# A short note on similarity of operator-valued multishifts

Soumitra Ghara  ,  Surjit Kumar  and  Shailesh Trivedi  Department of
Mathematics
Indian Institute of Technology Kharagpur, Midnapore-721302, India
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>Department of Mathematics
Indian Institute of Technology Madras, Chennai 600036, India
<EMAIL_ADDRESS>Department of Mathematics
Birla Institute of Technology and Science, Pilani 333031, India
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Abstract.

A complete characterization of the similarity between two operator-valued
multishifts with invertible operator weights is obtained purely in terms of
operator weights. This generalizes several existing results of the unitary
equivalence of two (multi)shifts. Further, we utilize the aforementioned
similarity criteria to determine the similarity between two tuples of
operators of multiplication by the coordinate functions on certain reproducing
kernel Hilbert spaces determined by diagonal kernels.

###### Key words and phrases:

operator-valued multishift, similarity, unitary equivalence, vector-valued
reproducing kernel Hilbert spaces.

###### 2020 Mathematics Subject Classification:

Primary 47A13, 47B37, Secondary 46E40, 47B32

The work of the first author is supported by INSPIRE Faculty Fellowship (Ref.
No. DST/INSPIRE/04/2021/002555). Support for the work of the second author was
provided in the form of the SERB-MATRICS grant (Ref. No. MTR/2022/000457) and
NFIG grant (Ref. No. RF/21-22/1540/MA/NFIG/008991) of IIT Madras. The work of
the third author was supported by Inspire Faculty Fellowship (Ref. No.
DST/INSPIRE/04/2018/000338), OPERA (FR/SCM/03-Nov-2022/MATH) funded by BITS
Pilani and SERB-SRG (SRG/2023/000641-G)

##  1\. Introduction

This article is another installment in the continuation of the study of
operator-valued multishift initiated in [5, 6]. In this short note, we focus
our attention on the similarity of two operator-valued multishifts with
invertible operator weights. We give a necessary and sufficient condition
under which two operator-valued multishifts with invertible operator weights
are similar. This condition is given purely in terms of the moments of the
respective operator-valued multishifts. To explain the main result of this
short note, we proceed by briefly recalling the preliminaries about the
operator-valued multishifts. The reader is referred to [5] and [6] for a
detailed overview of the operator-valued multishifts.

Let $\mathbb{N}$ denote the set of non-negative integers. Let $d$ be a
positive integer and $H$ be a complex separable Hilbert space. We define
$\ell^{2}_{H}(\mathbb{N}^{d}):=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}H$
as the orthogonal direct sum of $\mbox{Card}(\mathbb{N}^{d})$ many copies of
$H$. Then $\ell^{2}_{H}(\mathbb{N}^{d})$ is a Hilbert space equipped with the
following inner product:

$\displaystyle\langle{x},\,{y}\rangle:=\sum_{\alpha\in\mathbb{N}^{d}}\langle{x_{\alpha}},\,{y_{\alpha}}\rangle_{{}_{H}},\quad
x=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}x_{\alpha},\
y=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}y_{\alpha}\in\ell^{2}_{H}(\mathbb{N}^{d}).$

Let $\big{\\{}{A^{(j)}_{\alpha}}:\alpha\in\mathbb{N}^{d},\
j=1,\ldots,d\big{\\}}$ be a multisequence of bounded invertible operators on
$H$ with the following conditions: (i)
$\sup_{\alpha\in\mathbb{N}^{d}}\|A^{(j)}_{\alpha}\|<\infty$ for all
$j=1,\ldots,d$ and (ii)
$A^{(i)}_{\alpha+\varepsilon_{j}}A^{(j)}_{\alpha}=A^{(j)}_{\alpha+\varepsilon_{i}}A^{(i)}_{\alpha}\
\mbox{for all }\alpha\in\mathbb{N}^{d}$ and $i,j=1,\ldots,d$, where
$\varepsilon_{j}$ is the $d$-tuple in $\mathbb{N}^{d}$ in which $1$ is at the
$j$-th place and $0$ elsewhere. Then an operator-valued multishift on
$\ell^{2}_{H}(\mathbb{N}^{d})$ is the commuting $d$-tuple
$T=(T_{1},\ldots,T_{d})$ of bounded linear operators $T_{1},\ldots,T_{d}$ on
$\ell^{2}_{H}(\mathbb{N}^{d})$ defined as

$\displaystyle
T_{j}(\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}x_{\alpha}):=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}A^{(j)}_{\alpha-\varepsilon_{j}}x_{\alpha-\varepsilon_{j}},\quad\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}x_{\alpha}\in\ell^{2}_{H}(\mathbb{N}^{d}),\
j=1,\ldots,d.$

If for $\alpha=(\alpha_{1},\ldots,\alpha_{d})\in\mathbb{N}^{d}$,
$\alpha_{j}=0$, then we interpret $A^{(j)}_{\alpha-\varepsilon_{j}}$ as a zero
operator and $x_{\alpha-\varepsilon_{j}}$ as a zero vector. By an $H$-valued
formal power series we mean the series
$\displaystyle\sum_{\alpha\in\mathbb{N}^{d}}x_{\alpha}z^{\alpha}$,
$x_{\alpha}\in H$, without regard to convergence at any point
$z\in\mathbb{C}^{d}$. For a given multisequence
$\mathscr{B}=\big{\\{}B_{\alpha}:\alpha\in\mathbb{N}^{d}\big{\\}}$ of bounded
invertible operators on $H$, the Hilbert space $\mathcal{H}^{2}(\mathscr{B})$
of $H$-valued formal power series is defined as

$\displaystyle\mathcal{H}^{2}(\mathscr{B}):=\left\\{\sum_{\alpha\in\mathbb{N}^{d}}x_{\alpha}z^{\alpha}:\sum_{\alpha\in\mathbb{N}^{d}}\|B_{\alpha}x_{\alpha}\|^{2}<\infty\right\\}$

endowed with the following inner product: For
$f(z)=\sum_{\alpha\in\mathbb{N}^{d}}x_{\alpha}z^{\alpha}$ and
$g(z)=\sum_{\alpha\in\mathbb{N}^{d}}y_{\alpha}z^{\alpha}$ in
$\mathcal{H}^{2}(\mathscr{B})$,

$\displaystyle\langle{f(z)},\,{g(z)}\rangle_{\\!{}_{\mathcal{H}^{2}(\mathscr{B})}}:=\sum_{\alpha\in\mathbb{N}^{d}}\langle{B_{\alpha}x_{\alpha}},\,{B_{\alpha}y_{\alpha}}\rangle_{\\!{}_{H}}.$

Note that $H$-valued polynomials are dense in $\mathcal{H}^{2}(\mathscr{B})$.
It follows from [6, Theorem 5.4] that an operator-valued multishift on
$\ell^{2}_{H}(\mathbb{N}^{d})$ with invertible operator weights is unitarily
equivalent to the commuting $d$-tuple
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ of
(bounded) operators of multiplication by the coordinate function $z_{j}$ on
$\mathcal{H}^{2}(\mathscr{B})$, where
$\mathscr{B}=\big{\\{}B_{\alpha}:\alpha\in\mathbb{N}^{d}\big{\\}}$ is given by
[6, Proposition 2.3]. In view of this equivalence, we work with the
$\mathcal{H}^{2}(\mathscr{B})$ model of an operator-valued multishift in the
rest of this paper. The main result of this paper is stated as follows:

######  Theorem 1.1.

Let $H$ and $\tilde{H}$ be two complex separable Hilbert spaces and
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$,
$\tilde{\mathscr{B}}=\\{\tilde{B}_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ be
two multisequences of bounded invertible operators on $H$ and $\tilde{H}$,
respectively. Suppose that
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ is bounded
on $\mathcal{H}^{2}(\mathscr{B})$ and on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$. Then $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\mathscr{B})$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$ if and only if there exist a bounded
invertible operator $C:\tilde{H}\to H$ and positive constants $m_{1},m_{2}$
such that

$\displaystyle
m_{1}C^{*}B^{*}_{\alpha}B_{\alpha}C\leqslant{\tilde{B}_{\alpha}}^{*}\tilde{B}_{\alpha}\leqslant
m_{2}C^{*}B^{*}_{\alpha}B_{\alpha}C\quad\mbox{for all
}\alpha\in\mathbb{N}^{d}.$ (1)

Even for the simplest scenario where $d=1$, the criteria for the similarity
between two operator-valued unilateral weighted shifts weren’t previously
understood solely in terms of their moments. Instead, the similarity of these
shifts was characterized by examining the intertwiner rather than relying on
their moments, see [8, Lemma 2.1].

The result of Theorem 1.1 generalizes [7, Lemma 2.2] and [10, Theorem
$2^{\prime}$(b)]. The proof and the consequences of this theorem are presented
in the next section. All the notaions and symbols used in the proof of Theorem
1.1 are standard in the literature and have their usual meaning.

##  2\. Proof of Theorem 1.1

We begin by setting the following notation.

$\displaystyle\mathcal{H}_{\alpha}:=\\{xz^{\alpha}:x\in
H\\},\quad\alpha\in\mathbb{N}^{d}.$ (2)

If $\alpha_{j}=0$ for some $\alpha\in\mathbb{N}^{d}$, then we interpret
$\mathcal{H}_{\alpha-\varepsilon_{j}}$ as $\\{0\\}$. Note that

$\displaystyle\mathcal{H}^{2}(\mathscr{B})=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}\mathcal{H}_{\alpha}.$
(3)

The following well-known result (see [6, Theorem 5.4]) will be used in the
proof of the main result of this section.

######  Lemma 2.1.

Let $H$ be a complex separable Hilbert space and
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ be a multisequence of
bounded invertible operators on $H$. Let $\mathcal{H}^{2}(\mathscr{B})$ be the
Hilbert space of $H$-valued formal power series. Suppose that
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ is bounded
on $\mathcal{H}^{2}(\mathscr{B})$. Then the following statements are true:

  1. (i)

For all $x\in H$, $\alpha\in\mathbb{N}^{d}$ and $j=1,\ldots,d,$ we have

$\displaystyle\mathscr{M}^{*}_{z_{j}}(xz^{\alpha})=(B_{\alpha-\varepsilon_{j}}^{*}B_{\alpha-\varepsilon_{j}})^{-1}B_{\alpha}^{*}B_{\alpha}xz^{\alpha-\varepsilon_{j}},\
$ (4)

where $B_{\alpha-\varepsilon_{j}}$ is interpreted as a zero operator if
$\alpha_{j}=0$ for some $\alpha\in\mathbb{N}^{d}$.

  2. (ii)

The linear map
$P_{\mathcal{H}_{\alpha+\varepsilon_{j}}}\mathscr{M}_{z_{j}}|_{\mathcal{H}_{\alpha}}:\mathcal{H}_{\alpha}\rightarrow\mathcal{H}_{\alpha+\varepsilon_{j}}$
is bounded and invertible for all $\alpha\in\mathbb{N}^{d}$ and
$j=1,\ldots,d$, where for a closed subspace $\mathcal{E}$ of
$\mathcal{H}^{2}(\mathscr{B})$, $P_{\mathcal{E}}$ denotes the orthogonal
projection of $\mathcal{H}^{2}(\mathscr{B})$ onto $\mathcal{E}$.

We are now in a position to present the proof of the main result.

###### Proof of Theorem 1.1.

Suppose that $\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\mathscr{B})$ is similar to
$\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$. Let
$X:\mathcal{H}^{2}(\mathscr{B})\rightarrow\mathcal{H}^{2}(\tilde{\mathscr{B}})$
be an invertible operator such that
$X\mathscr{M}_{z_{j}}=\mathscr{M}_{z_{j}}X$ for all $j=1,\ldots,d$. For each
$\alpha\in\mathbb{N}^{d}$, let $\mathcal{H}_{\alpha}$ and
$\tilde{\mathcal{H}}_{\alpha}$ be the respective subspaces of
$\mathcal{H}^{2}(\mathscr{B})$ and $\mathcal{H}^{2}(\tilde{\mathscr{B}})$ as
defined by (2). In view of the decomposition (3), we get

$\displaystyle
P_{\tilde{\mathcal{H}}_{\alpha}}X\mathscr{M}_{z_{j}}|_{\mathcal{H}_{\beta}}=P_{\tilde{\mathcal{H}}_{\alpha}}\mathscr{M}_{z_{j}}X|_{\mathcal{H}_{\beta}}\mbox{
for all }\alpha,\beta\in\mathbb{N}^{d}\mbox{ and }j=1,\ldots,d.$

Consequently, it yields that

$\displaystyle\sum_{\gamma\in\mathbb{N}^{d}}P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\gamma}}P_{\mathcal{H}_{\gamma}}\mathscr{M}_{z_{j}}|_{\mathcal{H}_{\beta}}=\sum_{\gamma\in\mathbb{N}^{d}}P_{\tilde{\mathcal{H}}_{\alpha}}\mathscr{M}_{z_{j}}|_{\tilde{\mathcal{H}}_{\gamma}}P_{\tilde{\mathcal{H}}_{\gamma}}X|_{\mathcal{H}_{\beta}}\quad\mbox{for
all }\alpha,\beta\in\mathbb{N}^{d}.$ (5)

Both the series in (5) converge in the strong operator topology. Further, note
that for any $\alpha,\beta\in\mathbb{N}^{d}$,
$P_{\mathcal{H}_{\alpha}}\mathscr{M}_{z_{j}}|_{\mathcal{H}_{\beta}}=0$ if
$\beta\neq\alpha-\varepsilon_{j}$. Using this in (5), we get

$\displaystyle
P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\beta+\varepsilon_{j}}}P_{\mathcal{H}_{\beta+\varepsilon_{j}}}\mathscr{M}_{z_{j}}|_{\mathcal{H}_{\beta}}$
$\displaystyle=$ $\displaystyle
P_{\tilde{\mathcal{H}}_{\alpha}}\mathscr{M}_{z_{j}}|_{\tilde{\mathcal{H}}_{\alpha-\varepsilon_{j}}}P_{\tilde{\mathcal{H}}_{\alpha-\varepsilon_{j}}}X|_{\mathcal{H}_{\beta}}$
(6) $\displaystyle\mbox{for all }\ \alpha,\beta\in\mathbb{N}^{d}\mbox{ and
}j=1,\ldots,d.$

Putting $\alpha=0$ in (6) and using Lemma 2.1, we get

$\displaystyle
P_{\tilde{\mathcal{H}}_{0}}X|_{\mathcal{H}_{\beta+\varepsilon_{j}}}=0\ \mbox{
for all }\ \beta\in\mathbb{N}^{d}\ \mbox{ and }\ j=1,\ldots,d.$ (7)

Further, putting $\alpha=\beta+\varepsilon_{j}$ in (6), we get that

$\displaystyle
P_{\tilde{\mathcal{H}}_{\beta+\varepsilon_{j}}}X|_{\mathcal{H}_{\beta+\varepsilon_{j}}}$
$\displaystyle=$ $\displaystyle
P_{\tilde{\mathcal{H}}_{\beta+\varepsilon_{j}}}\mathscr{M}_{z_{j}}|_{\tilde{\mathcal{H}}_{\beta}}P_{\tilde{\mathcal{H}}_{\beta}}X|_{\mathcal{H}_{\beta}}(P_{\mathcal{H}_{\beta+\varepsilon_{j}}}\mathscr{M}_{z_{j}}|_{\mathcal{H}_{\beta}})^{-1}$
$\displaystyle\mbox{ for all }\ \beta\in\mathbb{N}^{d}\mbox{ and
}j=1,\ldots,d.$

Using recursion in the above expression, we get

$\displaystyle
P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\alpha}}=P_{\tilde{\mathcal{H}}_{\alpha}}\mathscr{M}^{\alpha}_{z}|_{\tilde{\mathcal{H}}_{0}}P_{\tilde{\mathcal{H}}_{0}}X|_{\mathcal{H}_{0}}(P_{\mathcal{H}_{\alpha}}\mathscr{M}^{\alpha}_{z}|_{\mathcal{H}_{0}})^{-1}\
\mbox{ for all }\ \alpha\in\mathbb{N}^{d}.$ (8)

Note that
$P_{\tilde{\mathcal{H}}_{0}}X|_{\mathcal{H}_{0}}:\mathcal{H}_{0}\rightarrow\tilde{\mathcal{H}}_{0}$
is an invertible operator. Indeed,

$\displaystyle
I_{\tilde{\mathcal{H}}_{0}}=P_{\tilde{\mathcal{H}}_{0}}XX^{-1}|_{\tilde{\mathcal{H}}_{0}}=\sum_{\alpha\in\mathbb{N}^{d}}P_{\tilde{\mathcal{H}}_{0}}X|_{\mathcal{H}_{\alpha}}P_{\mathcal{H}_{\alpha}}X^{-1}|_{\tilde{\mathcal{H}}_{0}}\overset{\eqref{X-0-alpha}}{=}P_{\tilde{\mathcal{H}}_{0}}X|_{\mathcal{H}_{0}}P_{\mathcal{H}_{0}}X^{-1}|_{\tilde{\mathcal{H}}_{0}}.$

Letting $C:=P_{\mathcal{H}_{0}}X^{-1}|_{\tilde{\mathcal{H}}_{0}}$, we deduce
from (8) that for all $\alpha\in\mathbb{N}^{d}$,

$\displaystyle
P_{\tilde{\mathcal{H}}_{0}}\mathscr{M}^{*\alpha}_{z}\mathscr{M}^{\alpha}_{z}|_{\tilde{\mathcal{H}}_{0}}=C^{*}(P_{\mathcal{H}_{\alpha}}\mathscr{M}^{\alpha}_{z}|_{\mathcal{H}_{0}})^{*}(P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\alpha}})^{*}(P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\alpha}})(P_{\mathcal{H}_{\alpha}}\mathscr{M}^{\alpha}_{z}|_{\mathcal{H}_{0}})C.$
(9)

Through repeated applications of (4), we obtain that

$P_{\mathcal{H}_{0}}\mathscr{M}^{*\alpha}_{z}\mathscr{M}^{\alpha}_{z}|_{\mathcal{H}_{0}}=B^{*}_{\alpha}B_{\alpha}\mbox{
for all }\alpha\in\mathbb{N}^{d}.$

Also, observe that
$(P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\alpha}})^{*}(P_{\tilde{\mathcal{H}}_{\alpha}}X|_{\mathcal{H}_{\alpha}})\leqslant\|X\|^{2}I.$
Substituting these values in (9), we get

$\displaystyle{\tilde{B}_{\alpha}}^{*}\tilde{B}_{\alpha}\leqslant\|X\|^{2}C^{*}B_{\alpha}^{*}B_{\alpha}C\mbox{
for all }\alpha\in\mathbb{N}^{d}.$ (10)

Similar arguments for $X^{-1}$ yield that

$\displaystyle\frac{1}{\|X^{-1}\|^{2}}C^{*}B_{\alpha}^{*}B_{\alpha}C\leqslant{\tilde{B}_{\alpha}}^{*}\tilde{B}_{\alpha}\quad\mbox{for
all }\alpha\in\mathbb{N}^{d}.$ (11)

Combining (10) and (11), we get

$\displaystyle\frac{1}{\|X^{-1}\|^{2}}C^{*}B_{\alpha}^{*}B_{\alpha}C\leqslant{\tilde{B}_{\alpha}}^{*}\tilde{B}_{\alpha}\leqslant\|X\|^{2}C^{*}B_{\alpha}^{*}B_{\alpha}C\mbox{
for all }\alpha\in\mathbb{N}^{d},$

establishing (1).

Conversely, suppose that there exist a bounded invertible operator $C$ on $H$
and positive constants $m_{1},m_{2}$ such that (1) holds. Define
$X:\mathcal{H}^{2}(\mathscr{B})\rightarrow\mathcal{H}^{2}(\tilde{\mathscr{B}})$
as

$\displaystyle
X\Big{(}\sum_{\alpha\in\mathbb{N}^{d}}x_{\alpha}z^{\alpha}\Big{)}=\sum_{\alpha\in\mathbb{N}^{d}}(C^{-1}x_{\alpha})z^{\alpha},\quad\sum_{\alpha\in\mathbb{N}^{d}}x_{\alpha}z^{\alpha}\in\mathcal{H}^{2}(\mathscr{B}).$

Note that $X$ is a block diagonal operator with respect to the decomposition
$\mathcal{H}^{2}(\mathscr{B})=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}\mathcal{H}_{\alpha}$.
Let
$X=\displaystyle\mathrel{\mathop{\oplus}_{\alpha\in\mathbb{N}^{d}}}X_{\alpha}$,
where $X_{\alpha}:\mathcal{H}_{\alpha}\rightarrow\tilde{\mathcal{H}}_{\alpha}$
is given by $X_{\alpha}(xz^{\alpha})=(C^{-1}x)z^{\alpha}$ for all
$\alpha\in\mathbb{N}^{d}$ and $x\in H$. Then (1) implies that
$\sqrt{m_{1}}\|xz^{\alpha}\|\leqslant\|X_{\alpha}(xz^{\alpha})\|\leqslant\sqrt{m_{2}}\|xz^{\alpha}\|$
for all $\alpha\in\mathbb{N}^{d}$ and $x\in H$. Thus $X$ is a bounded
invertible linear map. Further, the verification of the fact that
$X\mathscr{M}_{z_{j}}=\mathscr{M}_{z_{j}}X$ for all $j=1,\ldots,d$, is
obvious. This completes the proof. ∎

The proof of the preceding theorem shows that the similarity between the
multiplication operator tuples $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\mathscr{B})$ and on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$
can be characterized through an intertwining diagonal invertible operator (cf.
[8, Lemma 2.1]).

######  Corollary 2.2.

Let $H$, $\tilde{H}$ be two complex separable Hilbert spaces and
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$,
$\tilde{\mathscr{B}}=\\{\tilde{B}_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ be
two multisequences of invertible operators on $H$ and $\tilde{H}$,
respectively. Suppose that
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ is bounded
on $\mathcal{H}^{2}(\mathscr{B})$ and on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$. Then $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\mathscr{B})$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$ if and only if there exists a bounded
invertible operator $C:\tilde{H}\to H$ such that the map
$X:\mathcal{H}^{2}(\mathscr{B})\rightarrow\mathcal{H}^{2}(\tilde{\mathscr{B}})$,
given by $Xf(z)=C^{-1}f(z)$ for all $f\in\mathcal{H}^{2}(\mathscr{B})$, is
bounded and invertible.

###### Proof.

If $\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\mathscr{B})$ is similar to
$\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$, then by the proof
of the preceding theorem we can construct a bounded invertible operator
$X:\mathcal{H}^{2}(\mathscr{B})\rightarrow\mathcal{H}^{2}(\tilde{\mathscr{B}})$
given by $Xf(z)=C^{-1}f(z)$ for all $f\in\mathcal{H}^{2}(\mathscr{B})$.
Clearly, $X$ satisfies $X\mathscr{M}_{z_{j}}=\mathscr{M}_{z_{j}}X$ for all
$j=1,\ldots,d$. The converse part is straight forward. ∎

The following corollary generalizes as well as provides an alternate
verification of [6, Theorem 5.8]. It also recovers all the criteria of unitary
equivalence established in [8] and [10].

######  Corollary 2.3.

Let $H$, $\tilde{H}$ be two complex separable Hilbert spaces and
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$,
$\tilde{\mathscr{B}}=\\{\tilde{B}_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ be
two multisequences of invertible operators on $H$ and $\tilde{H}$,
respectively. Suppose that
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ is bounded
on $\mathcal{H}^{2}(\mathscr{B})$ and on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$. Then the following statements are
equivalent:

  1. (i)

$\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\mathscr{B})$ is unitarily equivalent to
$\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$.

  2. (ii)

There exists a unitary operator $V:\tilde{H}\to H$ such that

$\displaystyle{\tilde{B}_{\alpha}}^{*}\tilde{B}_{\alpha}=V^{*}B^{*}_{\alpha}B_{\alpha}V\quad\mbox{for
all }\alpha\in\mathbb{N}^{d}.$

  3. (iii)

There exists a unitary operator $V:\tilde{H}\to H$ such that the map
$X:\mathcal{H}^{2}(\mathscr{B})\rightarrow\mathcal{H}^{2}(\tilde{\mathscr{B}})$,
given by $Xf(z)=V^{*}f(z)$ for all $f\in\mathcal{H}^{2}(\mathscr{B})$, is
unitary.

###### Proof.

The equivalence of (i) and (ii) is immediate from the proof of Theorem 1.1,
whereas that of (i) and (iii) goes along the lines of the proof of the
preceding corollary. ∎

##  3\. Applications to reproducing kernel Hilbert space

In this section, we apply Theorem 1.1 to deduce the similarity between two
commuting tuples of multiplication operators on certain reproducing kernel
Hilbert spaces with diagonal kernels in which the polynomials are dense. We
proceed by recalling a few basic facts about such reproducing kernel Hilbert
spaces (for more details, see [10, 2, 9]).

Let $\Omega$ be a connected open subset of $\mathbb{C}^{d}$ containing 0 and
$E$ be a complex separable Hilbert space. Let $\mathcal{H}(\kappa)$ be the
reproducing kernel Hilbert space of $E$-valued holomorphic functions defined
on $\Omega$, where the reproducing kernel $\kappa$ is of the form

$\displaystyle\kappa(z,w)=\sum_{\alpha\in\mathbb{N}^{d}}C_{\alpha}z^{\alpha}\bar{w}^{\alpha},\
z,w\in\Omega;$

with $C_{\alpha}$ being bounded positive invertible operators on $E$. By a
diagonal kernel we mean a reproducing kernel $\kappa$ of the above form.
Moreover, if $\mathscr{M}_{z_{j}}$ is bounded on $\mathcal{H}(\kappa)$ for all
$j=1,\ldots,d$, then
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ on
$\mathcal{H}(\kappa)$ is unitarily equivalent to $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\mathscr{B})$, where
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ with
$B_{\alpha}=C_{\alpha}^{-1/2}$ for all $\alpha\in\mathbb{N}^{d}$.

The following result shows that perturbing finitely many operator coefficients
of $\kappa$ does not affect the similarity of $\mathscr{M}_{z}$.

######  Proposition 3.1.

Let $E$ be a complex separable Hilbert space and $\Omega$ be a connected open
subset of $\mathbb{C}^{d}$ containing 0. Let $\mathcal{H}(\kappa)$ and
$\mathcal{H}(\tilde{\kappa})$ be the reproducing kernel Hilbert spaces of
$E$-valued holomorphic functions defined on $\Omega$, where $\kappa$ and
$\tilde{\kappa}$ are given by

$\displaystyle\kappa(z,w)$ $\displaystyle=$
$\displaystyle\sum_{\alpha\in\mathbb{N}^{d}}C_{\alpha}z^{\alpha}\bar{w}^{\alpha},\quad
z,w\in\Omega;$ $\displaystyle\tilde{\kappa}(z,w)$ $\displaystyle=$
$\displaystyle\sum_{\underset{|\alpha|\leqslant
n}{\alpha\in\mathbb{N}^{d}}}D_{\alpha}z^{\alpha}\bar{w}^{\alpha}+\sum_{\underset{|\alpha|\geqslant
n+1}{\alpha\in\mathbb{N}^{d}}}C_{\alpha}z^{\alpha}\bar{w}^{\alpha},\quad
z,w\in\Omega;$

with $C_{\alpha}$, $D_{\alpha}$ being positive invertible operators on $E$. If
$\mathscr{M}_{z}$ is bounded on $\mathcal{H}(\kappa)$, then the following
statements are true:

  * (i)

$\mathscr{M}_{z}$ is bounded on $\mathcal{H}(\tilde{\kappa})$.

  * (ii)

$\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$.

###### Proof.

To see (i), let $j\in\\{1,\ldots,d\\}$. Note that $\mathscr{M}_{z_{j}}$ is
bounded on $\mathcal{H}(\kappa)$ if and only if there exists a positive
constant $\delta$ such that $(\delta-z_{j}\bar{w}_{j})\kappa(z,w)$ is a
positive definite kernel (refer to [9, Theorem 6.28]). It follows from [2,
Lemma 4.1(c)] that $\delta C_{\alpha}-C_{\alpha-\varepsilon_{j}}\geqslant 0$
for all $\alpha\in\mathbb{N}^{d}$. Consequently, we get that
$\mathscr{M}_{z_{j}}$ is bounded on $\mathcal{H}(\kappa)$ if and only if
$\sup_{\alpha\in\mathbb{N}^{d}}\|C_{\alpha}^{-1/2}C_{\alpha-\varepsilon_{j}}C_{\alpha}^{-1/2}\|<\infty$.
Here we follow the convention that $C_{\alpha-\varepsilon_{j}}=0$ if
$\alpha_{j}=0$. This establishes that the boundedness of $\mathscr{M}_{z}$ on
$\mathcal{H}(\kappa)$ is equivalent to that of $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$.

For the proof of (ii), suppose that $\mathscr{M}_{z}$ is bounded on
$\mathcal{H}(\kappa)$. Note that $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is
unitarily equivalent to $\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\mathscr{B})$
and $\mathscr{M}_{z}$ on $\mathcal{H}(\tilde{\kappa})$ is unitarily equivalent
to $\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$, where
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$,
$\tilde{\mathscr{B}}=\\{\tilde{B}_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ with

$B_{\alpha}=C_{\alpha}^{-1/2}\mbox{ for all }\alpha\in\mathbb{N}^{d}\
\mbox{~{}and~{}}\;\tilde{B}_{\alpha}=\begin{cases}D_{\alpha}^{-1/2}&\mbox{~{}if~{}}|\alpha|\leqslant
n,\\\ C_{\alpha}^{-1/2}&\mbox{~{}if~{}}|\alpha|\geqslant n+1.\end{cases}$

By choosing

$\displaystyle c_{1}$ $\displaystyle=$
$\displaystyle\min\left\\{\frac{1}{\|C_{\alpha}^{-1/2}D_{\alpha}C_{\alpha}^{-1/2}\|}:|\alpha|\leqslant
n\right\\}\mbox{ and}$ $\displaystyle c_{2}$ $\displaystyle=$
$\displaystyle\max\Big{\\{}\|C_{\alpha}^{1/2}D_{\alpha}^{-1}C_{\alpha}^{1/2}\|:|\alpha|\leqslant
n\Big{\\}},$

we obtain that

$\displaystyle c_{1}C_{\alpha}^{-1}\leqslant D_{\alpha}^{-1}\leqslant
c_{2}C_{\alpha}^{-1}\mbox{ for all }\alpha\in\mathbb{N}^{d}\mbox{ with
}|\alpha|\leqslant n.$

Now if we take $m_{1}=\min\\{1,c_{1}\\}$, $m_{2}=\max\\{1,c_{2}\\}$ and $C=I$,
then Theorem 1.1 establishes the similarity between $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\mathscr{B})$ and $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$. This completes the proof. ∎

In the preceding proposition, $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is
not unitarily equivalent to $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$, in general (see Corollary 2.3). The following
proposition provides another class of reproducing kernel Hilbert spaces in
which the similarity between two tuples of multiplication operators can be
deduced through Theorem 1.1.

######  Proposition 3.2.

Let $\lambda,\ \mu,\ \tilde{\lambda},\ \tilde{\mu}$ be $($strictly$)$ positive
real numbers not necessarily distinct. Let $\mathcal{H}(\kappa)$ and
$\mathcal{H}(\tilde{\kappa})$ be the reproducing kernel Hilbert spaces of
$\mathbb{C}^{2}$-valued holomorphic functions defined on the open unit
$($euclidean$)$ ball $\mathbb{B}_{d}$ of $\mathbb{C}^{d}$, where $\kappa$ and
$\tilde{\kappa}$ are given by

$\displaystyle\kappa(z,w)$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}\frac{1}{(1-\langle{z},\,{w}\rangle)^{\lambda}}&0\\\
&\vspace{-.3cm}\\\
0&\frac{1}{(1-\langle{z},\,{w}\rangle)^{\mu}}\end{matrix}\right),\quad
z,w\in\mathbb{B}_{d};$ $\displaystyle\tilde{\kappa}(z,w)$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}\frac{1}{(1-\langle{z},\,{w}\rangle)^{\tilde{\lambda}}}&0\\\
&\vspace{-.3cm}\\\
0&\frac{1}{(1-\langle{z},\,{w}\rangle)^{\tilde{\mu}}}\end{matrix}\right),\quad
z,w\in\mathbb{B}_{d}.$

Then $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is similar to
$\mathscr{M}_{z}$ on $\mathcal{H}(\tilde{\kappa})$ if and only if
$\\{\lambda,\mu\\}=\\{\tilde{\lambda},\tilde{\mu}\\}$.

###### Proof.

Routine calculations show $\mathscr{M}_{z}$ is bounded on
$\mathcal{H}(\kappa)$ and on $\mathcal{H}(\tilde{\kappa})$. Further,

$\displaystyle\kappa(z,w)$ $\displaystyle=$
$\displaystyle\sum_{\alpha\in\mathbb{N}^{d}}\frac{1}{\alpha!}\left(\begin{matrix}(\lambda)_{|\alpha|}&0\\\
&\vspace{-.3cm}\\\
0&(\mu)_{|\alpha|}\end{matrix}\right)z^{\alpha}\bar{w}^{\alpha},\quad
z,w\in\mathbb{B}_{d};$ $\displaystyle\tilde{\kappa}(z,w)$ $\displaystyle=$
$\displaystyle\sum_{\alpha\in\mathbb{N}^{d}}\frac{1}{\alpha!}\left(\begin{matrix}(\tilde{\lambda})_{|\alpha|}&0\\\
&\vspace{-.3cm}\\\
0&(\tilde{\mu})_{|\alpha|}\end{matrix}\right)z^{\alpha}\bar{w}^{\alpha},\quad
z,w\in\mathbb{B}_{d},$

where for $x>0$ and $n\in\mathbb{N}$,
$(x)_{n}:=\frac{\Gamma(x+n)}{\Gamma(x)}$. Thus $\mathscr{M}_{z}$ on
$\mathcal{H}(\kappa)$ is unitarily equivalent to $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\mathscr{B})$ and $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$ is unitarily equivalent to $\mathscr{M}_{z}$ on
$\mathcal{H}^{2}(\tilde{\mathscr{B}})$, where
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$,
$\tilde{\mathscr{B}}=\\{\tilde{B}_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ with

$\displaystyle(B^{*}_{\alpha}B_{\alpha})^{-1}=\frac{1}{\alpha!}\left(\begin{matrix}(\lambda)_{|\alpha|}&0\\\
&\vspace{-.3cm}\\\ 0&(\mu)_{|\alpha|}\end{matrix}\right)\ \mbox{ and }\
(\tilde{B}^{*}_{\alpha}\tilde{B}_{\alpha})^{-1}=\frac{1}{\alpha!}\left(\begin{matrix}(\tilde{\lambda})_{|\alpha|}&0\\\
&\vspace{-.3cm}\\\ 0&(\tilde{\mu})_{|\alpha|}\end{matrix}\right)$

for all $\alpha\in\mathbb{N}^{d}$.

Suppose that $\\{\lambda,\mu\\}=\\{\tilde{\lambda},\tilde{\mu}\\}$. If
$\lambda=\tilde{\lambda}$ and $\mu=\tilde{\mu}$, then clearly
$\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$. Let $\lambda=\tilde{\mu}$ and
$\mu=\tilde{\lambda}$. Then take $m_{1}=m_{2}=1$ and
$C=\left(\begin{matrix}0&1\\\ &\vspace{-.3cm}\\\ 1&0\end{matrix}\right)$. It
now immediately follows from Theorem 1.1 that $\mathscr{M}_{z}$ on
$\mathcal{H}(\kappa)$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$.

Conversely, suppose that $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is similar
to $\mathscr{M}_{z}$ on $\mathcal{H}(\tilde{\kappa})$. Hence,
$\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\mathscr{B})$ is similar to
$\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$. Therefore, by
Theorem 1.1, there exists an invertible matrix
$C=\left(\begin{matrix}c_{11}&c_{12}\\\ &\vspace{-.3cm}\\\
c_{21}&c_{22}\end{matrix}\right)$ and positive constants $m_{1}$, $m_{2}$ such
that for all $n\in\mathbb{N}$,

$\displaystyle m_{1}\left(\begin{matrix}(\lambda)_{n}&0\\\ &\vspace{-.3cm}\\\
0&(\mu)_{n}\end{matrix}\right)\leqslant
C\left(\begin{matrix}(\tilde{\lambda})_{n}&0\\\ &\vspace{-.3cm}\\\
0&(\tilde{\mu})_{n}\end{matrix}\right)C^{*}\leqslant
m_{2}\left(\begin{matrix}(\lambda)_{n}&0\\\ &\vspace{-.3cm}\\\
0&(\mu)_{n}\end{matrix}\right).$ (12)

The last inequality of (12) gives that

$\displaystyle\left.\begin{array}[]{cc}&|c_{11}|^{2}(\tilde{\lambda})_{n}+|c_{12}|^{2}(\tilde{\mu})_{n}\leqslant
m_{2}(\lambda)_{n},\\\
&|c_{21}|^{2}(\tilde{\lambda})_{n}+|c_{22}|^{2}(\tilde{\mu})_{n}\leqslant
m_{2}(\mu)_{n},\end{array}\right\\}n\in\mathbb{N}.$ (15)

From the first inequality of (12), we get

$\displaystyle m_{1}C^{-1}\left(\begin{matrix}(\lambda)_{n}&0\\\
&\vspace{-.3cm}\\\
0&(\mu)_{n}\end{matrix}\right)C^{*-1}\leqslant\left(\begin{matrix}(\tilde{\lambda})_{n}&0\\\
&\vspace{-.3cm}\\\ 0&(\tilde{\mu})_{n}\end{matrix}\right)\mbox{ for all
}n\in\mathbb{N}.$

This gives that

$\displaystyle\left.\begin{array}[]{cc}&\frac{m_{1}}{|\det(C)|^{2}}\big{(}|c_{22}|^{2}(\lambda)_{n}+|c_{12}|^{2}(\mu)_{n}\big{)}\leqslant(\tilde{\lambda})_{n},\\\
&\frac{m_{1}}{|\det(C)|^{2}}\big{(}|c_{21}|^{2}(\lambda)_{n}+|c_{11}|^{2}(\mu)_{n}\big{)}\leqslant(\tilde{\mu})_{n},\end{array}\right\\}n\in\mathbb{N}.$
(18)

Now we consider the following two cases:
Case I: Both $c_{11}$ and $c_{22}$ are non-zero.
Then by (15) and (18), we get

$\displaystyle\left.\begin{array}[]{cc}&\frac{m_{1}}{|\det(C)|^{2}}|c_{22}|^{2}(\lambda)_{n}\leqslant(\tilde{\lambda})_{n}\leqslant\frac{m_{2}}{|c_{11}|^{2}}(\lambda)_{n},\\\
&\frac{m_{1}}{|\det(C)|^{2}}|c_{11}|^{2}(\mu)_{n}\leqslant(\tilde{\mu})_{n}\leqslant\frac{m_{2}}{|c_{22}|^{2}}(\mu)_{n},\end{array}\right\\}n\in\mathbb{N}.$

Since $\frac{(\tilde{\lambda})_{n}}{(\lambda)_{n}}\sim
n^{\tilde{\lambda}-\lambda}$ as $n\to\infty$ (see [1, p. 257 (6.1.46)]), it
follows from above that $\lambda=\tilde{\lambda}$. By a similar argument, we
get $\mu=\tilde{\mu}$.
Case II: One of $c_{11}$ and $c_{22}$ is zero.
In this case, both $c_{12}$ and $c_{21}$ must be non-zero as $C$ is
invertible. Hence, again by (15) and (18), we obtain

$\displaystyle\left.\begin{array}[]{cc}&\frac{m_{1}}{|\det(C)|^{2}}|c_{21}|^{2}(\lambda)_{n}\leqslant(\tilde{\mu})_{n}\leqslant\frac{m_{2}}{|c_{12}|^{2}}(\lambda)_{n},\\\
&\frac{m_{1}}{|\det(C)|^{2}}|c_{12}|^{2}(\mu)_{n}\leqslant(\tilde{\lambda})_{n}\leqslant\frac{m_{2}}{|c_{21}|^{2}}(\mu)_{n},\end{array}\right\\}n\in\mathbb{N}.$

Again, using the argument as in the Case I, we get that $\tilde{\lambda}=\mu$
and $\tilde{\mu}=\lambda$. This completes the proof. ∎

We conclude this paper by characterizing the similarity between two $d$-tuples
of $\mathcal{U}(d)$-homogeneous multiplication operators acting on the
reproducing kernel Hilbert spaces determined by $n\times n$ matrix-valued
diagonal kernels defined on the open unit ball in $\mathbb{C}^{d}$.

Let $\mathcal{U}(d)$ denote the group of $d\times d$ unitary matrices and $E$
be a complex separable Hilbert space. Recall that a commuting $d$-tuple
$\mathscr{M}_{z}=(\mathscr{M}_{z_{1}},\ldots,\mathscr{M}_{z_{d}})$ of
multiplication operators on a reproducing kernel Hilbert space
$\mathcal{H}(\kappa)$ of $E$-valued holomorphic functions defined on the open
unit ball $\mathbb{B}_{d}$ of $\mathbb{C}^{d}$ is said to be
$\mathcal{U}(d)$-homogeneous (also known as spherical, see [3]) if for every
$u=(u_{ij})\in\mathcal{U}(d)$ there is a unitary operator $\Gamma(u)$ on
$\mathcal{H}(\kappa)$ such that
$\Gamma(u)\mathscr{M}_{z_{j}}=(u\cdot\mathscr{M}_{z})_{j}\Gamma(u)$ for all
$j=1,\ldots,d$, where

$(u\cdot\mathscr{M}_{z})_{j}=\sum_{k=1}^{d}u_{jk}\mathscr{M}_{z_{k}},\quad
j=1,\ldots,d.$

It follows from [4, Lemma 2.3] that $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$
is $\mathcal{U}(d)$-homogeneous if and only if $\kappa$ is quasi-invariant
under $\mathcal{U}(d)$ in the sense that for each $u\in\mathcal{U}(d)$, there
exists a unitary operator $J_{u}$ on $E$ such that
$\kappa(z,w)=J_{u}\,\kappa(u\cdot z,u\cdot w)\,J_{u}^{*}$ for all
$z,w\in\mathbb{B}_{d}$. In the case when $\kappa$ is scalar-valued, it turns
out that $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is
$\mathcal{U}(d)$-homogeneous if and only if $\kappa$ is
$\mathcal{U}(d)$-invariant (see [3, Theorem 2.12]). In view of this, note that
$\mathscr{M}_{z}$ appearing in the Proposition 3.2 are
$\mathcal{U}(d)$-homogeneous. If $\mathscr{M}_{z}$ is
$\mathcal{U}(d)$-homogeneous on $\mathcal{H}(\kappa)$ with $\kappa$ being a
diagonal kernel, then by [4, Theorem 4.10], the kernel $\kappa$ has the form

$\displaystyle\kappa(z,w)=\sum_{m\in\mathbb{N}}A_{m}\langle{z},\,{w}\rangle^{m},\quad
z,w\in\mathbb{B}_{d},$

where $\\{A_{m}\\}_{m\in\mathbb{N}}$ is a sequence of positive definite
$n\times n$ matrices.

######  Proposition 3.3.

Let $n$ be a positive integer. For the sequences
$\\{A_{m}\\}_{m\in\mathbb{N}}$ and $\\{\tilde{A}_{m}\\}_{m\in\mathbb{N}}$ of
$n\times n$ positive invertible matrices, let $\kappa$ and $\tilde{\kappa}$ be
the reproducing kernels given by

$\displaystyle\kappa(z,w)$ $\displaystyle=$
$\displaystyle\sum_{m\in\mathbb{N}}A_{m}\langle{z},\,{w}\rangle^{m},\quad
z,w\in\mathbb{B}_{d},$ $\displaystyle\tilde{\kappa}(z,w)$ $\displaystyle=$
$\displaystyle\sum_{m\in\mathbb{N}}\tilde{A}_{m}\langle{z},\,{w}\rangle^{m},\quad
z,w\in\mathbb{B}_{d}.$

Let $\mathcal{H}(\kappa)$ and $\mathcal{H}(\tilde{\kappa})$ be the reproducing
kernel Hilbert spaces of $\mathbb{C}^{n}$-valued holomorphic functions defined
on the open unit ball $\mathbb{B}_{d}$ of $\mathbb{C}^{d}$, determined by
$\kappa$ and $\tilde{\kappa}$ respectively. Suppose that $\mathscr{M}_{z}$ is
bounded on $\mathcal{H}(\kappa)$ and on $\mathcal{H}(\tilde{\kappa})$. Then
$\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$ if and only if there exist positive constants
$m_{1},m_{2}$ and an invertible $n\times n$ matrix $C$ such that

$\displaystyle m_{1}C^{*}A_{m}C\leqslant\tilde{A}_{m}\leqslant
m_{2}C^{*}A_{m}C\quad\mbox{for all }m\in\mathbb{N},$

###### Proof.

It is easy to see that

$\displaystyle\kappa(z,w)=\sum_{\alpha\in\mathbb{N}^{d}}C_{\alpha}z^{\alpha}\bar{w}^{\alpha}\quad\mbox{
and
}\quad\tilde{\kappa}(z,w)=\sum_{\alpha\in\mathbb{N}^{d}}\tilde{C}_{\alpha}z^{\alpha}\bar{w}^{\alpha},\
\ z,w\in\mathbb{B}_{d},$

where $C_{\alpha}=\frac{|\alpha|!}{\alpha!}A_{|\alpha|}$ and
$\tilde{C}_{\alpha}=\frac{|\alpha|!}{\alpha!}\tilde{A}_{|\alpha|}$ for all
$\alpha\in\mathbb{N}^{d}$. Since $\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is
unitarily equivalent to $\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\mathscr{B})$
and $\mathscr{M}_{z}$ on $\mathcal{H}(\tilde{\kappa})$ is unitarily equivalent
to $\mathscr{M}_{z}$ on $\mathcal{H}^{2}(\tilde{\mathscr{B}})$, where
$\mathscr{B}=\\{B_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$,
$\tilde{\mathscr{B}}=\\{\tilde{B}_{\alpha}\\}_{\alpha\in\mathbb{N}^{d}}$ with
$B_{\alpha}^{*}B_{\alpha}=C_{\alpha}^{-1}$ and
$\tilde{B}_{\alpha}^{*}\tilde{B}_{\alpha}=\tilde{C}_{\alpha}^{-1}$ for all
$\alpha\in\mathbb{N}^{d}$, it now immediately follows from Theorem 1.1 that
$\mathscr{M}_{z}$ on $\mathcal{H}(\kappa)$ is similar to $\mathscr{M}_{z}$ on
$\mathcal{H}(\tilde{\kappa})$ if and only if there exist positive constants
$m_{1},m_{2}$ and an invertible $n\times n$ matrix $C$ such that

$\displaystyle m_{1}C^{*}A_{|\alpha|}C\leqslant\tilde{A}_{|\alpha|}\leqslant
m_{2}C^{*}A_{|\alpha|}C\quad\mbox{for all }\alpha\in\mathbb{N}^{d}.$

This completes the proof. ∎

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