# Operator moment dilations as block operators

B. V. Rajarama Bhat  Statistics and Mathematics Unit, Indian Statistical
Institute, R. V. College Post, Bangalore 560059, India<EMAIL_ADDRESS>,
Anindya Ghatak  Statistics and Mathematics Unit, Indian Statistical Institute,
R. V. College Post, Bangalore 560059, India<EMAIL_ADDRESS>and
Santhosh Kumar Pamula  Department of Mathematical Sciences, Indian Institute
of Science Education and Research (IISER) Mohali, SAS Nagar, Manauli Post,
Punjab 140306, India<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{H}$ be a complex Hilbert space and let
$\big{\\{}A_{n}\big{\\}}_{n\geq 1}$ be a sequence of bounded linear operators
on $\mathcal{H}$. Then a bounded operator $B$ on a Hilbert space
$\mathcal{K}\supseteq\mathcal{H}$ is said to be a dilation of this sequence if

$A_{n}=P_{\mathcal{H}}B^{n}|_{\mathcal{H}}\;\text{for all}\;n\geq 1,$

where $P_{\mathcal{H}}$ is the projection of $\mathcal{K}$ onto $\mathcal{H}.$
The question of existence of dilation is a generalization of the classical
moment problem. We recall necessary and sufficient conditions for the
existence of self-adjoint, isometric and unitary dilations and present block
operator representations for these dilations. For instance, for self-adjoint
dilations one gets block tridiagonal representations similar to the classical
moment problem.

Given a positive invertible operator $A$, an operator $T$ is said to be in the
$\mathcal{C}_{A}$-class if the sequence
$\\{A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}:n\geq 1\\}$ admits a unitary
dilation. We identify a tractable collection of $\mathcal{C}_{A}$-class
operators for which isometric and unitary dilations can be written down
explicitly in block operator form. This includes the well-known
$\rho$-dilations for positive scalars. Here the special cases $\rho=1$ and
$\rho=2$ correspond to Schäffer representation for contractions and Ando
representation for operators with numerical radius not more than one
respectively.

###### Key words and phrases:

Moment problem, Dilation, Block operator, Poisson transform, Positive-definite
kernel, Toeplitz operator, Hankel operator

###### 2010 Mathematics Subject Classification:

46L07, 47A12, 47A20, 47A57, 47B35

##  1\. Introduction

Starting from the pioneering work of Sz.-Nagy ([25]), the dilation theory has
played a fundamental role in operator theory. This has been explored by many.
The early works mostly used techniques from classical function theory and
basic operator theory. Thanks to W. Arveson, V. Paulsen, and others ([28]) we
have new tools coming from the theory of completely positive maps. The
literature here is vast and so we simply refer to a recent survey of dilation
theory by Orr Shalit [35] and the references therein.

Let $\mathcal{H}$ be a complex Hilbert space and let
$\big{\\{}A_{n}\big{\\}}_{n\geq 1}$ be a sequence of bounded linear operators
on $\mathcal{H}$. Then a bounded operator $B$ on a Hilbert space
$\mathcal{K}\supseteq\mathcal{H}$ is said to be a dilation if

$A_{n}=P_{\mathcal{H}}B^{n}|_{\mathcal{H}}\;\text{for all}\;n\geq 1.$

Generally it is convenient to include $n=0$ in the sequence, where we would be
taking $A_{0}=I$ (the identity operator of the Hilbert space). The operator
moment problem is to determine conditions on the sequence of operators to
ensure existence of a dilation $B$ with prescribed property, for instance we
may demand $B$ to be unitary/isometry/self-adjoint/positive or we may require
the spectrum of $B$ to be contained in a given subset of complex plane.

Two special cases of this problem are very well-known. A famous result of
Sz.-Nagy tells us that if $A_{n}=A^{n},n\geq 0$ for some operator $A$, then it
admits a unitary dilation if and only if $A$ is a contraction.

Another special situation is when ${\mathcal{H}}$ is one dimensional so that
$A_{n}$ is a sequence of scalars and $B$ is a self-adjoint operator on
${\mathcal{K}}$. This amounts to requiring

$A_{n}=\langle u,B^{n}u\rangle=\int_{\sigma(B)}x^{n}d\mu(x)$

where $u$ is a unit vector in ${\mathcal{K}}$ and $\mu$ is the spectral
measure of $B$ coming from the vector state $\langle u,(\cdot)u\rangle.$ In
other words, this is the classical moment problem of seeking a measure with
specified moments.

The general operator moment problem is less well-known, but if we start
searching we find considerable literature scattered here and there. This
problem was already thought of by Sz.-Nagy ([26]). Some further references can
be found in [13, 21, 39]). Also, we refer an excellent book of V. I. Paulsen
[28], where the author discusses the dilation problem for operator valued
sequences, mostly through various exercises.

One of the main purpose of this article is to present some of the basic
results in the field in a unified way. Some of the early literature mentioned
above were inspired by classical moment problems and typically use similar
methods. Here in we mostly use modern techniques coming from the theory of
completely positive maps or we use the method of positive kernels to show the
existence of the dilation. Another aspect we emphasize is the existence of
dilations as block operator matrices, the kind got by Schäffer for Sz.-Nagy
dilation. This seems to be a fairly general phenomenon. This is very useful as
it helps us to visualize the dilation. For the powers of a contraction, the
Schäffer construction gives the dilation as a $2\times 2$ block operator
perturbation of the bilateral shift. In the last Section we have $4\times 4$
block operator perturbations of the bilateral shift appearing as dilation
operators.

Throughout the article, $\mathcal{H}$ denotes a complex Hilbert space and we
follow Physicists convention by considering the inner product
$\langle\cdot,\cdot\rangle$ on $\mathcal{H}$ as linear in the second variable
and anti-linear in the first variable. Also note that
$\mathcal{B}(\mathcal{H})$ denotes the space of all bounded linear operators
on $\mathcal{H}$ and $C(X)$ is the $C^{\ast}$-algebra of all continuous
functions on a compact Hausdorff space $X.$ In general, we denote
$C^{\ast}$-algebras by $\mathcal{A},\mathcal{B}$ etc. In particular, the
$C^{\ast}$ algebra of all $n\times n$ matrices with entries from $\mathcal{A}$
is denoted by $M_{n}(\mathcal{A})$. For a subset $M$ of $\mathcal{H}$, the
subspace $[M]:=\overline{\rm span}(M)$ is the smallest closed subspace of
$\mathcal{H}$ containing $M$.

######  Definition 1.1 (Dilation of operator sequences).

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of bounded linear operators on a
Hilbert space $\mathcal{H}$ with $A_{0}=I$. The operator sequence
$\\{A_{n}\\}_{n\geq 0}$ is said to admit a _dilation_ if there exist a bounded
linear operator $B$ on a Hilbert space $\mathcal{K}\supseteq\mathcal{H}$ such
that

$A_{n}=P_{\mathcal{H}}B^{n}\big{|}_{\mathcal{H}},\;\text{~{}for all~{}}\;n\geq
0.\;$ (1)

Then $B$ is called the dilation. A dilation is said to be positive/self-
adjoint/isometry/unitary if $B$ has that property. A positive/self-
adjoint/isometric dilation is said to be minimal if

${\mathcal{K}}=\overline{\rm
span}\\{B^{n}(\mathcal{H}):n\in{\mathbb{Z}_{+}}\\}.$ (2)

Some clarifications are in order. Here by convention for any operator $B$,
$B^{0}$ is taken as identity. So the condition
$A_{0}=P_{\mathcal{H}}B^{0}\big{|}_{\mathcal{H}}$ is superfluous once we take
$A_{0}=I.$ Therefore we are effectively just considering the dilation of
$\\{A_{n}\\}_{n\in{\mathbb{N}}}$. However, for minimality it is important to
include $n=0$ in Equation (2). In the case of unitary dilations, the
minimality condition in Equation (2) should be replaced by

${\mathcal{K}}=\overline{\rm span}\\{B^{n}(\mathcal{H}):n\in{\mathbb{Z}}\\}.$
(3)

######  Problem 1.2.

Given a sequence of operators we wish to consider dilations with prescribed
property of being positive, self-adjoint etc. Some natural questions that
arise are the following:

  1. (1)

Existence: What are necessary and sufficient conditions for existence of
dilations with prescribed property?

  2. (2)

Uniquenss: When a dilation with prescribed property exists, is it possible to
prove uniqueness of dilation up to unitary equivalence under the assumption of
minimality?

  3. (3)

Construction: Can we explicitly construct these dilations instead of just
abstractly proving their existence and uniqueness.

The question about uniqueness is easy to answer. For easy reference we state
it as a theorem.

######  Theorem 1.3.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of bounded operators on a Hilbert
space $\mathcal{H}$ admitting a self-adjoint/positive/isometric/unitary
dilation $B$ on a Hilbert space $\mathcal{K}\supseteq\mathcal{H}.$ Then the
given operator sequence admits a minimal dilation with the same prescribed
property. Moreover, such a minimal dilation is unique up to unitary
equivalence.

###### Proof.

Suppose $B$ is a self-adjoint dilation. Then $B$ restricted to
$\overline{\mbox{span}}\\{B^{n}h:h\in\mathcal{H},n\in\mathbb{Z}_{+}\\}$ is
again a self-adjoint dilation. Moreover, the inner products, $\langle
B^{m}g,B^{n}h\rangle,g,h\in\mathcal{H},m,n\in\mathbb{Z}_{+}$ are completely
determined by the given sequence $\\{A^{n}:n\in\mathbb{Z}_{+}\\}.$ This shows
the uniqueness of minimal dilation up to unitary equivalence. Clearly the same
statement holds for positive and isometric dilations. For unitary dilations we
change $\mathbb{Z}_{+}$ by $\mathbb{Z}$ and we have the analogous result. ∎

The first question is more delicate. It is obvious that given an arbitrary
operator valued sequence $\\{A_{n}\\}_{n\geq 0},$ there may not exist $B$ such
that dilation Equation (1) holds. Some necessary conditions follow easily. We
list the following:

  1. (1)

The sequence $\\{A_{n}\\}_{n\in{\mathbb{Z}}_{+}}$ should satisfy the growth
bound $\|A_{n}\|\leq M^{n}$ for some $M>0.$

  2. (2)

For $B$ to be positive (resp. self-adjoint), $A_{n}$’s should be positive
(resp. self-adjoint). For $B$ to be isometric or unitary, $A_{n}$’s should be
contractive.

The condition $(1)$ is natural as we are looking for dilations which are
bounded operators. The condition $(2)$ is also obvious.

We now briefly describe the plan of the article. We recall and summarize some
known answers to the first question of Problem 1.2 in Theorem 2.1, Theorem
2.4, Theorem 2.6, Theorem 3.1, and Theorem 4.1. To be more precise, in Theorem
2.1, we provide necessary and sufficient conditions of operator valued moment
sequence that admit self-adjoint dilation. This result was initially obtained
by Sz.-Nagy [26]. However, we provide a contemporary approach of dilation
using the theory of completely positive maps. In Theorem 2.1, and Theorem 2.4,
we have necessary and sufficient criteria for self-adjoint dilation problem in
terms of Hankel matrices. This can be treated as the operator analog of the
Hamburger moment problem. Finding necessary and sufficient conditions for an
operator sequence to have positive dilation referred to the Hausdorff moment
problem. In Theorem 2.6, we obtain necessary and sufficient conditions for
operator valued moment sequences to have positive dilation. In Theorem 3.1, we
have necessary and sufficient condition for an operator valued sequence to
have unitary dilation. This can be treated as an operator analog of Toeplitz
moment problem.

The main focus of this article is the third question. We try to obtain block
operator forms for various classes of dilations. In Theorem 2.9, we show that
if $\\{A_{n}\\}_{n\geq 0}$ admits self-adjoint dilation (say $B$) then $B$ has
a tri-diagonal form (see Theorem 2.8) whose blocks are given by the recursive
relation (see Section 2). In Theorem 3.2, we show that if $\\{A_{n}\\}_{n\geq
0}$ admits isometric dilation (say $V$), then $V$ has the form as in Equation
(27) and blocks of $V$ are given by the recursive relation described in
Section 3.

Then we focus on operator valued sequences of so called
$\mathcal{C}_{A}$-class that admit isometric dilations. We present several
necessary and sufficient conditions for an operator that belong to
$\mathcal{C}_{A}$-class (see Theorem 4.1 and Theorem 4.2). In the final
section, a special sub-class of $\mathcal{C}_{A}$-class operators for which we
can write down isometric and unitary dilations explicitly in block operator
form has been studied (see Theorem 5.3, and Theorem 5.5). Moreover, we
describe their minimal dilation spaces (see Proposition 5.4 and Remark 5.6).

##  2\. Self-adjoint and positive dilations

###  2.1. Classical moment problems

This is a well known topic in analysis. The subject has received considerable
attention over the years, beginning with the pioneering efforts of Stieltjes,
Riesz, Hamburger, Hausdorff and Kreĭn, followed by the works of Haviland,
Akhiezer, Fuglede, Berg, Atzmon, and many others (see for example [4, 32] and
references therein). For the convenience of the reader we recall a few basic
results from this theory, which are relevant for our current discussion.

Given a sequence $\\{m_{n}:n\geq 0\\}$ of real numbers, the moment problem
discusses the existence of a measure $\mu$ supported on a set
$K\subseteq\mathbb{R}$ such that

$m_{n}=\int_{K}x^{n}\;d\mu(x),~{}~{}\forall n\geq 0.$ (4)

Determining the existence of such a measure $\mu$ is known as $K$-moment
problem [33]. Three specific choices of $K$ stand out due to their natural
importance and for historical reasons.

For a real sequence $\\{m_{n}:\;n\geq 0\\},$ the _Hamburger moment problem_ is
determining as when does there exist a positive Radon measure $\mu$ on
$\mathbb{R}$ such that for all $n\geq 0,$ the integral
$\int\limits_{-\infty}^{\infty}x^{n}\;d\mu(x)$ converges and satisfies

$m_{n}=\int\limits_{-\infty}^{\infty}x^{n}\;d\mu(x).$

Similarly, the well known _Stieltjes moment problem_ and _Hausdorff moment
problem_ ask for existence of such measures $\mu$ (see Equation (4)) supported
on $[0,\infty)$ and $[0,1]$ respectively. Equivalent conditions for the
existence of solutions of these moment problems are very well known [4, 32]
(also see references therein). In fact, for each $n\geq 0$, the associated
Hankel matrices of the moment sequence are defined by

$\displaystyle H_{n}$
$\displaystyle=\begin{bmatrix}m_{0}&m_{1}&\cdots&m_{n}\\\
m_{1}&m_{2}&\cdots&m_{n+1}\\\ \vdots&\vdots&\ddots&\vdots\\\
m_{n}&m_{n+1}&\cdots&m_{2n}\end{bmatrix}_{(n+1)\times(n+1)},$ $\displaystyle
H^{(1)}_{n}$ $\displaystyle=\begin{bmatrix}m_{1}&m_{2}&\cdots&m_{n+1}\\\
m_{2}&m_{3}&\cdots&m_{n+2}\\\ \vdots&\vdots&\ddots&\vdots\\\
m_{n+1}&m_{n+2}&\cdots&m_{2n+1}\end{bmatrix}_{(n+1)\times(n+1)}.$

It is very well known that the Hamburger moment problem has a solution if and
only if the associated Hankel matrix

$\displaystyle H_{n}\geq 0\text{~{}for every~{}}n\geq 0.$

Further, the Stieltjes moment problem has a solution if and only if the
associated Hankel matrices

$\displaystyle H_{n}\geq 0\text{~{}and~{}}H_{n}^{(1)}\geq 0\text{~{}for
every~{}}n\geq 0.$

Furthermore, the Hausdorff moment problem has a solution if and only if the
sequence $\\{m_{n}:\;n\geq 0\\}$ is completely monotonic i.e.,

$(-1)^{k}(\Delta^{k}m)_{n}\geq 0\text{~{}for every }n,k\geq 0,$

where

$(\Delta^{k}m)_{n}=\sum\limits_{i=0}^{k}{k\choose i}(-1)^{i}m_{i+n}.$

One may also consider measures supported on subsets of the complex plane. For
instance see [3].

Coming to dilations of operator sequences, clearly the starting point is the
following famous theorem of Sz.-Nagy. Let $T\in\mathcal{B(H)}$ and consider
the operator valued sequence $\\{T^{n}\\}_{n\geq 0}$. This sequence admits a
minimal isometric or unitary dilation if and only if $T$ is a contraction.
Moreover, the minimal dilations are unique up to unitary equivalence. ( See
[25] or [28, Theorem 1.1]). We call them as dilations (or power dilations) of
$T$. In 1955, Schäffer [34] provided an explicit construction of minimal
isometric and unitary dilations as follows: Let
$D_{T}=(I-T^{*}T)^{\frac{1}{2}},D_{T^{*}}=(I-TT^{*})^{\frac{1}{2}}$ and
$\mathcal{D}_{T}:=\overline{\rm range}D_{T},$
$\mathcal{D}_{T^{*}}=\overline{\rm range}D_{T^{*}}$. Take

$\mathcal{K}:=\mathcal{H}\oplus\mathcal{D}_{T}\oplus\mathcal{D}_{T}\oplus\cdots.$

Then

$V=\begin{bmatrix}T&0&0&\cdots\\\ D_{T}&0&0&\cdots\\\ 0&I&0&\cdots\\\
0&0&I&\cdots\\\ \vdots&\vdots&\ddots&\ddots\end{bmatrix}$

on $\mathcal{K}$ is a minimal isometric dilation of $T$. Take
$\mathcal{L}=\cdots\oplus\mathcal{D}_{T^{*}}\oplus\mathcal{D}_{T^{*}}\oplus\mathcal{H}\oplus\mathcal{D}_{T}\oplus\mathcal{D}_{T}\oplus\cdots$
and define $U$ on $\mathcal{L}$ by

$U=\left(\begin{array}[]{ccccccc}&&&&&&\\\ \ddots&&&&&&\\\ &I&&&&&\\\
&&I&&&&\\\ &&&D_{T^{*}}&{\bf T}&&\\\ &&&-T^{*}&D_{T}&&\\\ &&&&&I&\\\
&&&&&&\ddots\end{array}\right).$

The bold font indicates the location of operator $T$ from $\mathcal{H}$ to
$\mathcal{H}$ and $0$ entries are not displayed. In this article we provide
Schäffer type constructions for several operator valued moment sequences.

###  2.2. Self-adjoint dilations

In 1952, Sz-Nagy obtained the following necessary and sufficient condition for
a sequence of operators to admit self-adjoint dilation with spectrum contained
in a given compact set (see Theorem 2.1). For reader’s convenience we present
this result using the theory of completely positive maps. This method seems to
have become a standard method to prove existence of operator dilations.

######  Theorem 2.1.

[26] Let $X\subseteq\mathbb{R}$ be a compact set. Let $\\{A_{n}:\;n\geq 0\\}$
be a sequence of bounded self-adjoint operators on a Hilbert space
$\mathcal{H}$ with $A_{0}=I$. It admits a self-adjoint operator dilation $B$
with $\sigma(B)\subseteq X$ if and only if

$c_{0}+c_{1}A_{1}+\cdots+c_{n}A_{n}\geq 0,\;$ (5)

whenever the complex polynomial $c_{0}+c_{1}x+\cdots+c_{n}x^{n}\geq 0\;$ for
all $x\in X.$

###### Proof.

Suppose Equation (5) is satisfied. Then by the functional calculus,

$\langle g,\;(c_{0}+c_{1}A_{1}+c_{2}A_{n}+\cdots+c_{n}A_{n})g\rangle=\langle
g,(c_{0}+c_{1}B+c_{2}B^{2}+\cdots+c_{n}B^{n})g\rangle\geq 0,$

for every $g\in\mathcal{H}$ whenever $\sum\limits_{i=0}^{n}c_{i}x^{i}$ is
positive on $X$. To prove the converse, let us define a map $\varphi\colon
C(X)\to\mathcal{B}(\mathcal{H})$ by

$\varphi(x^{n})=A_{n},~{}~{}\text{ for all }n\geq 0,$

where $C(X)$ is the $C^{*}$-algebra of continuous functions on $X$. Let
$\mathcal{P}(X)$ denote the algebra of all polynomials over $X$, then
$\mathcal{P}(X)$ is a $\ast$-subalgebra of $C(X)$, it separates points of $X$
and hence $\mathcal{P}(X)$ is dense in $C(X)$ by Stone-Weierstrass theorem.
From the hypothesis, it follows that $\varphi(p)\geq 0$ whenever
$p\in\mathcal{P}(X)$ is positive. So $\varphi$ can be extended to a positive
map on $C(X)$. Thus $\varphi$ is completely positive as $C(X)$ is a
commutative $C^{\ast}$-algebra. Therefore by Stinespring’s theorem, there is a
Hilbert space $\mathcal{K}$, an isometry $V\colon\mathcal{H}\to\mathcal{K}$
and a unital $\ast$-homomorphism $\pi\colon C(X)\to\mathcal{B}(\mathcal{K})$
such that

$\varphi(a)=V^{\ast}\pi(a)V,\;\text{for all}\;a\in\mathcal{A}.$

Let $\pi(x)=B$. Then $B^{\ast}=\pi(x)^{\ast}=\pi(x^{\ast})=\pi(x)=B$ since
$X\subset\mathbb{R}$. This implies that

$A_{n}=\varphi(x^{n})=V^{\ast}\pi(x^{n})V=V^{\ast}\pi(x)^{n}V=V^{\ast}B^{n}V=P_{V(\mathcal{H})}B^{n}\big{|}_{V(\mathcal{H})},~{}~{}\text{
for all }n\geq 0.$

Identifying $h\in\mathcal{H}$ with $Vh\in V(\mathcal{H})$ the proof is
complete. ∎

######  Remark 2.2.

It follows from the proof of Theorem 2.1 that if a sequence $\\{A_{n}:\;n\geq
0\\}$ admits self-adjoint dilation then

$\displaystyle
A_{2}-A_{1}^{2}=\varphi(x^{2})-\varphi(x)^{2}=V^{\ast}\pi(x)\big{[}I-VV^{\ast}\big{]}\pi(x)V\geq
0,$

since $V$ is an isometry.

The existence of self-adjoint dilation for an operator sequence is also linked
with positivity of the corresponding Hankel matrix. To see this we begin with
the following observation.

######  Lemma 2.3.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence in $\mathcal{B}(\mathcal{H})$ with
$A_{0}=I$. Then $\sum\limits_{i,j=0}^{n}X_{i}^{\ast}A_{i+j}X_{j}\geq 0$ for
any $X_{0},X_{1},\cdots,X_{n}\in\mathcal{B}(\mathcal{H}),n\geq 0$ if and only
if the associated Hankel matrix

$H_{n}:=\begin{bmatrix}I&A_{1}&A_{2}&\cdots&A_{n}\\\
A_{1}&A_{2}&A_{3}&\cdots&A_{n+1}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
A_{n}&A_{n+1}&A_{n+2}&\cdots&A_{2n}\end{bmatrix}\geq 0,\;\;\text{for
all}\;\;n\geq 0.$

###### Proof.

Suppose that $H_{n}\geq 0$ for all $n\geq 0$ and let
$X_{0},X_{1},\cdots,X_{n}\in\mathcal{B}(\mathcal{H})$. For every
$g\in\mathcal{H}$, we see that

$\displaystyle\big{\langle}g,\;\Big{(}\sum\limits_{i,j=0}^{n}X_{i}^{\ast}A_{i+j}X_{j}\Big{)}g\big{\rangle}$
$\displaystyle=\sum\limits_{i,j=0}^{n}\langle
X_{i}{g},\;A_{i+j}X_{j}{g}\rangle=\langle\widetilde{g},H_{n}\widetilde{g}\rangle\geq
0,$

where $\widetilde{g}=\begin{bmatrix}X_{0}g\\\ \vdots\\\ X_{n}g\end{bmatrix}$.
To prove the converse, choose and fix $g\in\mathcal{H}$ with $\|g\|=1.$ Now
for $h_{0},h_{1},\cdots,h_{n}\in\mathcal{H}$, take $X_{i}=|h_{i}\rangle\langle
g|\;(0\leq i\leq n)$. Then

$\displaystyle\Big{\langle}\begin{bmatrix}h_{0}\\\ \vdots\\\
h_{n}\end{bmatrix},\;H_{n}\begin{bmatrix}h_{0}\\\ \vdots\\\
h_{n}\end{bmatrix}\Big{\rangle}$ $\displaystyle=\sum\limits_{i,j=0}^{n}\langle
h_{i},\;A_{i+j}h_{j}\rangle$
$\displaystyle=\Big{\langle}g,\;\Big{(}\sum\limits_{i,j=0}^{n}|h_{i}\rangle\langle
g|^{\ast}A_{i+j}|h_{j}\rangle\langle g|\Big{)}g\Big{\rangle}$
$\displaystyle=\langle g,\sum_{i,j=0}^{n}X_{i}^{*}A_{i+j}X_{j}g\rangle$
$\displaystyle\geq 0.$

∎

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of self-adjoint operators acting on
a Hilbert space $\mathcal{H}$. Consider the associated Hankel matrices:

$H_{n}:=\begin{bmatrix}I&A_{1}&A_{2}&\cdots&A_{n}\\\
A_{1}&A_{2}&A_{3}&\cdots&A_{n+1}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
A_{n}&A_{n+1}&A_{n+2}&\cdots&A_{2n}\end{bmatrix},H^{(2)}_{n}:=\begin{bmatrix}A_{2}&A_{3}&A_{4}&\cdots&A_{n+2}\\\
A_{3}&A_{4}&A_{5}&\cdots&A_{n+3}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
A_{n+2}&A_{n+3}&A_{n+4}&\cdots&A_{2n+2}\end{bmatrix}.$

In the following theorem we use both the method of completely positive maps as
well as that of positive kernels.

######  Theorem 2.4.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of self-adjoint operators with
$A_{0}=I$ and $\|A_{n}\|\leq 1$ for all $n$. Then it admits a self-adjoint
contraction dilation if and only if $H_{n}\geq 0$ and $H^{(2)}_{n}\leq H_{n}$
for each $n.$

###### Proof.

Suppose there exists a self-adjoint contraction dilation $B$ on a Hilbert
space $\mathcal{K}\supseteq\mathcal{H}$. Then by the proof of Theorem 2.1,
there exists a completely positive map $\varphi:C[-1,1]\to\mathcal{B(H)}$ such
that $\varphi(x^{m})=A_{m}$ for each $m\geq 0.$ Consider the element $L\in
M_{n+1}(C[-1,1])$,where

$L=\begin{bmatrix}1&x&x^{2}&\cdots&x^{n}\\\ 0&0&0&\cdots&0\\\
0&0&0&\cdots&0\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
0&0&0&\cdots&0\end{bmatrix}.$

Then $L^{*}L\geq 0$ in $M_{n+1}(C[-1,1])$. Since
$H_{n}=\varphi_{n+1}(L^{*}L),$ and $\varphi$ is completely positive,
$H_{n}\geq 0.$ We define $G\in M_{n+1}(C[-1,1])$ by

$G=\begin{bmatrix}\sqrt{1-x^{2}}&x\sqrt{1-x^{2}}&\cdots&x^{n}\sqrt{1-x^{2}}\\\
0&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\cdots&\vdots\\\
0&0&\cdots&0\end{bmatrix}.$

Then $G^{*}G\geq 0$ in $M_{n+1}(C[-1,1])$. Since
$H_{n}-H^{(2)}_{n}=\varphi_{n+1}(G^{*}G),$ and $\varphi$ is completely
positive, $H_{n}-H^{(2)}_{n}\geq 0.$

Conversely, assume that $H_{n}\geq 0$ and $H^{(2)}_{n}\leq H_{n}$ for each
$n.$ Let $M:=\mathbb{Z}_{+}\times\mathcal{H}$. Define a map $k\colon M\times
M\to\mathbb{C}$ by

$k((m,g),(n,h))=\langle g,\;A_{m+n}h\rangle,\;\;\text{for
every}\;m,n\in\mathbb{Z}_{+},\;g,h\in\mathcal{H}.$ (6)

Since $H_{n}\geq 0$ for all $n\geq 0,$ $k$ is a positive definite kernel. Let
$V$ be a vector space of all complex functions on $M$ which is zero except for
finitely many points of $M$. Since $k$ is positive definite, $V$ is a semi-
inner product space with respect to:

$\langle\xi,\eta\rangle:=\sum\limits_{x,y\in
M}\overline{\xi(x)}\eta(y)k(x,y)\;\text{for every}\;\xi,\eta\in V.$

Let us take $\mathcal{N}=\\{\xi\in V:\;\langle\xi,\xi\rangle=0\\}$. Then by
Cauchy-Schwarz inequality,
$\mathcal{N}=\\{\xi:\;\langle\xi,\eta\rangle=0,\;\text{for every}\;\eta\in
V\\}$ is subspace of $V$. Take $\mathcal{K}$ as the Hilbert space obtained by
the completion of the quotient space $V/\mathcal{N}$. Define $\lambda\colon
M\to\mathcal{K}$ by

$\lambda(m,g)=\delta_{(m,g)}+\mathcal{N},\;\text{for all}\;(m,g)\in M.$

Since $\langle\lambda(0,g),\lambda(0,h)\rangle=\langle g,h\rangle$ for every
$g,h\in\mathcal{H}$, we see that $\mathcal{H}$ can be identified as a subspace
of $\mathcal{K}$ via the map $g\mapsto\lambda(0,g).$ Moreover,
$\mathcal{K}=\overline{\rm span}\\{\lambda(m,g):\;m\geq
0,\;g\in\mathcal{H}\\}$. Define $B(\lambda(m,g))=\lambda(m+1,g)$ for every
$(m,g)\in M$. Then

$\displaystyle\|B\Big{(}\sum\limits_{i=0}^{n}{c_{i}}\lambda(m_{i},g_{i})\Big{)}\|^{2}$
$\displaystyle=\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle\lambda(m_{i}+1,g_{i}),\;\lambda(m_{j}+1,g_{j})\rangle$
$\displaystyle=\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle
g_{i},\;A_{m_{i}+m_{j}+2}(g_{j})\rangle$
$\displaystyle\leq\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle
g_{i},\;A_{m_{i}+m_{j}}(g_{j})\rangle$
$\displaystyle=\big{\|}\sum\limits_{i=0}^{n}{c_{i}}\lambda(m_{i},g_{i})\big{\|}^{2}.$

This implies that $B$ extends to a linear contraction. It is easy to see that
it is self-adjoint. Moreover, for every $g,h\in\mathcal{H}$,

$\displaystyle\langle\lambda(0,g),P_{\mathcal{H}}B^{n}|_{\mathcal{H}}\lambda(0,h)\rangle=\langle\lambda(0,g),\;\lambda(n,h)\rangle=\langle
g,A_{n}h\rangle.$

Hence $B$ is a self-adjoint dilation of $\\{A_{n}\\}_{n\geq 0}.$ ∎

If a sequence $\\{A_{n}:n\geq 0\\}$ with $A_{0}=I$ admits a self-adjoint
dilation and suppose $A_{2}=A_{1}^{2}$. Then from the characterizations above
it is possible to see that $A_{n}=A_{1}^{n}$ for every $n\geq 1$. In fact,
much stronger results are known now (see [27] for more details).

###  2.3. Positive dilations

Here it is convenient to have the following standard definition.

######  Definition 2.5.

A sequence $\\{A_{n}\\}_{n\geq 0}$ of bounded operators on a Hilbert space
$\mathcal{H}$ is called completely monotone if $(-1)^{k}(\Delta^{k}A)_{n}\geq
0\text{ for every }n,k\geq 0,$ where

$(\Delta^{k}A)_{n}=\sum\limits_{i=0}^{k}{k\choose i}(-1)^{i}A_{i+n}.$

In the classical setup, the Hausdorff moment problem has a solution if and
only if the given sequence is completely monotone. The same result holds true
for operator sequences also. We prove the result here by defining an
appropriate positive definite kernel.

######  Theorem 2.6.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of positive operators in
$\mathcal{B}(\mathcal{H})$ with $A_{0}=I$. Then it is completely monotone if
and only if it admits a positive contraction dilation.

###### Proof.

Though the existence of such positive contraction is clear from the Theorem
2.1 when $X=[0,1]$, we provide an explicit proof for the construction of $B$.
Suppose that $\\{A_{n}:\;n\geq 0\\}$ is completely monotone and let
$M:=\mathbb{Z}_{+}\times\mathcal{H}$. Define the map $k\colon M\times
M\to\mathbb{C}$ by

$k((m,g),(n,h))=\langle g,\;A_{m+n}h\rangle,\;\;\text{for
every}\;m,n\in\mathbb{Z}_{+},\;g,h\in\mathcal{H}.$ (7)

Now, we show that $k$ is a positive definite kernel. Equivalently, it is
enough to show that

$H_{n}=\begin{bmatrix}I&A_{1}&A_{2}&\cdots&A_{n}\\\
A_{1}&A_{2}&A_{3}&\cdots&A_{n+1}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
A_{n}&A_{n+1}&A_{n+2}&\cdots&A_{2n}\end{bmatrix}\geq 0\;\text{for
every}\;n\geq 1.$

Let us define the map $\Phi(x^{n})=A_{n}$ for every $n$. By mathematical
induction it can be shown that every polynomial $q(x)$ of degree $n$ with real
coefficients over $[0,1]$ can be written using Bernstein polynomial (see [4,
Page 76]) as follows:

$\sum\limits_{i=0}^{N}{N\choose
i}x^{i}(1-x)^{N-i}q\Big{(}\frac{i}{N}\Big{)}=q(x)+\sum\limits_{j=1}^{n-1}\frac{u_{j}(x)}{N^{j}}\;\text{for
every}\;N\geq 1,$ (8)

where $u_{j}(x)$ is polynomial of degree less than or equal to $n$ and is
independent of $N.$ Since the sequence is completely monotone, for every
$k,n\geq 0$, we see that

$\Phi(x^{n}(1-x)^{k})=\Phi\Big{(}\sum\limits_{i=0}^{k}{k\choose
i}(-1)^{i}x^{i+n}\Big{)}=\sum\limits_{i=0}^{k}{k\choose
i}(-1)^{i}\Phi(x^{i+n})=\sum\limits_{i=0}^{k}{k\choose i}(-1)^{i}A_{i+n}\geq
0.$

If $q(x)$ is a positive polynomial over $[0,1]$, it follows from Equation (8)
that

$\displaystyle\Phi(q(x))$ $\displaystyle=\sum\limits_{i=0}^{N}{N\choose
i}\;\Phi(x^{i}(1-x)^{N-i})q\Big{(}\frac{i}{N}\Big{)}-\sum\limits_{j=1}^{n-1}\frac{\Phi(u_{j}(x))}{N^{j}}$
$\displaystyle\geq 0-\sum\limits_{j=1}^{n-1}\frac{\Phi(u_{j}(x))}{N^{j}}$
$\displaystyle\rightarrow 0,\;\;\;\text{as}\;N\to\infty.$

This shows that $\Phi$ is a positive map. Since $\pm q\leq\|q\|_{\infty},$ it
implies that $\Phi(q)\leq\|q\|_{\infty}\Phi(1)$ and thus $\Phi$ is continuous
on the space of polynomials over $[0,1].$ Therefore $\Phi$ is a positive map
on $C([0,1])$ and so it is completely positive map. For every $n\geq 0$, we
have $H_{n}=\Phi_{n+1}(L^{\ast}L)\geq 0,$ where

$L=\begin{bmatrix}1&x&x^{2}&\cdots&x^{n}\\\ 0&0&0&\cdots&0\\\
0&0&0&\cdots&0\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
0&0&0&\cdots&0\end{bmatrix}.$

Hence $k$ is a positive definite kernel. Furthermore, if we denote (for $n\geq
0$)

$H^{(1)}_{n}:=\begin{bmatrix}A_{1}&A_{2}&A_{3}&\cdots&A_{n+1}\\\
A_{2}&A_{3}&A_{4}&\cdots&A_{n+2}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
A_{n+1}&A_{n+2}&A_{n+3}&\cdots&A_{2n+1}\end{bmatrix};\;H^{(2)}_{n}:=\begin{bmatrix}A_{2}&A_{3}&A_{4}&\cdots&A_{n+2}\\\
A_{3}&A_{4}&A_{5}&\cdots&A_{n+3}\\\ \vdots&\vdots&\vdots&\cdots&\vdots\\\
A_{n+2}&A_{n+3}&A_{n+4}&\cdots&A_{2n+2}\end{bmatrix}$ (9)

then

$H_{n}^{(1)}=\Phi_{n+1}(F^{\ast}F)\geq
0\;\text{and}\;H_{n}-H^{(2)}_{n}=\Phi_{n+1}(G^{\ast}G)\geq 0,\;$ (10)

where

$F=\begin{bmatrix}x^{\frac{1}{2}}&x^{\frac{3}{2}}&\cdots&x^{\frac{2n+1}{2}}\\\
0&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\cdots&\vdots\\\
0&0&\cdots&0\end{bmatrix},\;G=\begin{bmatrix}(1-x^{2})^{\frac{1}{2}}&x(1-x^{2})^{\frac{1}{2}}&\cdots&x^{n}(1-x^{2})^{\frac{1}{2}}\\\
0&0&\cdots&0\\\ 0&0&\cdots&0\\\ \vdots&\vdots&\cdots&\vdots\\\
0&0&\cdots&0\end{bmatrix}.$

Let $V$ be a vector space of all complex functions on $M$ which is zero except
for finitely many points of $M$. Then $V$ is a semi-inner product space with
respect to:

$\langle\xi,\eta\rangle:=\sum\limits_{x,y\in
M}\overline{\xi(x)}\eta(y)k(x,y)\;\text{for every}\;\xi,\eta\in V.$

Let us take $\mathcal{N}=\\{\xi\in V:\;\langle\xi,\xi\rangle=0\\}$. Thus by
Cauchy-Schwarz inequality, it follows that
$\mathcal{N}=\\{\xi:\;\langle\xi,\eta\rangle=0,\;\text{for every}\;\eta\in
V\\}$. Take $\mathcal{K}$ as the Hilbert space obtained by the completion of
the quotient space $V/\mathcal{N}$. Define $\lambda\colon M\to\mathcal{K}$ by

$\lambda(m,g)=\delta_{(m,g)}+\mathcal{N},\;\text{for all}\;(m,g)\in M.$

Since $\langle\lambda(0,g),\lambda(0,g)\rangle=\|g\|^{2}$ for every
$g\in\mathcal{H}$, we see that $\mathcal{H}$ can be identified as a subspace
of $\mathcal{K}$ via the map $g\mapsto\lambda(0,g).$ Moreover,
$\mathcal{K}=\overline{\rm span}\\{\lambda(m,g):\;m\geq
0,\;g\in\mathcal{H}\\}$. Define $B(\lambda(m,g))=\lambda(m+1,g)$ for every
$(m,g)\in M$. Then

$\displaystyle\Big{\langle}\sum\limits_{i=0}^{n}c_{i}\lambda(m_{i},g_{i}),\;B\Big{(}\sum\limits_{i=0}^{n}c_{i}\lambda(m_{i},g_{i})\Big{)}\Big{\rangle}$
$\displaystyle=\Big{\langle}\sum\limits_{i=0}^{n}c_{i}\lambda(m_{i},g_{i}),\;\sum\limits_{i=0}^{n}c_{i}\lambda(m_{i}+1,g_{i})\Big{\rangle}$
$\displaystyle=\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle
g_{i},\;A_{m_{i}+m_{j}+1}(g_{j})\rangle$ $\displaystyle\geq 0\;[\text{~{}since
$H^{(1)}_{n}\geq 0$ by Equation~{}}\eqref{Equation: Hankel positive
definite}]\ $

and

$\displaystyle\|B\Big{(}\sum\limits_{i=0}^{n}{c_{i}}\lambda(m_{i},g_{i})\Big{)}\|^{2}$
$\displaystyle=\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle\lambda(m_{i}+1,g_{i}),\;\lambda(m_{j}+1,g_{j})\rangle$
$\displaystyle=\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle
g_{i},\;A_{m_{i}+m_{j}+2}(g_{j})\rangle$
$\displaystyle\leq\sum\limits_{i,j=0}^{n}\overline{c_{i}}c_{j}\langle
g_{i},\;A_{m_{i}+m_{j}}(g_{j})\rangle\;[\text{~{}since $H_{n}\geq H^{(2)}_{n}$
by Equation~{}}\eqref{Equation: Hankel positive definite}]$
$\displaystyle=\big{\|}\sum\limits_{i=0}^{n}{c_{i}}\lambda(m_{i},g_{i})\big{\|}^{2}.$

This implies that $B$ defines a contractive and positive operator. Finally,
for every $g,h\in\mathcal{H}$, we see that

$\displaystyle\langle\lambda(0,g),P_{\mathcal{H}}B^{n}|_{\mathcal{H}}\lambda(0,h)\rangle=\langle\lambda(0,g),\;\lambda(n,h)\rangle=\langle
g,A_{n}h\rangle.$

Therefore, every monotone sequence admits a contractive positive dilation.
Conversely, suppose that there is a positive contraction $B$ satisfying,
$A_{n}=P_{\mathcal{H}}B^{n}\big{|}_{\mathcal{H}}$ for $n\geq 0$ then

$\displaystyle\big{\langle}g,\;\sum\limits_{i=0}^{k}{k\choose
i}(-1)^{i}A_{i+n}(g)\big{\rangle}$
$\displaystyle=\big{\langle}g,\;\sum\limits_{i=0}^{k}{k\choose
i}(-1)^{i}B^{i+n}(g)\big{\rangle}$
$\displaystyle=\big{\langle}g,\;B^{n}(I-B)^{k}(g)\rangle$
$\displaystyle=\langle(I-B)^{\frac{k}{2}}g,\;B^{n}(I-B)^{\frac{k}{2}}g\rangle$
$\displaystyle\geq 0$

for every $k,n\geq 0.$ Therefore, $\\{A_{n}\\}_{n\geq 0}$ is a completely
monotone sequence. ∎

###  2.4. Concrete self-adjoint dilations

Now we turn our discussion to a concrete construction of self-adjoint
dilation. Before that, let us recall some known facts from the literature that
a bounded self-adjoint operator say $B$ defined on a separable Hilbert space
with a unit cyclic vector $v$ can be represented by a tridiagonal matrix with
respect to the basis obtained by Gram-Schmidt process from the set
$\\{v,Bv,B^{2}v,\cdots\\}$. Suppose the matrix $B$ is expressed as,

$B=\begin{bmatrix}a_{0}&b_{0}&0&0&\cdots\\\ {b_{0}}&a_{1}&b_{1}&0&\ldots\\\
0&{b_{1}}&a_{2}&b_{2}&\ldots\\\ 0&0&{b_{2}}&a_{3}&\ddots\\\
\vdots&\vdots&\ddots&\ddots&\ddots\\\ \end{bmatrix}$ (11)

then it is related to a family of monic orthogonal polynomials given by
$p_{0}(x)=1,p_{1}(x)=(x-a_{0})$,

$p_{n}(x)=(x-a_{n-1})p_{n-1}(x)-b_{n-1}^{2}p_{n-2}(x),\;\text{for all}\;n\geq
2$ (12)

Note that here equation (12) is obtained by expanding the determinant of $(xI-
B_{n})$ using the last row, where $B_{n}$ is the $n\times n$ truncated matrix
of $B$. More information on this can be found in the well known classic on
Hilbert space linear transformations [37]. Some applications of this idea to
quantum theory with worked out examples can be seen in [12].

Let $\mu(\cdot)=\|E(\cdot)v\|^{2}$ be the probability measure defined on the
Borel $\sigma-$field of the real line, where $E$ is the spectral measure
associated to $B$. Clearly it is supported on the spectrum of $B$. Then
$\\{p_{n}(x):\;n\geq 0\\}$ is an orthonormal basis of $L^{2}(\mu)$ (for
details, see [37]). Moreover, $m$-th moment of $\mu$ is given by

$\displaystyle\langle
v,\;B^{m}v\rangle=\int\limits_{-\infty}^{\infty}\lambda^{m}\;d\langle
v,\;E_{B}(\lambda)v\rangle,\;\text{for all}\;m\geq 0.$

Conversely, given any compactly supported probability measure $\mu$ on the
real line we can take $B$ as the operator ‘multiplication by $x$’, on
$L^{2}(\mu)$ and cyclic vector $v$ as the constant function $1$. We can
observe the tridiagonal form of $B$ on the basis normalized orthogonal
polynomials. The coefficients $\\{a_{n},b_{n}:n\geq 0\\}$ are known as Jacobi
parameters of the measure $\mu.$ Here $B$ is a self-adjoint dilation of the
moment sequence of the probability measure. This motivates us to construct
such tridiagonal operator matrix $B$ for a self-adjoint dilation of an
operator sequence $\\{A_{n}:\;n\geq 0\\}.$ Such tri-diagonal blocks, known as
generalized Jacobi 3-diagonal relations are well known in quantum theory (See
[1]).

######  Lemma 2.7.

Let $B$ be a bounded operator on some Hilbert space ${\mathcal{K}}$. Let
${\mathcal{H}}$ be a closed subspace of ${\mathcal{K}}.$ Assume

${\mathcal{K}}=\overline{\mbox{s}pan}\\{B^{n}h:h\in{\mathcal{H}},n\in{\mathbb{Z}}_{+}\\}.$
(13)

Then ${\mathcal{K}}$ decomposes as a direct sum of Hilbert spaces

${\mathcal{K}}={\mathcal{H}}_{0}\oplus{\mathcal{H}}_{1}\oplus{\mathcal{H}}_{2}\oplus\cdots$

where ${\mathcal{H}}_{0}={\mathcal{H}}$ and with respect to this decomposition
the operator $B$ has the ‘upper Hessenberg’ form:

$B=\begin{bmatrix}B_{00}&B_{01}&B_{02}&B_{03}&\cdots\\\
B_{10}&B_{11}&B_{12}&B_{13}&\cdots\\\ 0&B_{21}&B_{22}&B_{23}&\cdots\\\
0&0&B_{32}&B_{33}&\cdots\\\ \vdots&\vdots&\vdots&\ddots&\ddots\\\
\end{bmatrix}$

where $\mathcal{H}_{n}=\overline{B_{n(n-1)}(\mathcal{H}_{n-1})}$ for every
$n$. Conversely any such $B$ satisfies Equation (13).

###### Proof.

Take ${\mathcal{H}}_{n]}=\overline{\rm span}\\{B^{m}h:0\leq m\leq
n-1,h\in{\mathcal{H}}\\}$ and
${\mathcal{H}}_{n}={\mathcal{H}}_{n]}\bigcap{{\mathcal{H}}_{(n-1)]}^{\perp}}$
for $n\geq 2$ with ${\mathcal{H}}_{1]}={\mathcal{H}}_{1}.$ Then clearly
${\mathcal{K}}=\oplus_{n\geq 1}{\mathcal{H}}_{n},$ and as
$B({\mathcal{H}}_{n]})\subseteq{\mathcal{H}}_{(n+1)]}$,
$B({\mathcal{H}}_{n})\subseteq\oplus_{m=0}^{n+1}{\mathcal{H}}_{m+1}$.
Consequently, the operator $B$ has the form described above. The range and
condition and the converse statements are easy to see. ∎

######  Theorem 2.8.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of self-adjoint operators in
$\mathcal{B}(\mathcal{H})$ with $A_{0}=I$, admitting a minimal self-adjoint
dilation $B$ in $\mathcal{B}(\mathcal{K})$ for some Hilbert space
$\mathcal{K}$. Then the space
$\mathcal{K}=\mathcal{H}_{0}\oplus\mathcal{H}_{1}\oplus\cdots$ (
$\mathcal{H}_{0}=\mathcal{H})$, so that the operator $B$ has the tridiagonal
form:

$B=\begin{bmatrix}B_{00}&B_{01}^{\ast}&0&0&\cdots\\\
B_{10}&B_{11}&B_{21}^{\ast}&0&\cdots\\\
0&B_{21}&B_{22}&B_{32}^{\ast}&\cdots\\\ 0&0&B_{32}&B_{33}&\cdots\\\
\vdots&\vdots&\vdots&\ddots&\ddots\\\ \end{bmatrix},$ (14)

and $\mathcal{H}_{n}=\overline{B_{n(n-1)}(\mathcal{H}_{n-1})}$ for all $n\geq
1.$

###### Proof.

From the previous lemma, $B$ has upper Hessenberg form. But since $B$ is self-
adjoint, $B_{ij}=B_{ji}^{*}=0$ for $j>(i+1).$

∎

Consider the set up of this theorem. We try to determine the blocks using the
sequence $\\{A_{n}\\}_{n\geq 0}$ and the dilation property. It is done
recursively and to do this we need to invert some operators. In general, it is
quite possible that some of these operators are not invertible and we may have
to modify the construction. For the moment we assume invertibility of
concerned operators as and when required.

Since $B$ is a self-adjoint dilation for the moment sequence
$\\{A_{n}\\}_{n\geq 0}$ we have $(B^{n})_{00}=A_{n}$ for every $n\geq 0$.
Clearly, $B_{00}=A_{1}$ and $B_{10}$ is obtained as,

$\displaystyle
A_{2}=(B^{2})_{00}=B_{00}^{2}+B_{10}^{\ast}B_{10}=A_{1}^{2}+B_{10}^{\ast}B_{10}.$

That is, $B_{10}^{\ast}B_{10}=A_{2}-A_{1}^{2}$. Since $(A_{2}-A_{1}^{2})\geq
0$ (see Remark 2.2), we can choose $B_{10}=(A_{2}-A_{1}^{2})^{1/2}.$ Firstly,
we explain the process to compute the diagonal block $B_{11}$ and then
establish the recurrence relation to obtain diagonal and lower diagonal
blocks. From the Equation (14), we have

$\displaystyle A_{3}=(B^{3})_{00}=\sum\limits_{0\leq r_{1},r_{2}\leq
2}B_{0r_{1}}B_{r_{1}r_{2}}B_{r_{2}0}=\sum\limits_{0\leq r_{1},r_{2}\leq
1}B_{0r_{1}}B_{r_{1}r_{2}}B_{r_{2}0},$

since $B_{0r_{1}}B_{r_{1}r_{2}}B_{r_{2}0}=0$ when either $r_{1}$ or $r_{2}$ is
2. This implies that

$\displaystyle A_{3}$
$\displaystyle=B_{00}^{3}+B_{00}B_{01}B_{10}+B_{01}B_{10}B_{00}+B_{01}B_{11}B_{10}$
$\displaystyle=B_{00}^{3}+B_{00}B^{\ast}_{10}B_{10}+B^{\ast}_{10}B_{10}B_{00}+B^{\ast}_{10}B_{11}B_{10}.$

By substituting the first column information we get

$B_{11}=(A_{2}-A_{1}^{2})^{-1/2}\Big{[}(A_{3}-A_{1}^{3})-A_{1}(A_{2}-A_{1}^{2})-(A_{2}-A_{1}^{2})A_{1}\Big{]}(A_{2}-A_{1}^{2})^{-1/2}.$

The formulae for diagonal and off-diagonal blocks are as follows. Firstly,
note that each diagonal block is a self-adjoint operator and it is computed by
the compression of odd powers of $B$. Suppose that $(n-1)$ columns of $B$ are
known then $B_{nn}$ is computed as follows:

$\displaystyle A_{2n-1}$ $\displaystyle=$ $\displaystyle(B^{2n-1})_{00}$
$\displaystyle=$ $\displaystyle\sum\limits_{0\leq
r_{1},r_{2},\ldots,r_{2(n-1)}\leq 2n-2}B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2(n-1)}0}$ $\displaystyle=$
$\displaystyle\sum\limits_{\underset{\&\;(r_{n-1},r_{n})\neq(n,n)}{0\leq
r_{1},r_{2},\ldots,r_{2(n-1)}\leq 2n-2}}B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2(n-1)}0}+B_{01}\cdots B_{(n-1)n}B_{nn}B_{n(n-1)}\cdots B_{10}$
$\displaystyle=$
$\displaystyle\sum\limits_{\underset{\&\;(r_{n-1},r_{n})\neq(n,n)}{0\leq
r_{1},r_{2},\ldots,r_{2(n-1)}\leq 2n-2}}B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2(n-1)}0}+B_{10}^{\ast}\cdots B^{\ast}_{n(n-1)}B_{nn}B_{n(n-1)}\cdots
B_{10}.$

This implies that

$\displaystyle
B_{nn}=\Big{(}\prod\limits_{i=1}^{n}B^{\ast}_{i(i-1)}\Big{)}^{-1}\Big{[}A_{2n-1}\;-\sum\limits_{\underset{\&\;(r_{n-1},r_{n})\neq(n,n)}{1\leq
r_{1},r_{2},\ldots,r_{2(n-1)}\leq 2n-2}}B_{1r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2(n-1)}1}\Big{]}{\Big{(}\prod\limits_{i=1}^{n}B^{\ast}_{i(i-1)}\Big{)}^{\ast}}^{-1}.$

Now notice that the lower diagonal block in the first column is given by
$B_{21}=(A_{2}-A_{1}^{2})^{1/2}$. These lower diagonal blocks can be obtained
by the compression of even powers of $B.$ Suppose that $(n-1)$ columns of $B$
are known then $B_{(n+1)n}$ is obtained as below:

$\displaystyle=$ $\displaystyle A_{2n}=(B^{2n})_{00}$ $\displaystyle=$
$\displaystyle\sum\limits_{0\leq r_{1},r_{2},\ldots,r_{2n-1}\leq
2n-1}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!B_{1r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2n-1}0}$ $\displaystyle=$
$\displaystyle\sum\limits_{\underset{\&\;r_{n}\neq n}{0\leq
r_{1},r_{2},\ldots,r_{2n-1}\leq
2n-1}}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2n-1}0}+B_{01}\cdots
B_{(n-2)(n-1)}B_{(n-1)n}B_{n(n-1)}B_{(n-1)(n-2)}\cdots B_{10}$
$\displaystyle=$ $\displaystyle\sum\limits_{\underset{\&\;r_{n}\neq n}{0\leq
r_{1},r_{2},\ldots,r_{2n-1}\leq
2n-1}}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2n-1}0}+B_{10}^{\ast}\cdots
B^{\ast}_{(n-1)(n-2)}B^{\ast}_{n(n-1)}B_{n(n-1)}B_{(n-1)(n-2)}\cdots B_{10}$

This implies that

$\displaystyle
B^{\ast}_{n(n-1)}B_{n(n-1)}=\Big{(}\prod\limits_{i=1}^{n-1}B^{\ast}_{i(i-1)}\Big{)}^{-1}\Big{[}A_{2n}\;-\sum\limits_{\underset{\&\;r_{n}\neq
n}{0\leq r_{1},r_{2},\ldots,r_{2n-1}\leq
2n-1}}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2n-1}0}\Big{]}{\Big{(}\prod\limits_{i=1}^{n-1}B^{\ast}_{i(i-1)}\Big{)}^{\ast}}^{-1}.$

In this case, we can choose

$B_{n(n-1)}=\Big{\\{}\Big{(}\prod\limits_{i=1}^{n-1}B^{\ast}_{i(i-1)}\Big{)}^{-1}\Big{[}A_{2n}\;-\sum\limits_{\underset{\&\;r_{n}\neq
n}{0\leq r_{1},r_{2},\ldots,r_{2n-1}\leq
2n-1}}\\!\\!\\!\\!\\!\\!\\!\\!\\!\\!B_{0r_{1}}B_{r_{1}r_{2}}\cdots
B_{r_{2n-1}0}\Big{]}{\Big{(}\prod\limits_{i=1}^{n-1}B^{\ast}_{i(i-1)}\Big{)}^{\ast}}^{-1}\Big{\\}}^{1/2}.$

With these computations and Lemma 2.7, we get the following result.

######  Theorem 2.9.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of self-adjoint operators in
$\mathcal{B}(\mathcal{H})$ with $A_{0}=I$, admitting a self-adjoint dilation.
If the inverses appearing in the recurrence relations above are well-defined
bounded operators, then these formulae provide a self-adjoint minimal dilation
with block tridiagonal form as above.

######  Example 2.10.

Let $T\in\mathcal{B(H)}$ be a self-adjoint operator. We define a moment
sequence $\\{A_{n}\\}_{n\geq 0}$ by

$\displaystyle A_{n}=$
$\displaystyle\left\\{\begin{array}[]{cc}2^{\frac{n-2}{2}}T^{n}&\text{~{}if~{}}n\text{
is even}\\\ &\\\ 0&\text{~{}if~{}}n\text{ is odd}~{}~{}~{}~{}~{}~{}.\\\
\end{array}\right.$

Then $V$ acting on $\mathcal{H}\oplus\mathcal{H}\oplus\mathcal{H}$,

$\displaystyle V=\begin{bmatrix}0&{T}&0\\\ T&0&T\\\ 0&T&0\end{bmatrix}$

is a self-adjoint dilation of $\\{A_{n}:n\geq 0\\}.$

This example can be generalized as follows. Suppose $\\{m_{n}:n\geq 0\\}$ is a
moment sequence of a compactly supported probability measure $\mu$ on
$\mathbb{R}.$ Let $T$ be a bounded self-adjoint operator on a Hilbert space
$\mathcal{H}$. Then $\\{m_{n}T^{n}:n\geq 0\\}$ is a operator moment sequence
admitting self-adjoint dilation. Take the dilation space
$\mathcal{K}=\mathcal{H}\otimes L^{2}(\mu)$ with $\mathcal{H}$ identified as a
subspace by identifying $h\in\mathcal{H}$ with $h\otimes 1$, where $1$ is the
constant function 1 in $L^{2}(\mu).$ Let $B$ be the tri-diagonal form of
multiplication by ‘$x$’ operator on $L^{2}(\mu)$ with cyclic vector $1$ as in
Equation (11). Then $T\otimes B$ is a self-adjoint dilation of
$\\{m_{n}T^{n}:n\geq 0\\}$:

$T\otimes B=\begin{bmatrix}a_{0}T&b_{0}T&0&0&\cdots\\\
{b_{0}}T&a_{1}T&b_{1}T&0&\ldots\\\ 0&{b_{1}}T&a_{2}T&b_{2}T&\ldots\\\
0&0&{b_{2}}T&a_{3}T&\ddots\\\ \vdots&\vdots&\ddots&\ddots&\ddots\\\
\end{bmatrix}.$

An operator $T\in\mathcal{B(H)}$ is called _quasinormal_ if
$T(T^{*}T)=(T^{*}T)T.$ We recall that $T$ is quasinormal if and only if
$(T^{*}T)^{k}=T^{*k}T^{k}$ for all $k\in\mathbb{Z}_{+}$ [27].

######  Example 2.11.

Let $T\in\mathcal{B(H)}$ be a quasinormal operator. Then the moment sequence
$\\{A_{n}\\}_{n\geq 0}$ defined by

$\displaystyle A_{n}=$
$\displaystyle\left\\{\begin{array}[]{cc}2^{\frac{n-2}{2}}T^{*\frac{n}{2}}T^{\frac{n}{2}}&\text{~{}if~{}}n\text{
is even}\\\ &\\\ 0&\text{~{}if~{}}n\text{ is odd}{}{}{}{}{}{}\\\
\end{array}\right.$

admits a self-adjoint dilation. In fact, $V$ acting on
$\mathcal{H}\oplus\mathcal{H}\oplus\mathcal{H}$ defined by

$\displaystyle V=\begin{bmatrix}0&T^{*}&0\\\ T&0&T^{*}\\\ 0&T&0\end{bmatrix}$

is a self-adjoint dilation of $\\{A_{n}:n\geq 0\\}.$

Like before, this example also can be generalized to have self-adjoint
dilations for operator moment sequences
$\\{1,0,m_{2}T^{*}T,0,m_{4}(T^{*})^{2}T^{2},0,\ldots\\}$, where
$\\{m_{n}:n\geq 0\\}$ is moment sequence of a compactly supported probability
measure $\mu$ on $\mathbb{R}$, which is symmetric around $0$. Such a symmetry
ensures that odd moments and also diagonal Jacobi parameters of the measure
are all equal to 0.

##  3\. Unitary and isometric dilations

In this section, we mainly discuss necessary and sufficient conditions on
operator sequences to admit unitary or isometric dilations. Let
$\\{A_{n}\\}_{n\geq 0}$ be a sequence in $\mathcal{B}(\mathcal{H})$ with
$A_{0}=I$. Then $\\{A_{n}\\}_{n\geq 0}$ is said to admit a unitary
(respectively isometric) dilation if there is a Hilbert space $\mathcal{K}$
containing $\mathcal{H}$ and a unitary (respectively unitary)
$U\in\mathcal{B}(\mathcal{K})$ such that

$A_{n}=P_{\mathcal{H}}U^{n}|_{\mathcal{H}},\;\text{for all }\;n\geq 0.$ (15)

Obviously every unitary dilation is an isometric dilation. Sz.-Nagy dilation
theorem implies in particular that every isometry has a power dilation to a
unitary. Therefore, if a sequence admits an isometric dilation it also admits
a unitary dilation. Consequently, an operator sequence admits an isometric
dilation if and only if it admits a unitary dilation. However, there is a
difference in the notion of minimality. With notation as above, a unitary
dilation $U$ acting on $\mathcal{K}$ is minimal if

$\mathcal{K}=\overline{\mbox{span}}\\{U^{n}h:n\in\mathbb{Z},h\in\mathcal{H}\\}.$

On the other hand, an isometric dilation $U$ acting on $\mathcal{K}$ is
minimal if

$\mathcal{K}=\overline{\mbox{span}}\\{U^{n}h:n\in\mathbb{Z}_{+},h\in\mathcal{H}\\}.$

The classical Szegő kernel and Poisson kernel [20] (also see [15] for more
details on Poisson kernel) are denoted and defined by

$\displaystyle S(w,z)$
$\displaystyle=\frac{1}{1-\overline{w}z}{~{}for~{}all~{}}z,w\in\mathbb{D}.$
$\displaystyle P_{r}(\theta)$
$\displaystyle=\frac{1-r^{2}}{1-2r\cos{\theta}+r^{2}}{~{}for~{}all~{}}0\leq
r<1,0\leq\theta\leq 2\pi.$

A Poisson kernel can be written using the Szegő kernel:

$\displaystyle P_{r_{1}r_{2}}(\theta-t)$
$\displaystyle=S(w,z)+\overline{S(w,z)}-1\text{ for all
}z,w\in\mathbb{D}\text{ and }z=r_{1}e^{i\theta},w=r_{2}e^{it}.$

Moreover, $P_{r}(\theta)\geq 0$ for all $0\leq r<1$ and $0\leq\theta\leq
2\pi.$

Keeping Szegő and Poisson kernels in mind, we define an operator valued kernel
function for a sequence of operators. Let ${\bf A}=\\{A_{n}\\}_{n\geq 0}$ be a
sequence of contractions in $\mathcal{B(H)}$. The associated Szegő kernel
function $S_{\bf A}:\mathbb{D}\to\mathcal{B(H)}$ is defined as

$\displaystyle S_{\bf
A}(z)=\sum_{n=0}^{\infty}z^{n}A^{*}_{n}\text{~{}for~{}all~{}}z\in\mathbb{D}.$
(16)

We define the associated Poisson kernel function by

$\displaystyle P_{\bf A}(z)=S_{\bf A}(z)+S_{\bf A}(z)^{*}-I\text{~{}for
all~{}}z\in\mathbb{D}.$ (17)

F. H. Vasilescu introduced operator valued Poisson kernel functions [40] for a
$d$-tuples of operators using defect operators to study Holomorphic functional
calculus. Such Poisson kernel functions can not be extended to operator valued
sequences as we do not have semigroup property (i.e. $A_{n}A_{m}=A_{n+m}$ may
not hold). In the next result, we discuss necessary and sufficient criteria
for isometric and unitary dilations in terms of the Poisson kernel.

######  Theorem 3.1.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence in $\mathcal{B}(\mathcal{H})$ with
$A_{0}=I$. Then the following are equivalent:

(i) $\\{A_{n}\\}_{n\geq 0}$ admits a unitary/isometric dilation.

(ii)

$\sum_{\ell,k=0}^{n}\bar{c_{\ell}}c_{k}A_{k-\ell}\geq 0,$ (18)

where $A_{-n}:=A^{*}_{n}$ for $n\in{\mathbb{N}}.$

(iii) $P_{\bf A}(z)\geq 0$ in $\mathcal{{B(H)}}$ for all $z\in\mathbb{D}.$

Before coming to the proof of this result, we point out that there is a subtle
point here. The condition in Equation (18) is a priori weaker than having
complete positivity of the $B({\mathcal{H}})$ valued kernel
$K:{\mathbb{Z}}\times{\mathbb{Z}}\to B({\mathcal{H}})$ defined by:

$K(k,\ell)=A_{(\ell-k)},~{}~{}k,\ell\in{\mathbb{Z}}$ (19)

as that would mean,

$\sum_{\ell,k=0}^{n}\langle g_{k},A_{\ell-k}g_{l}\rangle\geq 0,\;\text{for
all}\;g_{0},g_{1},\ldots,g_{n}\in{\mathcal{H}},n\in{\mathbb{N}}.$

###### Proof.

$(i)\implies(ii)$ Suppose assume that $(i)$ holds true. Then the Equation(15)
holds true. For every $N\geq 0$ and $g\in\mathcal{H}$, we see that

$\displaystyle\Big{\langle}g,\;\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}A_{(k-\ell)}g\Big{\rangle}$
$\displaystyle=\Big{\langle}\begin{bmatrix}c_{0}g\\\ c_{1}g\\\ \vdots\\\
c_{n}g\end{bmatrix},\;\begin{bmatrix}I&U&\cdots&U^{N}\\\
U^{\ast}&I&\cdots&U^{N-1}\\\ \vdots&\vdots&\ddots&\vdots\\\
{U^{\ast}}^{N}&{U^{\ast}}^{N-1}&\cdots&I\end{bmatrix}\begin{bmatrix}c_{0}g\\\
c_{1}g\\\ \vdots\\\ c_{n}g\end{bmatrix}\Big{\rangle}$
$\displaystyle=\Big{\langle}\sum\limits_{\ell=0}^{N}c_{\ell}U^{\ell}g,\;\sum\limits_{\ell=0}^{N}c_{\ell}U^{\ell}g\Big{\rangle}$
$\displaystyle\geq 0.$

$(ii)\implies(i)$ Consider the operator system
$\mathcal{S}=\Big{\\{}\sum\limits_{k=-N}^{N}c_{k}e^{ik\theta}:\;N\geq
0\Big{\\}}\subset C(\mathbb{T})$. Define
$\Phi\colon\mathcal{S}\to\mathcal{B}(\mathcal{H})$ by
$\Phi(e^{in\theta})=A_{n}$ for every $n\in\mathbb{Z}$. Suppose that
$f\in\mathcal{S}$ is strictly positive then by Fejér-Riesz theorem (see [28,
Lemma 2.5]), we see that
$f(e^{i\theta})=\sum\limits_{i,j=0}^{N}\overline{c_{\ell}}c_{k}e^{i(k-\ell)\theta}$
for some $c_{0},c_{1},\ldots,c_{N}$ and hence

$\Phi(f(e^{i\theta}))=\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\Phi(e^{i(k-\ell)\theta})=\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}A_{k-\ell}\geq
0.$

Further, if $g\in\mathcal{S}$ is non-negative, then $g+\epsilon\cdot 1$ is
strictly positive for every $\epsilon>0$ and so $\Phi(g+\epsilon\cdot
1)=\Phi(g)+\epsilon I\geq 0$ for every $\epsilon>0$. Thus $\Phi(g)\geq 0,$ it
follows that $\Phi$ is a positive map and hence it is bounded. By Stone-
Weierstrass theorem, $\mathcal{S}$ is dense in $C(\mathbb{T})$. So $\Phi$ can
be extended to a positive map on $C(\mathbb{T})$ by [28, Exercise 2.2]. Again
we denote it by $\Phi$. Therefore, it follows from [28, Theorem 3.11] that
$\Phi$ is completely positive map on $C(\mathbb{T})$.

Now by Stinespring’s theorem there exists a triple $({\mathcal{K}},\pi,V)$
where ${\mathcal{K}}$ is a Hilbert space, $\pi:C({\mathbb{T}})\to
B({\mathcal{K}})$ is a unital $*$-homomorphism,
$V\colon{\mathcal{H}}\to{\mathcal{K}}$ a bounded linear map satisfying

$\Phi(f)=V^{*}\pi(f)V,~{}~{}\forall f\in C({\mathbb{T}}).$ (20)

Since $\Phi$ is unitary, $V$ is an isometry and we may consider
${\mathcal{H}}$ as a subspace of ${\mathcal{K}}$ by identifying
$h\in{\mathcal{H}}$ with $Vh$ in ${\mathcal{K}}$. Then the Equation (20) reads
as

$\Phi(f)=P_{\mathcal{H}}\pi(f)|_{\mathcal{H}},f\in C({\mathbb{T}}).$

Since the function $f_{0}(z)=z$ is a unitary, $\pi(f_{0})$ is a unitary in
${\mathcal{B}(K)}$. Taking $U=\pi(f_{0})$ and considering
$f_{0}^{n}(z)=z^{n},$ for $n\geq 0$, we get

$A_{n}=P_{\mathcal{H}}U^{n}|_{\mathcal{H}},n\geq 0.$

$(i)\implies(iii)$ Suppose $\\{A_{n}:n\geq 0\\}$ admits a unitary dilation.
Then there is a unitary operator $U\in\mathcal{B(K)}$ for some Hilbert space
$\mathcal{K}\supseteq\mathcal{H}$ such that

$\displaystyle A_{n}=$ $\displaystyle
P_{\mathcal{H}}U^{n}|_{\mathcal{H}}\text{~{}for all~{}}n\geq 0.$ (21)

First notice that for each $z\in\mathbb{D},$ $(1-zU^{*})$ is invertible and
$(1-zU^{*})^{-1}=1+zU^{*}+z^{2}U^{*2}+\cdots$ and

$\|I+zU^{*}+z^{2}U^{*2}+\cdots\|\leq 1+|z|+|z|^{2}+\cdots=\frac{1}{1-|z|}.$

Now a simple computation ensures that
$(I-zU^{*})^{-1}+(I-\overline{z}U)^{-1}-I=(1-|z|^{2})(I-\overline{z}U)(I-\overline{z}U)^{*}.$
Also, it is immediate to see that
$(I-zU^{*})^{-1}+(I-\overline{z}U)^{-1}-I\geq 0$ as $|z|<1.$ It implies that
$P_{\mathcal{H}}\Big{(}(I-zU^{*})^{-1}+(I-\overline{z}U)^{-1}-I\Big{)}|_{\mathcal{H}}\geq
0.$ Using the hypothesis of dilation Equation (21), we see that

$\displaystyle\|I\|+\|zA^{*}_{1}\|+\|z^{2}A^{*}_{2}\|+\cdots+\|z^{n}A_{n}^{*}\|\leq$
$\displaystyle
1+|z|\|A^{*}_{1}\|+|z^{2}|\|A^{*}_{2}\|+\cdots+|z^{n}|\|A_{n}^{*}\|$
$\displaystyle\leq$ $\displaystyle
1+|z|\|U^{*}\|+|z^{2}|\|U^{*2}\|+\cdots+|z^{n}|\|U^{*n}\|$ $\displaystyle=$
$\displaystyle 1+|z|+|z|^{2}+\cdots+|z|^{n}$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-|z|}\text{~{}for all~{}}z\in\mathbb{D}.$

Thus the series $\sum_{n=0}^{\infty}z^{n}A^{*}_{n}$ converges in norm
absolutely for all $z\in\mathbb{D}.$ Therefore the map $S_{\bf
A}(z):\mathbb{D}\to\mathcal{B(H)}$ is well defined. Further, we compute that

$\displaystyle S_{\bf A}(z)+S_{\bf A}(z)^{*}-I$
$\displaystyle=\sum_{n=0}^{\infty}z^{n}A^{*}_{n}+\sum_{m=0}^{\infty}\overline{z}^{m}A^{m}-I$
$\displaystyle=\sum_{n=0}^{\infty}z^{n}P_{\mathcal{H}}U^{*n}|_{\mathcal{H}}+\sum_{m=0}^{\infty}\overline{z}^{m}P_{\mathcal{H}}U^{m}|_{\mathcal{H}}-I$
$\displaystyle=P_{\mathcal{H}}\big{(}\sum_{n=0}^{\infty}z^{n}U^{*n}\big{)}|_{\mathcal{H}}+P_{\mathcal{H}}\big{(}\sum_{m=0}^{\infty}\overline{z}^{m}U^{m}\big{)}|_{\mathcal{H}}-I$
$\displaystyle=P_{\mathcal{H}}\Big{(}(I-zU^{*})^{-1}+(I-\overline{z}U)^{-1}-I\Big{)}|_{\mathcal{H}}.$

Since
$P_{\mathcal{H}}\Big{(}(I-zU^{*})^{-1}+(I-\overline{z}U)^{-1}-I\Big{)}|_{\mathcal{H}}\geq
0,$ therefore we have $P_{\bf A}(z)\geq 0$ for all $z\in\mathbb{D}.$

$(iii)\implies(i)$ Assume that $P_{\bf A}(z)\geq 0$ for all $z\in\mathbb{D}.$
Let $0\leq r<1,$ then for each $l\in\mathbb{Z},$ we have

$\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}e^{il\theta}P_{\bf
A}(re^{i\theta})d\theta$
$\displaystyle=\frac{1}{2\pi}\int_{0}^{2\pi}e^{il\theta}\big{(}\sum_{n=0}^{\infty}r^{n}e^{in\theta}A^{*}_{n}+\sum_{m=0}^{\infty}r^{m}e^{-im\theta}A_{m}\big{)}d\theta$
$\displaystyle=\left\\{\begin{array}[]{cc}r^{l}A_{l}&\text{if}\;l\geq 0\\\
&\\\ r^{-l}A_{-l}^{*}&\text{if}\;l<0.\\\ \end{array}\right.$ (25)

Let us define a map $\varphi$ from an operator system $\mathcal{S}$ to
$\mathcal{B(H)}$ by

$\varphi\big{(}\sum_{n=0}^{N}p_{n}e^{in\theta}+\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\big{)}=\sum_{n=0}^{N}p_{n}A_{n}+\sum_{m=0}^{N}\overline{q_{m}}A_{m}^{*}$

for all
$\sum_{n=0}^{N}p_{n}e^{in\theta}+\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\in\mathcal{S}.$
Now it is enough to show that the map
$\varphi:\mathcal{S}\to\mathcal{B}(\mathcal{H})$ is a positive map. Now, we
observe the following

$\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\Big{(}\sum_{n=0}^{N}p_{n}e^{in\theta}\Big{)}P_{\bf
A}(re^{i\theta})d\theta$
$\displaystyle=\sum_{n=0}^{N}p_{n}\frac{1}{2\pi}\int_{0}^{2\pi}e^{in\theta}P_{\bf
A}(re^{i\theta})d\theta$ $\displaystyle=\sum_{n=0}^{N}p_{n}r^{n}A_{n}.$

Similarly, we have

$\displaystyle\frac{1}{2\pi}\int_{0}^{2\pi}\Big{(}\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\Big{)}P_{\bf
A}(re^{i\theta})d\theta$
$\displaystyle=\sum_{m=0}^{N}\overline{q_{m}}r^{m}A_{m}^{*}.$

Combining all, we finally obtain that

$\displaystyle\varphi\big{(}\sum_{n=0}^{N}p_{n}e^{in\theta}+\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\big{)}$
$\displaystyle=\sum_{n=0}^{N}p_{n}A_{n}+\sum_{m=0}^{N}\overline{q_{m}}A_{m}^{*}$
$\displaystyle=\lim_{r\to
1-}\big{(}\sum_{n=0}^{N}p_{n}r^{n}A_{n}+\sum_{m=0}^{N}\overline{q_{m}}r^{m}A_{m}^{*}\big{)}$
$\displaystyle=\lim_{r\to
1-}\Big{(}\frac{1}{2\pi}\int_{0}^{2\pi}\big{(}\sum_{n=0}^{N}p_{n}e^{in\theta}+\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\big{)}P_{\bf
A}(re^{i\theta})d\theta\Big{)}.$ (26)

Notice that for all $0\leq\theta\leq 2\pi$ and $0\leq r<1,$ we have $P_{\bf
A}(re^{i\theta})\geq 0.$ It follows that
$\int_{0}^{2\pi}\big{(}\sum_{n=0}^{N}p_{n}e^{in\theta}+\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\big{)}P_{\bf
A}(re^{i\theta})d\theta\geq 0$ whenever
$\sum_{n=0}^{N}p_{n}e^{in\theta}+\sum_{m=0}^{N}\overline{q_{m}}e^{-im\theta}\geq
0.$ These completes the proof as $\varphi$ is a positive map. ∎

Let $V$ be minimal isometric dilation on some Hilbert space ${\mathcal{K}}$
for a sequence of contractions $\\{A_{n}\\}_{n\geq 0}$ on $\mathcal{H}.$ In
particular, ${\mathcal{K}}=\overline{\rm span}\\{V^{n}\mathcal{H}:n\geq 0\\}$.
Then by Lemma 2.7, ${\mathcal{K}}$ decomposes as

${\mathcal{K}}={\mathcal{H}}_{0}\oplus{\mathcal{H}}_{1}\oplus\cdots,$

where ${\mathcal{H}}_{0}={\mathcal{H}}$ and with respect to this decomposition
the operator $V$ has the block form:

$V=\begin{bmatrix}V_{00}&V_{01}&V_{02}&\cdots&V_{0n}&\cdots\\\
V_{10}&V_{11}&V_{12}&\cdots&V_{1n}&\cdots\\\
0&V_{21}&V_{22}&\cdots&V_{2n}&\cdots\\\ 0&0&V_{32}&\cdots&V_{3n}&\cdots\\\
\vdots&\ddots&\ddots&\vdots&\vdots&\vdots\\\ \end{bmatrix}.$ (27)

We wish to construct $V_{ij}$’s using the given sequence $\\{A_{n}\\}.$ Recall
that ${\mathcal{H}}_{n]}=\overline{\mbox{s}pan}\\{V^{m}h:0\leq m\leq
n-1,h\in{\mathcal{H}}\\}$ and
${\mathcal{H}}_{n}={\mathcal{H}}_{n]}\bigcap{{\mathcal{H}}_{(n-1)]}^{\perp}}$
for $n\geq 1$ with ${\mathcal{H}}_{0]}={\mathcal{H}}_{0}.$

We try to determine the blocks using the sequence $\\{A_{n}\\}_{n\geq 0}$ and
the dilation property. Like in the case of self-adjoint dilation we assume
invertibility of concerned operators as and when required. Since $V$ is an
isometric dilation for the moment sequence $\\{A_{n}\\}_{n\geq 0}$ we have
$P_{\mathcal{H}}V^{n}\big{|}_{\mathcal{H}}=A_{n}$. In other words,
$(V^{n})_{00}=A_{n},\;\text{for}\;n\geq 0.$ Let us compute the first column of
$V$. Clearly, $V_{00}=A_{1}$ and since $V$ is an isometry, we get

$I=(V^{\ast}V)_{00}=V_{00}^{\ast}V_{00}+V_{10}^{\ast}V_{10}=A_{1}^{\ast}A_{1}+V_{10}^{\ast}V_{10}.$

One can choose that $V_{10}=(I-A_{1}^{\ast}A_{1})^{1/2}$ as $A_{1}$ is a
contraction. The first row of $V$ is given by the following recursive
relation: For every $n\in\mathbb{N},$ we have

$\displaystyle A_{n}$ $\displaystyle=(V^{n})_{00}$
$\displaystyle=\sum\limits_{0\leq r_{1},r_{2},\ldots,r_{(n-1)}\leq
n-1}V_{0r_{1}}V_{r_{1}r_{2}}\cdots V_{r_{(n-1)}0}$
$\displaystyle=\sum\limits_{0\leq r_{1},r_{2},\cdots,r_{n-1}\leq
n-1\;\&\;r_{1}\neq n-1}\\!\\!\\!\\!\\!\\!V_{0r_{1}}V_{r_{1}r_{2}}\cdots
V_{r_{(n-1)}0}+\sum\limits_{0\leq r_{2},\cdots,r_{n-1}\leq
n-1}V_{0(n-1)}V_{(n-1)r_{2}}\cdots V_{r_{(n-1)}0}.$

Since $V_{ij}=0$ whenever $i>j+1$, it implies that

$\sum\limits_{0\leq r_{2},\cdots,r_{n-1}\leq
n-1}V_{0(n-1)}V_{(n-1)r_{2}}\cdots
V_{r_{(n-1)}0}=V_{0(n-1)}V_{(n-1)(n-2)}V_{(n-2)(n-3)}\cdots V_{10}.$

Therefore,

$V_{0n}=\Big{[}A_{n}-\sum\limits_{0\leq r_{1},r_{2},\cdots,r_{n-1}\leq
n-1\;\&\;r_{1}\neq n-1}\\!\\!\\!\\!\\!\\!V_{0r_{1}}V_{r_{1}r_{2}}\cdots
V_{r_{(n-1)}0}\Big{]}\Big{(}V_{(n-1)(n-2)}V_{(n-2)(n-3)}\cdots
V_{10}\Big{)}^{-1}.$

Next, we use the isometric property of $V$ to obtain expression for the
$n^{\text{th}}$ column of $V$. For a fixed $n\in\mathbb{N}$, suppose the first
$(n-1)$ columns are known then $V_{1n}$ is given by considering the inner
product of the first column with the $n^{\text{th}}$ column which is zero.
That is, $V_{00}^{\ast}V_{0n}+V_{10}^{\ast}V_{1n}=0,$ and therefore

$\displaystyle V_{1n}$
$\displaystyle=-(I-A_{1}^{\ast}A_{1})^{-1/2}A^{\ast}_{1}V_{0n}.$

Indeed $V_{kn}$ for $0<k\leq n$ is given by the inner product of the
$(k-1)^{\text{th}}$ column with the $n^{\text{th}}$ column which is again
zero. That is, $\sum\limits_{i=0}^{k}V^{\ast}_{i(k-1)}V_{in}=0.$ Equivalently,
$V^{\ast}_{k(k-1)}V_{kn}+\sum\limits_{i=0}^{k-1}V^{\ast}_{i(k-1)}V_{in}=0.$
This implies that

$V_{kn}=-\big{(}V_{k(k-1)}^{\ast}\big{)}^{-1}\sum\limits_{i=0}^{k-1}V_{i(k-1)}^{\ast}V_{in}.$

For $k=n+1$, we have $\sum\limits_{i=0}^{n+1}V_{in}^{\ast}V_{in}=I$ and
equivalently,
$V_{(n+1)n}^{\ast}V_{(n+1)n}+\sum\limits_{i=0}^{n}V_{in}^{\ast}V_{in}=I.$ It
implies that

$\displaystyle V_{(n+1)n}^{\ast}V_{(n+1)n}$
$\displaystyle=I-\sum\limits_{i=0}^{n}V_{in}^{\ast}V_{in}.$

One can choose that

$V_{(n+1)n}=\big{[}I-\sum\limits_{i=0}^{n}V_{in}^{\ast}V_{in}\big{]}^{1/2}.$

As a result, the recurrence relations are given by
$V_{00}=A_{1},\;V_{10}=(I-A_{1}^{\ast}A_{1})^{1/2}$,

$V_{0n}=\Big{[}A_{n}-\sum\limits_{0\leq r_{1},r_{2},\cdots,r_{n-1}\leq
n-1\;\&\;r_{1}\neq n-1}V_{0r_{1}}V_{r_{1}r_{2}}\cdots
V_{r_{(n-1)}0}\Big{]}\Big{(}V_{(n-1)(n-2)}V_{(n-2)(n-3)}\cdots
V_{10}\Big{)}^{-1}$

and

$V_{kn}=\left\\{\begin{array}[]{cc}-\big{(}V_{k(k-1)}^{\ast}\big{)}^{-1}\sum\limits_{i=0}^{k-1}V_{i(k-1)}^{\ast}V_{in}&\mbox{
if\; $0<k\leq n$}\\\ &\\\
\Big{[}I-\sum\limits_{i=0}^{n}V_{in}^{\ast}V_{in}\Big{]}^{1/2}&\mbox{if \;
$k=n+1$}\end{array}\right.$

for $n>0.$

The computations above lead to the following result.

######  Theorem 3.2.

Let $\\{A_{n}\\}_{n\geq 0}$ be a sequence of contractions on $\mathcal{H}$ and
$A_{0}=I$ admitting an isometric dilation. If the inverses appearing in the
recurrence relations above are well defined bounded operators, then these
formulae provide a minimal isometric dilation with the blocks of $V$ described
as above.

Now we look at unitary dilations.

######  Theorem 3.3.

Suppose $U$ is a unitary on a Hilbert space $\mathcal{K}$ and $\mathcal{H}$ is
a closed subspace of $\mathcal{K}$, satisfying $\mathcal{K}=\overline{\rm
span}\\{U^{n}h:h\in\mathcal{H},n\in\mathbb{Z}\\}.$ Then ${\mathcal{K}}$
decomposes as

${\mathcal{K}}=\cdots\oplus{\mathcal{H}}_{-1}\oplus{\mathcal{H}}_{0}\oplus{\mathcal{H}}_{1}\oplus\cdots,$

where ${\mathcal{H}}_{0}={\mathcal{H}}$ so that with respect to this
decomposition $U$ has the form:

$U=\left[\begin{array}[]{@{}ccccc|ccccccc@{}}&\ddots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\\\
&&I&0&0&0&0&0&\\\ &\ldots&0&I&0&0&0&0&\ldots\\\
\hline\cr&\ldots&0&0&U_{0(-1)}&U_{00}&U_{01}&U_{02}&\ldots\\\
&\ldots&0&0&U_{1(-1)}&U_{10}&U_{11}&U_{12}&\ldots\\\
&\ldots&0&0&U_{2(-1)}&0&U_{21}&U_{22}&\ldots\\\
&\ldots&0&0&U_{3(-1)}&0&0&U_{32}&\ldots\\\
&&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\ddots\\\ \end{array}\right].$

###### Proof.

Take ${\mathcal{H}}_{0]}={\mathcal{H}}_{0}$ and
${\mathcal{H}}_{n]}=\overline{\mbox{span}}\\{U^{m}h:0\leq m\leq
n-1,h\in{\mathcal{H}}\\}$ and
${\mathcal{H}}_{n}={\mathcal{H}}_{n]}\bigcap{{\mathcal{H}}_{(n-1)]}^{\perp}}$
for $n\geq 1.$
$\mathcal{K}_{+}=\bigvee^{\infty}\limits_{n=0}U^{n}\mathcal{H},$ and
$\mathcal{K}_{-}=\mathcal{K}^{\perp}_{+}.$ Then clearly
${\mathcal{K}}_{+}=\oplus_{n\geq 0}{\mathcal{H}}_{n}$ and
$U({\mathcal{H}}_{n]})\subseteq{\mathcal{H}}_{(n+1)]}.$ It implies that $n$-th
column of the block matrix satisfies the desired property for all $n\geq 0.$
Notice that $U^{*}(\mathcal{K}_{-})\subseteq\mathcal{K}_{-}$ as
$\mathcal{K}_{-}=\mathcal{K}^{\perp}_{+}$ and
$U(\mathcal{K}_{+})\subseteq\mathcal{K}_{+}.$ We claim that

$\bigcap_{n=0}^{\infty}U^{*n}(\mathcal{K}_{-})=\\{0\\}.$

Let $x\in\bigcap\limits_{n=0}^{\infty}U^{*n}(\mathcal{K}_{-})=\\{0\\}.$ Let
$n\geq 0,$ and $h\in\mathcal{H}$ notice that $x=U^{*n}x_{n}$ for some
$x_{n}\in\mathcal{K}_{-}.$ Then $\langle x,U^{*n}h\rangle=\langle
U^{*n}x_{n},U^{*n}h\rangle=\langle x_{n},h\rangle=0,$ and $\langle
x,U^{m}h\rangle=0$ for all $m\geq 0.$ It follows that $x=0.$ Hence
$U^{*}:\mathcal{K}_{-}\to\mathcal{K}_{-}$ is a shift with wandering subspace

$\mathcal{H}_{-1}:=\mathcal{K}_{-}\ominus U^{*}\mathcal{K}_{-}.$

Therefore $\mathcal{K}_{-}$ can be decomposed as

$\mathcal{K}_{-1}=\cdots\oplus U^{*2}\mathcal{H}_{-1}\oplus
U^{*}\mathcal{H}_{-1}\oplus\mathcal{H}_{-1}.$

Now, it is enough to see that $U(\mathcal{H}_{-1})\subseteq\mathcal{K}_{+}.$
Consider $x\in\mathcal{H}_{-1},$ then for any $y\in\mathcal{K}_{-},$ we see
that $\langle Ux,y\rangle=\langle x,U^{*}y\rangle=0,$ as $x\perp
U^{*}\mathcal{K}_{-}.$ This completes the proof. ∎

##  4\. $\mathcal{C}_{A}$-class operators

Inspired by Sz.-Nagy’s dilation theorem for contractions, Berger and
Stamfli[11], obtained the following interesting theorem. Suppose
$T\in\mathcal{B(H)}$. Then the numerical radius $w(T)\leq 1$ if and only if
the operator sequence $\\{\frac{1}{2}T^{n},\;n\geq 1\\}$ admits a unitary
dilation. This led to the following definition. Fix $\rho>0$. Then a bounded
operator $T\in\mathcal{B(H)}$ is said to be in $\mathcal{C}_{\rho}$-class if
there is a unitary operator $U$ on some Hilbert space
$\mathcal{K}\supset\mathcal{H}$ such that $T^{n}=\rho
P_{\mathcal{H}}U^{n}\big{|}_{\mathcal{H}}$ for $n\geq 1$ (see page 43 of
[25]). Thanks to Sz.-Nagy’s dilation theorem, $\mathcal{C}_{1}$-class is
precisely the set of all contractions in $\mathcal{B(H)}$. Similarly from the
Berger-Stamfli’s Theorem [11] we know that $\mathcal{C}_{2}$-class is the set
of all operators in $\mathcal{B(H)}$ whose numerical radius is less than equal
to one. An operator $T$ is in $\mathcal{C}_{\rho}$-class (see Theorem 11.1 of
[25]) if and only if

$(\rho-2)\|(I-zT)h\|^{2}+2Re\big{\langle}h,\;(I-zT)h\big{\rangle}\geq
0,\;\text{for }\;|z|\leq 1,\;h\in\mathcal{H}.$ (28)

M. A. Dritschel, H. J. Woerdeman [23] have developed a model theory for the
$\mathcal{C}_{2}$-class. Over the past few decades there has been an extensive
study of $\mathcal{C}_{\rho}$-class operators. To mention some of them we
refer [5, 10, 11, 22, 17, 18, 24] and references therein. Generalizing this
notion in a natural way H. Langer introduced the $\mathcal{C}_{A}$-class. Let
$A\in\mathcal{B(H)}$ be a positive invertible operator. Then an operator
$T\in\mathcal{B(H)}$ is said to be of $\mathcal{C}_{A}$-class if the operator
valued sequence $\\{A_{n}\\}_{n\geq 1}$, where

$\displaystyle A_{n}:=A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}},n\geq 1,$ (29)

admits unitary dilation. The only reference we could find regarding this
notion is [25]. In particular we couldn’t find part (iii) of the following
Theorem in the literature.

######  Theorem 4.1.

Let $A\in\mathcal{B(H)}$ be a positive invertible operator and the operator
$T\in\mathcal{B(H)}.$ Then the following are equivalent:

  * (i)

$T\in\mathcal{C}_{A}$-class.

  * (ii)

For every $N\in\mathbb{N}$ and $c_{1},\cdots,c_{N}\in\mathbb{C}$, we have

$\sum_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\zeta_{A}(k-\ell)\geq
0\text{~{}for each~{}}N\in\mathbb{N},$ (30)

where $\zeta_{A}(n)$ is defined as follows:

$\displaystyle\zeta_{A}(n):=\left\\{\begin{array}[]{ll}A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}&{\mbox{i}f}~{}n>0;\\\
I&\mbox{if}~{}n=0;\\\
A^{-\frac{1}{2}}T^{*-n}A^{-\frac{1}{2}}&\mbox{if}~{}n<0.\\\
\end{array}\right.$

  * (iii)

$T$ satisfies the following:

$\displaystyle\left\langle(I-zT)h,(A-2I)(I-zT)h\right\rangle+2{\rm Re}\langle
h,(I-zT)h\rangle\geq 0\text{~{}~{}~{}for all~{}}|z|<1,h\in\mathcal{H}.$ (31)

###### Proof.

In view of Theorem 3.1, it is immediate to see (i) and (ii) are equivalent.
Now, we claim that (i) and (iii) are equivalent.

(i)$\implies$ (iii): Let ${\bf A}=\\{A_{n}\\}_{n\geq 0}$, where
$A_{n}=A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}$ for $n\geq 1$ and $A_{0}=I.$
Assume that $T\in\mathcal{C}_{A}.$ Then by Theorem 3.1, $P_{\bf
A}(\overline{z})\geq 0$ for all $z\in\mathbb{D}.$ In other words,

$(I+zA^{-\frac{1}{2}}TA^{-\frac{1}{2}}+z^{2}A^{-\frac{1}{2}}T^{2}A^{-\frac{1}{2}}+\ldots\infty)+(I+\overline{z}A^{-\frac{1}{2}}T^{*}A^{-\frac{1}{2}}+\overline{z}^{2}A^{-\frac{1}{2}}T^{*2}A^{-\frac{1}{2}}+\ldots\infty)-I\geq
0.$

It follow that

$(I-zT)^{-1}+(I-\overline{z}T^{*})^{-1}+A-2I\geq 0.$

Let $h\in\mathcal{H},$ then for all $z\in\mathbb{D},$ we have

$\displaystyle\left\langle(I-zT)h,\big{(}(I-zT)^{-1}+(I-\overline{z}T^{*})^{-1}\big{)}(I-zT)h\right\rangle+\left\langle(I-zT),(A-2I)(I-zT)h\right\rangle\geq
0.$

Since
$\left\langle(I-zT)h,\big{(}(I-zT)^{-1}+(I-\overline{z}T^{*})^{-1}\big{)}(I-zT)h\right\rangle=2Re\langle(I-zT)h,h\rangle,$
therefore

$\left\langle(I-zT)h,(A-2I)(I-zT)h\right\rangle+2{\rm
Re}\langle(I-zT)h,h\rangle\geq 0\text{~{}~{}~{}for
all~{}}|z|<1,h\in\mathcal{H}.$

(iii)$\implies$ (i): Assume that
$\left\langle(I-zT)h,(A-2I)(I-zT)h\right\rangle+2{\rm
Re}\langle(I-zT)h,h\rangle\geq 0\text{~{}~{}~{}for
all~{}}|z|<1,h\in\mathcal{H}.$ Then by reverse computation, we can show that
$P_{\bf A}(\overline{z})\geq 0$ for all $z\in\mathbb{D}.$ Hence by Theorem
3.1, $T\in\mathcal{C}_{A}.$ ∎

In the next result, employing standard techniques we provide another necessary
and sufficient criterion in terms of positive maps. As an immediate
application of Theorem 4.2, we recover a result of V. Istrăţescu [16] that if
$0\leq B\leq A,$ and $A,B$ are invertible, then
$\mathcal{C}_{B}\subseteq\mathcal{C}_{A}.$

######  Theorem 4.2.

Let $A\in\mathcal{B(H)}$ be a positive invertible operator. Then an operator
$T\in\mathcal{B(H)}$ is in $\mathcal{C}_{A}$-class if and only if the map
$\varphi:\mathcal{S}\subseteq C(\mathbb{T})\to\mathcal{B(H)}$ defined by

$\displaystyle\varphi_{A}\big{(}p(e^{i\theta})+\overline{q(e^{i\theta})}\big{)}:=p(T)+q(T)^{*}+(A-I)\big{(}p(0)+\overline{q(0)}\big{)}I.$

is a positive map.

###### Proof.

Let $T\in{\mathcal{C}}_{A}.$ Now, let
$f(e^{i\theta})=\sum\limits_{n=-N}^{N}a_{n}e^{in\theta}\in\mathcal{S}$ be
strictly positive. Then applying Riesz-Fejer theorem, we see that
$f(e^{i\theta})=\sum\limits_{0\leq\ell,k\leq
N}\overline{c_{\ell}}c_{k}e^{i(k-\ell)\theta}$ for some
$c_{0},\ldots,c_{N}\in\mathbb{C}.$ This implies that

$\displaystyle\varphi_{A}(f(e^{i\theta}))$
$\displaystyle=\sum\limits_{\ell=0}^{N}|c_{\ell}|^{2}+\sum\limits_{0\leq\ell\neq
k\leq
N}\overline{c_{\ell}}c_{k}{T_{(k-\ell)}}+(A-I)\sum\limits_{\ell=0}^{N}|c_{\ell}|^{2}$
$\displaystyle=\sum\limits_{\ell=0}^{N}|c_{\ell}|^{2}A+\sum_{0\leq l\neq k\leq
N}\overline{c_{\ell}}c_{k}{T_{(k-\ell)}}$
$\displaystyle=A^{\frac{1}{2}}\big{(}\sum\limits_{\ell=0}^{N}|c_{\ell}|^{2}I+\sum_{0\leq\ell\neq
k\leq
N}\overline{c_{\ell}}c_{k}A^{-\frac{1}{2}}{T_{(k-\ell)}}A^{-\frac{1}{2}}\big{)}A^{\frac{1}{2}}$
$\displaystyle=A^{\frac{1}{2}}\Big{(}\sum_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\zeta_{A}(k-\ell)\Big{)}A^{\frac{1}{2}}.$

As $T\in\mathcal{C}_{A},$ by Theorem 4.1, we have
$\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\zeta_{A}(k-\ell)\geq 0.$
Therefore $\varphi_{A}(f(z))\geq 0$ whenever $f(z)$ is strictly positively
element in $\mathcal{S}.$ Now, let $g\in\mathcal{S}$ be positive. Then
$g+\varepsilon\cdot 1$ is strictly positive for any $\epsilon>0.$ Then
$\varphi_{A}(g)+\varepsilon A=\varphi_{A}(g+\varepsilon\cdot 1)\geq 0$ for all
$\varepsilon>0.$ Taking limit $\epsilon\to 0,$ we obtain $\varphi_{A}(g)\geq
0.$ Hence the map $\varphi_{A}$ is positive.

Conversely, let $\varphi_{A}$ be a positive map. In view of Theorem 4.1, it is
enough to prove that
$\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\zeta_{A}(k-\ell)\geq 0.$
Notice that
$\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\zeta_{A}(k-l)=A^{-\frac{1}{2}}\varphi_{A}(\sum\limits_{0\leq\ell,k\leq
N}\overline{c_{\ell}}c_{k}e^{i(k-\ell)\theta})A^{-\frac{1}{2}}.$ Since
$\varphi_{A}$ is positive and $\sum\limits_{0\leq\ell,k\leq
N}\overline{c_{\ell}}c_{k}e^{i(k-\ell)\theta}\geq 0,$ therefore
$\sum\limits_{\ell,k=0}^{N}\overline{c_{\ell}}c_{k}\zeta_{A}(k-l)\geq 0.$ This
completes the proof. ∎

##  5\. Concrete isometric and unitary dilations for a subclass of
$\mathcal{C}_{A}$-class operators

Let $A\in\mathcal{B(H)}$ be a positive invertible operator. Recall that an
operator $T$ is in $\mathcal{C}_{A}$-class means that the sequence
$\\{A_{n}\\}_{n\geq 1}$ given by

$\displaystyle A_{n}:=A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}},n\geq 1$ (32)

admits unitary dilation. We do not know how to write down the dilation of
$\mathcal{C}_{A}$-class operators in general. Here we do it for a subclass.

First we write down an isometric dilation for
$\\{A^{-\frac{1}{2}}T^{n}A^{\frac{1}{2}}\\}_{n\geq 1}$. Let
$A,C\in\mathcal{B(H)}$ be commuting pair such that $A\geq 0$ with $\|C\|\leq
1.$ Suppose,

$T=A[I+A(A-2I)C^{*}C]^{-\frac{1}{2}}(I-C^{*}C)^{\frac{1}{2}}C.$

Here $I+A(A-2I)C^{*}C\geq 0$ as $C$ is a contraction and
$I+A(A-2I)C^{*}C=I-C^{*}C+(A-I)^{2}C^{*}C$. We are assuming that it is
invertible. In such a case, we will see that $T$ is in $\mathcal{C}_{A}$-class
and we can explicitly write down isometric and unitary dilations of $T$.

Consider $A,C$ as above. It is convenient to have some notation. Take
$B=(I+A(A-2I)C^{*}C)^{-\frac{1}{2}},D=(I-C^{*}C)^{\frac{1}{2}},D_{*}=(I-CC^{*})^{\frac{1}{2}}.$
Therefore, $T=ABDC.$ Now we take the following sequence $\\{T_{n}\\}_{n\geq
0}$ of bounded operators defined by

$\displaystyle
T_{0}=I,\;\text{and}\;T_{n}=A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}=A^{n-1}(BDC)^{n}\text{
for all }n\geq 1.$ (33)

Our aim is to show that the sequence $\\{T_{n}\\}_{n\geq 0}$ admits an
isometric dilation. In other words, there exists an isometry $V$ on some
Hilbert space $\mathcal{K}\supseteq\mathcal{H}$ such that
$T_{n}=P_{\mathcal{H}}V^{n}|_{\mathcal{H}}$ for all $n\geq 0.$ In addition, we
want to find an $V$ with explicit block structure.

We observe that since $A$ is positive and commutes with $C$, it also commutes
with $C^{*}$. It follows that $A$ commutes with also $B,D,D_{*}$. Consequently
we get $BD=DB$. On the other hand, we see that

$D^{2}_{\ast}C=(I-CC^{\ast})C=C(I-C^{\ast}C)=CD^{2}.$

As both $D$ and $D_{\ast}$ are positive operators, $D_{\ast}C=CD.$
Furthermore,

$\displaystyle B^{2}(D^{2}+(A-I)^{2}C^{*}C)$
$\displaystyle=(I+A(A-2I)C^{\ast}C)^{-1}\big{(}I-C^{\ast}C+A(A-I)C^{\ast}C-(A-I)C^{\ast}C\big{)}$
$\displaystyle=(I+A(A-2I)C^{\ast}C)^{-1}\big{(}I-A(A-I)C^{\ast}C-AC^{\ast}C\big{)}$
$\displaystyle=(I+A(A-2I)C^{\ast}C)^{-1}(I+A(A-2I)C^{\ast}C)$
$\displaystyle=I.$

Therefore,

$\displaystyle(BD)^{2}=I-B^{2}(A-I)^{2}C^{*}C.$ (34)

We first obtain a partial isometric dilation on
$\mathcal{H}\oplus\mathcal{H}.$

######  Lemma 5.1.

Define $R$ on $\mathcal{H}\oplus\mathcal{H}$ by:

$\displaystyle R=\begin{bmatrix}BDC&BDD_{*}\\\
(A-I)CBC&(A-I)CBD_{*}\end{bmatrix}$ (35)

Then $R$ is a partial isometry and

$T_{n}=P_{\mathcal{H}}R^{n}P_{\mathcal{H}}\text{ for all }n\geq 1.$

###### Proof.

Making use of the commutation relations observed above, by direct computation
we see that

$R^{*}R=\begin{bmatrix}C^{*}C&C^{*}D_{*}\\\ D_{*}C&D^{2}_{*}\end{bmatrix}.$

Clearly, $R^{*}R$ is a self-adjoint operator and

$\displaystyle(R^{*}R)^{2}$ $\displaystyle=\begin{bmatrix}C^{*}C&C^{*}D_{*}\\\
D_{*}C&D^{2}_{*}\end{bmatrix}\begin{bmatrix}C^{*}C&C^{*}D_{*}\\\
D_{*}C&D^{2}_{*}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}(C^{*}C)^{2}+C^{*}D^{2}_{*}C&C^{*}CC^{*}D_{*}+C^{*}D_{*}D_{*}^{2}\\\
D_{*}CC^{*}C+D^{2}_{*}D_{*}C&D_{*}CC^{*}D_{*}+D_{*}^{4}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}C^{*}C+C^{*}(1-CC^{*})C&C^{*}CC^{*}D_{*}+C^{*}(1-CC^{*})D_{*}\\\
D_{*}CC^{*}C+D_{*}(1-CC^{*})C&D_{*}^{2}CC^{*}+D_{*}^{2}(1-CC^{*})\end{bmatrix}$
$\displaystyle=\begin{bmatrix}C^{*}C&C^{*}D_{*}\\\
D_{*}C&D^{2}_{*}\end{bmatrix}.$

Therefore $R^{*}R$ is a projection and $R$ is a partial isometry. Next, we
compute $R^{2}$ as follows:

$\displaystyle R^{2}$ $\displaystyle=\begin{bmatrix}BDC&BDD_{*}\\\
(A-1)CBC&(A-1)CBD_{*}\end{bmatrix}\begin{bmatrix}BDC&BDD_{*}\\\
(A-1)CBC&(A-1)CBD_{*}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}(BDC)^{2}+(A-1)BDD_{*}CBC&*\\\
(A-1)CBCBDC+(A-1)^{2}CBD_{*}CBC&*\end{bmatrix}$
$\displaystyle=\begin{bmatrix}(BDC)^{2}+(A-1)BDCDCBC&*\\\
(A-1)CBCBDC+(A-1)^{2}CBCDBC&*\end{bmatrix}$
$\displaystyle=\begin{bmatrix}A(BDC)^{2}&*\\\ A(A-1)CBCBDC&*\end{bmatrix}.$

Therefore,
$P_{\mathcal{H}}R^{2}|P_{\mathcal{H}}=A(BDC)^{2}=A^{-\frac{1}{2}}T^{2}A^{-\frac{1}{2}}.$
Hence by induction, we conclude that
$P_{\mathcal{H}}R^{n}|_{\mathcal{H}}=A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}.$ ∎

Let $D_{R}=(1-R^{*}R)^{\frac{1}{2}}$ be the defect operator associate the
partial isometry. Since $(1-R^{*}R)$ is a projection,
$(1-R^{*}R)^{\frac{1}{2}}=(1-R^{*}R).$ Therefore,
$D_{R}=\begin{bmatrix}D^{2}&-C^{*}D_{*}\\\ -D_{*}C&CC^{*}\end{bmatrix}.$

######  Remark 5.2.

Using the Schäffer construction of $R,$ and Proposition 5.1, we can provide
directly explicit form of isometric dilation of the moment sequence
$\\{A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}\\}$ associated to the
$\mathcal{C}_{A}$-class. Let
$\tilde{V}:\ell^{2}(\mathcal{H}\oplus\mathcal{H})\to\ell^{2}(\mathcal{H}\oplus\mathcal{H})$
be the isometry given by

$\tilde{V}=\left[\begin{array}[]{@{}c|ccccccccc@{}}R&0&0&0&0&\ldots\\\
D_{R}&0&0&0&0&\ldots\\\ \hline\cr 0&I&0&0&0&\ldots\\\ 0&0&I&0&0&\ldots\\\
0&0&0&I&0&\ldots\\\ \vdots&\vdots&\vdots&\vdots&\ddots&\ddots\\\
\end{array}\right],$

Recalling Schäffer construction of $R$, it is clear that
$P_{\mathcal{H}}\tilde{V}^{n}|_{\mathcal{H}}=P_{\mathcal{H}}R^{n}|_{\mathcal{H}}=A^{-\frac{1}{2}}T^{n}A^{-\frac{1}{2}}.$

Here is an alternative construction of an isometric dilation which is simpler
looking.

######  Theorem 5.3.

Take $\mathcal{K}=\mathcal{H}^{\oplus 3}\oplus\ell^{2}(\mathcal{H}).$ Consider
$V$ on $\mathcal{K}$ defined by

$V=\left[\begin{array}[]{@{}ccc|cccccccc@{}}BDC&BDD_{*}&0&0&0&0&\ldots\\\
(A-I)CBC&(A-I)CBD_{*}&0&0&0&0&\ldots\\\ D&-C^{*}&0&0&0&0&\ldots\\\ \hline\cr
0&0&I&0&0&0&\ldots\\\ 0&0&0&I&0&0&\ldots\\\ 0&0&0&0&I&0&\ldots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\ddots\\\ \end{array}\right].$

Then $V$ is an isometric dilation of $\\{A_{n}\\}_{n\geq 0}.$

###### Proof.

The dilation property follows from the Lemma 5.1. The isometric property of
$V$ is clear once we observe that

$D_{R}=(I-R^{*}R)=\left[\begin{array}[]{c}D\\\
-C\end{array}\right]\left[\begin{array}[]{cc}D&-C^{*}\end{array}\right].$

∎

In the last theorem there is no claim of minimality. To get the minimal
isometric dilation we need to identify the minimal dilation space
$\bigvee\limits_{n=0}^{\infty}V^{n}\mathcal{H}.$ This we do here as a
proposition.

######  Proposition 5.4.

Let $V$ be the minimal isometric dilation defined in Theorem 5.3. Then the
minimal dilation space $\bigvee\limits_{n=0}^{\infty}V^{n}\mathcal{H}$ is of
the form

$\bigvee\limits_{n=0}^{\infty}V^{n}\mathcal{H}=\bigoplus_{n=0}^{\infty}\mathcal{H}_{n},$

where
$\mathcal{H}_{0}:=\mathcal{H},\mathcal{H}_{1}:=V\mathcal{H}\ominus\mathcal{H}.$
Moreover,

$\mathcal{H}_{1}=\\{(0\oplus(A-I)CBCh\oplus
Dh\oplus\cdots):h\in\mathcal{H}\\}$

and

$\mathcal{H}_{n}:=V^{n}\mathcal{H}\ominus\bigvee\limits_{m=0}^{n-1}V^{m}\mathcal{H}=\\{(0\oplus(n~{}\mbox{copies}~{})\oplus
0\oplus(I-A)BCPh\oplus Dh\oplus 0\cdots):h\in\mathcal{H}\\}$

for $n\geq 2,$ where $P$ is the projection onto $\ker(I-A)BDC.$

###### Proof.

First note that $\mathcal{H}\bigvee
V\mathcal{H}=\mathcal{H}\oplus(V\mathcal{H}\ominus\mathcal{H}).$ Let
$h\in\mathcal{H},$ then notice that

$\displaystyle Vh$ $\displaystyle=\big{(}BDCh\oplus(A-I)CBCh\oplus Dh\oplus
0\oplus\cdots\big{)},$ $\displaystyle V^{2}h$
$\displaystyle=\big{(}A(BDC)^{2}h\oplus A(A-I)CBCBDCh\oplus
DBDCh-(A-I)C^{*}CBCh\oplus Dh\oplus 0\oplus\cdots\big{)}.$

Then it is immediate to see that
$V\mathcal{H}\ominus\mathcal{H}=\\{(0\oplus(A-I)CBCh\oplus Dh\oplus 0\oplus
0\oplus\cdots):h\in\mathcal{H}\\}.$ Next, we shall find
$V^{2}\mathcal{H}\ominus(\mathcal{H}\bigvee V\mathcal{H}).$ First, we write
$V^{2}h$ as

$\displaystyle V^{2}h$ $\displaystyle=\big{(}A(BDC)^{2}h\oplus
0\oplus\cdots\big{)}+\big{(}0\oplus(A-I)CBC(ABDCh)\oplus DABDCh\oplus
0\oplus\cdots\big{)}$ $\displaystyle+\big{(}0\oplus 0\oplus(I-A)BCh\oplus
Dh\oplus 0\oplus\cdots\big{)}.$

Obviously, the first and second terms belong to $\mathcal{H}$ and
$V\mathcal{H}\ominus\mathcal{H}$ respectively. We want to understand the last
term of this equation. We claim that

$V^{2}\mathcal{H}\ominus(\mathcal{H}\vee V\mathcal{H})=\big{\\{}\big{(}0\oplus
0\oplus(I-A)BCPh\oplus Dh\oplus
0\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}}.$

Notice that $\big{\langle}\big{(}0\oplus(A-I)CBCg\oplus Dg\oplus 0\oplus
0\oplus\cdots\big{)},\big{(}0\oplus 0\oplus(I-A)BCPh\oplus Dh\oplus
0\oplus\cdots\big{)}\big{\rangle}=0$ as $(I-A)BDCPh=0.$ Therefore, we have

$\big{\\{}\big{(}0\oplus 0\oplus(I-A)BCPh\oplus Dh\oplus
0\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}}\subseteq
V^{2}\mathcal{H}\ominus(\mathcal{H}\vee V\mathcal{H}).$

Now, it is sufficient to prove that $\big{(}0\oplus 0\oplus(I-A)BC(I-P)h\oplus
0\oplus\cdots\big{)}\in V\mathcal{H}\ominus\mathcal{H}$ for all
$h\in\mathcal{H}.$ To see this, suppose

$\displaystyle\big{\langle}\big{(}0\oplus 0\oplus(I-A)BC(I-P)h\oplus
0\oplus\cdots\big{)},\big{(}0\oplus(A-I)CBCg\oplus Dg\oplus
0\oplus\cdots\big{)}\big{\rangle}=0,$ (36)

for all $g\in\mathcal{H}.$ This implies that for all $g\in\mathcal{H},$

$0=\big{\langle}(I-A)BC(I-P)h,Dg\big{\rangle}=\big{\langle}(I-A)BDC(I-P)h,g\big{\rangle}=\big{\langle}(I-A)BDCh,g\big{\rangle},$

as $(I-A)BDC(I-P)h=0.$ It follows that $(I-A)BDCh=0$ and $h\in{\rm range}{P}.$
Therefore, we have $\big{(}0\oplus 0\oplus(I-A)BC(I-P)h\oplus
0\oplus\cdots\big{)}=0$ under the condition defined Equation (36).
Consequently, we get $\big{(}0\oplus 0\oplus(I-A)BC(I-P)h\oplus
0\oplus\cdots\big{)}\in V\mathcal{H}\ominus\mathcal{H}$ for all
$h\in\mathcal{H}.$ Hence, we have $\big{\\{}\big{(}0\oplus
0\oplus(I-A)BCPh\oplus Dh\oplus
0\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}}=V^{2}\mathcal{H}\ominus(\mathcal{H}\vee
V\mathcal{H}).$ Now it is easy to see that

$\big{\\{}\big{(}0\oplus 0\oplus 0\oplus(I-A)BCPh\oplus Dh\oplus
0\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}}=V^{3}\mathcal{H}\ominus(\mathcal{H}\vee
V\mathcal{H}\vee V^{2}\mathcal{H}).$

By induction, we can prove that

$\big{\\{}\big{(}0\oplus(n~{}\mbox{copies}~{})\oplus(I-A)BCPh\oplus Dh\oplus
0\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}}=V^{n}\mathcal{H}\ominus(\bigvee\limits_{m=0}^{n-1}V^{m}\mathcal{H}).$

∎

Now we explicitly write down a unitary dilation for the sequence
$\\{T_{n}\\}_{n\geq 0}$ as a

######  Theorem 5.5.

Let $\mathcal{K}=\ell^{2}(\mathcal{H})\oplus\mathcal{H}^{\oplus
3}\oplus\ell^{2}(\mathcal{H})$ and $U$ acting on $\mathcal{K}$ is defined by

$\displaystyle U=\left[\begin{array}[]{@{}ccc|cccc|cccc@{}}\ddots&&&&&&&&&\\\
&I&&&&&&&&\\\ \hline\cr&&I&0&0&0&0&&&\\\ &&&-(A-I)BC^{*}&{\bf
BDC}&BDD_{*}&0&&&\\\ &&&B_{*}D_{*}&(A-I)CBC&(A-I)CBD_{*}&0&&&\\\
&&&0&D&-C^{*}&0&&&\\\ \hline\cr&&&0&0&0&I&&&\\\ &&&&&&&I&&\\\
&&&&&&&&\ddots&\\\ \end{array}\right].$ (46)

where $BDC$ that appear in the bold font is the $(00)$ entry of $U,$ and the
$B_{*}$ is given by

$B_{*}=(I+A(A-2I)CC^{*})^{-\frac{1}{2}}.$

Then $U$ is a unitary dilation of $\\{T_{n}\\}_{n\geq 0}.$

###### Proof.

Consider the block operator matrix $M$ defined by:

$\displaystyle M=\begin{bmatrix}&0&0&0&0\\\ &-(A-I)BC^{*}&{\bf
BDC}&BDD_{*}&0\\\ &B_{*}D_{*}&(A-I)CBC&(A-I)CBD_{*}&0\\\
&0&D&-C^{*}&0\end{bmatrix}.$ (47)

It is straight forward to verify that

$M^{*}M=\begin{bmatrix}I&0&0&0\\\ 0&I&0&0\\\ 0&0&I&0\\\ 0&0&0&0\\\
\end{bmatrix},~{}~{}~{}~{}MM^{*}=\begin{bmatrix}0&0&0&0\\\ 0&I&0&0\\\
0&0&I&0\\\ 0&0&0&I\\\ \end{bmatrix}.$

This clearly implies that $U^{*}U=UU^{*}=I.$ Finally, we obtain the desired
claim

$\displaystyle
P_{\mathcal{H}}U^{n}|_{\mathcal{H}}=P_{\mathcal{H}}M^{n}|_{\mathcal{H}}=T_{n}\text{
for all }n\geq 0.$

These completes the proof. ∎

As we discussed earlier, minimal dilation space of the unitary dilation is of
the form $\mathcal{K}=\bigvee^{\infty}\limits_{n=-\infty}U^{n}\mathcal{H}.$
The minimal dilation space can be written as
$\mathcal{K}_{+}\bigvee\mathcal{K}^{-}=\mathcal{K},$ where
$\mathcal{K}_{+}=\bigvee^{\infty}\limits_{n=0}U^{n}\mathcal{H},$ and
$\mathcal{K}^{-}=\bigvee^{-\infty}\limits_{n=0}U^{n}\mathcal{H}.$ Since
$U|_{\mathcal{K}_{+}}$ is the isometry $V$ constructed before, the
decomposition of $\mathcal{K}_{+}$ is clear.

######  Remark 5.6.

$\mathcal{K}_{-}$ can be decomposed as

$\mathcal{K}_{-}=\bigoplus_{n=1}^{\infty}\mathcal{H}_{-n}$

$\text{ where
}\mathcal{H}_{-1}:=U^{*}\mathcal{H}\ominus\mathcal{H},\mathcal{H}_{-n}:=U^{*n}\mathcal{H}\ominus\bigvee\limits_{m=0}^{n-1}U^{*m}\mathcal{H}\text{
for }n\geq 2.$ Moreover,

$\mathcal{H}_{-1}:=\big{\\{}\big{(}\cdots\oplus-(A-I)CBh\oplus\textbf{0}\oplus
D_{*}DBh\oplus 0\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}},$
$\mathcal{H}_{-n}=\big{\\{}\big{(}\cdots\oplus-(A-I)CBCh\oplus
B^{-\frac{1}{2}}_{*}DBQh\oplus
0\oplus(-n+2)~{}\mbox{times}~{}\oplus\textbf{0}\oplus\cdots\big{)}:h\in\mathcal{H}\big{\\}}$

and $Q$ is the projection onto $\ker(A-1)B^{\frac{1}{2}}C^{*}DB.$

###  5.1. Special case ($\mathcal{C}_{\rho}$-class)

The case when $A$ is a positive scalar $\rho,$ corresponds to the
$\mathcal{C}_{\rho}$\- class of operators. This class is very rich and widely
studied. However, as far as we know explicit block operator description of
isometric and unitary dilations of operators this class is not found in the
literature except for $\rho=2$. An operator $T$ is in $\mathcal{C}_{2}$ class
if and only if $w(T)\leq 1,$ where $w(T)$ denote the numerical range of $T$.
Unitary dilation of this class was exhibited by T. Ando [6] and coincides with
the following construction (with $\rho=2$). We recall from Durzst [14] that
$T\in\mathcal{C}_{\rho}(\rho>0)$ if and only if $T$ is of the form:

$T=\rho(1+\rho(\rho-2)C^{*}C)^{-\frac{1}{2}}DC,$

where $\rho>0,C$ is a contraction and $D=(I-C^{*}C)^{\frac{1}{2}}.$ In this
case we may write down isometric and unitary dilations as above. We observe
that
$B=(I+\rho(\rho-2)C^{*}C)^{-\frac{1}{2}},B_{*}=(I+\rho(\rho-2)CC^{*})^{-\frac{1}{2}}$
and $T_{n}=\rho^{n-1}(BDC)^{n}=\frac{(\rho
BDC)^{n}}{\rho}=\frac{T^{n}}{\rho}.$ Moreover, for the case $\rho=2,$ being
$B=B^{*}=I,$ the operators $B,B^{*}$ will not play any role in unitary and
isometric dilation of $\mathcal{C}_{2}$-class.

######  Question 5.7.

How to write block decompositions for isometric and unitary dilations of
general $\mathcal{C}_{A}$-class operators?

Acknowledgements: The first author is funded by the J C Bose Fellowship
JBR/2021/000024 of SERB(India). The second author is supported by the NBHM
postdoctoral fellowship, Department of Atomic Energy (DAE), Government of
India (File No. 0204/1(4)/2022/ R&D-II/1198). The third author is funded by
the Startup Research Grant (File No. SRG/2022/001795) of SERB, India. We also
acknowledge Stat-Math Unit of the Indian Statistical Institute Bangalore
Centre for providing excellent research environment.

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