# Fixed points and normal automorphisms of the unit ball of bounded operators
on $\mathbb{C}^{n}$

Rachna Aggarwal, Krishnendu Gongopadhyay and Mukund Madhav Mishra  Department
of Mathematics, University of Delhi, Delhi, India<EMAIL_ADDRESS>Department of Mathematical Sciences, Indian Institute of Science Education and
Research (IISER) Mohali, Knowledge City, S.A.S. Nagar, Sector 81, P.O. Manauli
140306, India<EMAIL_ADDRESS>Department of Mathematics, Hans
Raj College, University of Delhi, Delhi, India<EMAIL_ADDRESS>
###### Abstract.

We examine the group of isometries of the open unit ball of a complex Banach
space of certain bounded linear operators equipped with the Carathéodory
metric. Therein we obtain a charactrization of the normal isometries in terms
of their special type of fixed points.

###### Key words and phrases:

unit ball, Caratan domain, automorphism group, normal isometry, bounded linear
operators

###### 2020 Mathematics Subject Classification:

Primary 32M15; Secondary 47B02, 47B15, 47B91

##  1\. Introduction

Classical group $U(n,1)$ has been extensively studied in the literature from
various perspectives. For instance [1], [4], [16], [6] and many more. A
natural generalization of the unitary group $U(n,1)$ is the pseudo-unitary
group $U(p,q)$. For its definition and details, see ([15]), for instance. It
exhibits exhibit substantial contrast in its formulative theory due to the
presence of a higher rank component. These groups have been explored at
certain places like [15] [5], [18], [12], [13] and [11]. Some infinite
dimensional versions have also been discussed, for instance, in [3], [8], [7]
and [9].Also Popescu has studied hyperbolic geometry on non commutative unit
balls in infinite dimensional set up, see for instance, [18].

In this article, we consider certain infinite dimensional extension of these
pseudo-unitary groups and investigate one of their significant subclass from
the fixed point classification perspective. The subclass under consideration
is the class of normal elements. If we consider holomorphic isometry on the
open unit ball of complex Banach space of bounded linear operators between
Hilbert spaces then it can be represented by a linear operator on a Hilbert
space. We reserve the phrases ‘normal isometry’ and ‘non-unitary normal
isometry’ for an isometry whose linear representation is normal and non-
unitary normal as an operator. A bounded linear operator on a Hilbert space is
called normal if it commutes with its Hilbert adjoint and unitary if its
Hilbert adjoint coincides with its inverse. An operator is non-unitary normal
if it is not unitary and a normal operator. The central result of this article
is Theorem 1.4 which characterizes a non-unitary normal isometry on the basis
of a special class of its fixed points. Before describing our main result, we
make the following set up below.

Let $K$ and $H$ be complex Hilbert spaces and $B(K,H)$ be the complex Banach
space of all bounded linear operators from $K$ to $H$ equipped with the
operator norm. If $K=H$, this will be abridged to $B(H)$. Let $\mathcal{B}$ be
the open unit ball of $B(K,H)$ equipped with the Carathéodory metric. It is an
infinite dimensional extension of the Cartan domain of type 1, i.e. Let $U(H)$
denote the group of unitary operators on $H$. The holomorphic automorphisms of
$\mathcal{B}$ has an action on the unit ball of $\mathcal{B}(K,H)$ by
operator-valued Möbius transformations. This is an operator valued analogue of
the classical Möbius transformations acting on the Poincaré disc. The
automorphism group of $\mathcal{B}$ as operator-valued linear fractional
transformations was determined independently by Greenfield and Wallach [7],
Harris [8] and Kaup [9] respectively in different contexts. The general form
of a holomorphic automorphism of $\mathcal{B}$ also follows from these works.
In [3], Franzoni used a different approach that works also in infinite
dimensions and provided a linear representation of the group of holomorphic
automorphisms. Considering this linear group as our framework, we mainly work
with the assumption $K=\mathbb{C}^{n}$. In the following subsection, we
describe the main result of this article.

###  1.1. Main result of the article

We begin by providing the machinery which is needed to understand the
statement of the main result. We shall follow Franzoni, [3] in our treatment
to recall the basics. Let $\mathcal{G}$ denote the group of all bijective
linear operators defined on $H\oplus K$ which leave the Hermitian form
$\mathcal{S}$ invariant, where $\mathcal{S}:(H\oplus K)\times(H\oplus
K)\rightarrow\mathbb{C}$ is defined as

$\mathcal{S}\big{(}(h_{1},k_{1}),(h_{2},k_{2})\big{)}=\left<h_{1},h_{2}\right>-\left<k_{1},k_{2}\right>.$

From now on, throughout this subsection, $K=\mathbb{C}^{n}$ unless stated
otherwise and the sets $\\{1,\,2,\,...,\,l\\}$ and $\\{1,\,2,\,...,\,k_{i}\\}$
will be denoted by the symbols $I$ and $J_{i}$ respectively where $i\in I$.

A general element of $\mathcal{G}$ is $T=\left[{\begin{array}[]{cc}BU&CV\\\
C^{*}U&EV\\\ \end{array}}\right]$ where

  * •

$U\in U(H)$, $V\in U(\mathbb{C}^{n})$, $B\in B(H)$, $E\in B(\mathbb{C}^{n})$
and $C\in B(\mathbb{C}^{n},H)$.

  * •

$E$ decomposes $\mathbb{C}^{n}$ orthogonally into eigenspaces as
$\mathbb{C}^{n}=\overset{l}{\underset{i=1}{\oplus}}{K_{i}}\oplus{K^{\prime}}^{\perp}$
where $K^{\prime}=\overset{l}{\underset{i=1}{\oplus}}{K_{i}}$.
$E\restriction_{K_{i}}=a_{i}\,I$ and $E\restriction_{{K^{\prime}}^{\perp}}=I$,
$a_{i}>1$, $i\in I$.
Let $dim\,(K^{\prime})=k$ and $dim\,(K_{i})=k_{i}$, hence
$\sum\limits_{i=1}^{l}{k_{i}}=k$. Let $\beta_{i}=\\{e_{i_{j}},\,\,j\in
J_{i}\\}$ be an orthonormal basis of $K_{i}$, $i\in I$.

  * •

$C(e_{i_{j}})=\xi_{i_{j}}$ where $\xi_{i_{j}}(\neq 0)\in H$ for every $i\in I$
and $j\in J_{i}$ and distinct ${\xi_{i_{j}}}^{\prime}s$ are mutually
orthogonal making $C\restriction_{K^{\prime}}$ a bijective operator and
$C\restriction_{{K^{\prime}}^{\perp}}\equiv 0$. Hence
$dim\,(ran\,C)=k=dim\,(K^{\prime})$ where $ran\,C=$ Range of $C$.

  * •

$\|\xi_{i_{j}}\|=\|\xi_{i_{j^{\prime}}}\|$ for an arbitrarily fixed $i\in I$
and $j,\,j^{\prime}\in J_{i}.$ Let
$\|\xi_{i_{j}}\|=\|\xi_{i_{j^{\prime}}}\|=\delta_{i}$ giving
${a_{i}}^{2}=1+{\delta_{i}}^{2},\,\,i\in I.$

  * •

$C^{*}(\xi_{i_{j}})={\delta_{i}}^{2}\,e_{i_{j}}$ and $C^{*}\equiv 0$ on
${(ran\,C)}^{\perp}$, for every $i\in I$ and $j\in J_{i}$.

  * •

Let $M_{i}=span\,\\{\xi_{i_{j}},\,\,j\in J_{i}\\}$. Then
$ran\,C=\overset{l}{\underset{i=1}{\oplus}}{M_{i}}$,
$dim\,(M_{i})=dim\,(K_{i})=k_{i}$, $i\in I$. The operator $B$ decomposes $H$
orthogonally into eigenspaces as
$H=\overset{l}{\underset{i=1}{\oplus}}{M_{i}}\oplus{(ran\,C)}^{\perp}$ where
$B\restriction_{M_{i}}=a_{i}\,I$ and $B\restriction_{{(ran\,C)}^{\perp}}=I$.
Let us denote an orthogonal basis $\\{\xi_{i_{j}},\,\,j\in J_{i}\\}$ of
$M_{i}$, $i\in I$ by ${\beta_{i}}$.

######  Proposition 1.1.

Let $K$ be any arbitrary Hilbert space. An isometry
$T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ is normal if and only if $UB=BU$, $UC=CV$
and $VE=EV$.

######  Proposition 1.2.

For $K=\mathbb{C}^{n}$, unitary elements in $\mathcal{G}$ are of the form
$\left[{\begin{array}[]{cc}U&0\\\ 0&V\\\ \end{array}}\right]$ where $U\in
U(H)$ and $V\in U(\mathbb{C}^{n})$.

######  Proposition 1.3.

For $K=\mathbb{C}^{n}$, Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$. Then $T$ is non-unitary normal if and only
if

(1.1) $\displaystyle U(M_{i})$ $\displaystyle=M_{i},$ (1.2) $\displaystyle
V(K_{i})$ $\displaystyle=K_{i},\,\,\text{and}$ (1.3)
$\displaystyle{[U\restriction_{M_{i}}]}_{{\beta_{i}}^{\prime}}$
$\displaystyle={[V\restriction_{K_{i}}]}_{{\beta_{i}}}.$

######  Definition 1.

An operator $F\in B(\mathbb{C}^{n},H)$ will be called a fixed point of
$T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ if there exists a basis
$\\{\boldsymbol{\mathrm{z_{r}}},\,\,r=1,\,2,\,...,\,n\\}$ of $\mathbb{C}^{n}$
such that $\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\ \end{array}}\right],\,\,\,\,r=1,\,2,\,...,\,n$
are eigenvectors of $T$.

######  Definition 2.

The functions $F_{\theta}\in B(\mathbb{C}^{n},H)$ defined below will be called
of generic type.
Let $S=\\{(\epsilon_{m})_{m=1}^{q},\,\,\epsilon_{m}\in\\{1,-1\\}\\}$ be a
collection of $q$-tuples where $q\leq n$. For
$\theta=(\epsilon_{m})_{m=1}^{q}\in S$, define $F_{\theta}\in
B(\mathbb{C}^{n},H)$ as

$\displaystyle F_{\theta}(\boldsymbol{\mathrm{z}_{m}})$
$\displaystyle=\epsilon_{m}\,y_{m},\,\,\,\,m=1,\,2,\,...,\,q,$ $\displaystyle
F_{\theta}(\boldsymbol{\mathrm{z}_{s}})$
$\displaystyle=0,\,\,\,\,s=q+1,\,...,\,n$

where $\\{\boldsymbol{\mathrm{z}_{r}},\,\,r=1,\,2,\,...,\,n\\}$ is an
orthonormal basis of $\mathcal{C}^{n}$ and $\\{y_{m},\,\,m=1,\,2,\,...,\,q\\}$
are non-zero vectors in $H$.
If $\left[{\begin{array}[]{c}F_{\theta}(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\
\end{array}}\right],\,\,\,\,r=1,\,2,\,...,\,n,\,\theta\in S$ are eigenvectors
of $T$ then in view of Definition 1, ${F_{\theta}}^{\prime}s$ are called fixed
points of generic type.

Note that in Definition 2, $|S|=2^{q}$. Hence $|\\{F_{\theta},\,\,\theta\in
S\\}|=2^{q}$. Now we state the main result.

######  Theorem 1.4.

Let an isometry $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ be such that $dim\,(ran\,C)=k$. Then $T$ is
non-unitary normal if and only if it possesses exactly $2^{k}$ fixed points of
the generic type. The generic fixed points lie on the boundary of the unit
ball ${\mathcal{B}}$.

Structure of the article. Section 2 reviews the work of [8] and [3] by
discussing the general form of a holomorphic automorphism of $\mathcal{B}$ and
its linear representation. Section 3 discusses the general form of an isometry
in $\mathcal{G}$ along with the description of its normal, self-adjoint and
unitary elements for a general Hilbert space $K$. Section 4 investigates
general form, normal and unitary elements under the finite dimensionality
assumption on $K$. Section 5 focuses on normal isometries and explores their
spectral properties along with their fixed point characterization. We prove
our main result in this section.

##  2\. Holomorphic automorphisms of $\mathcal{B}$

As mentioned above, let $\mathcal{B}$ be the open unit ball in $B(K,H)$. A
biholomorphic map or a holomorphic automorphism of $\mathcal{B}$ is a
bijective holomorphic mapping from $\mathcal{B}$ to $\mathcal{B}$ having
holomorphic inverse. Let $\overline{\mathcal{B}}$ denote the closure of
$\mathcal{B}$ in the operator norm and $\partial{\mathcal{B}}$, the boundary
of $\overline{\mathcal{B}}$ in $B(K,H)$. For $T\in B(H)$, $\sigma(T)$ and
$\sigma_{pt}(T)$ denote the spectrum and the collection of eigenvalues of $T$,
respectively. Since biholomorphic automrphisms turn out to be isometries for
the Carathéodory metric (cf. [2] Proposition. IV.1.6), throughout this
article, the word isometry would either mean a biholomorphic automorphism of
$\mathcal{B}$ or its linear representation. As Harris [8] and Franzoni [3]
give the general form of a holomorphic isometry acting on $\mathcal{B}$ and
discuss its linear representation, we begin with a quick review of their work
which forms the basis of our study.

######  Remark 1.

([14], chapter 1, (3.4)) For a contraction $T\in B(H)$, an application of
continuous functional calculus tells that

$T(I-T^{*}T)^{1/2}=(I-TT^{*})^{1/2}T.$

For $B\in\mathcal{B}$, the map $T_{B}:\mathcal{B}\rightarrow B(K,H)$ defined
by

(2.1) $\displaystyle\hskip
28.45274ptT_{B}(A)=(I-BB^{*})^{-1/2}(A+B)(I+B^{*}A)^{-1}(I-B^{*}B)^{1/2}$

is a holomorphic automorphism of $\mathcal{B}$ such that $T_{B}(0)=B$ (see
Remark 1) and ${T_{B}}^{-1}=T_{-B}$. $T_{B}$ is called the Möbius
transformation ([8], Theorem 2).

######  Theorem 2.1.

([3], Theorem 1.6) Every biholomorphic mapping
$h:\mathcal{B}\rightarrow\mathcal{B}$ such that $h(0)=B$ is of the form

$h=T_{B}\circ L,\,\,\,\,B\in\mathcal{B}$

where $L$ is a norm preserving surjective linear map of $B(K,H)$ and $T_{B}$
is a Möbius transformation (2.1) defined on $\mathcal{B}$.

So far there has been no restriction on the dimensions of $K$ and $H$.
From now onwards we will assume $H$ and $K$ have different dimensions, i.e.
they have orthonormal bases of different cardinalities.

Let $K$ and $H$ be Hilbert spaces having different dimensions. Then for a norm
preserving surjective linear isometry $L$ defined on $B(K,H)$, there exist
$U\in U(H)$ and $V\in U(K)$ such that

(2.2) $\displaystyle L(A)=UAV,\,\,\,\,\,\,\,\forall A\in B(K,H),$

(cf. [3], Theorem 4.6). $L$ is the generalization of a unitary operator to
Banach spaces. Collection of all the biholomorphic mappings of the form $h$
defined in Theorem 2.1 forms a group which will be denoted by
$Aut(\mathcal{B})$. This group acts transitively on $\mathcal{B}$ (see [8],
Corollary 2).
For $A,B\in\mathcal{B}$, the Carathéodory distance $C_{D}$ ([3], Sec. 3) is
defined as

$C_{D}(A,B)=\rho\bigg{(}0,\|T_{-B}(A)\|\bigg{)}=tanh^{-1}\bigg{(}\|T_{-B}(A)\|\bigg{)}$

where $\rho$ is the Poincaré metric on the open unit ball in $\mathbb{C}$ and
$T_{-B}$ is the inverse of Möbius transformation $T_{B}$ defined in (2.1). For
definition and details on Carathéodory metric, refer [2].
A linear representation of $Aut(\mathcal{B})$ due to Franzoni ([3], section 5)
is as follows.

Consider the Hermitian form $\mathcal{S}:(H\oplus K)\times(H\oplus
K)\rightarrow\mathbb{C}$ defined as

$\mathcal{S}\big{(}(h_{1},k_{1}),(h_{2},k_{2})\big{)}=\left<h_{1},h_{2}\right>-\left<k_{1},k_{2}\right>.$

Let $\mathcal{G}$ be the group of all bijective linear transformations defined
on $H\oplus K$ leaving $\mathcal{S}$ invariant. The general form of an element
of $\mathcal{G}$ is $T=\left[{\begin{array}[]{cc}B&C\\\ D&E\\\
\end{array}}\right]$ where $B:H\rightarrow H$, $C:K\rightarrow H$,
$D:H\rightarrow K$ and $E:K\rightarrow K$ are bounded linear operators
satisfying

(2.3) $\displaystyle B^{*}B-D^{*}D$ $\displaystyle=I,$ $\displaystyle
E^{*}E-C^{*}C$ $\displaystyle=I,$ $\displaystyle C^{*}B-E^{*}D$
$\displaystyle=0,$ $\displaystyle BB^{*}-CC^{*}$ $\displaystyle=I,$
$\displaystyle EE^{*}-DD^{*}$ $\displaystyle=I,\,\,\text{and}$ $\displaystyle
DB^{*}-EC^{*}$ $\displaystyle=0.$

The above equations imply that $B$ and $E$ are invertible operators (see [3],
Lemma 5.1).

######  Remark 2.

For $T=\left[{\begin{array}[]{cc}B&C\\\ D&E\\\
\end{array}}\right]\in\mathcal{G}$,
$T^{-1}=\left[{\begin{array}[]{cc}B^{*}&-D^{*}\\\ -C^{*}&E^{*}\\\
\end{array}}\right]$, (see [3], page 63).

The following theorem shows the identification of $Aut(\mathcal{B})$ with
$\mathcal{G}$. It has been proved in [3] but here we give the detailed version
of the proof.

######  Theorem 2.2.

([3], Theorem 5.3) Let $H$ and $K$ be Hilbert spaces of different dimensions.
The map $\psi:\mathcal{G}\rightarrow Aut(\mathcal{B})$ defined as
$T\mapsto\psi(T)$, where $T=\left[{\begin{array}[]{cc}B&C\\\ D&E\\\
\end{array}}\right]$ and $\big{(}\psi(T)\big{)}(A)=(BA+C)(DA+E)^{-1}$,
$A\in\mathcal{B}$, is a surjective homomorphism.

###### Proof.

Well-definedness and homomorphicity of $\psi$ are self evident. Let
$h_{A}=T_{A}\circ L\in Aut(\mathcal{B})$, $A\in\mathcal{B}$. As $\psi$ is a
homomorphism, it is sufficient to provide pre-images of $T_{A}$ and $L$
separately under $\psi$. Notice that for $L\in Aut(\mathcal{B})$ of the form
$L(X)=UXV$ (cf. 2.2), $T_{1}=\left[{\begin{array}[]{cc}U&0\\\ 0&V^{-1}\\\
\end{array}}\right]$ does the job. Next, we need to find a suitable element
$T_{2}=\left[{\begin{array}[]{cc}B&C\\\ D&E\\\
\end{array}}\right]\in\mathcal{G}$ satisfying

$(BX+C)(DX+E)^{-1}=(I-AA^{*})^{-1/2}(X+A)(I+A^{*}X)^{-1}(I-A^{*}A)^{1/2}$

for all $X\in\mathcal{B}$. Putting $X\equiv 0$ in the above equation gives
$CE^{-1}=(I-AA^{*})^{-1/2}A(I-A^{*}A)^{1/2}=A$ by Remark 1, thus $C=AE$. Also
by (2.3), $E^{*}E-I=C^{*}C=E^{*}A^{*}AE$, i.e. $E^{*}E=E^{*}A^{*}AE+I$
implying $E^{*}=E^{*}A^{*}A+E^{-1}$ and $EE^{*}=EE^{*}A^{*}A+I$. Hence
$EE^{*}(I-A^{*}A)=I$ or $EE^{*}=(I-A^{*}A)^{-1}$. Choose $E$ to be the
positive square root of $(I-A^{*}A)^{-1}$. This gives $C=A(I-A^{*}A)^{-1/2}$.
Next the relation $BB^{*}=I+CC^{*}$ and Neumann Series expansion for the
operator $(I-A^{*}A)^{-1}$ yield $BB^{*}=(I-AA^{*})^{-1}$. Choose $B$ to be
the positive square root of the operator $(I-AA^{*})^{-1}$. Lastly
$DB^{*}=EC^{*}$ gives $D=(I-A^{*}A)^{-1/2}A^{*}$. Observe that

(2.4) $\displaystyle D=C^{*}.$

So using Remark 1,
$T_{2}=\left[{\begin{array}[]{cc}(I-AA^{*})^{-1/2}&{(I-AA^{*})^{-1/2}}A\\\
(I-A^{*}A)^{-1/2}A^{*}&(I-A^{*}A)^{-1/2}\\\ \end{array}}\right]$. ∎

######  Remark 3.

The elements in the centre $Z(\mathcal{G})$ of $\mathcal{G}$ are all the unit
constant multiples of identity ([3], page 65) and $ker(\psi)=Z(\mathcal{G})$,
thereby making $\widetilde{\psi}:\mathcal{G}/Z(\mathcal{G})\rightarrow
Aut(\mathcal{B})$ a surjective isomorphism.

##  3\. Normal isometries of $\mathcal{B}$

We begin by presenting the general form of an element of $\mathcal{G}$ which
is a direct application of the proof of Theorem 2.2.

######  Proposition 3.1.

A general element of $\mathcal{G}$ is represented as

$T=e^{i\theta}\left[{\begin{array}[]{cc}(I-AA^{*})^{-1/2}U&{(I-AA^{*})^{-1/2}}AV\\\
(I-A^{*}A)^{-1/2}A^{*}U&(I-A^{*}A)^{-1/2}V\\\ \end{array}}\right]$

where $\theta\in\mathbb{R}$, $A\in\mathcal{B}$, $U\in U(H)$ and $V\in U(K)$.

Observe that a general element $T$ of $\mathcal{G}$ in Proposition 3.1 is a
composition of a self-adjoint element
$\left[{\begin{array}[]{cc}(I-AA^{*})^{-1/2}&{(I-AA^{*})^{-1/2}}A\\\
(I-A^{*}A)^{-1/2}A^{*}&(I-A^{*}A)^{-1/2}\\\ \end{array}}\right]$ and a unitary
operator $e^{i\theta}\left[{\begin{array}[]{cc}U&0\\\ 0&V\\\
\end{array}}\right]$.

For the sake of simplicity, let $B=(I-AA^{*})^{-1/2}$,
$C={(I-AA^{*})^{-1/2}}A$ and $E=(I-A^{*}A)^{-1/2}$. So a general element of
$\mathcal{G}$ would be $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]$. Observe that the fact $B$ and $E$ are positive operators
simplifies (2.3) to

(3.1) $\displaystyle B^{2}-CC^{*}$ $\displaystyle=I,$ $\displaystyle
E^{2}-C^{*}C$ $\displaystyle=I,\,\,\text{and}$ $\displaystyle BC$
$\displaystyle=CE.$

We now discuss the characterising conditions for an isometry in $\mathcal{G}$
to be normal.

Proof of Proposition 1.1. We will compute the expressions of $TT^{*}$ and
$T^{*}T.\\\ $ $TT^{*}=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{cc}U^{-1}B&U^{-1}C\\\
V^{-1}C^{*}&V^{-1}E\\\
\end{array}}\right]=\left[{\begin{array}[]{cc}B^{2}+CC^{*}&BC+CE\\\
C^{*}B+EC^{*}&C^{*}C+E^{2}\\\ \end{array}}\right]$

and $T^{*}T=\left[{\begin{array}[]{cc}U^{-1}B&U^{-1}C\\\
V^{-1}C^{*}&V^{-1}E\\\ \end{array}}\right]\left[{\begin{array}[]{cc}BU&CV\\\
C^{*}U&EV\\\ \end{array}}\right]$

$=\left[{\begin{array}[]{cc}U^{-1}(B^{2}+CC^{*})U&U^{-1}(BC+CE)V\\\
V^{-1}(C^{*}B+EC^{*})U&V^{-1}(C^{*}C+E^{2})V\\\ \end{array}}\right]$. Equating
the respective entries of $TT^{*}$ and $T^{*}T$, we get

$\displaystyle U(B^{2}+CC^{*})$ $\displaystyle=(B^{2}+CC^{*})U,$
$\displaystyle U(BC+CE)$ $\displaystyle=(BC+CE)V,$ $\displaystyle
V(C^{*}B+EC^{*})$ $\displaystyle=(C^{*}B+EC^{*})U,\,\text{and}$ $\displaystyle
V(C^{*}C+E^{2})$ $\displaystyle=(C^{*}C+E^{2})V$

which in view of (3.1) simplifies to $UB^{2}=B^{2}U,\linebreak
UBC=BCV,\,\text{and}\,\,VE^{2}=E^{2}V.$ Hence $UB=BU$ and $VE=EV$. Also
$UBC=BUC=BCV$ implies $UC=CV$. Thus $T$ is normal if and only if $UB=BU$,
$UC=CV$ and $VE=EV.$

As a corollary, we present the form of self-adjoint isometries.

######  Corollary 3.1.1.

Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ be a normal isometry. Then $T$ is self-
adjoint if and only if both $U$ and $V$ are self-adjoint operators.

###### Proof.

$T$ is self-adjoint if and only if $T=T^{*}$, i.e.
$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]=\left[{\begin{array}[]{cc}U^{-1}B&U^{-1}C\\\
V^{-1}C^{*}&V^{-1}E\\\ \end{array}}\right]$ which on equating the respective
entries gives $BU=U^{-1}B,\,CV=U^{-1}C\,$ and $EV=V^{-1}E$. Now corollary
follows using Proposition 1.1 and the fact that a unitary operator is self-
adjoint if and only if it is an involution, i.e. order two element. ∎

######  Proposition 3.2.

Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$, then $T$ is unitary if and only if
$C\equiv 0$.

###### Proof.

For $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$, $T$ is unitary if and only if
$T^{*}=T^{-1}$ , i.e. $\left[{\begin{array}[]{cc}U^{-1}B&U^{-1}C\\\
V^{-1}C^{*}&V^{-1}E\\\
\end{array}}\right]=\left[{\begin{array}[]{cc}U^{-1}B&-U^{-1}C\\\
-V^{-1}C^{*}&V^{-1}E\\\ \end{array}}\right]$. It is easy to see that
respective entries are equal if and only if $C\equiv 0$. ∎

######  Corollary 3.2.1.

A unitary operator in $\mathcal{G}$ is of the form
$\left[{\begin{array}[]{cc}W&0\\\ 0&W^{\prime}\\\ \end{array}}\right]$ for
some $W\in U(H)$ and $W^{\prime}\in U(K)$.

###### Proof.

Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ be a unitary operator. Then
$T=\left[{\begin{array}[]{cc}BU&0\\\ 0&EV\\\ \end{array}}\right]$ where $B$
and $E$ are involution operators by (3.1). Hence $B$ and $E$ are unitary as
these are self-adjoint also. So $T=\left[{\begin{array}[]{cc}W&0\\\
0&W^{\prime}\\\ \end{array}}\right]$ where $W=BU$ and $W^{\prime}=EV$. ∎

##  4\. Normal isometries for $\mathcal{B}\subseteq B(\mathbb{C}^{n},H)$

From now onwards, we will assume $K=\mathbb{C}^{n}$ and a normal isometry
would mean non-unitary normal isometry.

Let us understand the general form of elements of $\mathcal{G}$ more
explicitly for finite dimensional $K$.

From the proof of Theorem 2.2, $E=(I-A^{*}A)^{-1/2}$ is a positive invertible
operator where $A\in\mathcal{B}$. Since
$\|A^{*}A\|<1,\,\sigma(A^{*}A)\in[0,1)$. This implies
$\sigma(I-A^{*}A)\in(0,1]$ and hence $\sigma(E)\in[1,\infty)$. Thus for
$T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$, $E$ and $B$ are positive invertible
operators with $E$ having eigenvalues $\geq$ 1\. Let $E$ decompose the space
$\mathbb{C}^{n}$ orthogonally into eigenspaces as
$\mathbb{C}^{n}=\overset{p}{\underset{i=1}{\oplus}}K_{i}$ such that
$E\restriction_{K_{i}}=a_{i}\,I$, $a_{i}\geq 1$.
Let $dim\,(K_{i})=k_{i}$ and let
$\beta_{i}=\\{e_{{i}_{j}},\,\,j=1,\,2,\,...,\,k_{i}\\}$ be an orthonormal
basis of $K_{i}$ for each $i=1,\,2,\,...,\,p$.
Let $C(e_{i_{j}})=\xi_{i_{j}}$. This implies
$C^{*}(h)=\sum\limits_{r=1}^{p}\sum\limits_{s=1}^{k_{r}}{\left<h,\xi_{r_{s}}\right>\,e_{r_{s}}}$.
The relation $E^{2}-C^{*}C=I$ in (3.1) yields

${a_{i}}^{2}\,e_{i_{j}}-\sum\limits_{r=1}^{p}\sum\limits_{s=1}^{k_{r}}{\left<\xi_{i_{j}},\xi_{r_{s}}\right>\,e_{r_{s}}}=e_{i_{j}}.$

Taking inner product with $e_{i_{j}}$ gives
${a_{i}}^{2}=1+\|\xi_{i_{j}}\|^{2}$ which further implies that for all
$j,\,j^{\prime}\in\\{1,\,2,\,...,\,k_{i}\\}$,
$\|\xi_{i_{j}}\|=\|\xi_{i_{j^{\prime}}}\|=\delta_{i}$ (say). Hence
${a_{i}}^{2}=1+{\delta_{i}}^{2}$ for all $i=1,\,2,\,...,\,p$.
Now taking inner product in the same equation with $e_{r_{s}}$, we get
$\left<\xi_{i_{j}},\xi_{r_{s}}\right>=0$, $\xi_{i_{j}}\neq\xi_{r_{s}}$, i.e.
non-zero distinct ${\xi_{i_{j}}}^{\prime}s$ are mutually orthogonal. Also some
of the ${\xi_{i_{j}}}^{\prime}s$ may be zero as well and if $\xi_{i_{j}}=0$
then $\xi_{i_{j^{\prime}}}=0$ for all
$j^{\prime}\in\\{1,\,2,\,...,\,k_{i}\\}$. This will make the corresponding
$a_{i}=1$.
Let us re-define the orthogonal decomposition of $\mathbb{C}^{n}$ via $E$ as
follows.
Out of $p$ number of eigenspaces of $E$ let the first $l$ number of
eigenspaces ($l\leq p$) be such that $C$ restricted to these is throughout
non-zero and it vanishes on the orthogonal complement. This makes the
eigensvalues of $E$ strictly bigger than 1 for these $l$ eigenspaces and $E=I$
on the orthogonal complement.
Let $K^{\prime}=\overset{l}{\underset{i=1}{\oplus}}{K_{i}}$. So,
$\mathbb{C}^{n}=K^{\prime}\oplus{K^{\prime}}^{\perp}$ where
$E\restriction_{K_{i}}=a_{i}\,I$, $a_{i}>1$, $i=1,\,2,\,...,\,l$ and
$E\restriction_{{K^{\prime}}^{\perp}}=I.$ Let $dim\,K^{\prime}=k$ which gives
$\sum\limits_{i=1}^{l}k_{i}=k$.
Hence $C\restriction_{{K^{\prime}}^{\perp}}\equiv 0$ and due to mutual
orthogonality of ${\xi_{i_{j}}}^{\prime}s$, $C\restriction_{K^{\prime}}$
becomes bijective where $C(e_{i_{j}})=\xi_{i_{j}}$, $i=1,\,2,\,...,\,l$ and
$j=1,\,2,\,...,\,k_{i}$. Hence
$C^{*}(\xi_{i_{j}})={\delta_{i}}^{2}\,e_{i_{j}},\,\,j=1,\,2,\,...,\,k_{i},\,\,i=1,\,2,\,...,\,l$
and $C^{*}\equiv 0$ on $(ran\,C)^{\perp}$ where $ran\,C$ denote the range of
$C$.
At last, the relation $B^{2}-CC^{*}=I$ yields
$B(\xi_{i_{j}})=a_{i}\,\xi_{i_{j}},\,\,j=1,\,2,\,...,\,k_{i},\,\,i=1,\,2,\,...,\,l$
and $B=I$ on $(ran\,C)^{\perp}.$
Let $M_{i}=span\,\\{\xi_{i_{j}},\,\,j=1,\,2,\,...,\,k_{i}\\}$ for each
$i=1,\,2,\,...,\,l$. So $\overset{l}{\underset{i=1}{\oplus}}{M_{i}}=ran\,C$
and $B$ decomposes $H$ orthogonally into eigenspaces as
$H=ran\,C\oplus{(ran\,C)}^{\perp}$ where $B\restriction_{M_{i}}=a_{i}\,I$.
Also $dim\,M_{i}=dim\,K_{i}=k_{i}$. Let
${\beta_{i}}^{\prime}=\\{\xi_{i_{j}},\,\,j=1,\,2,\,...,\,k_{i}\\}$ which is an
orthogonal basis of $M_{i}$ for each $i=1,\,2,\,...,\,l$.

In the rest of the article we reserve the notations $I$ and $J_{i}$ for the
sets $\\{1,\,2,\,...,\,l\\}$ and $\\{1,\,2,\,...,\,k_{i}\\}$, respectively
where $i\in I$.
As a summary of the preceding analysis, we have the following information.

######  Remark 4.

Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$.Then $E$ decomposes $\mathbb{C}^{n}$
orthogonally into eigenspaces as
$\mathbb{C}^{n}=\overset{l}{\underset{i=1}{\oplus}}{K_{i}}\oplus{K^{\prime}}^{\perp}$
where

(4.1) $\displaystyle
K^{\prime}=\overset{l}{\underset{i=1}{\oplus}}{K_{i}},\,\,\text{and}$ (4.2)
$\displaystyle E\restriction_{K_{i}}$ $\displaystyle=a_{i}\,I,\,\,a_{i}>1,$
$\displaystyle E\restriction_{{K^{\prime}}^{\perp}}$ $\displaystyle=I.$

Let $dim\,(K^{\prime})=k$ and $dim\,(K_{i})=k_{i}$ for each $i\in I$, hence

(4.3) $\displaystyle\sum\limits_{i=1}^{l}{k_{i}}=k.$ (4.4)
$\displaystyle\beta_{i}=\\{e_{i_{j}},\,\,j\in J_{i}\\}$

is an orthonormal basis of $K_{i}$ for each $i\in I$.

(4.5) $\displaystyle C(e_{i_{j}})$ $\displaystyle=\xi_{i_{j}}(\neq 0),\,\,j\in
J_{i},\,\,i\in I,$ $\displaystyle C$ $\displaystyle\equiv
0\,\,\text{on}\,\,{K^{\prime}}^{\perp}.$

As distinct ${\xi_{i_{j}}}^{\prime}s$ are mutually orthogonal,
$C\restriction_{K^{\prime}}$ is a bijective operator and
$dim\,(ran\,C)=k=dim\,K^{\prime}$.
Also $\|\xi_{i_{j}}\|=\|\xi_{i_{j^{\prime}}}\|$ for every $j,\,j^{\prime}\in
K_{i}$. Let

(4.6) $\displaystyle\|\xi_{i_{j}}\|=\delta_{i},\,\,j\in J_{i}.$

Hence

(4.7) $\displaystyle{a_{i}}^{2}=1+{\delta_{i}}^{2}.$ (4.8) $\displaystyle
C^{*}(\xi_{i_{j}})$ $\displaystyle={\delta_{i}}^{2}e_{i_{j}},\,\,j\in
J_{i},\,\,i\in I,$ $\displaystyle C^{*}$ $\displaystyle\equiv
0\,\,\text{on}\,\,(ran\,C)^{\perp}.$

Let $M_{i}=span\,\\{\xi_{i_{j}},\,\,j\in J_{i}\\}$ for each $i\in I$. Then

(4.9) $\displaystyle ran\,C=\overset{l}{\underset{i=1}{\oplus}}M_{i}$

where $dim\,(M_{i})=k_{i}=dim\,(K_{i})$.

(4.10) $\displaystyle{\beta_{i}}^{\prime}=\\{\xi_{i_{j}},\,\,j\in J_{i}\\}$

is an orthogonal basis of $M_{i}$ for each $i\in I$. $B$ decomposes $H$
orthogonally as
$H=\overset{l}{\underset{i=1}{\oplus}}M_{i}\oplus(ran\,C)^{\perp}$ where

(4.11) $\displaystyle B\restriction_{M_{i}}$ $\displaystyle=a_{i}\,I,\,\,j\in
J_{i},\,\,i\in I,$ $\displaystyle B$
$\displaystyle=I\,\,\text{on}\,\,(ran\,C)^{\perp}.$

Now we are ready to characterize non-unitary normal isometries for
$K=\mathbb{C}^{n}$.

Proof of Proposition 1.3. Let $T$ be a non-unitary normal isometry. Then in
view of Proposition 1.1, $UB=BU$ and $VE=EV$ which implies $U(M_{i})=M_{i}$
and $V(K_{i})=K_{i}$ using (4.11) and (4.2) respectively. Also
$UC(e_{i_{j}})=U(\xi_{i_{j}})$ by (4.5). As $U(M_{i})=M_{i}$, let
$U(\xi_{i_{j}})=\sum\limits_{p=1}^{k_{i}}{{\alpha_{j_{p}}}^{(i)}}{\xi_{i_{p}}}$
using (4.10) where ${{\alpha_{j_{p}}}^{(i)}}\in\mathbb{C}$. Similarly, as
$V(K_{i})=K_{i}$, let
$V(e_{i_{j}})=\sum\limits_{p=1}^{k_{i}}{{\gamma_{j_{p}}}^{(i)}}{e_{i_{p}}}$
using (4.4) where ${{\gamma_{j_{p}}}^{(i)}}\in\mathbb{C}$. This gives
$CV(e_{i_{j}})=\sum\limits_{p=1}^{k_{i}}{{\gamma_{j_{p}}}^{(i)}}{\xi_{i_{p}}}$.
As $UC=CV$, $UC(e_{i_{j}})=CV(e_{i_{j}})$ which implies
${{\alpha_{j_{p}}}^{(i)}}={{\gamma_{j_{p}}}^{(i)}}$ for all $i\in I,\,j,\,p\in
J_{i}$ using mutual orthogonality of distinct ${\xi_{i_{p}}}^{\prime}s$. Hence

(4.12)
$\displaystyle[U(\xi_{i_{j}})]_{{\beta_{i}}^{\prime}}=[V(e_{i_{j}})]_{{\beta_{i}}}$

and
${[U\restriction_{M_{i}}]}_{{\beta_{i}}^{\prime}}={[V\restriction_{K_{i}}]}_{{\beta_{i}}}$
for every $i\in I$.
Conversely, tracing back the steps of forward part together with the facts
that $B=I$ on $(ran\,C)^{\perp}$, $E=I$ on ${K^{\prime}}^{\perp}$ and $C\equiv
0$ on ${K^{\prime}}^{\perp}$ yield $UB=BU$, $VE=EV$ and $UC=CV$.

######  Corollary 4.0.1.

If $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ is a non-unitary normal isometry then
$T=\overset{l}{\underset{i=1}{\oplus}}{T_{i}}\oplus T^{\prime}\oplus
T^{\prime\prime}$ where $T_{i}=T\restriction_{M_{i}\oplus K_{i}}$, $i\in I$,
$T^{\prime}=U\restriction_{{(ran\,C)}^{\perp}}$ and
$T^{\prime\prime}=V\restriction_{{K^{\prime}}^{\perp}}$.

###### Proof.

Note that ${\beta_{i}}^{\prime}\cup\beta_{i}=\\{(\xi_{i_{j}},0),\,\,\,\,j\in
J_{i}\\}\cup\\{(0,e_{i_{j}}),\,\,\,\,j\in J_{i}\\}$ is a basis of $M_{i}\oplus
K_{i}$ which follows from (4.10) and (4.4). Now using (4.11), (4.8) and
(4.12), we get

$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{c}\xi_{i_{j}}\\\ 0\\\
\end{array}}\right]=\left[{\begin{array}[]{c}BU(\xi_{i_{j}})\\\
C^{*}U(\xi_{i_{j}})\\\
\end{array}}\right]=\left[{\begin{array}[]{c}a_{i}U(\xi_{i_{j}})\\\
{\delta_{i}}^{2}V(e_{i_{j}})\\\ \end{array}}\right]$ $\in M_{i}\oplus K_{i}$.

Similarly, the fact that $UC=CV$, (4.5) and (4.2) give

$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{c}0\\\ e_{i_{j}}\\\
\end{array}}\right]=\left[{\begin{array}[]{c}CV(e_{i_{j}})\\\ EV(e_{i_{j}})\\\
\end{array}}\right]=\left[{\begin{array}[]{c}U(\xi_{i_{j}})\\\
a_{i}V(e_{i_{j}})\\\ \end{array}}\right]\in M_{i}\oplus K_{i}$.

This implies $T(M_{i}\oplus K_{i})=M_{i}\oplus K_{i}$ as $T$ is one-one. For
$h\in(ran\,C)^{\perp}$, the fact that $ran\,C$ reduces $U$ together with
(4.11) and (4.8) yield

$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{c}h\\\ 0\\\
\end{array}}\right]=\left[{\begin{array}[]{c}U(h)\\\ 0\\\
\end{array}}\right]\in(ran\,C)^{\perp}$

implying $T((ran\,C)^{\perp})=(ran\,C)^{\perp}$. Let
$T^{\prime}=T\restriction_{(ran\,C)^{\perp}}=U\restriction_{(ran\,C)^{\perp}}$.
Also for $\boldsymbol{\mathrm{z}}\in{K^{\prime}}^{\perp}$, the fact that
$K^{\prime}$ reduces $V$ together with (4.5) and (4.2) imply
$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{c}0\\\ \boldsymbol{\mathrm{z}}\\\
\end{array}}\right]=\left[{\begin{array}[]{c}0\\\
V(\boldsymbol{\mathrm{z}})\\\ \end{array}}\right]\in{K^{\prime}}^{\perp}$
giving $T({K^{\prime}}^{\perp})={K^{\prime}}^{\perp}$. Let
$T^{\prime\prime}=T\restriction_{{K^{\prime}}^{\perp}}=V\restriction_{{K^{\prime}}^{\perp}}$.
Hence we have $T=\overset{l}{\underset{i=1}{\oplus}}{T_{i}}\oplus
T^{\prime}\oplus T^{\prime\prime}$. ∎

######  Corollary 4.0.2.

$[T_{i}]_{{\beta_{i}}^{\prime}\cup{\beta_{i}}}=\left[{\begin{array}[]{cc}a_{i}R_{i}&R_{i}\\\
{\delta_{i}}^{2}R_{i}&a_{i}R_{i}\\\ \end{array}}\right]_{2k_{i}\times 2k_{i}}$
where
$R_{i}={[U\restriction_{M_{i}}]}_{{\beta_{i}}^{\prime}}={[V\restriction_{K_{i}}]}_{{\beta_{i}}}$.

###### Proof.

Using the expressions of $T(\xi_{i_{j}},0)$ and $T(0,e_{i_{j}})$ from the
proof of Corollary 4.0.1 and the expressions of $U(\xi_{i_{j}})$ and
$V(e_{i_{j}})$ from the proof of Proposition 1.3 along with (4.12), we get

$[T_{i}]_{{\beta_{i}}^{\prime}\cup{\beta_{i}}}=[{t_{r_{s}}}^{(i)}]_{2k_{i}\times
2k_{i}}$

where $r$ is the number of rows and $s$ is the number of columns and

$\displaystyle{t_{r_{s}}}^{(i)}=\left\\{\begin{array}[]{llll}\left<T(\xi_{i_{s}},0),(\xi_{i_{r}},0)\right>&\mbox{if
}1\leq r,s\leq k_{i},\\\ \left<T(\xi_{i_{s}},0),(0,e_{i_{p}})\right>&\mbox{if
}r=k_{i}+p,\,1\leq p\leq k_{i}\,\,\mbox{and}\,\,1\leq s\leq k_{i},\\\
\left<T(0,e_{i_{p}}),(\xi_{i_{r}},0)\right>&\mbox{if }1\leq r\leq
k_{i}\,\,\mbox{and}\,\,s=k_{i}+p,\,1\leq p\leq k_{i},\\\
\left<T(0,e_{i_{p}}),(0,e_{i_{m}},0)\right>&\mbox{if
}r=k_{i}+m\,\,\mbox{and}\,\,s=k_{i}+p,\,1\leq m,p\leq
k_{i}.\end{array}\right.$

The above analysis in compact form gives
$[T_{i}]_{{\beta_{i}}^{\prime}\cup{\beta_{i}}}=\left[{\begin{array}[]{cc}a_{i}R_{i}&R_{i}\\\
{\delta_{i}}^{2}R_{i}&a_{i}R_{i}\\\ \end{array}}\right]_{2k_{i}\times 2k_{i}}$
where
$R_{i}={[U\restriction_{M_{i}}]}_{{\beta_{i}}^{\prime}}={[V\restriction_{K_{i}}]}_{{\beta_{i}}}$.

∎

Next we describe the unitary elements in $\mathcal{G}$ for $K=\mathbb{C}^{n}$.

Proof of Proposition 1.2. Let $T=\left[{\begin{array}[]{cc}BU&CV\\\
C^{*}U&EV\\\ \end{array}}\right]\in\mathcal{G}$ be a unitary operator. By
Proposition 3.2, $C\equiv 0$. This would make $(ran\,C)^{\perp}=H$ and hence
$B=I$ on $H$ by (4.11). Also as $\xi_{i_{j}}=0$ for all $i\in I$ and $j\in
J_{i}$, we have $a_{i}=1$ for every $i\in I$ using (4.6) and (4.7), hence
$E=I$ by (4.2). Thus $T=\left[{\begin{array}[]{cc}U&0\\\ 0&V\\\
\end{array}}\right]$.

##  5\. Fixed points of non-unitary normal isometries for
$\mathcal{B}\subseteq B(\mathbb{C}^{n},H)$

Due to a result of Kaup [10], every holomorphic automorphism on $\mathcal{B}$
is the restriction of a holomorphic map defined on some neighbourhood of
$\overline{\mathcal{B}}$ taking values in $B(K,H)$. What follows is a result
from [3] which guarantees existence of fixed points on
$\overline{\mathcal{B}}$.

######  Theorem 5.1 ([3], Corollary 2.4).

Every holomorphic automorphism of the open unit ball $\mathcal{B}$ in
$B(\mathbb{C}^{n},H)$ has a fixed point in $\overline{\mathcal{B}}$.

Also, to have a representation of fixed points for linear isometries, we need
to ensure that the expression $(BA+C){(DA+E)}^{-1}$ which appears in Theorem
2.2 is meaningful for $A\in\overline{\mathcal{B}}$. As (2.4) says $D=C^{*}$ ,
we will show that

######  Lemma 5.2.

$C^{*}A+E$ is invertible for $A\in\overline{\mathcal{B}}$.

###### Proof.

As $E$ is a positive invertible operator, for $\textbf{z}\,(\neq
0)\in\mathbb{C}^{n}$, (3.1) yields
$\left<E^{-1}(E^{2}-C^{*}C){E^{-1}}(\textbf{z}),\textbf{z}\right>=\left<E^{-1}{E^{-1}}(\textbf{z}),\textbf{z}\right>>0$.
In particular $\left<E^{-1}C^{*}C{E^{-1}}(\textbf{z}),\textbf{z}\right><1$ for
$\|\textbf{z}\|=1$. Since the unit sphere in $\mathbb{C}^{n}$ is a compact
set, we have
$\underset{\|\textbf{z}\|=1}{sup}\left<E^{-1}C^{*}{(E^{-1}C^{*})}^{*}(\textbf{z}),\textbf{z}\right><1$.
This means $\Big{\|}E^{-1}C^{*}\Big{\|}^{2}<1$ and
$\Big{\|}E^{-1}C^{*}A\Big{\|}<1$ for $A\in\overline{\mathcal{B}}$ which makes
$I+E^{-1}C^{*}A$ and hence $E+C^{*}A$ an invertible operator. ∎

Let us now have a representation of fixed points for linear isometries.

######  Proposition 5.3.

Let $h\in Aut(\mathcal{B})$ be defined as

$h(A)=(BUA+CV){(C^{*}UA+EV)}^{-1}$

and $T_{h}=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]$ be the corresponding linear isometry (cf. Theorem 2.2).
Let $\\{\boldsymbol{\mathrm{z}_{r}},\,\,r=1,\,2,\,...,\,n\\}$ be a basis of
$\mathbb{C}^{n}$ and $F\in B({\mathbb{C}}^{n},H)$ be such that
$\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\ \end{array}}\right]$ are eigenvectors of
$T_{h}$, $r=1,\,2,\,...,\,n$, then $F$ is a fixed point of $h$. Conversely, if
$F$ is a fixed point of $h$ and there exists an eigenbasis
$\\{\boldsymbol{\mathrm{z}_{r}},\,\,r=1,\,2,\,...,\,n\\}$ of $C^{*}UF+EV$,
then
$\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\ \end{array}}\right]$ are eigenvectors of
$T_{h}$.

###### Proof.

Let $\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\ \end{array}}\right]$ be eigenvectors of $T_{h}$
having eigenvalues $\lambda_{r}$ (say), i.e.
$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\
\end{array}}\right]=\lambda_{r}\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\ \end{array}}\right]$. This gives
$BU(F(\boldsymbol{\mathrm{z}_{r}}))+CV(\boldsymbol{\mathrm{z}_{r}})=F(C^{*}U(F(\boldsymbol{\mathrm{z}_{r}}))+EV(\boldsymbol{\mathrm{z}_{r}}))$
and hence $BUF+CV=F(C^{*}UF+EV)$ or $(BUF+CV)(C^{*}UF+EV)^{-1}=F$ as
$\\{\boldsymbol{\mathrm{z}_{r}},\,\,r=1,\,2,\,...,\,n\\}$ is a basis of
$\mathbb{C}^{n}$. Thus $F$ is a fixed point of $h$.
Conversely, Let $F$ be a fixed point of $h$ and ${\mu_{r}}^{\prime}s$ be the
eigenvalues corresponding to the eigenbasis
$\\{\boldsymbol{\mathrm{z}_{r}},\,\,r=1,\,2,\,...,\,n\\}$ of $C^{*}UF+EV$.
This gives
$\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\
\end{array}}\right]=\left[{\begin{array}[]{c}BU(F(\boldsymbol{\mathrm{z}_{r}}))+CV(\boldsymbol{\mathrm{z}_{r}})\\\
C^{*}U(F(\boldsymbol{\mathrm{z}_{r}}))+EV(\boldsymbol{\mathrm{z}_{r}})\\\
\end{array}}\right]$
$=\left[{\begin{array}[]{c}F(C^{*}U(F(\boldsymbol{\mathrm{z}_{r}}))+EV(\boldsymbol{\mathrm{z}_{r}}))\\\
C^{*}U(F(\boldsymbol{\mathrm{z}_{r}}))+EV(\boldsymbol{\mathrm{z}_{r}})\\\
\end{array}}\right]$ =
$\mu_{r}\left[{\begin{array}[]{c}F(\boldsymbol{\mathrm{z}_{r}})\\\
\boldsymbol{\mathrm{z}_{r}}\\\ \end{array}}\right],\,\,r=1,\,2,\,...,\,n$.
This completes the proof. ∎

The above proposition tells that the fixed points which are obtained via
eigenvectors form a subclass of collection of all the fixed points.

In view of Theorem 2.2, Fixed points of an isometry $T\in\mathcal{G}$ would
mean fixed points of its identification in $Aut(\mathcal{B})$ and from now
onwards, fixed points would mean the ones which are obtained through
eigenvectors only, (cf. Proposition 5.3).

In order to investigate the fixed points of a non-unitary normal isometry, let
us first examine its spectrum.

######  Lemma 5.4.

For the matrix $S_{i}=\left[{\begin{array}[]{cc}a_{i}I&I\\\
{\delta_{i}}^{2}I&a_{i}I\\\ \end{array}}\right]_{2k_{i}\times 2k_{i}}$ where
$I$ is a $k_{i}\times k_{i}$ identity matrix,
$\sigma(S_{i})=a_{i}\pm\delta_{i}$. The corresponding eigenspaces are
generated by the sets $\\{[\Delta_{j_{m}}]_{2k_{i}\times 1},\,\,\,\,j\in
J_{i}\\}$ where

$\displaystyle{\Delta_{j_{m}}}=\left\\{\begin{array}[]{llll}1&\mbox{if
}m=k_{i}+j,\\\ 0&\mbox{if }m=k_{i}+p,\,p\in J_{i}\setminus\\{j\\},\\\
\pm\dfrac{1}{\delta_{i}}&\mbox{if }j=m,\\\ 0&\mbox{if }m\in
J_{i}\setminus\\{j\\}.\end{array}\right.$

###### Proof.

For $\lambda_{i}\in\mathbb{C}$, consider the expression

$\left[{\begin{array}[]{cc}(a_{i}-\lambda_{i})I&I\\\
{\delta_{i}}^{2}I&(a_{i}-\lambda_{i})I\\\
\end{array}}\right]\left[{\begin{array}[]{c}r_{j}\\\
\end{array}}\right]_{2k_{i}\times 1}\equiv 0.$

This implies

$\displaystyle(a_{i}-\lambda_{i})r_{j}+r_{k_{i}+j}$ $\displaystyle=0,$
$\displaystyle{\delta_{i}}^{2}r_{j}+(a_{i}-\delta_{i})r_{k_{i}+j}$
$\displaystyle=0,$

$j\in J_{i}$. On simplification, the above two equations yield
$\lambda_{i}=a_{i}\pm\delta_{i}$ are the two eigenvalues of $S_{i}$ and
general eigenvectors are $\left[{\begin{array}[]{c}r_{j}\\\
\end{array}}\right]_{2k_{i}\times 1}$, respectively where for arbitrary values
of $r_{k_{i}+j}\in\mathbb{C}$, $r_{j}=\pm\dfrac{r_{k_{i}+j}}{\delta_{i}}$,
$j\in J_{i}$. The eigenspaces corresponding to the eigenvalues
$\lambda_{i}\pm\delta_{i}$ are

$E_{i}=span\,\Bigg{\\{}\Big{(}\underbrace{0,\,0,\,...,\,\underbrace{\dfrac{1}{\delta_{i}}}_{j^{th}\,\,\text{place}},0,...,0}_{\text{first}\,\,k_{i}\,\,\text{entries}},\underbrace{0,...,\underbrace{1}_{k_{i}+j^{th}\,\,\text{place}},0,...,0}_{\text{next}\,\,k_{i}\,\,\text{entries}}\Big{)},\,\,\,\,j\in
J_{i}\Bigg{\\}}$

and

${E_{i}}^{\prime}=span\,\Bigg{\\{}\Big{(}\underbrace{0,\,0,\,...,\,\underbrace{-\dfrac{1}{\delta_{i}}}_{j^{th}\,\,\text{place}},0,...,0}_{\text{first}\,\,k_{i}\,\,\text{entries}},\underbrace{0,...,\underbrace{1}_{k_{i}+j^{th}\,\,\text{place}},0,...,0}_{\text{next}\,\,k_{i}\,\,\text{entries}}\Big{)},\,\,\,\,j\in
J_{i}\Bigg{\\}}$

respectively which is expanded version of the conclusion of the statement of
lemma. ∎

######  Proposition 5.5.

Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ be a non-unitary normal isometry. Then

$\sigma(T)=\overset{l}{\underset{i=1}{\cup}}\\{\lambda_{i_{j}}(a_{i}\pm\delta_{i}),\,\,j\in
J_{i}\\}\cup\sigma(U\restriction_{(ran\,C)^{\perp}})\cup\sigma_{pt}(V\restriction_{{K^{\prime}}^{\perp}})$

where ${\lambda_{i_{j}}}$, $j\in J_{i}$ are eigenvalues of
$U\restriction_{M_{i}}$ (cf (1.3)) for each $i\in I$. The eigenvectors
corresponding to the eigenvalues $\lambda_{i_{j}}(a_{i}\pm\delta_{i})$ are
$\left(\pm\dfrac{1}{\delta_{i}}C(\boldsymbol{\mathrm{z_{i_{j}}}}),\boldsymbol{\mathrm{z_{i_{j}}}}\right)$,
respectively for $j\in J_{i}$ and $i\in I$ where the collection
$\\{\boldsymbol{\mathrm{z_{i_{j}}}},\,\,j\in J_{i}\\}$ is some orthonormal
basis of $K_{i}$ for each $i\in I$.

###### Proof.

In view of Corollary 4.0.1,
$\sigma(T)=\overset{l}{\underset{i=1}{\cup}}{\sigma_{pt}{(T_{i})}}\cup\sigma(U\restriction_{(ran\,C)^{\perp}})\cup\sigma_{pt}(V\restriction_{{K^{\prime}}^{\perp}})$
where $\sigma(U\restriction_{(ran\,C)^{\perp}})$ and
$\sigma_{pt}(V\restriction_{{K^{\prime}}^{\perp}})$ are contained in the unit
circle. We will show that for every $i\in I$,
$\sigma_{pt}{(T_{i})}=\\{{\lambda_{i_{j}}}(a_{i}\pm\delta_{i}),\,\,j\in
J_{i}\\}$.
For $i\in I$, consider $T_{i}=T\restriction_{M_{i}\oplus K_{i}}$. From
Corollary 4.0.2,

$\displaystyle[T_{i}]_{{\beta_{i}}^{\prime}\cup{\beta_{i}}}$
$\displaystyle=\left[{\begin{array}[]{cc}a_{i}R_{i}&R_{i}\\\
{\delta_{i}}^{2}R_{i}&a_{i}R_{i}\\\ \end{array}}\right]_{2k_{i}\times 2k_{i}}$
$\displaystyle=\left[{\begin{array}[]{cc}a_{i}{[I]}_{k_{i}\times
k_{i}}&{[I]}_{k_{i}\times k_{i}}\\\ {\delta_{i}}^{2}{[I]}_{k_{i}\times
k_{i}}&a_{i}{[I]}_{k_{i}\times k_{i}}\\\
\end{array}}\right]\left[{\begin{array}[]{cc}R_{i}&0\\\ 0&R_{i}\\\
\end{array}}\right]=S_{i}\,W_{i}$

where $S_{i}=\left[{\begin{array}[]{cc}a_{i}{[I]}_{k_{i}\times
k_{i}}&{[I]}_{k_{i}\times k_{i}}\\\ {\delta_{i}}^{2}{[I]}_{k_{i}\times
k_{i}}&a_{i}{[I]}_{k_{i}\times k_{i}}\\\ \end{array}}\right]$ and
$W_{i}=\left[{\begin{array}[]{cc}R_{i}&0\\\ 0&R_{i}\\\ \end{array}}\right]$.
From the proof of Lemma 5.4, it can be observed that $S_{i}$ is
diagonalizable. It has eigenvalues $a_{i}\pm\delta_{i}$ and $E_{i}$ and
${E_{i}}^{\prime}$ as corresponding eigenvectors. Also it is easy to see that
$S_{i}$ and $W_{i}$ commute. This implies $W(E_{i})=E_{i}$ and
$W({E_{i}}^{\prime})={E_{i}}^{\prime}$. Let
$\left\\{\left({c_{j_{p}}}^{(i)}\right)_{p=1}^{k_{i}},\,\,j\in J_{i}\right\\}$
be an orthonormal eigenbasis of $R_{i}$ having $\lambda_{i_{j}}$ as
eigenvalues, ${c_{j_{p}}}^{(i)}\in\mathbb{C}$. Hence
$\left\\{\left(\dfrac{\left({c_{j_{p}}}^{(i)}\right)_{p=1}^{k_{i}}}{\delta_{i}},\left({c_{j_{p}}}^{(i)}\right)_{p=1}^{k_{i}}\right),\,\,j\in
J_{i}\right\\}$ and
$\left\\{\left(-\dfrac{\left({c_{j_{p}}}^{(i)}\right)_{p=1}^{k_{i}}}{\delta_{i}},\left({c_{j_{p}}}^{(i)}\right)_{p=1}^{k_{i}}\right),\,\,j\in
J_{i}\right\\}$ would serve as eigenbases of $W_{i}\restriction_{{E_{i}}}$ and
$W_{i}\restriction_{{E_{i}}^{\prime}}$ respectively with common eigenvalues as
$\\{{\lambda_{i_{j}}},\,\,j\in J_{i}\\}$. As these are also eigenbases of
$T_{i}\restriction_{E_{i}}$ and $T_{i}\restriction_{{E_{i}}^{\prime}}$, union
of these eigenbases is eigenbasis of
$[T_{i}]_{{\beta_{i}}^{\prime}\cup{\beta_{i}}}$. Let

(5.1)
$\displaystyle\boldsymbol{\mathrm{z_{i_{j}}}}=\sum\limits_{p=1}^{k_{i}}{{c_{j_{p}}}^{(i)}}e_{i_{p}}$

where $\\{e_{i_{p}},\,\,p\in J_{i}\\}$ is as in (4.4). Hence
$\\{\boldsymbol{\mathrm{z_{i_{j}}}},\,\,j\in J_{i}\\}$ is an orthonormal basis
of $K_{i}$. Also

(5.2) $\displaystyle
C(\boldsymbol{\mathrm{z_{i_{j}}}})=\sum\limits_{p=1}^{k_{i}}{{c_{j_{p}}}^{(i)}}\xi_{i_{p}}$

by (4.5). This gives the eigenvectors corresponding to eigenvalues
$\lambda_{i_{j}}(a_{i}\pm\delta_{i})$ as
$\left(\pm\dfrac{1}{\delta_{i}}C(\boldsymbol{\mathrm{z_{i_{j}}}}),\boldsymbol{\mathrm{z_{i_{j}}}}\right),\,\,\,\,j\in
J_{i}$. ∎

It is easy to observe that eigenvalues $\lambda_{i_{j}}(a_{i}\pm\delta_{i})$
are distinct for distinct values of $i$. Also
$\sigma(T_{i}\restriction_{E_{i}})\cap\sigma(T_{i}\restriction_{{E_{i}}^{\prime}})=\phi$.

######  Remark 5.

Notice that for a non-unitary normal isometry
$T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$, $\left[{\begin{array}[]{c}h\\\
\boldsymbol{\mathrm{z}}\\\
\end{array}}\right]\in{(ran\,C)}^{\perp}\oplus{K^{\prime}}^{\perp}$ is an
eigenvector of $T$ if and only if there exists an eigenvalue $\mu$ such that
$U(h)=\mu h$ and $V(\boldsymbol{\mathrm{z}})=\mu\boldsymbol{\mathrm{z}}$.

Let us conclude the preceding analysis by exhibiting some special fixed points
of a non-unitary normal isometry.

######  Theorem 5.6.

Let $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ be a non-unitary normal isometry such that
$dim\,(ran\,C)=k$. Then it has exactly $2^{k}$ fixed points of the generic
type (cf. Definition 2) such that $\|F_{\theta}\|=1$ for all $\theta\in S$.

###### Proof.

In view of Definition 2, consider $2^{k}$ number of generic functions for
$q=k$ as follows. As $\sum\limits_{i=1}^{l}{k_{i}}=k$, re-define the $k$-tuple
$(\epsilon_{m})_{m=1}^{k}$ as $(\epsilon_{i_{j}})_{i\in I,\,\,j\in J_{i}}$,
orthonormal basis $\\{\boldsymbol{\mathrm{z_{m}}},\,\,m=1,\,2,\,...,\,k\\}$ as
$\\{\boldsymbol{\mathrm{z_{i_{j}}}},\,\,i\in I,\,\,j\in J_{i}\\}$ where the
collection $\\{\boldsymbol{\mathrm{z_{i_{j}}}}\\}$ is as in Proposition 5.5
and choose $\\{\boldsymbol{\mathrm{z_{s}}},\,\,s=k+1,\,....,\,n\\}$ to be an
orthonormal eigenbasis of $V\restriction_{{K^{\prime}}^{\perp}}$. Take
$\\{y_{m},\,\,m=1,\,2,\,...,\,k\\}=\left\\{\dfrac{C(\boldsymbol{\mathrm{z_{i_{j}}}})}{\delta_{i}},\,\,i\in
I,\,\,j\in J_{i}\right\\}$. Note that
$\dfrac{C(\boldsymbol{\mathrm{z_{i_{j}}}})}{\delta_{i}}\neq 0$ as
$C\restriction_{K^{\prime}}$ is a bijective operator. For generic functions
defined in this way, Proposition 5.5 tells that
$\left[\begin{array}[]{c}F_{\theta}(\boldsymbol{\mathrm{z_{i_{j}}}})\\\
\boldsymbol{\mathrm{z_{i_{j}}}}\\\ \end{array}\right]$ are eigenvectors of $T$
for each $\theta\in S$, $i\in I$ and $j\in J_{i}$. Also it is easy to see that
$\left[\begin{array}[]{c}0\\\ \boldsymbol{\mathrm{z}_{s}}\\\
\end{array}\right]$ are eigenvectors of $T$ for $s=k+1,\,...,\,n$. Thus in
view of Proposition 5.3, $T$ has exactly $2^{k}$ fixed points of the generic
type. Now we will see that $\|F_{\theta}\|=1$ for $\theta\in S$.
Equations (5.2) and (4.6) yield that the collection
$\left\\{\dfrac{C(\boldsymbol{\mathrm{z_{i_{j}}}})}{\delta_{i}},\,\,i\in
I,\,j\in J_{i}\right\\}$ is an orthonormal set and hence forms an orthonormal
basis of $ran\,C$. This makes the generic function
${F_{\theta}}\restriction_{K^{\prime}}$ a norm preserving surjective linear
map and hence $\|{F_{\theta}}\restriction_{K^{\prime}}\|=1$. Also
${F_{\theta}}\restriction_{{K^{\prime}}^{\perp}}\equiv 0$. Therefore
$\|F_{\theta}\|=\|{F_{\theta}}\restriction_{K^{\prime}}\|=1$. ∎

Another way of obtaining fixed points is subject to some condition. Let
$\sigma_{pt}(U\restriction_{{(ran\,C)}^{\perp}})\cap\sigma_{pt}(V\restriction_{{K^{\prime}}^{\perp}})\neq\phi$
and $\mu$ be a common eigenvalue. Define $F\in B(\mathbb{C}^{n},H)$ such that
$F\restriction_{K^{\prime}}={F_{\theta}}\restriction_{K^{\prime}}$ where
$F_{\theta}$ is as in Theorem 5.6. For ${K^{\prime}}^{\perp}$, $F$ may take
basis elements of the eigenspace of $V\restriction_{{K^{\prime}}^{\perp}}$ to
the values in the eigenspace of $U\restriction_{{(ran\,C)}^{\perp}}$
corresponding to the common eigenvalue $\mu$. Set $F$ to vanish on those
eigenspaces whose corresponding eigenvalues have empty intersection with
$\sigma_{pt}(U\restriction_{{(ran\,C)}^{\perp}})$. Such an $F$ is also a fixed
point of non-unitary normal isometry by Remark 5 and Proposition 5.3.

###  5.1. Proof of Theorem 1.4

######  Lemma 5.7.

Let an isometry $T=\left[{\begin{array}[]{cc}BU&CV\\\ C^{*}U&EV\\\
\end{array}}\right]\in\mathcal{G}$ be such that
$dim\,(ran\,C)=k$ and it possesses exactly $2^{q}$ fixed points of generic
type (cf. Definition 2). Then $q=k$ and $\boldsymbol{\mathrm{z}_{m}}\in K_{i}$
and $y_{m}$ lies in corresponding $M_{i}$ for every
$m\in\\{1,\,2,\,...,\,k\\}$ and some $i\in I$ depending on $m$.

###### Proof.

By hypothesis,
$\left[\begin{array}[]{c}F_{\theta}(\boldsymbol{\mathrm{z}_{m}})\\\
\boldsymbol{\mathrm{z}_{m}}\\\ \end{array}\right]$ are eigenvectors of $T$ for
every $\theta\in S$ and $m=1,\,2,\,...,\,q$, i.e.
$\left[\begin{array}[]{c}\epsilon_{m}\,y_{m}\\\ \boldsymbol{\mathrm{z}_{m}}\\\
\end{array}\right]=\left[\begin{array}[]{c}\pm y_{m}\\\
\boldsymbol{\mathrm{z}_{m}}\\\ \end{array}\right]$ are eigenvectors of $T$.
Let $\lambda_{m}$ and ${\lambda_{m}}^{\prime}$ be corresponding eigenvalues.
Expanding and simplifying the so formed expressions, we get

(5.3) $\displaystyle CV(\boldsymbol{\mathrm{z}_{m}})$
$\displaystyle=\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{2}y_{m}$ (5.4)
$\displaystyle BU(y_{m})$
$\displaystyle=\dfrac{\lambda_{m}+{\lambda_{m}}^{\prime}}{2}y_{m}$ (5.5)
$\displaystyle EV(\boldsymbol{\mathrm{z}_{m}})$
$\displaystyle=\dfrac{\lambda_{m}+{\lambda_{m}}^{\prime}}{2}\boldsymbol{\mathrm{z}_{m}},\,\,\,\,\text{and}$
$\displaystyle C^{*}U(y_{m})$
$\displaystyle=\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{2}\boldsymbol{\mathrm{z}_{m}}.$

The above equations further simplify to

(5.6) $\displaystyle(\lambda_{m}+{\lambda_{m}}^{\prime})\,C^{*}B^{-1}(y_{m})$
$\displaystyle=(\lambda_{m}-{\lambda_{m}}^{\prime})\,\boldsymbol{\mathrm{z}_{m}},\,\,\,\,\text{and}$
(5.7)
$\displaystyle(\lambda_{m}+{\lambda_{m}}^{\prime})\,CE^{-1}(\boldsymbol{\mathrm{z}_{m}})$
$\displaystyle=(\lambda_{m}-{\lambda_{m}}^{\prime})\,y_{m}.$

As $E$ is invertible, (5.5) yields $\lambda_{m}\neq{-\lambda_{m}}^{\prime}$.
In view of (4.1) and (4.9), each $\boldsymbol{\mathrm{z}_{m}}$ is a linear
combination of basis elements of
$\overset{l}{\underset{i=1}{\oplus}}{K_{i}}\oplus{K^{\prime}}^{\perp}$ and
each $y_{m}$ is a linear combination of basis elements of
$\overset{l}{\underset{i=1}{\oplus}}{M_{i}}\oplus{(ran\,C)}^{\perp}$
respectively. Thus for each $m\in\\{1,\,2,\,...,\,q\\}$, let

(5.8) $\displaystyle\boldsymbol{\mathrm{z}_{m}}$
$\displaystyle=\sum_{i=1}^{l}\sum_{j=1}^{k_{i}}{r^{(m)}_{{i}_{j}}}\,e_{{i}_{j}}+\boldsymbol{\mathrm{w_{m}}},\,\,\,\,\boldsymbol{\mathrm{w_{m}}}\in{K^{\prime}}^{\perp},\,\,\text{(by}\,\,(\ref{eq:24}))$
(5.9) $\displaystyle y_{m}$
$\displaystyle=\sum_{i=1}^{l}\sum_{j=1}^{k_{i}}{s^{(m)}_{{i}_{j}}}\,\xi_{{i}_{j}}+h_{m},\,\,\,\,h_{m}\in{(ran\,C)}^{\perp},\,\,\text{(by}\,\,(\ref{eq:27}))$

$r^{(m)}_{{i}_{j}},\,s^{(m)}_{{i}_{j}}\in\mathbb{C}$.
Substituting the above values of $\boldsymbol{\mathrm{z}_{m}}$ and $y_{m}$ in
(5.6) gives $\boldsymbol{\mathrm{w_{m}}}=0$, hence
$\boldsymbol{\mathrm{z}_{m}}\in K^{\prime}$. This also provides the following
relation between the coefficients $r^{(m)}_{{i}_{j}}$ and $s^{(m)}_{{i}_{j}}$

(5.10) $\displaystyle
s^{(m)}_{{i}_{j}}=\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{\lambda_{m}+{\lambda_{m}}^{\prime}}\,\dfrac{a_{i}}{{\delta_{i}}^{2}}\,r^{(m)}_{{i}_{j}}$

using (4.11) and (4.8). Similarly, using the values of
$\boldsymbol{\mathrm{z}_{m}}$, $y_{m}$ and (5.10) in (5.7) yield $h_{m}=0$,
hence $y_{m}\in ran\,C$. Comparing the coefficients in (5.7) and using (5.10)
yield

(5.11)
$\displaystyle\left(\dfrac{1}{a_{i}}-{\left(\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{\lambda_{m}+{\lambda_{m}}^{\prime}}\right)}^{2}\,\dfrac{a_{i}}{{\delta_{i}}^{2}}\right)\,r^{(m)}_{{i}_{j}}=0$

by (4.5) and (4.2). In view of (5.10), $\lambda_{m}={\lambda_{m}}^{\prime}$ is
a contradiction to the hypothesis that all ${y_{m}}^{\prime}s$ are non-zero as
$h_{m}=0$. So, $\lambda_{m}\neq\pm{\lambda_{m}}^{\prime}$.

We will see that $q=k$. Since
$\\{\boldsymbol{\mathrm{z}_{m}},\,\,m=1,\,2,\,...,q\\}\subseteq K^{\prime}$ is
linearly independent, we have $q\leq k$ as $dim\,(K^{\prime})=k$ by Remark 4.
Also, $\left[\begin{array}[]{c}0\\\ \boldsymbol{\mathrm{z_{s}}}\\\
\end{array}\right]$, $s=q+1,\,...,\,n$ are eigenvectors of $T$ as fixed points
obtained are of generic type. Let the corresponding eigenvalues be
$\lambda_{s}$. Expanding the concerned expression gives
$CV(\boldsymbol{\mathrm{z}_{s}})=0$. This means $dim\,(ker\,C)\,(=n-k)\geq
n-q$, i.e. $q\geq k$ hence $q=k$.

Next claim is that each $\boldsymbol{\mathrm{z}_{m}},\,y_{m}$ lies in some
$K_{i},\,M_{i}$. As $\boldsymbol{\mathrm{z}_{m}}\neq 0$, for every
$m\in\\{1,\,2,\,...,\,k\\}$, there exist atleast one $i\in I$ and $j\in J_{i}$
such that $r^{(m)}_{{i}_{j}}\neq 0$ which means
$\dfrac{1}{a_{i}}-{\left(\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{\lambda_{m}+{\lambda_{m}}^{\prime}}\right)}^{2}\,\dfrac{a_{i}}{{\delta_{i}}^{2}}=0$
or

(5.12)
$\displaystyle\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{\lambda_{m}+{\lambda_{m}}^{\prime}}=\pm\dfrac{{\delta_{i}}}{a_{i}}$

using (5.11) and hence

(5.13) $\displaystyle
s^{(m)}_{{i}_{j}}=\pm\dfrac{1}{\delta_{i}}\,r^{(m)}_{{i}_{j}}$

by (5.10). Suppose there exists $p\neq i\in\\{1,\,2,\,...,\,l\\}$ such that
$r^{(m)}_{{p}_{j}}\neq 0$ for some $j\in J_{p}\\}$. This means
$\dfrac{1}{a_{p}}-{\left(\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{\lambda_{m}+{\lambda_{m}}^{\prime}}\right)}^{2}\,\dfrac{a_{p}}{{\delta_{p}}^{2}}=0$
which further implies
$\dfrac{{\delta_{i}}^{2}}{(a_{i})^{2}}=\dfrac{{\delta_{p}}^{2}}{(a_{p})^{2}}$,
and $a_{i}=a_{p}$ by (4.7) giving a contradiction. Thus for each
$m\in\\{1,\,2,\,...,\,k\\}$, there exists $i\in I$ such that

(5.14)
$\displaystyle\boldsymbol{\mathrm{z}_{m}}=\sum\limits_{j=1}^{k_{i}}{r^{(m)}_{{i}_{j}}}\,e_{{i}_{j}}\in
K_{i}$

by (4.4). As (5.10) yields ${r_{i_{j}}}^{(m)}$ and ${s_{i_{j}}}^{(m)}$ vanish
together,

(5.15) $\displaystyle
y_{m}=\sum\limits_{j=1}^{k_{i}}{s^{(m)}_{{i}_{j}}}\,\xi_{{i}_{j}}\in M_{i}$

using (4.10). ∎

Proof of Theorem 1.4 Necessary part follows from Theorem 5.6. For the
sufficient part we will show $T$ is normal by using Proposition 1.1 together
with Lemma 5.7. First we will show that the set
$\\{y_{m},\,\,m=1,\,2,\,...,\,k\\}$ is linearly independent. As each
$\boldsymbol{\mathrm{z}_{m}}$ lies in exactly one $K_{i}$ by Lemma 5.7, we
have each $\boldsymbol{\mathrm{z}_{m}}$ is an eigenvector of $V$ by (5.5) and
(4.2), hence $K^{\prime}$ reduces $V$. Now bijectivity of
$C\restriction_{K^{\prime}}$ asserts linear independence of $\\{y_{m}\\}$’s
using (5.3). Again using $\boldsymbol{\mathrm{z}_{m}}$ is an eigenvector of
$V$ and (5.5), it follows that $EV=VE$. Similar argument applied to subspaces
of $ran\,C$ establishes $UB=BU$. Now we need to show $UC=CV$. Using (5.15) and
(5.13),
$y_{m}=\pm\dfrac{1}{\delta_{i}}\sum\limits_{j=1}^{k_{i}}{r^{(m)}_{{i}_{j}}}\,\xi_{{i}_{j}}=\pm\dfrac{1}{\delta_{i}}C(\boldsymbol{\mathrm{z}_{m}})$
or $C(\boldsymbol{\mathrm{z}_{m}})=\pm\delta_{i}\,y_{m}$ by (4.5). Now using
(5.4) and (4.11),
$UC(\boldsymbol{\mathrm{z}_{m}})=\pm\delta_{i}\left(\dfrac{\lambda_{m}+{\lambda_{m}}^{\prime}}{2a_{i}}\right)\,y_{m}$
$=\left(\dfrac{\lambda_{m}-{\lambda_{m}}^{\prime}}{2}\right)\,y_{m}=CV(\boldsymbol{\mathrm{z}_{m}})$
by (5.12) and (5.3). Hence $T$ is normal and by Theorem 5.6, its generic fixed
points lie on $\partial{\mathcal{B}}$. This completes the proof.

######  Remark 6.

When $dim\,(K)>1$, then there exist non-trivial orthogonal vectors amongst and
between the collection of time like and light like vectors. For example, if
$H=K=\mathbb{C}^{2}$ with standard orthonormal basis $\\{e_{1},\,e_{2}\\}$
then $(e_{1},\,e_{1})$ and $(e_{2},\,e_{2})$ are two light like orthogonal
vectors. $(0,\,e_{1})$ and $(0,\,e_{2})$ are two orthogonal time like vectors
and $(e_{1},\,e_{1})$ and $(0,\,e_{2})$ are two orthogonal light like and time
like vectors.

The following example discussed in [3], page 55 shows that in general
holomorphic automorphisms in a $C^{*}$-algebra may not possess fixed points.
Consider the $C^{*}$-algebra of all complex valued continuous functions
defined on $\overline{\Delta}=\\{x\in\mathbb{C}\,\,\,\,|\,\,\,\,|x|\leq 1\\}$
with $\|.\|_{\infty}$ norm. Let ${B}^{\prime}$ be its open unit ball. Then the
function $F\in Aut({B}^{\prime})$ defined by
$F(f)(x)=\dfrac{f(x)-\frac{1}{2}x}{1-\frac{1}{2}\overline{x}f(x)}$ doesnot
have a fixed point in $\overline{{B}^{\prime}}$.

### Acknowledgement

The authors acknowledge partial support from the TARE project grant
TAR/2019/000379. During the course of this work, Rachna acknowledges support
from CSIR-SRF grant.

Part of this work was carried out when two of the authors (R.A and M.M.M)
visited IISER Mohali. They acknowledge hospitality and support from IISER
Mohali during their stay there.

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