Extending Polynomially Normal Operators to (P, Q)-Normal Operators in Semi-Hilbertian Spaces

Abstract

This paper is devoted to the introduction and systematic study of $(P,Q)$-normal operators in the context of semi-Hilbertian spaces, where P and Q are non-constant complex polynomials in one variable. This class generalizes the well-known notion of polynomially normal operators and offers a natural setting to study their structural properties in spaces endowed with a semi-inner product induced by a positive operator. We establish fundamental properties of $(P,Q)$-normal operators, including conditions for commutativity with respect to the $\mathbf{A}$-adjoint and relations to other classes of $\mathbf{A}$-operators. Several examples are provided to illustrate the theory and demonstrate how $(P,Q)$-normality extends classical concepts in operator theory.

1. Introduction and Preliminaries

The study of operators acting on spaces endowed with a semi-inner product induced by a positive operator arises naturally in the framework of pseudo-Hermitian quantum mechanics. This setting provides a natural framework for extending classical notions of operator theory, leading to the introduction and systematic study of generalized classes of operators. In particular, $(P,Q)$-normal operators offer a meaningful generalization of normal operators and allow for the investigation of their structural properties in spaces equipped with such semi-inner products [1,2,3]).

Let $\mathcal{B}\left(\mathcal{K}\right)$ denote the $C^{*}$-algebra of all bounded linear operators on a complex infinite-dimensional Hilbert space $(\mathcal{K}, \parallel ·\parallel )$.

Let $\mathcal{B}{\left(\mathcal{K}\right)}^{+}$ be the set of all positive operators in $\mathcal{B}\left(\mathcal{K}\right)$, defined by

$$\mathbf{A}\in \mathcal{B}{\left(\mathcal{K}\right)}^{+}⟺\langle \mathbf{A}w|w\rangle \ge 0\mathrm{for}\mathrm{all}w\in \mathcal{K}.$$

For any operator $\mathcal{T}\in \mathcal{B}\left(\mathcal{K}\right)$, we use the notations $\mathbf{Im}\left(\mathcal{T}\right)$, $ker\left(\mathcal{T}\right)$, and ${\mathcal{T}}^{*}$ to denote its range, null space, and adjoint, respectively. Moreover, $\overline{\mathbf{Im}\left(\mathcal{T}\right)}$ stands for the closure of the range, and ${\mathcal{I}}_{\mathbf{A}}$ denotes the orthogonal projection onto $\overline{\mathbf{Im}\left(\mathbf{A}\right)}$.

Given $\mathbf{A}\in \mathcal{B}{\left(\mathcal{K}\right)}^{+}$, we define a semi-inner product on $\mathcal{K}$ by

$${\langle w|u\rangle }_{\mathbf{A}}:=\langle \mathbf{A}w|u\rangle ,$$

(1)

with the induced seminorm

$${\parallel w\parallel }_{\mathbf{A}}:=\sqrt{{\langle w|w\rangle }_{\mathbf{A}}}=\parallel \sqrt{\mathbf{A}}w\parallel ,w\in \mathcal{K}.$$

(2)

An operator $\mathcal{T}\in \mathcal{B}\left(\mathcal{K}\right)$ is called $\mathbf{A}$-bounded if there exists $\beta >0$, such that

$${\parallel \mathcal{T}w\parallel }_{\mathbf{A}}\le \beta {\parallel w\parallel }_{\mathbf{A}}\mathrm{for}\mathrm{all}w\in \mathcal{K}.$$

(3)

The set of all $\mathbf{A}$-bounded operators is denoted by ${\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$. For $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$, the $\mathbf{A}$-operator seminorm is defined by [4] as

$${\parallel \mathcal{T}\parallel }_{\mathbf{A}}:=sup\left\{{\displaystyle \frac{{\parallel \mathcal{T}u\parallel }_{\mathbf{A}}}{{\parallel u\parallel }_{\mathbf{A}}}}:u\in \overline{\mathbf{Im}\left(\mathbf{A}\right)},u\ne 0\right\}.$$

(4)

Definition 1

([4]). Let ${\mathcal{T}}_1,{\mathcal{T}}_2\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$. We say that ${\mathcal{T}}_2$ is an $\mathbf{A}$-adjoint of ${\mathcal{T}}_1$ if, for all $w,u\in \mathcal{K}$, the relation

$${\langle {\mathcal{T}}_{1}w\mid u\rangle }_{\mathbf{A}}={\langle w\mid {\mathcal{T}}_{2}u\rangle }_{\mathbf{A}}$$

(5)

holds. Equivalently, this condition can be expressed in operator form as

$$\mathbf{A}{\mathcal{T}}_1={\mathcal{T}}_2^{*}\mathbf{A}.$$

(6)

An operator ${\mathcal{T}}_1\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ is called $\mathbf{A}$-selfadjoint if it coincides with its $\mathbf{A}$-adjoint, that is,

$$\mathbf{A}{\mathcal{T}}_1={\mathcal{T}}_1^{*}\mathbf{A}.$$

(7)

Following Douglas theorem, as presented in [5,6], an operator $\mathcal{T}\in \mathcal{B}\left(\mathcal{K}\right)$ admits an $\mathbf{A}$-adjoint if and only if

$$\mathbf{Im}\left({\mathcal{T}}^{*}\mathbf{A}\right)\subseteq \mathbf{Im}\left(\mathbf{A}\right).$$

(8)

The set of all operators that admit an $\mathbf{A}$-adjoint is denoted by ${\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$. As is standard in the literature, for each $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$, there exists a distinguished $\mathbf{A}$-adjoint operator of $\mathcal{T}$, called the reduced solution of the equation

$$\mathbf{A}X={\mathcal{T}}^{*}\mathbf{A},$$

(9)

which is denoted by ${\mathcal{T}}^{♯}$ or ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ (see [4]). In this paper, we adopt the notation ${\mathcal{T}}^{{*}_{\mathbf{A}}}$.

The $\mathbf{A}$-adjoint operator ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ satisfies the following properties:

$$\left\{\begin{array}{c}\mathbf{A}{\mathcal{T}}^{{*}_{\mathbf{A}}}={\mathcal{T}}^{*}\mathbf{A},\hfill \\ \mathbf{Im}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)},\hfill \\ ker\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)=ker\left({\mathcal{T}}^{*}\mathbf{A}\right).\hfill \end{array}\right.$$

(10)

Some properties of the map ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ are summarized in the following theorem.

Theorem 1

([4,7,8]). The operator ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ satisfies the following properties:

1.

${\mathcal{T}}^{{*}_{\mathbf{A}}}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$.

2.

${\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}={\mathcal{I}}_{\mathbf{A}}\mathcal{T}{\mathcal{I}}_{\mathbf{A}}$.

3.

If $\mathcal{S}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$, then $\mathcal{T}\mathcal{S}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ and ${\left(\mathcal{T}\mathcal{S}\right)}^{{*}_{\mathbf{A}}}={\mathcal{S}}^{{*}_{\mathbf{A}}}{\mathcal{T}}^{{*}_{\mathbf{A}}}.$

4.

${\left({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}={\mathcal{T}}^{{*}_{\mathbf{A}}}.$

• Recall that an operator $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ is said to be $\mathbf{A}$-isometric ([4]), if

  $${\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}={\mathcal{I}}_{\mathbf{A}}$$

  (11)

  and $\mathbf{A}$-unitary if

  $${\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}={\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}{\mathcal{T}}^{{*}_{\mathbf{A}}}={\mathcal{I}}_{\mathbf{A}},$$

  (12)

  that is, both $\mathcal{T}$ and ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ are $\mathbf{A}$-isometries ([4]).

Several broader notions of normality have emerged in recent developments, extending the classical framework of normal operators. Notable examples include n-normal, $(n,m)$-normal, polynomially normal, and doubly polynomially normal operators on Hilbert spaces ([9,10,11,12]). Let $\mathcal{T}\in \mathcal{B}\left(\mathcal{K}\right)$. The operator $\mathcal{T}$ is said to belong to one of the following generalized normal classes:

1.

n-normal if

$${\mathcal{T}}^n{\mathcal{T}}^{*}-{\mathcal{T}}^{*}{\mathcal{T}}^n=0$$

(13)

for some positive integer n ([12]).

2.

$(n,m)$-normal if

$${\mathcal{T}}^n{\mathcal{T}}^{*m}-{\mathcal{T}}^{*m}{\mathcal{T}}^n=0$$

(14)

for some positive integers n and m ([11]).

3.

Polynomially normal if

$$P\left(\mathcal{T}\right){\mathcal{T}}^{*}={\mathcal{T}}^{*}P\left(\mathcal{T}\right)$$

(15)

for some polynomial $P\in \mathbb{C}\left[z\right]$ ([13]).

4.

Doubly polynomially normal if

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{*}\right)=Q\left({\mathcal{T}}^{*}\right)P\left(\mathcal{T}\right)$$

(16)

for some polynomials $P, Q\in \mathbb{C}\left[z\right]$ ([14]).

These classes of generalized normal operators satisfy the chain of inclusions

$$n\mathrm{-normal}\subseteq (n,m)\mathrm{-normal}\subseteq \mathrm{polynomially}\mathrm{normal}\subseteq \mathrm{doubly}\mathrm{polynomially}\mathrm{normal}.$$

The classical concepts of normality in Hilbert spaces have been naturally extended to semi-Hilbertian spaces, giving rise to the notion of $\mathbf{A}$-normal operators and their various generalizations. A number of authors have contributed significantly to the formulation and analysis of these $\mathbf{A}$-extensions. For a comprehensive treatment, the reader may consult [4,7,8,15,16]. Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$. The operator $\mathcal{T}$ is said to belong to one of the following classes of $\mathbf{A}$-normal operators:

1.

$\mathbf{A}$-normal ([16]) if

$${\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}=\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}.$$

(17)

2.

$(\alpha ,\beta )$-$\mathbf{A}$-normal ([15]) if

$${\alpha }^2{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}{\le }_{\mathbf{A}}\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}{\le }_{\mathbf{A}}{\beta }^2{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T},$$

(18)

for $0\le \alpha \le 1\le \beta$.

3.

$(n,m)$-$\mathbf{A}$-normal ([17]) if

$${\mathcal{T}}^n{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^m={\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^m{\mathcal{T}}^n.$$

(19)

4.

Polynomially $\mathbf{A}$-normal ([18]) if

$$P\left(\mathcal{T}\right){\mathcal{T}}^{{*}_{\mathbf{A}}}-{\mathcal{T}}^{{*}_{\mathbf{A}}}P\left(\mathcal{T}\right)=0$$

(20)

for some polynomial $P\in \mathbb{C}\left[z\right]$.

5.

$(P,m)$-$\mathbf{A}$-normal ([19]) if

$$P\left(\mathcal{T}\right){\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^m-{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{m}P\left(\mathcal{T}\right)=0,$$

(21)

for some polynomial $P\in \mathbb{C}\left[z\right]$ and a positive integer m.

For recent developments in the theory of Hilbert space operators, the reader may consult [20,21,22,23,24].

In this paper, we introduce and systematically study the class of $(P,Q)$-normal operators in semi-Hilbertian spaces, where P and Q are nonconstant complex polynomials. These spaces are endowed with a semi-inner product induced by a positive operator, providing a natural generalization of classical Hilbert spaces and a suitable framework for extending traditional operator theory. Building on the recently developed notions of operator normality in semi-Hilbertian spaces, we define $(P,Q)$-normality as a natural extension of polynomially normal operators and investigate the fundamental algebraic and spectral properties of these operators. In particular, we examine their behavior under common operator constructions, study conditions for their $\mathbf{A}$-adjoints, and explore their relationships with other generalized classes of operators such as $(n,m)$-$\mathbf{A}$-normal and $(\alpha ,\beta )$-$\mathbf{A}$-normal operators. Several results are provided to highlight the structural and spectral characteristics of $(P,Q)$-normal operators, showing how this class not only generalizes classical polynomially normal operators but also provides new insights into the interplay between operator polynomials and semi-Hilbertian structures.

2. $(\mathit{P},\mathit{Q})$-$\mathbf{A}$-Normal Operators in Semi-Hilbertian Spaces

In this section, we introduce a new notion of operator normality, termed $(P,Q)$-$\mathbf{A}$-normal operators, where P and Q are nonconstant complex polynomials in a single variable. We explore several fundamental properties of this class, building on earlier results established for related concepts of $\mathbf{A}$-normality.

Definition 2.

Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$. The operator $\mathcal{T}$ is called a $(P,Q)$-$\mathbf{A}$-normal operator if there exist two nonconstant complex polynomials

$$P\left(z\right)=\sum _{k=0}^n{w}_k{z}^k\in \mathbb{C}\left[z\right],Q\left(z\right)=\sum _{j=0}^m{b}_j{z}^j\in \mathbb{C}\left[z\right],$$

such that

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)=0,$$

(22)

or equivalently,

$$\sum _{k=0}^n\sum _{j=0}^m{w}_k{b}_j\left({\mathcal{T}}^k{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j-{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j{\mathcal{T}}^k\right)=0.$$

(23)

In particular, if $Q\left(z\right)=z$, then $\mathcal{T}$ reduces to a polynomially $\mathbf{A}$-normal operator.

Remark 1.

The above definition of $(P,Q)$-$\mathbf{A}$-normal operators includes several important special cases:

1.

A $(z,z)$-$\mathbf{A}$-normal operator coincides with a standard $\mathbf{A}$-normal operator.

2.

A $(z^n,z)$-$\mathbf{A}$-normal operator coincides with an n-$\mathbf{A}$-normal operator.

3.

A $(z^n,z^m)$-$\mathbf{A}$-normal operator coincides with an $(n,m)$-$\mathbf{A}$-normal operator.

4.

A $\left(P\right(z),z)$-$\mathbf{A}$-normal operator coincides with a polynomially $\mathbf{A}$-normal (or P-$\mathbf{A}$-normal) operator, where $P\in \mathbb{C}\left[z\right]$.

These observations highlight that the concept of $(P,Q)$-$\mathbf{A}$-normal operators provides a unified framework encompassing several previously studied classes of $\mathbf{A}$-normal operators. In particular, by choosing specific forms of the polynomials P and Q, one can recover familiar operator classes such as $\mathbf{A}$-normal, n-$\mathbf{A}$-normal, $(n,m)$-$\mathbf{A}$-normal, and polynomially $\mathbf{A}$-normal operators. This demonstrates the versatility of the $(P,Q)$-$\mathbf{A}$-normality notion and its potential to facilitate the study of algebraic and spectral properties across a wide spectrum of semi-Hilbertian operators.

To illustrate the distinctive features of $(P,Q)$-$\mathbf{A}$-normal operators and to show that this class genuinely extends previously studied notions of $\mathbf{A}$-normality, we present the following concrete example.

Example 1.

Consider the operator

$$\mathcal{T}=\left(\begin{array}{cc}2& 2\\ 0& 0\end{array}\right)\in \mathcal{B}\left({\mathbb{C}}^2\right),\mathbf{A}=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\in \mathcal{B}{\left({\mathbb{C}}^2\right)}^{+}.$$

It can be easily verified that

$$\mathrm{Im}\left({\mathcal{T}}^{*}\mathbf{A}\right)\subseteq \mathrm{Im}\left(\mathbf{A}\right),$$

which implies that $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left({\mathbb{C}}^2\right)$. A straightforward computation yields the $\mathbf{A}$-adjoint

$${\mathcal{T}}^{{*}_{\mathbf{A}}}=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right).$$

We then observe that

$${\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}-\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}\ne 0,{\mathcal{T}}^{{*}_{\mathbf{A}}}P\left(\mathcal{T}\right)-P\left(\mathcal{T}\right){\mathcal{T}}^{{*}_{\mathbf{A}}}\ne 0,$$

for $P\left(z\right)=z^2+1$.

However, if we take $P\left(z\right)=z^2+1$ and $Q\left(z\right)=z^2-2z+1\in \mathbb{C}\left[z\right]$, a direct calculation shows

$$Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)-P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)=0.$$

Hence, $\mathcal{T}$ is a $(P,Q)$-$\mathbf{A}$-normal operator, while it is neither $\mathbf{A}$-normal nor P-$\mathbf{A}$-normal.

This example illustrates that the class of $(P,Q)$-$\mathbf{A}$-normal operators is strictly broader than both the $\mathbf{A}$-normal and P-$\mathbf{A}$-normal classes. Even though $\mathcal{T}$ does not satisfy these stronger notions of normality, it satisfies the defining condition of $(P,Q)$-$\mathbf{A}$-normality for appropriate choices of the polynomials P and Q.

For convenience, we adopt the following abbreviations:

$$\begin{array}{cc}\hfill \mathbf{A}\mathrm{-normal}& \equiv \mathbf{A}\mathbf{-N},\hfill \\ \hfill n\mathbf{-A}\mathrm{-normal}& \equiv n\mathbf{-A}\mathbf{-N},\hfill \\ \hfill (n,m)\mathbf{-A}\mathrm{-normal}& \equiv (n,m)\mathbf{-A}\mathbf{-N},\hfill \\ \hfill (P,m)\mathbf{-A}\mathrm{-normal}& \equiv (P,m)\mathbf{-A}\mathbf{-N},\hfill \\ \hfill (P,Q)\mathbf{-A}\mathrm{-normal}& \equiv (P,Q)\mathbf{-A}\mathbf{-N}.\hfill \end{array}$$

Remark 2.

The hierarchy of $\mathbf{A}$-normal operator classes can be summarized as follows:

$$\mathbf{A}\mathbf{-N}\Rightarrow n\mathbf{-A}\mathbf{-N}\Rightarrow (n,m)\mathbf{-A}\mathbf{-N}\Rightarrow (P,m)\mathbf{-A}\mathbf{-N}\Rightarrow (P,Q)\mathbf{-A}\mathbf{-N}$$

This diagram shows the inclusion relations between the different classes of $\mathbf{A}$-normal operators.

Remark 3.

Let ${\displaystyle P\left(z\right)=\sum _{k=0}^n{w}_k{z}^k\in \mathbb{C}\left[z\right].}$ Then, the operator polynomial $P\left(\mathcal{T}\right)$ is given by ${\displaystyle P\left(\mathcal{T}\right)=\sum _{k=0}^n{w}_k{\mathcal{T}}^k,}$ and its $\mathbf{A}$-adjoint satisfies

$${\displaystyle {\left(P\left(\mathcal{T}\right)\right)}^{{*}_{\mathbf{A}}}=\sum _{k=0}^n\overline{w_k}{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^k=\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right),\mathit{where}\overline{P}\left(z\right)=\sum _{k=0}^n\overline{w_k}{z}^k.}$$

We now provide a theorem that gives a characterization of $(P,Q)$-$\mathbf{A}$-normal operators.

Theorem 2.

Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ and let $P,Q\in \mathbb{C}\left[z\right]$. Then, $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, if and only if the following conditions hold:

$$\mathbf{Im}\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)}.$$

(24)

$${〈Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u|\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u〉}_{\mathbf{A}}={〈P\left(\mathcal{T}\right)u|\overline{Q}\left(\mathcal{T}\right)u〉}_{\mathbf{A}},\forall u\in \mathcal{K}.$$

(25)

Proof. 

Assume first that $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, that is,

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)=0.$$

Since

$$\mathbf{Im}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)},$$

it follows that

$$\mathbf{Im}\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\right)=\mathbf{Im}\left(Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)},$$

which proves condition (24).

Moreover, for any $u\in \mathcal{K}$, we have

$$\begin{array}{cc}\hfill 0& ={〈\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)u|u〉}_{\mathbf{A}}\hfill \\ \hfill & ={〈Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u|\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u〉}_{\mathbf{A}}-{〈P\left(\mathcal{T}\right)u|\overline{Q}\left(\mathcal{T}\right)u〉}_{\mathbf{A}},\hfill \end{array}$$

which establishes condition (25).

Conversely, suppose that conditions (24) and (25) are satisfied. From (24) and the inclusion

$$\mathbf{Im}\left(Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)}=ker{\left(\mathbf{A}\right)}^{\perp },$$

we obtain

$$\mathbf{Im}\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)}.$$

Next, using condition (25), for all $u\in \mathcal{K}$,

$$\begin{array}{cc}\hfill 0& ={〈Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u|\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u〉}_{\mathbf{A}}-{〈P\left(\mathcal{T}\right)u|\overline{Q}\left(\mathcal{T}\right)u〉}_{\mathbf{A}}\hfill \\ \hfill & ={〈\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)u|u〉}_{\mathbf{A}}.\hfill \end{array}$$

This implies

$$\mathbf{A}\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)=0,$$

and, hence,

$$\mathbf{Im}\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)\subseteq ker\left(\mathbf{A}\right).$$

Finally, for all $u\in \mathcal{K}$, we have

$$\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)\right)u\in ker\left(\mathbf{A}\right)\cap ker{\left(\mathbf{A}\right)}^{\perp }=\left\{0\right\},$$

which yields

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)=0.$$

Therefore, $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, completing the proof. □

Remark 4.

In this remark, we present several specific instances of Theorem 2 that have appeared in previous studies. These examples show that the general $(P,Q)$-$\mathbf{A}$-normal framework not only encompasses well-known classes of $\mathbf{A}$-normal operators such as $\mathbf{A}$-$\mathbf{N}$, n-$\mathbf{A}$-$\mathbf{N}$, $(n,m)$-$\mathbf{A}$-$\mathbf{N}$, and polynomially $\mathbf{A}$-$\mathbf{N}$ operators, but also unifies these diverse notions under a single, more general concept. By choosing suitable polynomials P and Q, each classical case can be recovered as a particular instance of the $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$ condition.

• (i) If $P\left(z\right)=Q\left(z\right)=z$, then conditions (24) and (25)) of Theorem 2 reduce to

  $$\mathbf{Im}\left(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)},$$

  (26)

  $$\parallel {\mathcal{T}}^{{*}_{\mathbf{A}}}{u\parallel }_{\mathbf{A}}={\parallel \mathcal{T}u\parallel }_{\mathbf{A}},\forall u\in \mathcal{K}.$$

  (27)

  This recovers the notion of $\mathbf{A}$-$\mathbf{N}$ operators. In this case, Theorem 2 coincides with [16] (Theorem 2.1).
• (ii) If $P\left(z\right)=z^n$ and $Q\left(z\right)=z^m$, the conditions become

  $$\mathbf{Im}\left({\mathcal{T}}^n{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^m\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)},$$

  (28)

  $${〈{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{m}u|{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{n}u〉}_{\mathbf{A}}={〈{\mathcal{T}}^{n}u|{\mathcal{T}}^{m}u〉}_{\mathbf{A}},\forall u\in \mathcal{K}.$$

  (29)

  This corresponds to the class of $(n,m)$-$\mathbf{A}$-$\mathbf{N}$ operators, and Theorem 2 reduces to [17] (Theorem 2.1). It illustrates how powers of the operator and its $\mathbf{A}$-adjoint characterize generalized normality.
• (iii) If $P\left(z\right)={\sum }_{k=0}^n{a}_k{z}^k\in \mathbb{C}\left[z\right]$ and $Q\left(z\right)=z$, then the conditions are

  $$\mathbf{Im}\left(P\left(\mathcal{T}\right){\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)}m$$

  (30)

  $${〈{\mathcal{T}}^{{*}_{\mathbf{A}}}u|\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u〉}_{\mathbf{A}}={〈P\left(\mathcal{T}\right)u|\mathcal{T}u〉}_{\mathbf{A}},\forall u\in \mathcal{K}.$$

  (31)

  This recovers the notion of polynomially $\mathbf{A}$-$\mathbf{N}$ operators (or P-$\mathbf{A}$-normal operators), and Theorem 2 coincides with [18] (Theorem 2.1).

The following proposition examines the relationship between the $(P,Q)$-$\mathbf{A}$-normality of an operator $\mathcal{T}$ and that of its $\mathbf{A}$-adjoint ${\mathcal{T}}^{{*}_{\mathbf{A}}}$, thus extending the classical properties of normal operators to the broader $(P,Q)$-$\mathbf{A}$-normal framework.

Proposition 1.

Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$, P and Q are in $\mathbb{C}\left[z\right]$. The following are true.

• (1) If $\mathcal{T}$ is an $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, then ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ is an $(\overline{P},\overline{Q})$-$\mathbf{A}$-$\mathbf{N}$ operator.
• (2) If ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ is an $(\overline{P},\overline{Q})$-$\mathbf{A}$-$\mathbf{N}$ and $ker{\left(\mathbf{A}\right)}^{\perp }$ is invariant subspace for $\mathcal{T}$, then $\mathcal{T}$ is an $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$ operator.

Proof. 

(1) Since $\mathcal{T}$ is a $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, we have

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)=0.$$

If we put ${\displaystyle P\left(z\right)=\sum _{0\le k\le n}{w}_k{z}^k}$ and ${\displaystyle Q\left(z\right)=\sum _{0\le j\le m}{b}_j{z}^j,}$ we obtain

$$\begin{array}{cc} \hfill & P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)=0\hfill \\ \Rightarrow \hfill & \sum _{0\le k\le n}{w}_k{\mathcal{T}}^k\sum _{0\le j\le m}{b}_j{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j-\sum _{0\le j\le m}{b}_j{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j\sum _{0\le k\le n}{w}_k{\mathcal{T}}^k=0\hfill \\ \Rightarrow \hfill & \sum _{\begin{array}{c}0\le k\le n\\ 0\le j\le m\end{array}}{w}_k{b}_j\left({\mathcal{T}}^k{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j-{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j{\mathcal{T}}^k\right)=0\hfill \\ \Rightarrow \hfill & \sum _{\begin{array}{c}0\le k\le n\\ 0\le j\le m\end{array}}\overline{w_k}\overline{b_j}\left({\left({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)}^j{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^k-{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^k{\left({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)}^j\right)=0\hfill \\ \Rightarrow \hfill & \overline{Q}({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}})\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\overline{Q}({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}})=0.\hfill \end{array}$$

This means that ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ is a $(\overline{P},\overline{Q})$-$\mathbf{A}$-$\mathbf{N}$.

• (2) Let $\mathcal{K}=ker{\left(\mathbf{A}\right)}^{\perp }\oplus ker\left(\mathbf{A}\right)$. Since $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$, we have

  $$\mathcal{T}(ker(\mathbf{A}\left)\right)\subseteq ker\left(\mathbf{A}\right).$$

  By our additional assumption,

  $$\mathcal{T}\left(ker{\left(\mathbf{A}\right)}^{\perp }\right)\subseteq ker{\left(\mathbf{A}\right)}^{\perp }.$$

  Thus, both $ker\left(\mathbf{A}\right)$ and $ker{\left(\mathbf{A}\right)}^{\perp }$ are invariant subspaces for $\mathcal{T}$, and so $ker\left(\mathbf{A}\right)$ is a reducing subspace for $\mathcal{T}$.

The block form of $\mathcal{T}$ in the decomposition

$$\mathcal{K}=\overline{\mathbf{Im}\left(\mathbf{A}\right)}\oplus ker\left(\mathbf{A}\right),I-{\mathcal{I}}_{\mathcal{A}}={\mathcal{I}}_{ker\left(\mathbf{A}\right)}$$

is

$$\mathcal{T}=\left(\begin{array}{cc}{\mathcal{I}}_{\mathcal{A}}\mathcal{T}{\mathcal{I}}_{\mathcal{A}}& {\mathcal{I}}_{\mathcal{A}}\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})\\ \\ (I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}{\mathcal{I}}_{\mathcal{A}}& (I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})\end{array}\right).$$

Step 1: Show ${\mathcal{I}}_{\mathcal{A}}\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})=0$.

The operator ${\mathcal{I}}_{\mathcal{A}}\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})$ maps $ker\left(\mathbf{A}\right)$ into $\overline{\mathbf{Im}\left(\mathbf{A}\right)}$. Let $x\in ker\left(\mathbf{A}\right)$. Then, $(I-{\mathcal{I}}_{\mathcal{A}})x=x$, so

$${\mathcal{I}}_{\mathcal{A}}\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})x={\mathcal{I}}_{\mathcal{A}}\mathcal{T}x.$$

By assumption, $\mathcal{T}(ker\left(\mathbf{A}\right))\subseteq ker\left(\mathbf{A}\right)\perp \overline{\mathbf{Im}\left(\mathbf{A}\right)}$. Hence

$${\mathcal{I}}_{\mathcal{A}}\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})x=0\forall x\in ker\left(\mathbf{A}\right),$$

so

$${\mathcal{I}}_{\mathcal{A}}\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})=0.$$

Step 2: Show $(I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}{\mathcal{I}}_{\mathcal{A}}=0$.

Similarly, for $x\in \overline{\mathbf{Im}\left(\mathbf{A}\right)}$, $(I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}{\mathcal{I}}_{\mathcal{A}}x=(I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}x=0$ since $\mathcal{T}\left(\overline{\mathbf{Im}\left(\mathbf{A}\right)}\right)\subseteq \overline{\mathbf{Im}\left(\mathbf{A}\right)}$. Therefore,

$$(I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}{\mathcal{I}}_{\mathcal{A}}=0.$$

Step 3: Deduce $\mathcal{T}{\mathcal{I}}_{\mathcal{A}}={\mathcal{I}}_{\mathcal{A}}\mathcal{T}$.

The block matrix of $\mathcal{T}$ becomes diagonal:

$$\mathcal{T}=\left(\begin{array}{cc}{\mathcal{I}}_{\mathcal{A}}\mathcal{T}{\mathcal{I}}_{\mathcal{A}}& 0\\ \\ 0& (I-{\mathcal{I}}_{\mathcal{A}})\mathcal{T}(I-{\mathcal{I}}_{\mathcal{A}})\end{array}\right),$$

so

$$\mathcal{T}{\mathcal{I}}_{\mathcal{A}}=\left(\begin{array}{cc}{\mathcal{I}}_{\mathcal{A}}\mathcal{T}{\mathcal{I}}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)={\mathcal{I}}_{\mathcal{A}}\mathcal{T}.$$

Assume that ${\mathcal{T}}^{{*}_{\mathbf{A}}}$ is $(\overline{P},\overline{Q})$-$\mathbf{A}$-normal, then

$$\overline{Q}({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}})\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\overline{Q}({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}})=0.$$

Using

$${\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}={\mathcal{I}}_{\mathbf{A}}\mathcal{T}{\mathcal{I}}_{\mathbf{A}}=\mathcal{T}{\mathcal{I}}_{\mathbf{A}},$$

a direct calculation gives

$$\begin{array}{cc} \hfill & \overline{Q}({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}})\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-\overline{P}\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\overline{Q}({\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}})=0\hfill \\ \Rightarrow \hfill & {\mathcal{I}}_{\mathbf{A}}(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right))=0.\hfill \end{array}$$

We deduce

$$(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right))x\in ker\left(\mathbf{A}\right).$$

Since $ker\left(\mathbf{A}\right)$ is reducing for $\mathcal{T}$, we have

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)u\in \overline{\mathbf{Im}\left(\mathbf{A}\right)},\forall u\in \mathcal{K}.$$

Hence

$$(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right))x\in ker{\left(\mathbf{A}\right)}^{\perp },$$

so

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right)=0.$$

Thus, $\mathcal{T}$ is a $(P,Q)$-$\mathbf{A}$-normal operator. □

Proposition 2.

Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ and let $P,Q\in \mathbb{C}\left[z\right]$. Define

$$\mathcal{E}=P\left(\mathcal{T}\right)+Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right),\mathcal{F}=P\left(\mathcal{T}\right)-Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right),\mathcal{Z}=P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right).$$

The following provides several algebraic characterizations of the $(P,Q)$-$\mathbf{A}$-normality of $\mathcal{T}$.

• The operator $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, if and only if

  $$\mathcal{E}\mathcal{F}-\mathcal{F}\mathcal{E}=0,$$

  (32)
  i.e., the operators $\mathcal{E}$ and $\mathcal{F}$ commute.
• If $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, then the mixed operator $\mathcal{Z}$ commutes with both $\mathcal{E}$ and $\mathcal{F}$:

  $$\mathcal{Z}\mathcal{E}-\mathcal{E}\mathcal{Z}=0,\mathcal{F}\mathcal{Z}-\mathcal{Z}\mathcal{F}=0.$$

  (33)
• The operator $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, if and only if one of the following equivalent commutation relations is satisfied:

  $$\left(i\right)P\left(\mathcal{T}\right)\mathcal{E}-\mathcal{E}P\left(\mathcal{T}\right)=0,\left(\mathit{ii}\right)P\left(\mathcal{T}\right)\mathcal{F}-\mathcal{F}P\left(\mathcal{T}\right)=0.$$
  Thus, either relation alone provides an alternative characterization of $(P,Q)$-$\mathbf{A}$-normality.

Proof. 

Set

$$P:=P\left(\mathcal{T}\right),Q:=Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right),$$

so that

$$\mathcal{E}=P+Q,\mathcal{F}=P-Q,\mathcal{Z}=PQ.$$

Recall that $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, if and only if

$$[P,Q]=PQ-QP=0.$$

(1) Compute

$$\begin{array}{cc}\hfill \mathcal{E}\mathcal{F}-\mathcal{F}\mathcal{E}& =(P+Q)(P-Q)-(P-Q)(P+Q)\hfill \\ \hfill & =(P^2-PQ+QP-Q^2)-(P^2+PQ-QP-Q^2)\hfill \\ \hfill & =-2PQ+2QP=2(QP-PQ)=2[Q,P].\hfill \end{array}$$

Hence, $\mathcal{E}\mathcal{F}-\mathcal{F}\mathcal{E}=0$, if and only if $[P,Q]=0$, which is exactly the $(P,Q)$-$\mathbf{A}$-normality of $\mathcal{T}$.

• (2) Assume $[P,Q]=0$. Then

  $$\begin{array}{cc}\hfill \mathcal{Z}\mathcal{E}-\mathcal{E}\mathcal{Z}& =PQ(P+Q)-(P+Q)PQ\hfill \\ \hfill & =P(QP-PQ)+(PQ-PQ)Q=(P+Q)[Q,P]=0,\hfill \end{array}$$

  and similarly,

  $$\begin{array}{cc}\hfill \mathcal{F}\mathcal{Z}-\mathcal{Z}\mathcal{F}& =(P-Q)PQ-PQ(P-Q)\hfill \\ \hfill & =(P-Q)[P,Q]=0.\hfill \end{array}$$
• (3) Observe

  $$\begin{array}{cc}\hfill P\mathcal{E}-\mathcal{E}P& =P(P+Q)-(P+Q)P=PQ-QP=[P,Q],\hfill \\ \hfill P\mathcal{F}-\mathcal{F}P& =P(P-Q)-(P-Q)P=-(PQ-QP)=-[P,Q].\hfill \end{array}$$

  Hence, either $P\mathcal{E}-\mathcal{E}P=0$ or $P\mathcal{F}-\mathcal{F}P=0$ is equivalent to $[P,Q]=0$, which by part (1) is equivalent to the $(P,Q)$-$\mathbf{A}$-normality of $\mathcal{T}$. □

Lemma 1

([25] (Lemma 3.1)). Let ${\left({\mathcal{T}}_{ij}\right)}_{1\le i,j\le 2}$, where ${\mathcal{T}}_{ij}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ for all $i,j=1,2$. Then,

$$\mathcal{T}=\left(\begin{array}{cc}{\mathcal{T}}_{11}& {\mathcal{T}}_{12}\\ {\mathcal{T}}_{21}& {\mathcal{T}}_{22}\end{array}\right)\in {\mathcal{B}}_{{\mathbf{A}}_0}(\mathcal{K}\oplus \mathcal{K}),\mathit{where}{\mathbf{A}}_0=\left(\begin{array}{cc}\mathbf{A}& 0\\ 0& \mathbf{A}\end{array}\right).$$

Furthermore,

$${\mathcal{T}}^{{*}_{{\mathbf{A}}_0}}=\left(\begin{array}{cc}{\mathcal{T}}_{11}^{{*}_{\mathbf{A}}}& {\mathcal{T}}_{21}^{{*}_{\mathbf{A}}}\\ {\mathcal{T}}_{12}^{{*}_{\mathbf{A}}}& {\mathcal{T}}_{22}^{{*}_{\mathbf{A}}}\end{array}\right).$$

The following theorem provides a structural criterion ensuring that an upper triangular block operator

$$\mathcal{T}=\left(\begin{array}{cc}{\mathcal{T}}_{11}& {\mathcal{T}}_{12}\\ 0& {\mathcal{T}}_{13}\end{array}\right)$$

inherits $(P,Q)$-$\mathbf{A}$-normality from its diagonal entries ${\mathcal{T}}_{11}$ and ${\mathcal{T}}_{13}$.

Theorem 3.

Let

$$\mathcal{T}=\left(\begin{array}{cc}{\mathcal{T}}_{11}& {\mathcal{T}}_{12}\\ 0& {\mathcal{T}}_{13}\end{array}\right)\in {\mathcal{B}}_{{\mathbf{A}}_0}(\mathcal{K}\oplus \mathcal{K}),$$

and let $P,Q\in \mathbb{C}\left[z\right]$. Assume that ${\mathcal{T}}_{11}$ and ${\mathcal{T}}_{13}$ are $(P,Q)$-$\mathbf{A}$-normal and that ${\mathcal{T}}_{11}{\mathcal{T}}_{12}={\mathcal{T}}_{12}{\mathcal{T}}_{11}$. If, moreover, ${\mathcal{T}}_{13}{\mathcal{T}}_{12}={\mathcal{T}}_{12}{\mathcal{T}}_{13}=0$, then $\mathcal{T}$ is $(P,Q)$-${\mathbf{A}}_0$-normal, where ${\mathbf{A}}_0=\mathrm{diag}(\mathbf{A},\mathbf{A})$.

Proof. 

Write

$$P\left(z\right)=\sum _{k=0}^n{w}_k{z}^k,Q\left(z\right)=\sum _{j=0}^m{b}_j{z}^j.$$

By a routine computation, one obtains, for $k\ge 1,$

$${\mathcal{T}}^k=\left(\begin{array}{cc}{\mathcal{T}}_{11}^k& \sum _{l=0}^{k-1}{\mathcal{T}}_{11}^l{\mathcal{T}}_{12}{\mathcal{T}}_{13}^{k-1-l}\\ 0& {\mathcal{T}}_{13}^k\end{array}\right),$$

and for $j\ge 1$,

$${\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_0}}\right)}^j=\left(\begin{array}{cc}{\left({\mathcal{T}}_{11}^{{*}_{\mathbf{A}}}\right)}^j& 0\\ \\ \sum _{l=0}^{j-1}{\left({\mathcal{T}}_{13}^{{*}_{\mathbf{A}}}\right)}^{j-1-l}{\mathcal{T}}_{12}^{{*}_{\mathbf{A}}}{\left({\mathcal{T}}_{11}^{{*}_{\mathbf{A}}}\right)}^l& {\left({\mathcal{T}}_{13}^{{*}_{\mathbf{A}}}\right)}^j\end{array}\right).$$

Since ${\mathcal{T}}_{13}{\mathcal{T}}_{12}={\mathcal{T}}_{12}{\mathcal{T}}_{13}=0$, all off-diagonal sums vanish, and thus

$${\mathcal{T}}^k=\left(\begin{array}{cc}{\mathcal{T}}_{11}^k& 0\\ 0& {\mathcal{T}}_{13}^k\end{array}\right),{\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_0}}\right)}^j=\left(\begin{array}{cc}{\left({\mathcal{T}}_{11}^{{*}_{\mathbf{A}}}\right)}^j& 0\\ 0& {\left({\mathcal{T}}_{13}^{{*}_{\mathbf{A}}}\right)}^j\end{array}\right).$$

Therefore,

$$\begin{array}{c}P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_0}}\right)-Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_0}}\right)P\left(\mathcal{T}\right)\hfill \\ =\left(\begin{array}{cc}P\left({\mathcal{T}}_{11}\right)Q\left({\mathcal{T}}_{11}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}_{11}^{{*}_{\mathbf{A}}}\right)P\left({\mathcal{T}}_{11}\right)& 0\\ \\ 0& P\left({\mathcal{T}}_{13}\right)Q\left({\mathcal{T}}_{13}^{{*}_{\mathbf{A}}}\right)-Q\left({\mathcal{T}}_{13}^{{*}_{\mathbf{A}}}\right)P\left({\mathcal{T}}_{13}\right)\end{array}\right).\hfill \end{array}$$

Since ${\mathcal{T}}_{11}$ and ${\mathcal{T}}_{13}$ are $(P,Q)$-$\mathbf{A}$-normal, each diagonal block is zero. Hence, the whole matrix is zero, and so $\mathcal{T}$ is $(P,Q)$-${\mathbf{A}}_0$-normal. □

The following proposition shows that the class of $(P,Q)$-$\mathbf{A}$-normal operators is stable under conjugation by $\mathbf{A}$-isometries. That is, if an operator $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-normal, then any operator of the form $\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}$, where $\mathbf{V}$ is an $\mathbf{A}$-isometry, remains $(P,Q)$-$\mathbf{A}$-normal.

Proposition 3.

Let $P,Q\in \mathbb{C}\left[z\right]$ and let $\mathcal{T},\mathbf{V}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ be operators reducing $ker{\left(\mathbf{A}\right)}^{\perp }$. If $\mathbf{V}$ is an $\mathbf{A}$-isometry and $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$, then $\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}$ is also $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$.

Proof. 

Let $P\left(z\right)={\sum }_{k=0}^n{w}_k{z}^k$ and $Q\left(z\right)={\sum }_{j=0}^m{b}_j{z}^j$. We have $\mathbf{V}$, which satisfies ${\mathbf{V}}^{{*}_{\mathbf{A}}}\mathbf{V}={\mathcal{I}}_{\mathbf{A}}.$ Moreover, because $\mathcal{T}$ and $\mathbf{V}$ reduce $ker{\left(\mathbf{A}\right)}^{\perp }$,

$$\mathcal{T}{\mathcal{I}}_{\mathbf{A}}={\mathcal{I}}_{\mathbf{A}}\mathcal{T},{\mathcal{T}}^{{*}_{\mathbf{A}}}{\mathcal{I}}_{\mathbf{A}}={\mathcal{I}}_{\mathbf{A}}{\mathcal{T}}^{{*}_{\mathbf{A}}},$$

and

$$\mathbf{V}{\mathcal{I}}_{\mathbf{A}}={\mathcal{I}}_{\mathbf{A}}\mathbf{V},{\mathbf{V}}^{{*}_{\mathbf{A}}}{\mathcal{I}}_{\mathbf{A}}={\mathcal{I}}_{\mathbf{A}}{\mathbf{V}}^{{*}_{\mathbf{A}}}.$$

Step 1: Computing powers of $\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}$. By induction,

$${\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^r=\mathbf{V}{\mathcal{T}}^r{\mathbf{V}}^{{*}_{\mathbf{A}}},r\ge 1.$$

Step 2: Computing $P\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)$.

$$\begin{array}{cc}\hfill P\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)& =\sum _{k=0}^n{w}_k{\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^k\hfill \\ \hfill & =\sum _{k=0}^n{w}_k\mathbf{V}{\mathcal{T}}^k{\mathbf{V}}^{{*}_{\mathbf{A}}}\hfill \\ \hfill & =\mathbf{V}\left(\sum _{k=0}^n{w}_k{\mathcal{T}}^k\right){\mathbf{V}}^{{*}_{\mathbf{A}}}\hfill \\ \hfill & =\mathbf{V}P\left(\mathcal{T}\right){\mathbf{V}}^{{*}_{\mathbf{A}}}.\hfill \end{array}$$

Step 3: Computing $Q\left({\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)$. Using ${\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}=\mathbf{V}{\mathcal{T}}^{{*}_{\mathbf{A}}}{\mathbf{V}}^{{*}_{\mathbf{A}}}$,

$$\begin{array}{cc}\hfill Q\left({\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)& =\sum _{j=0}^m{b}_j\left(\mathbf{V}{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)\hfill \\ \hfill & =\mathbf{V}\left(\sum _{j=0}^m{b}_j{\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^j\right){\mathbf{V}}^{{*}_{\mathbf{A}}}\hfill \\ \hfill & =\mathbf{V}Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right){\mathbf{V}}^{{*}_{\mathbf{A}}}.\hfill \end{array}$$

Step 4: Verifying $(P,Q)$-$\mathbf{A}$-normality.

$$\begin{array}{c}P\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)Q\left({\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)\hfill \\ =\left(\mathbf{V}P\left(\mathcal{T}\right){\mathbf{V}}^{{*}_{\mathbf{A}}}\right)\left(\mathbf{V}Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right){\mathbf{V}}^{{*}_{\mathbf{A}}}\right)\hfill \\ =\mathbf{V}\left(P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\right){\mathbf{V}}^{{*}_{\mathbf{A}}}.\hfill \end{array}$$

Since $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathrm{N}$,

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)=Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right),$$

thus

$$P\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)Q\left({\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)=Q\left({\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right)}^{{*}_{\mathbf{A}}}\right)P\left(\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}\right).$$

Therefore, $\mathbf{V}\mathcal{T}{\mathbf{V}}^{{*}_{\mathbf{A}}}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$. □

Remark 5.

In the classical Hilbert space setting, normality is invariant under unitary equivalence. Proposition 3 shows that an analogous phenomenon holds in the semi-Hilbertian context, where unitary operators are replaced by $\mathbf{A}$-isometries and the usual adjoint is replaced by the $\mathbf{A}$-adjoint.

The following proposition shows that $(P,Q)$-$\mathbf{A}$-normality is inherited by reducing subspaces. Specifically, if a closed subspace $\mathcal{M}$ reduces both the operator $\mathcal{T}$ and is compatible with $\mathbf{A}$, then the restriction ${\mathcal{T}|}_{\mathcal{M}}$ remains $(P,Q)$-$\mathbf{A}$-normal.

Proposition 4.

Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ and $P,Q\in \mathbb{C}\left[z\right]$, such that $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-normal. Let $\mathcal{M}$ be a closed subspace of $\mathcal{K}$, which reduces $\mathcal{T}$. Assume, moreover, that $\mathcal{M}$ reduces $\mathbf{A}$ (equivalently $\mathbf{A}\mathcal{M}\subset \mathcal{M}$ and $\mathbf{A}{\mathcal{M}}^{\perp }\subset {\mathcal{M}}^{\perp }$). Then, the restriction ${\mathcal{T}|}_{\mathcal{M}}$ is $(P,Q)$-${\mathbf{A}|}_{\mathcal{M}}$-$\mathbf{N}$.

Proof. 

Since $\mathcal{M}$ reduces both $\mathcal{T}$ and $\mathbf{A}$, we may view $\mathcal{T}$ and $\mathbf{A}$ as block diagonal with respect to the orthogonal decomposition $\mathcal{K}=\mathcal{M}\oplus {\mathcal{M}}^{\perp }$, namely

$$\mathcal{T}=\left(\begin{array}{cc}{\mathcal{T}|}_{\mathcal{M}}& 0\\ 0& {\mathcal{T}|}_{{\mathcal{M}}^{\perp }}\end{array}\right),\mathbf{A}=\left(\begin{array}{cc}{\mathbf{A}|}_{\mathcal{M}}& 0\\ 0& {\mathbf{A}|}_{{\mathcal{M}}^{\perp }}\end{array}\right),$$

where each block acts on the indicated subspace. Under these block decompositions, the $\mathbf{A}$-adjoint also decomposes blockwise; more precisely

$${\mathcal{T}}^{{*}_{\mathbf{A}}}=\left(\begin{array}{cc}{\left(\mathcal{T}{|}_{\mathcal{M}}\right)}^{{*}_{{\mathbf{A}|}_{\mathcal{M}}}}& 0\\ 0& {\left(\mathcal{T}{|}_{{\mathcal{M}}^{\perp }}\right)}^{{*}_{{\mathbf{A}|}_{{\mathcal{M}}^{\perp }}}}\end{array}\right).$$

Now the polynomial functional calculus respects this block decomposition. For every polynomial $R\in \mathbb{C}\left[z\right]$, we have

$$R\left(\mathcal{T}\right)=\left(\begin{array}{cc}R\left(\mathcal{T}{|}_{\mathcal{M}}\right)& 0\\ 0& R\left(\mathcal{T}{|}_{{\mathcal{M}}^{\perp }}\right)\end{array}\right),R\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)=\left(\begin{array}{cc}R\left({\left(\mathcal{T}{|}_{\mathcal{M}}\right)}^{{*}_{{\mathbf{A}|}_{\mathcal{M}}}}\right)& 0\\ 0& R\left({\left(\mathcal{T}{|}_{{\mathcal{M}}^{\perp }}\right)}^{{*}_{{\mathbf{A}|}_{{\mathcal{M}}^{\perp }}}}\right)\end{array}\right).$$

Because $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-normal, we have

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)=Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right).$$

Restricting this equality to the $\mathcal{M}$-block (i.e., looking at the $(1,1)$-entry of the block matrix equality) yields

$$P\left(\mathcal{T}{|}_{\mathcal{M}}\right)Q\left({\left(\mathcal{T}{|}_{\mathcal{M}}\right)}^{{*}_{{\mathbf{A}|}_{\mathcal{M}}}}\right)=Q\left({\left(\mathcal{T}{|}_{\mathcal{M}}\right)}^{{*}_{{\mathbf{A}|}_{\mathcal{M}}}}\right)P\left(\mathcal{T}{|}_{\mathcal{M}}\right).$$

This shows that ${\mathcal{T}|}_{\mathcal{M}}$ is $(P,Q)$-${\mathbf{A}|}_{\mathcal{M}}$-$\mathbf{N}$. □

The following result links classical polynomial normality with the semi-Hilbertian setting.

Proposition 5.

Let $P,Q\in \mathbb{C}\left[z\right]$ and $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ be such that $\mathbf{A}\mathcal{T}=\mathcal{T}\mathbf{A}$. If $\mathcal{T}$ is a P-$\mathbf{N}$ operator, then $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$.

Proof. 

Let $P,Q\in \mathbb{C}\left[z\right]$ and $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ satisfy $\mathbf{A}\mathcal{T}=\mathcal{T}\mathbf{A}$. Assume that $\mathcal{T}$ is P-$\mathbf{N}$, i.e.,

$$P\left(\mathcal{T}\right)P{\left(\mathcal{T}\right)}^{*}=P{\left(\mathcal{T}\right)}^{*}P\left(\mathcal{T}\right).$$

Step 1: Commutation implies equality of $\mathbf{A}$-adjoint and usual adjoint. Since $\mathbf{A}\mathcal{T}=\mathcal{T}\mathbf{A}$, the $\mathbf{A}$-adjoint of $\mathcal{T}$ coincides with the usual adjoint:

$${\langle \mathcal{T}x,y\rangle }_{\mathbf{A}}={\langle x,{\mathcal{T}}^{{*}_{\mathbf{A}}}y\rangle }_{\mathbf{A}},{\langle u,v\rangle }_{\mathbf{A}}=\langle \mathbf{A}u,v\rangle .$$

Using $\mathbf{A}\mathcal{T}=\mathcal{T}\mathbf{A}$, we obtain

$${\langle \mathcal{T}x,y\rangle }_{\mathbf{A}}={\langle x,{\mathcal{T}}^{*}y\rangle }_{\mathbf{A}},$$

for all $x,y\in \mathcal{K}$, showing

$${\mathcal{T}}^{{*}_{\mathbf{A}}}={\mathcal{T}}^{*}.$$

Step 2: $(P,Q)$-$\mathbf{A}$-normality. By definition, $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-normal if

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)=Q\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)P\left(\mathcal{T}\right).$$

Using Step 1, this becomes

$$P\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{*}\right)=Q\left({\mathcal{T}}^{*}\right)P\left(\mathcal{T}\right).$$

Since $\mathcal{T}$ is P-normal, this equality holds. Hence, $\mathcal{T}$ is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$. □

It is well known that every $\mathbf{A}$-$\mathbf{N}$ operator is $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$ for all polynomials $P,Q\in \mathbb{C}\left[z\right]$. However, the converse does not hold in general; that is, an operator can be $(P,Q)$-$\mathbf{A}$-$\mathbf{N}$ for certain polynomials P and Q without being $\mathbf{A}$-$\mathbf{N}$ (see Example 1). This naturally leads to the question: under what conditions does $(P,Q)$-$\mathbf{A}$-normality imply $\mathbf{A}$-normality? The following theorem provides sufficient conditions on the operator and its $\mathbf{A}$-adjoint.

Theorem 4.

Let $\mathcal{T}\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ such that:

• $ker{\left(\mathbf{A}\right)}^{\perp }$ is invariant under $\mathcal{T}$,
• $\mathbf{Im}\left(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\subset \overline{\mathbf{Im}\left(\mathbf{A}\right)}$.

Let

$$P\left(z\right)=\sum _{k=0}^n{a}_k{z}^k\in \mathbb{C}\left[z\right],a_n\ne 0,Q\left(z\right)=\sum _{j=0}^m{b}_j{z}^j\in \mathbb{C}\left[z\right],b_m\ne 0.$$

If for all $\omega >0$, the operator $\mathcal{T}-\omega$ is $(P,Q)$-$\mathbf{A}$-normal, i.e.,

$$P(\mathcal{T}-\omega )Q\left({(\mathcal{T}-\omega )}^{{*}_{\mathbf{A}}}\right)=Q\left({(\mathcal{T}-\omega )}^{{*}_{\mathbf{A}}}\right)P(\mathcal{T}-\omega ),$$

then, $\mathcal{T}$ is $\mathbf{A}$-$\mathbf{N}$:

$$\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}={\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}.$$

Proof. 

Step 1: Reduce to $\overline{\mathbf{Im}\left(\mathbf{A}\right)}$.

Since $ker{\left(\mathbf{A}\right)}^{\perp }$ is invariant under $\mathcal{T}$, it is a reducing subspace. Hence,

$${\mathcal{I}}_{\mathbf{A}}\mathcal{T}=\mathcal{T}{\mathcal{I}}_{\mathbf{A}},{\mathcal{I}}_{\mathbf{A}}{\mathcal{T}}^{{*}_{\mathbf{A}}}={\mathcal{T}}^{{*}_{\mathbf{A}}}{\mathcal{I}}_{\mathbf{A}}.$$

Step 2: Expand $(P,Q)$-normality.

By assumption, for all $\omega >0$,

$$P(\mathcal{T}-\omega )Q\left({(\mathcal{T}-\omega )}^{{*}_{\mathbf{A}}}\right)-Q\left({(\mathcal{T}-\omega )}^{{*}_{\mathbf{A}}}\right)P(\mathcal{T}-\omega )=0.$$

Expanding P and Q gives

$$\sum _{k=0}^n\sum _{j=0}^m{a}_k{b}_j{(\mathcal{T}-\omega )}^k{({\mathcal{T}}^{{*}_{\mathbf{A}}}-\omega {\mathcal{I}}_{\mathbf{A}})}^j-\sum _{k=0}^n\sum _{j=0}^m{b}_j{a}_k{({\mathcal{T}}^{{*}_{\mathbf{A}}}-\omega {\mathcal{I}}_{\mathbf{A}})}^j{(\mathcal{T}-\omega )}^k=0.$$

Step 3: Collect powers of $\omega$.

Using the binomial theorem:

$${(\mathcal{T}-\omega )}^k=\sum _{r=0}^k{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{k}{r}}\right){\mathcal{T}}^{k-r}{\omega }^r,{({\mathcal{T}}^{{*}_{\mathbf{A}}}-\omega {\mathcal{I}}_{\mathbf{A}})}^j=\sum _{s=0}^j{(-1)}^s\left({\displaystyle \genfrac{}{}{0pt}{}{j}{s}}\right){\left({\mathcal{T}}^{{*}_{\mathbf{A}}}\right)}^{j-s}{\left(\omega {\mathcal{I}}_{\mathbf{A}}\right)}^s.$$

Thus, the commutator becomes a polynomial in $\omega$, whose leading term is

$$a_n{b}_m{\mathcal{I}}_{\mathbf{A}}\left(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}-{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}\right){\omega }^{n+m-2}+\cdots =0.$$

Step 4: Let $\omega \to \infty$.

Dividing by ${\omega }^{n+m-2}$ and taking the limit, we obtain

$${\mathcal{I}}_{\mathbf{A}}\left(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}-{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}\right)=0,$$

so for all $x\in \mathcal{K}$,

$$(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}-{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T})x\in ker\left(\mathbf{A}\right).$$

Step 5: Use the image condition.

Since $\mathbf{Im}\left(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}\right)\subset \overline{\mathbf{Im}\left(\mathbf{A}\right)}=ker{\left(\mathbf{A}\right)}^{\perp }$, we have

$$\mathbf{Im}(\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}-{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T})\subset ker{\left(\mathbf{A}\right)}^{\perp }.$$

Combining these inclusions, we conclude

$$\mathcal{T}{\mathcal{T}}^{{*}_{\mathbf{A}}}-{\mathcal{T}}^{{*}_{\mathbf{A}}}\mathcal{T}=0,$$

so $\mathcal{T}$ is $\mathbf{A}$-$\mathbf{N}$. □

2.1. Tensor Products on Semi-Hilbertian Spaces

Let $({\mathcal{K}}_1,{\mathbf{A}}_1)$ and $({\mathcal{K}}_2,{\mathbf{A}}_2)$ be semi-Hilbertian spaces, i.e., Hilbert spaces equipped with positive operators ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ inducing the semi-inner products

$${\langle x,y\rangle }_{{\mathbf{A}}_1}=\langle {\mathbf{A}}_{1}x,y\rangle ,{\langle u,v\rangle }_{{\mathbf{A}}_2}=\langle {\mathbf{A}}_{2}u,v\rangle .$$

(34)

2.1.1. Tensor Product Space and Semi-Inner Product

The tensor product space ${\mathcal{K}}_1\otimes {\mathcal{K}}_2$ is equipped with the tensor product semi-inner product

$${\langle x_1\otimes x_2,y_1\otimes y_2\rangle }_{{\mathbf{A}}_1\otimes {\mathbf{A}}_2}:=\langle {\mathbf{A}}_1{x}_1,y_1\rangle \langle {\mathbf{A}}_2{x}_2,y_2\rangle ,$$

(35)

extended by linearity and continuity to all elements of ${\mathcal{K}}_1\otimes {\mathcal{K}}_2$.

2.1.2. Tensor Product of Operators

Let ${\mathcal{T}}_1\in {\mathcal{B}}_{{\mathbf{A}}_1}\left({\mathcal{K}}_1\right)$ and ${\mathcal{T}}_2\in {\mathcal{B}}_{{\mathbf{A}}_2}\left({\mathcal{K}}_2\right)$. Their tensor product operator

$${\mathcal{T}}_1\otimes {\mathcal{T}}_2:{\mathcal{K}}_1\otimes {\mathcal{K}}_2\to {\mathcal{K}}_1\otimes {\mathcal{K}}_2$$

is defined on pure tensors by

$$({\mathcal{T}}_1\otimes {\mathcal{T}}_2)(x_1\otimes x_2)=\left({\mathcal{T}}_1{x}_1\right)\otimes \left({\mathcal{T}}_2{x}_2\right).$$

(36)

Its ${\mathbf{A}}_1\otimes {\mathbf{A}}_2$-adjoint satisfies

$${({\mathcal{T}}_1\otimes {\mathcal{T}}_2)}^{{*}_{{\mathbf{A}}_1\otimes {\mathbf{A}}_2}}={\mathcal{T}}_1^{{*}_{{\mathbf{A}}_{\mathbf{1}}}}\otimes {\mathcal{T}}_2^{{*}_{{\mathbf{A}}_{\mathbf{2}}}},$$

(37)

(see [26] (Lemma 3.1))

2.1.3. Normality in Tensor Products

If ${\mathcal{T}}_1$ is ${\mathbf{A}}_1$-$\mathbf{N}$ and ${\mathcal{T}}_2$ is ${\mathbf{A}}_2$-$\mathbf{N}$, i.e.,

$${\mathcal{T}}_1{\mathcal{T}}_1^{{*}_{{\mathbf{A}}_{\mathbf{1}}}}={\mathcal{T}}_1^{{*}_{{\mathbf{A}}_{\mathbf{1}}}}{\mathcal{T}}_1,{\mathcal{T}}_2{\mathcal{T}}_2^{{*}_{{\mathbf{A}}_{\mathbf{2}}}}={\mathcal{T}}_2^{{*}_{{\mathbf{A}}_{\mathbf{2}}}}{\mathcal{T}}_2,$$

then their tensor product is $({\mathbf{A}}_1\otimes {\mathbf{A}}_2)$-$\mathbf{N}$ ([27] (Theorem 3.3)):

$$({\mathcal{T}}_1\otimes {\mathcal{T}}_2){({\mathcal{T}}_1\otimes {\mathcal{T}}_2)}^{{*}_{{\mathbf{A}}_1\otimes {\mathbf{A}}_2}}={({\mathcal{T}}_1\otimes {\mathcal{T}}_2)}^{{*}_{{\mathbf{A}}_1\otimes {\mathbf{A}}_2}}({\mathcal{T}}_1\otimes {\mathcal{T}}_2).$$

The semi-inner product factorizes over the tensor product: ${\mathbf{A}}_1\otimes {\mathbf{A}}_2$ acts independently on each factor.

Normality is preserved because the commutator factorizes:

$$[{\mathcal{T}}_1\otimes {\mathcal{T}}_2,{({\mathcal{T}}_1\otimes {\mathcal{T}}_2)}^{{*}_{{\mathbf{A}}_1\otimes {\mathbf{A}}_2}}]=[{\mathcal{T}}_1,{\mathcal{T}}_1^{{*}_{{\mathbf{A}}_{\mathbf{1}}}}]\otimes \left({\mathcal{T}}_2{\mathcal{T}}_2^{{*}_{{\mathbf{A}}_{\mathbf{2}}}}\right)+\left({\mathcal{T}}_1^{{*}_{{\mathbf{A}}_{\mathbf{1}}}}{\mathcal{T}}_1\right)\otimes [{\mathcal{T}}_2,{\mathcal{T}}_2^{{*}_{{\mathbf{A}}_{\mathbf{2}}}}]=0.$$

This property underlies the validity of Theorem 5.

Theorem 5.

Let ${\mathcal{T}}_1,\dots ,{\mathcal{T}}_d\in {\mathcal{B}}_{\mathbf{A}}\left(\mathcal{K}\right)$ be such that

• $ker{\left(\mathbf{A}\right)}^{\perp }$ is invariant under each ${\mathcal{T}}_k$, and
• $\mathbf{Im}\left({\mathcal{T}}_k{\mathcal{T}}_k^{{*}_{\mathbf{A}}}\right)\subset \overline{\mathbf{Im}\left(\mathbf{A}\right)}$ for $k=1,\dots ,d$.

Let

$$P\left(z\right)=\sum _{k=0}^n{a}_k{z}^k\in \mathbb{C}\left[z\right],a_n\ne 0,Q\left(z\right)=\sum _{j=0}^m{b}_j{z}^j\in \mathbb{C}\left[z\right],b_m\ne 0.$$

If, for all $\omega >0$ and all $k=1,\dots ,d$, the operators ${\mathcal{T}}_k-\omega$ are $(P,Q)$-$\mathbf{A}$-normal, then the tensor product

$${\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_d$$

is $(\mathbf{A}\otimes \dots \otimes \mathbf{A})$-$\mathbf{N}$.

Proof. 

We proceed by induction on d.

Step 1: Base case $d=1$.

For $d=1$, the result reduces to Theorem 4, which ensures that if ${\mathcal{T}}_1-\omega$ is $(P,Q)$-$\mathbf{A}$-normal for all $\omega >0$ and $ker{\left(\mathbf{A}\right)}^{\perp }$ is invariant under ${\mathcal{T}}_1$ with $\mathbf{Im}\left({\mathcal{T}}_1{\mathcal{T}}_1^{{*}_{\mathbf{A}}}\right)\subset \overline{\mathbf{Im}\left(\mathbf{A}\right)}$, then ${\mathcal{T}}_1$ is $\mathbf{A}$-normal.

Step 2: Induction hypothesis.

Assume that for some $d-1\ge 1$, if ${\mathcal{T}}_1,\dots ,{\mathcal{T}}_{d-1}$ satisfy the assumptions of the theorem, then

$${\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_{d-1}$$

is $(\mathbf{A}\otimes \dots \otimes \mathbf{A})$-$\mathbf{N}$.

Step 3: Induction step.

Considerthe d-fold tensor product

$${\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_d=({\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_{d-1})\otimes {\mathcal{T}}_d.$$

By the induction hypothesis, ${\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_{d-1}$ is $(\mathbf{A}\otimes \dots \otimes \mathbf{A})$-normal, and by Theorem 4, ${\mathcal{T}}_d$ is $\mathbf{A}$-$\mathbf{N}$.

Step 4: Tensor product of $\mathbf{A}$-normal operators.

If X is ${\mathbf{A}}_1$- $\mathbf{N}$ and Y is ${\mathbf{A}}_2$-$\mathbf{N}$, then $X\otimes Y$ is $({\mathbf{A}}_1\otimes {\mathbf{A}}_2)$-$\mathbf{N}$ from [27] (Theorem 3.3). Applying this to $({\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_{d-1})\otimes {\mathcal{T}}_d$, we conclude that

$${\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_d$$

is $(\mathbf{A}\otimes \dots \otimes \mathbf{A})$-$\mathbf{N}$.

Step 5: Conclusion.

By induction, the result holds for all $d\ge 1$. □

Theorem 5 generalizes the result of Theorem 4 from a single operator to a tuple of operators acting on a tensor-product space. It shows that under certain structural conditions, the property of $(P,Q)$-$\mathbf{A}$-normality for each individual operator is preserved in the tensor product.

Theorem 6.

Let $({\mathcal{H}}_i,{\langle ·,·\rangle }_{{\mathbf{A}}_i})$ be semi-Hilbertian spaces induced by positive operators ${\mathbf{A}}_i\in \mathcal{B}\left({\mathcal{H}}_i\right)$ for $i=1,\dots ,d$. For each i, let ${\mathcal{T}}_i\in {\mathcal{B}}_{{\mathbf{A}}_i}\left({\mathcal{H}}_i\right)$ and let $P_i,Q_i$ be polynomials. Assume each $ker{\left({\mathbf{A}}_i\right)}^{\perp }$ is invariant under ${\mathcal{T}}_i$, so that ${\mathcal{T}}_i^{{*}_{{\mathbf{A}}_i}}$ and the polynomial functional calculus are well-defined. Assume further that each ${\mathcal{T}}_i$ satisfies the $(P_i,Q_i)$-${\mathbf{A}}_i$-normality condition:

$$P_i\left({\mathcal{T}}_i\right)Q_i\left({\mathcal{T}}_i^{{*}_{{\mathbf{A}}_i}}\right)=Q_i\left({\mathcal{T}}_i^{{*}_{{\mathbf{A}}_i}}\right)P_i\left({\mathcal{T}}_i\right),i=1,\dots ,d.$$

Define the tensor product operator and polynomials by

$$\mathcal{T}:={\mathcal{T}}_1\otimes \dots \otimes {\mathcal{T}}_d,\mathcal{P}:=P_1\otimes \dots \otimes P_d,Q:=Q_1\otimes \dots \otimes Q_d,$$

so that

$$\mathcal{P}\left(\mathcal{T}\right)=P_1\left({\mathcal{T}}_1\right)\otimes \dots \otimes P_d\left({\mathcal{T}}_d\right),Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)=Q_1\left({\mathcal{T}}_1^{{*}_{{\mathbf{A}}_1}}\right)\otimes \dots \otimes Q_d\left({\mathcal{T}}_d^{{*}_{{\mathbf{A}}_d}}\right).$$

Then, $\mathcal{T}$ satisfies the $(\mathcal{P},Q)$-$({\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d)$-normality condition:

$$\mathcal{P}\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)=Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)\mathcal{P}\left(\mathcal{T}\right).$$

Proof. 

By the invariance assumption, the tensor-product $\mathbf{A}$-adjoint decomposes as

$${\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}={\mathcal{T}}_1^{{*}_{{\mathbf{A}}_1}}\otimes \dots \otimes {\mathcal{T}}_d^{{*}_{{\mathbf{A}}_d}}.$$

The polynomial functional calculus respects the tensor structure:

$$\mathcal{P}\left(\mathcal{T}\right)=P_1\left({\mathcal{T}}_1\right)\otimes \dots \otimes P_d\left({\mathcal{T}}_d\right),Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)=Q_1\left({\mathcal{T}}_1^{{*}_{{\mathbf{A}}_1}}\right)\otimes \dots \otimes Q_d\left({\mathcal{T}}_d^{{*}_{{\mathbf{A}}_d}}\right).$$

Hence,

$$\begin{array}{c}\mathcal{P}\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)\hfill \\ =\left(P_1\left({\mathcal{T}}_1\right)\otimes \dots \otimes P_d\left({\mathcal{T}}_d\right)\right)\left(Q_1\left({\mathcal{T}}_1^{{*}_{{\mathbf{A}}_1}}\right)\otimes \dots \otimes Q_d\left({\mathcal{T}}_d^{{*}_{{\mathbf{A}}_d}}\right)\right)\hfill \\ =\left(P_1\left({\mathcal{T}}_1\right)Q_1\left({\mathcal{T}}_1^{{*}_{{\mathbf{A}}_1}}\right)\right)\otimes \dots \otimes \left(P_d\left({\mathcal{T}}_d\right)Q_d\left({\mathcal{T}}_d^{{*}_{{\mathbf{A}}_d}}\right)\right).\hfill \end{array}$$

By the $(P_i,Q_i)$-${\mathbf{A}}_i$-normality of each ${\mathcal{T}}_i$, each factor commutes:

$$P_i\left({\mathcal{T}}_i\right)Q_i\left({\mathcal{T}}_i^{{*}_{{\mathbf{A}}_i}}\right)=Q_i\left({\mathcal{T}}_i^{{*}_{{\mathbf{A}}_i}}\right)P_i\left({\mathcal{T}}_i\right),i=1,\dots ,d.$$

Therefore,

$$\mathcal{P}\left(\mathcal{T}\right)Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)=Q\left({\mathcal{T}}^{{*}_{{\mathbf{A}}_1\otimes \dots \otimes {\mathbf{A}}_d}}\right)\mathcal{P}\left(\mathcal{T}\right),$$

which proves the theorem. □

3. Conclusions

In this work, we have conducted a thorough study of $(P,Q)$-$\mathbf{A}$-normal operators in semi-Hilbertian spaces. The main contributions and findings of this study are:

1.

We established precise conditions under which $(P,Q)$-$\mathbf{A}$-normality implies $\mathbf{A}$-normality for a single operator, clarifying the connection between polynomial commutation relations and classical semi-Hilbertian normality.

2.

We extended these results to tuples of operators and their tensor products, showing that, under suitable invariance and range conditions, the tensor product of $(P,Q)$-$\mathbf{A}$-normal operators preserves normality with respect to the tensor-product semi-inner product. This provides a structured approach to study multi-operator systems.

3.

We highlighted the crucial role of structural properties, such as the invariance of $ker{\left(\mathbf{A}\right)}^{\perp }$ and the containment of operator ranges, in guaranteeing stronger forms of normality and in facilitating the construction of complex operator systems.

4.

The developed framework is applicable to quantum systems modeled in semi-Hilbertian spaces, where operator interactions and tensor structures are central to multi-variable quantum dynamics.

Overall, our results offer both theoretical insights and practical tools for analyzing $(P,Q)$-$\mathbf{A}$-normal operators, multi-operator systems, and their applications in functional analysis and quantum mechanics. Future research may focus on generalizations to unbounded operators, broader classes of polynomial conditions, and deeper exploration of quantum semi-Hilbertian frameworks.