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\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}

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