j A_i \in \mathcal{C} \;(i, j = 1, 2,\ldots, n)$. Then by the Gelfand theorem (see \cite[Theorem 4.4]{TAK}) there is a *-isomorphism $\pi$ of $\mathcal C$ to the commutative C*-algebra $C(\Omega)$ of continuus functions on a compact set $\Omega$. Here the adjoint $f^*$ of a function $f \in C(\Omega)$ is determined by \begin{equation}\label{2.1} f^*(\omega)=\overline{f(\omega)}\;\;\; (\omega\in\Omega). \end{equation} Therefore we can write $f^* = \overline{f}$. Notice further that positivity of a $C(\Omega)$-matrix $[f_{i,j} ]^n_{i,j=1}$ is equivalent to saying that for every $\omega \in \Omega$ the numerical matrix $[f_{i,j}(\omega)]^n_{i,j=1}$ is positive semi-definite in the usual sense. \newline Now let \[f_{i,j} \equiv \pi(A^*_j A_i)\;\,\,(i, j = 1, 2,\ldots , n)\] Then by \eqref{2.1} \[f_{j,i} = \pi(A^*_i A_j) = \pi(A^*_j A_i)^* = \overline{f_{i,j}} .\] Then since \[[A^*_i A_j ]^n_{i,j=1} = [A_1,A_2, ...., A_n]^*\cdot[A_1,A_2, ...., A_n] \geq 0\] it follows that $[f_{j,i}]^n_{i,j=1} \geq 0$. Therefore for any $\omega\in\Omega$ \[[f_{i,j}(\omega)]^n_{i,j=1} = [\overline{f_{j,i}(\omega)}]^n_{i,j=1} \geq 0.\] Recall the positivity theorem for \emph{Schur product} (or Hadamard product) (see \cite[Theorem 5.2.1]{H-J/T}) that for two numerical $n \times n$ matrices \begin{equation} [\alpha_{i,j} ]^n_{i,j=1} \geq 0 \;\text{ and }\; [\beta_{i,j} ]^n_{i,j=1} \geq 0\; \Longrightarrow\; [\alpha_{i,j}\beta_{i,j} ]^n_{i,j=1} \geq 0. \end{equation} Then since \[[(I - A^*_j A_i)^{-1}]^n_{i,j=1} =\sum_{k=0}^{\infty}[(A^*_j A_i)^k]^n_{i,j=1},\] and \[[\pi((A^*_j A_i)^k)]^n_{i,j=1} = [f^k_{i,j} ]^n_{i,j=1},\] it follows from the Schur product theorem (2.2) that \[[(A^*_j A_i)^k]^n_{i,j=1} \geq0 \;\;(k = 1, 2,\ldots ),\] consequently $[(I - A^*_j A_i)^{-1}]^n_{i,j=1} \geq 0$. \end{proof} \end{theorem} In a similar way we can prove \begin{theorem}\label{th2.2} Let $A_j\,\,(j = 1, 2,\ldots , n)$ be strict contractions. If the products $A_iA^*_j \;(i, j = 1,\ldots , n)$ are commuting normal operators, $\textbf{G}_n(A_1,A_2,\ldots ,A_n)\geq 0$. \end{theorem} \begin{remark} Positivity of $\textbf{G}_3(A_1,A_2,A_3)$ in Example \ref{ex1.3} follows from Theorem \ref{th2.2}. \end{remark} In the linear systems theory (see \cite[Chapter 10]{Z-D-G}), for a time-invariant linear system with a state-space realization matrix {\scriptsize $\left[ \begin{array}{cc} B_{1,1} & B_{1,2} \\ B_{2,1} & B_{2,2} \\ \end{array} \right] $} it is common to consider the operator-valued function, called the \emph{transfer function}, defined as \[\zeta\longmapsto B_{2,2} + B_{2,1}(\zeta I - B_{1,1})^{-1}B_{1,2}\] for complex numbers $\zeta$ for which $\zeta I- B_{1,1}$ are invertible. In operator theory, however, it is more convenient to consider a linear-fractional transformation $\Theta(\zeta)$ defined as \[\Theta(\zeta) = B_{2,2} + \zeta B_{2,1}(I - \zeta B_{1,1})^{-1}B_{1,2}.\] (See \cite[Chapter 6]{N-F}) Extending the variable from a number $\zeta$ to an operator $Z$, let us define a map $\Theta(Z)$ as \begin{equation}\label{eq2.3} \Theta(Z) = B_{2,2} + B_{2,1}Z(I - B_{1,1}Z)^{-1}B_{1,2}. \end{equation} For a contraction $B$, define its \emph{defect operator} $D_B$ as \begin{equation}\label{eq2.4} D_B = (I - B^*B)^{1/2}. \end{equation} The following relations are immediate from definition (2.4) \begin{equation}\label{eq2.5} BD_B = D_{B^*}B,\;\;\text{ and }\;B^*D_{B^*} = D_BB^*, \end{equation} and for any strict contraction $Z$ the operators $I-B^*Z$ and $I-ZB^*$ are invertible and the following relation holds \begin{equation}\label{eq2.6} Z(I - B^*Z)^{-1} = (I - ZB^*)^{-1}Z. \end{equation} \begin{lemma}\label{lem2.4} When $B$ is a strict contraction, the operator-matrix {\scriptsize $ \left[ \begin{array}{cc} B^* & D_B \\ -D_{B^*}& B \\ \end{array} \right]$} is unitary, and the map \[\Theta(Z) = B - D_{B^*}Z(I - B^*Z)^{-1}D_B = B - D_{B^*}(I - ZB^*)^{-1}ZD_B\] satisfies the following relations that for any strict contractios $Z,W$ \[I - \Theta(Z)^*\Theta(W) = D_B(I - Z^*B)^{-1}(I - Z^*W)(I - B^*W)^{-1}D_B.\] \end{lemma} \begin{proof} The proof of unitarity is immediate from \eqref{eq2.5} and omitted. Now since \begin{eqnarray*} \Theta(Z)^*\Theta(W) &=& B^*B - D_B(I - Z^*B)^{-1}Z^*D_{B^*}B - B^*D_{B^*}W(I - B^*W)^{-1}D_B\\ &&+D_B(I - Z^*B)^{-1}Z^*(I - BB^*)W(I - B^*W)^{-1}D_B, \end{eqnarray*} by \eqref{eq2.5} and \eqref{eq2.6} we can see \begin{eqnarray*} I - \Theta(Z)^*\Theta(W) &=& D_B\{I + (I - Z^*B)^{-1}Z^*B + B^*W(I - B^*W)^{-1}\\ &&- (I - Z^*B)^{-1}(I - BB^*)W(I - B^*W)^{-1}\}D_B\\ &=& D_B(I - Z^*B)^{-1}\{(I - Z^*B)(I - B^*W) + Z^*B(I - B^*W)\\ &&+ (I - Z^*B)B^*W - Z^*(I - BB^*)W \}(I - B^*W)^{-1}D_B\\ &=& D_B(I - Z^*B)^{-1}(I - Z^*W)(I - B^*W)^{-1}D_B. \end{eqnarray*} \end{proof} Given a complex number $\beta$ with $|\beta| < 1$, the M\"obius transformation at $\beta$ \[M_\beta(\zeta)\equiv\frac{\beta-\zeta}{1-\overline{\beta}\zeta}\] is a conformal map of the open unit disc of the complex plane, which maps 0 to $\beta$ and $\beta$ to 0, and is involutive, that is, $M_\beta\left(M_\beta(\zeta)\right)=\zeta$. The following is an analogy for the case of the open unit disc $\mathcal D$ of strict contractions. \begin{proposition}\label{prop2.5} For a strict contraction $B$, the M\"obius map $\Theta_B(\cdot)$ at $B$, defined by \[\Theta_B(Z) \equiv D^{-1}_{B^*}(B - Z)(I - B^*Z)^{-1}D_B,\] is an involutive map of the open unit disc $\mathcal D$, that is, \[\Theta_B(\Theta_B(Z)) = Z \;\,\,(Z \in \mathcal D).\] \end{proposition} It is clear from the definition that $\Theta_B(Z)$ is \emph{holomorphic} with respect to the operator variable $Z$. Since $\Theta(\cdot)$ is involutive, its inverse is also holomorphic. Therefore $\Theta_B(\cdot)$ becomes a \emph{biholomorphic} map of the open unit disc $\mathcal D$ of strict contractions, and is considered as a natural generalization of the M\"obius transformation on the open unit disc of the complex plane. \begin{proof} First let us show the map $\Theta_B(\cdot)$ is nothing but the linear-fractinal transformation $\Theta(\cdot)$ of the unitary operator-matrix {\scriptsize $\left[ \begin{array}{cc} B^* & D_B \\ -D_{B^*} & B \\ \end{array} \right] $.} In fact, by definition and \eqref{eq2.5} \begin{eqnarray*} \Theta(Z) &=& B - D_{B^*}Z(I - B^*Z)^{-1}D_B \\ &=& D^{-1}_{B^*}\left\{D_{B^*}BD^{-1}_B (I - B^*Z) - (I - BB^*)Z\right\}(I - B^*Z)^{-1}D_B\\ &=& D_{B^*}^{-1}(B - Z)(I - B^*Z)^{-1}D_B = \Theta_B(Z). \end{eqnarray*} Next $\Theta_B(\cdot)$ maps the open unit disc $\mathcal D$ to itself. In fact, by Lemma \ref{lem2.4} \[I - \Theta_B(Z)^*\Theta_B(Z) = D_B(I - Z^*B)^{-1}(I - Z^*Z)(I - B^*Z)^{-1}D_B >0\;\,\,(Z \in \mathcal{D}).\] Finally the involutivity follows from the following two relations: \[B - \Theta(Z) = D_{B^*}Z(I - B^*Z)^{-1}D_B\] and \begin{eqnarray*} I - B^*\Theta(Z) &=& I - B^*B + B^*D_{B^*}Z(I - B^*Z)^{-1}D_B\\ &=& D^2_B+ D_BB^*Z(I - B^*Z)^{-1}D_B\\ &=& D_B\{I + B^*Z(I - B^*Z)^{-1}\}D_B = D_B(I - B^*Z)^{-1}D_B. \end{eqnarray*} \end{proof} \begin{corollary} If an operator-matrix $[B_{i,j} ]^2_{i,j=1}$ with $\|B_{2,2}\| < 1$ is unitary, then the map \[\Theta(Z) \equiv B_{2,2} + B_{2,1}Z(I - B_{1,1}Z)^{-1}B_{1,2}\] is a biholomorphic map of the open unit disc $\mathcal D$ of strict contractions. \end{corollary} \begin{proof} Let $B = B_{2,2}$. Then it is easy to see from unitarity that there are unitary $U, V$ such that \[B_{1,1} = UB^*V,\; B_{1,2} = UD_B\; \text{ and }\; B_{2,1} = -D^*_B V.\] Then we have \[\Theta(Z) = \Theta_B(VZU)\,\,(Z \in \mathcal D),\] where $\Theta_B(\cdot)$ is the M\"obius map at $B$. Finally since $Z \longmapsto VZU$ is a biholomorphic map of $\mathcal D$, the assertion follows from Proposition \ref{prop2.5}. \end{proof} The following is the main result of this paper. \begin{theorem}\label{th2.7} Let $B$ be a strict contraction, and $\Theta_B(\cdot)$ the M\"obius map at $B$ on the open unit disc $\mathcal D$ of strict contractions. Then for any $A_i\in \mathcal{D}\,\,(i =1, 2,\ldots , n)$ \[\textbf{H}_n(A_1,A_2,\ldots ,A_n) \geq 0 \;\text{ implies } \; \textbf{H}_n\left(\Theta_B(A_1),\Theta_B(A_2), \ldots,\Theta_B(A_n)\right) \geq 0.\] \end{theorem} \begin{proof} Since by Lemma \ref{lem2.4} \[\left(I - \Theta_B(A_j)^*\Theta_B(A_i)\right)^{-1}= D^{-1}_B (I - B^*A_i)(I - A^*_j A_i)^{-1}(I - A^*_j B)D^{-1}_B ,\] we have \[\textbf{H}_n\left(\Theta_B(A_1),\Theta_B(A_2),\ldots ,\Theta_B(A_n)\right)= \textbf{D} \cdot \textbf{H}_n(A_1,A_2,\ldots ,A_n)\cdot \textbf{D}^*\] where \[\textbf{D} = \text{diag} \left(D^{-1}_B(I - B^*A_1),D^{-1}_B (I - B^*A_2),\ldots ,D^{-1}_B (I - B^*A_n)\right).\] This identity proves the assertion. \end{proof} \begin{remark} It is not clear whether or not \[\textbf{G}_n(A_1,A_2,\ldots ,A_n) \geq 0 \;\text{ implies } \; \textbf{G}_n\left(\Theta_B(A_1),\Theta_B(A_2), \ldots,\Theta_B(A_n)\right) \geq 0.\] \end{remark} \begin{remark} In Introduction we stated that $\textbf{H}_2(A,B) \geq 0$ is valid for any strict contraction $A,B$. Let us show that this result is included in the combination of Theorem \ref{th2.2} and Theorem \ref{th2.7}. In fact, consider the M\"obius map $\Theta_B(\cdot)$ at $B$. Then by Proposition \ref{prop2.5} $A = \Theta_B(\tilde{A})$ where $\tilde{A} = \Theta_B(A)$ and $B = \Theta_B(0)$ and by Theorem \ref{th2.2} $\textbf{H}_2(\tilde{A}, 0) \geq 0$. Then apply Theorem \ref{th2.7} to see $\textbf{H}_2(A,B) \geq 0$. \end{remark} \textbf{Acknowledgement} The author would like to thank Professor F. Zhang for the paper \cite{X-X-Z} before publication and useful comments on the original version of the present paper. %------------------------------------------------------- \bibliographystyle{amsplain} \begin{thebibliography}{99} \bibitem{AND} T. Ando, \textit{Hua-Marcus inequalities}, Linear Multilinear Alg. \textbf{8} (1980), 347--352. \bibitem{H-J} R. Horn and Ch. Johnson, \textit{Matrix Analysis}, Cambridge Univ. Press, 1985. \bibitem{H-J/T} R. Horn and Ch. 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