A Structural Study of Generalized ${\displaystyle [\mathit{m},\mathcal{C}]}$-Symmetric Extension Operators

Abstract

This manuscript introduces and investigates a new class of operators, termedn-quasi-$[m,\mathcal{C}]$-symmetric operators, which generalize and extend the existing notions of $[m,\mathcal{C}]$-symmetric and n-quasi-$[m,\mathcal{C}]$-isometric operators. Specifically, given a conjugation $\mathcal{C}$ on a Hilbert space, an operator $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is said to be n-quasi-$[m,\mathcal{C}]$-symmetric if it satisfies the relation${\mathcal{Q}}^{\ast n}\left({\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\mathcal{C}{\mathcal{Q}}^{m-j}\mathcal{C}{\mathcal{Q}}^j\right){\mathcal{Q}}^n=0$. Our study systematically explores the algebraic properties and structural characterization of n-quasi-$[m,\mathcal{C}]$-symmetric operators through matrix representations, providing a deeper understanding of their internal structure. Moreover, we establish sufficient conditions under which the powers and products of such operators inherit the n-quasi-$[m,\mathcal{C}]$-symmetric property. Additionally, we investigate the tensor products of n-quasi-$[m,\mathcal{C}]$-symmetric operators. Finally, we identify conditions that distinguish n-quasi-$[m,\mathcal{C}]$-symmetric operators from n-quasi-$[m-1;\mathcal{C}]$-symmetric operators.

1. Introduction

The set $\mathcal{B}\left(\mathcal{K}\right)$ refers to the set of all bounded linear operators on a complex Hilbert space $\mathcal{K}$. For every $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$, the range of $\mathcal{Q}$ is denoted by $\mathcal{R}\left(\mathcal{Q}\right)$, the null space of $\mathcal{Q}$ is denoted by $ker\left(\mathcal{Q}\right)$, and the adjoint of $\mathcal{Q}$ is denoted by ${\mathcal{Q}}^{\ast }$. A subspace $\mathcal{M}\subset \mathcal{K}$ is invariant for $\mathcal{Q}$ (or $\mathcal{Q}$-invariant) if $\mathcal{Q}\mathcal{M}=\{\mathcal{Q}\omega ,\omega \in \mathcal{K}\}\subset \mathcal{M}$.

The norm of $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is quantity $\parallel \mathcal{Q}\parallel$ defined by

$$\parallel \mathcal{Q}\parallel =sup\{\parallel \mathcal{Q}\omega \parallel ,\parallel \omega \parallel \le 1\}.$$

A conjugation $\mathcal{C}:\mathcal{K}⟶\mathcal{K}$ is a conjugate-linear operator that satisfies ${\mathcal{C}}^2=I$ and is isometric (i.e., $\langle \mathcal{C}\omega \mid \mathcal{C}\psi \rangle =\langle \psi \mid \omega \rangle (\forall \omega ,\psi \in \mathcal{K})$). We denote by $\mathcal{C}\left[\mathcal{K}\right]$ the set of all conjugation operators $\mathcal{C}:\mathcal{K}⟶\mathcal{K}$. If $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$, then the following properties hold:

(i) $\mathcal{C}\mathcal{Q}\mathcal{C}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\parallel \mathcal{C}\mathcal{Q}\mathcal{C}\parallel =\parallel \mathcal{Q}\parallel$ for all $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$,

(ii) ${\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right)}^k=\mathcal{C}{\mathcal{Q}}^k\mathcal{C}$,

(iii) ${\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right)}^{\ast }=\mathcal{C}{\mathcal{Q}}^{\ast }\mathcal{C}$ for $k=0,1,\cdots ,$ (see [1,2] for more details).

Let $\mathcal{Q},\mathcal{U}\in \mathcal{B}\left(\mathcal{K}\right)$. In [3], Helton studied the operator ${\mathbf{J}}_{\mathcal{Q},\mathcal{U}}:\mathcal{B}\left(\mathcal{K}\right)\to \mathcal{B}\left(\mathcal{K}\right)$, defined by

$${\mathbf{J}}_{\mathcal{Q},\mathcal{U}}\left(X\right)=\mathcal{Q}X-X\mathcal{U}.$$

Then,

$${\mathbf{J}}_{\mathcal{Q},\mathcal{U}}^m\left(I\right)=\sum _{0\le k\le m}^m{(-1)}^{m-k}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\mathcal{Q}}^k{\mathcal{U}}^{m-k}.$$

Let $\mathcal{Q},\mathcal{U}\in \mathcal{B}\left(\mathcal{K}\right)$. If ${\mathbf{J}}_{\mathcal{Q},\mathcal{U}}^m\left(I\right)=0$ for some $m\ge 1$, $\mathcal{U}$ belongs to the Helton class of $\mathcal{Q}$, denoted by $\mathcal{U}\in {\mathrm{Helton}}_m\left(\mathcal{Q}\right)$. In Helton’s work [3], m-symmetries refer to a family of operators that satisfy the specific condition

$${\mathbf{J}}_{\mathcal{Q}}^m\left(I\right)={\mathbf{J}}_{{\mathcal{Q}}^{\ast },\mathcal{Q}}^m\left(I\right)=0.$$

$\mathcal{Q}$ is defined as m-symmetric if it satisfies

$${\displaystyle {\displaystyle \sum _{0\le k\le m}}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right){\mathcal{Q}}^{\ast k}{\mathcal{Q}}^{m-k}=0.}$$

(1)

In what follows, m and n will represent positive integers. An operator $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is defined as an $(m,\mathcal{C})$-symmetric operator [4,5] if there exists $\mathcal{C}\in \mathcal{C}\left(\mathcal{K}\right)$ such that

$${\displaystyle {\displaystyle \sum _{0\le j\le m}}{(-1)}^{m-j}\left(\genfrac{}{}{0pt}{}{m}{j}\right){\mathcal{Q}}^{\ast j}\mathcal{C}{\mathcal{Q}}^{m-j}\mathcal{C}=0,}$$

(2)

for some positive integer m. It is apparent that the $\mathcal{C}$-symmetric operator $(\mathcal{C}\mathcal{Q}\mathcal{C}={\mathcal{Q}}^{\ast })$ is $(m,\mathcal{C})$-symmetric. For further information on a class of operators associated with complex symmetric operators, see [1,6,7,8,9,10,11,12,13,14,15,16,17]. The set of $(m,C)$-symmetric operators is surprisingly large.

According to [18], $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is called an $[m,\mathcal{C}]$-isometric operator for some $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$ if

$${\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\mathcal{C}{\mathcal{Q}}^{\ast m-j}\mathcal{C}{\mathcal{Q}}^{m-j}=0.$$

Building on the definitions of $(m,\mathcal{C})$-symmetry, $(m,\mathcal{C})$-isometry, and $[m,C]$-isometry, the authors in [19] defined $[m,\mathcal{C}]$-symmetric $\mathcal{Q}$ as follows: An operator $\mathcal{Q}$ is said to be an $[m,\mathcal{C}]$-symmetric operator if, for some $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$,

$${\mathsf{\Psi }}_m\left(\mathcal{Q};\mathcal{C}\right):={\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\mathcal{C}{\mathcal{Q}}^{m-j}\mathcal{C}{\mathcal{Q}}^j=0,$$

(3)

for some $m\ge 1$. It was observed that

$${\mathsf{\Psi }}_{k+1}(\mathcal{Q};\mathcal{C})=\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right)-{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right)\mathcal{Q}.$$

(4)

Hence, if $\mathcal{Q}$ is an $[m,\mathcal{C}]$-symmetric operator, then $\mathcal{Q}$ is a $[k,\mathcal{C}]$-symmetric operator for all positive integers $k\ge m$. The interested reader is referred to [19,20].

The concept of n-quasi-$(m;C)$-isometric operators, introduced by the authors in [21], serves as a generalization of $(m;C)$-isometric operators. An operator $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is called an n-quasi-$(m,\mathcal{C})$-isometric operator if there exists some $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$ such that

$${\mathcal{Q}}^{\ast n}\left({\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right){\mathcal{Q}}^{\ast m-j}\mathcal{C}{\mathcal{Q}}^{m-j}\mathcal{C}\right){\mathcal{Q}}^n=0.$$

In [22], Shen studied n-quasi-$[m,\mathcal{C}]$-isometric operators, which serve as a generalization of $[m,\mathcal{C}]$-isometric operators. $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is called an n-quasi-$[m,\mathcal{C}]$-isometric operator if there exists $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$, for which

$${\mathcal{Q}}^{\ast n}\left({\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\mathcal{C}{\mathcal{Q}}^{m-j}{\mathcal{Q}}^{m-j}\mathcal{C}\right){\mathcal{Q}}^n=0.$$

For additional information on this family of operators, we invite readers to refer to [23].

Based on the works that have been published about n-quasi-$(m,\mathcal{C})$-isometries and n-quasi-$[m,\mathcal{C}]$-isometries, in this research, we were interested in the investigation of n-quasi-$[m,\mathcal{C}]$-symmetric operators, which extend the class of $[m,\mathcal{C}]$-symmetries.

This paper will proceed as follows: Section 2 starts with a few remarks and examples that aim to provide clarity on the concept of n-quasi-$[m,\mathcal{C}]$-symmetric operators. Certain properties were derived by taking advantage of the specific operator matrix representation linked to them. The main concepts to be discussed are powers, invariance, product, and tensor product. We show that, if $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n_1,m_1)}\left(\mathcal{K}\right)$ and $\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(n_2,m_2)}\left(\mathcal{K}\right)$, then $\mathcal{Q}\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(max\{n_1,n_2\},m_1+m_2-1)}\left(\mathcal{K}\right)$ under suitable conditions. Moreover, we examined the perturbation of a member $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$ by a nilpotent operator $\mathbf{T}$ of order q under suitable conditions. Section 3 deals with the case of n-quasi strict $[m,\mathcal{C}]$-symmetric operators. We show that, given an operator $\mathcal{Q}$, which is a strict n-quasi-$[m,\mathcal{C}]$-symmetry for some conjugation $\mathcal{C}$ such that $\left[{\mathcal{Q}}^{\ast },\mathcal{C}\mathcal{Q}\mathcal{C}\right]=0$, then the set $\left\{{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n,k=0,1,\cdots ,m-1\right\}$ is linearly independent. Moreover, if $\mathcal{R}\left(\mathcal{Q}\right)$ is dense, then the sets

$$\left\{{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m-k},k=0,1,\cdots ,m-1\right\}$$

and

$$\left\{{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+k},k=0,1,\cdots ,m-1\right\}$$

are linearly independent. These results enable us to show that, if ${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}\otimes \mathcal{C}]$-symmetry, then ${\mathcal{Q}}_1$ is an n-quasi strict-$[m_1,\mathcal{C}]$-symmetry if, and only if, ${\mathcal{Q}}_2$ is an n-quasi strict-$[m_2,\mathcal{C}]$.

2. Basic Properties

This section develops the algebraic framework and operator identities defining the new class; explores matrix representations to study their structural properties; establishes sufficient conditions under which powers, products, and tensor products preserve the n-quasi-[$m,C$]-symmetric property; examines perturbations by nilpotent operators; presents several illustrative examples; and provides results distinguishing these operators from closely related classes (e.g., [$m-1,C$]-symmetric operators).

Example 1.

Let $\mathcal{C}\in \mathcal{C}\left[{\mathcal{K}}_0\right]$ for ${\mathcal{K}}_0={\mathbb{C}}^3$ be defined by $\mathcal{C}(x_1,x_2,x_3)=(\overline{x_3},-\overline{x_2},\overline{x_1})$, and consider $\mathcal{Q}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right)$ on ${\mathbb{C}}^3$. A basic calculation proves that $\mathcal{C}\mathcal{Q}\mathcal{C}-\mathcal{Q}\ne 0$, and ${\mathcal{Q}}^{\ast }\left(\mathcal{C}\mathcal{Q}\mathcal{C}-\mathcal{Q}\right)\mathcal{Q}=0$.

Definition 1.

A member $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ is called an n-quasi-$[m,\mathcal{C}]$-symmetric operator if

$${\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_m\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n:={\mathcal{Q}}^{\ast n}\left({\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\mathcal{C}{\mathcal{Q}}^{m-j}\mathcal{C}{\mathcal{Q}}^j\right){\mathcal{Q}}^n=0,$$

(5)

for some $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$ and positive integers n and m.

From now on, ${\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ will represent the set of n-quasi-$[m,C]$-symmetric operators.

Remark 1.

It is not difficult to prove that Statement (5) is equivalent to the following statement:

$$\mathcal{P}\left({\displaystyle \sum _{0\le j\le m}}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\mathcal{C}{\mathcal{Q}}^{m-j}\mathcal{C}{\mathcal{Q}}^j\right)\mathcal{P}=0,$$

(6)

where $\mathcal{P}$ denotes the projection on $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}$.

Remark 2.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$.

(i)

If $\mathcal{Q}$ is an $[m,\mathcal{C}]$-symmetric operator, then $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right].$

(ii)

$\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]⟺\mathcal{C}\mathcal{Q}\mathcal{C}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right].$

(iii)

If $\mathcal{Q}$ has a dense range, and $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$, then $\mathcal{Q}$ is an $[m,\mathcal{C}]$- symmetric operator.

(iv)

If $\mathcal{C}\mathcal{Q}=\mathcal{Q}\mathcal{C}$, then $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ if, and only if, $\mathcal{Q}\in {\left[\mathcal{QS}\right]}_{(n,m)}\left[\mathcal{K}\right].$

(v)

A 1-quasi-$[m,\mathcal{C}]$-symmetric operator is often referred to as a quasi-$[m,\mathcal{C}]$-symmetric operator.

As stated earlier, all $[m,\mathcal{C}]$-symmetric operators are in ${\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$. The next example illustrates a member of ${\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$ that does not qualify as an $[m,\mathcal{C}]$-symmetric operator.

Example 2.

From Example 1, we have an operator $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(1,1)}\left[{\mathcal{K}}_0\right]$ that does not qualify as a $[1,\mathcal{C}]$-symmetric operator.

Example 3.

Let $\mathcal{C}\in \mathcal{C}\left[{\mathbb{C}}^3\right]$ be defined by $\mathcal{C}(x_1,x_2,x_3)=(\overline{x_3},-\overline{x_2},\overline{x_1})$, and consider $\mathcal{Q}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)$ on ${\mathbb{C}}^3$. A simple calculation shows that ${\mathcal{Q}}^{\ast }\left(\mathcal{C}\mathcal{Q}\mathcal{C}-\mathcal{Q}\right)\mathcal{Q}\ne 0$ and ${\mathcal{Q}}^{\ast 2}\left(\mathcal{C}\mathcal{Q}\mathcal{C}-\mathcal{Q}\right){\mathcal{Q}}^2=0$. This shows that $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(2,1)}\left[{\mathbb{C}}^3\right]$, but $\mathcal{Q}\notin {\left[\mathcal{QSC}\right]}_{(1,1)}\left[{\mathbb{C}}^3\right]$.

Remark 3.

Observe that $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$; then, $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n+1,m)}\left[\mathcal{K}\right]$. It is not generally true that the converse holds. For instance, by choosing $\mathcal{Q}$ and $\mathcal{C}$, as in Example 3, it is easily shown that $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(2,1)}\left[{\mathbb{C}}^3\right]$, but $\mathcal{Q}\notin {\left[\mathcal{QSC}\right]}_{(1,1)}\left[{\mathbb{C}}^3\right]$.

In the following theorem, we establish that, under certain assumptions on $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$, for $n\ge 2$, it becomes ${\left[\mathcal{QSC}\right]}_{(1,m)}\left[\mathcal{K}\right].$

Proposition 1.

Let $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$. If $ker\left({\mathcal{Q}}^{(\ast r)}\right)=ker\left({\mathcal{Q}}^{\ast (r+1}\right)$ for some integer $r\in \{1,\cdots ,n-1\},$ then $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(r,m)}\left[\mathcal{K}\right]$.

Proof. 

Since $ker\left({\mathcal{Q}}^{\ast r}\right)=ker\left({\mathcal{Q}}^{\ast (r+1)}\right)$, it follows that $ker\left({\mathcal{Q}}^{\ast r}\right)=ker\left({\mathcal{Q}}^{\ast n}\right)$. Consequently, if ${\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_m\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0$, then ${\mathcal{Q}}^{\ast r}{\mathsf{\Psi }}_m\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^r=0$. The statement of the proposition follows directly as a consequence of this fact. □

Proposition 2.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\mathcal{M}$ be a closed subspace $\mathcal{Q}$-invariant of $\mathcal{K}$ and let $\mathcal{C}={\mathcal{C}}_1\oplus {\mathcal{C}}_2\in \mathcal{C}\left[\mathcal{K}\right]$, where ${\mathcal{C}}_1\in \mathcal{C}\left[\mathcal{M}\right]$ and ${\mathcal{C}}_2\in \mathcal{C}\left[{\mathcal{M}}^{\perp }\right]$. Assume that $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$; then, ${\mathcal{Q}}_1=\mathcal{Q}{|}_{\mathcal{M}}\in {\left[{\mathcal{QSC}}_1\right]}_{(n,m)}\left[\mathcal{M}\right]$.

Proof. 

Let us consider

$$\mathcal{Q}=\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)\mathrm{on}\mathcal{K}=\mathcal{M}\oplus {\mathcal{M}}^{\perp }.$$

According to $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$, we obtain

$$\begin{array}{cc}& {\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^{\ast n}\left\{{\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\left(\begin{array}{cc}{\mathcal{C}}_1& 0\\ 0& {\mathcal{C}}_2\end{array}\right){\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^{m-k}\left(\begin{array}{cc}{\mathcal{C}}_1& 0\\ 0& {\mathcal{C}}_2\end{array}\right){\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^k\right\}\hfill \\ & \times {\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^n\hfill \\ =& 0,\hfill \end{array}$$

and it is easy to see that

$${\mathcal{Q}}_1^{\ast n}\left({\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\mathcal{C}}_1{\mathcal{Q}}_1^{m-k}{\mathcal{C}}_1{\mathcal{Q}}_1^k\right){\mathcal{Q}}_1^n=0.$$

Thus, ${\mathcal{Q}}_1\in {\left[{\mathcal{QSC}}_1\right]}_{(n,m)}\left[\mathcal{M}\right]$. □

We give the following characterization of members of ${\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$.

Theorem 1.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\mathcal{C}={\mathcal{C}}_1\oplus {\mathcal{C}}_2\in \mathcal{C}\left[\mathcal{K}\right]$, where ${\mathcal{C}}_1\in \mathcal{C}\left[\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\right]$ and ${\mathcal{C}}_2\in \mathcal{C}[ker\left({{\mathcal{Q}}^{\ast }}^n\right)]$. Assume that ${\mathcal{Q}}^n\ne 0$ and $\overline{\mathcal{R}\left(\mathcal{Q}\right)}\ne \mathcal{K}.$ The following statements are equivalent:

(1)

$\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$,

(2)

$\mathcal{Q}=\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)$ on $\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({{\mathcal{Q}}^{\ast }}^n\right)$, where ${\mathcal{Q}}_1$ is an $[m,{\mathcal{C}}_1]$-symmetric operator on $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}$. In addition, we have ${\mathcal{Q}}_3^n=0$ and $\sigma \left(\mathcal{Q}\right)=\sigma \left({\mathcal{Q}}_1\right)\cup \left\{0\right\}$, where $\sigma \left(\mathcal{Q}\right)$ is the spectrum of $\mathcal{Q}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$. Consider the following matrix representations:

$$\mathcal{Q}=\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right),\mathcal{C}=\left(\begin{array}{cc}{\mathcal{C}}_1& 0\\ 0& {\mathcal{C}}_2\end{array}\right)\mathrm{on}\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({{\mathcal{Q}}^{\ast }}^n\right).$$

Let $\mathcal{P}$ be the orthogonal projection of $\mathcal{K}$ onto $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}.$ Since $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$, we obtain from Remark 1 that

$$\mathcal{P}\left({\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\mathcal{C}{\mathcal{Q}}^{m-k}\mathcal{C}{\mathcal{Q}}^k\right)\mathcal{P}=0.$$

This yields

$${\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){\mathcal{C}}_1{\mathcal{Q}}_1^{m-k}{\mathcal{C}}_1{\mathcal{Q}}_1^k=0.$$

Hence, ${\mathcal{Q}}_1$ is an $[m,{\mathcal{C}}_1]$-symmetric operator on $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}$.

Let $\psi ={\psi }_1\oplus {\psi }_2\in \overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({{\mathcal{Q}}^{\ast }}^n\right)=\mathcal{K}$. If $\psi \in ker\left({{\mathcal{Q}}^{\ast }}^n\right)$, then

$$\begin{array}{ccc}\hfill \langle {{\mathcal{Q}}_3}^n\psi \mid \psi \rangle & =& \langle (I-\mathcal{P}){\mathcal{Q}}^n\psi \mid \psi \rangle \hfill \\ & =& \langle {\mathcal{Q}}^n\psi \mid (I-\mathcal{P})\psi \rangle \hfill \\ & =& \langle {\mathcal{Q}}^n\psi \mid \psi \rangle \hfill \\ & =& \langle \psi \mid {\mathcal{Q}}^{\ast n}\psi \rangle \hfill \\ & =& 0.\hfill \end{array}$$

Hence, ${{\mathcal{Q}}_3}^n=0$. From [24] (Corollary 7), we obtain $\sigma \left(\mathcal{Q}\right)\cup \mathcal{S}=\sigma \left({\mathcal{Q}}_1\right)\cup \sigma \left({\mathcal{Q}}_3\right)$, where $\mathcal{S}$ is the union of some of the holes in $\sigma \left(\mathcal{Q}\right)$, which is a subset of $\sigma \left({\mathcal{Q}}_1\right)\cap \sigma \left({\mathcal{Q}}_3\right)$. Moreover, $\sigma \left({\mathcal{Q}}_3\right)=\left\{0\right\}$, and $\sigma \left({\mathcal{Q}}_1\right)\cap \sigma \left({\mathcal{Q}}_3\right)$ has no interior points. So, we have $\sigma \left(\mathcal{Q}\right)=\sigma \left({\mathcal{Q}}_1\right)\cup \sigma \left({\mathcal{Q}}_3\right)=\sigma \left({\mathcal{Q}}_1\right)\cup \left\{0\right\},$ by [24] (Corollary 8).

$\left(2\right)\Rightarrow \left(1\right)$ Let us assume that

$$\mathcal{Q}=\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)\mathrm{on}\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({\mathcal{Q}}^{\ast n}\right),$$

where ${\mathcal{Q}}_1$ is $[m,{\mathcal{C}}_1]$-symmetric, and ${\mathcal{Q}}_3^n=0$. Since

$${\mathcal{Q}}^n=\left(\begin{array}{cc}{\mathcal{Q}}_1^n& {\displaystyle {\displaystyle \sum _{0\le j\le n-1}}{\mathcal{Q}}_1^j{\mathcal{Q}}_2{\mathcal{Q}}_3^{n-1-j}}\\ 0& 0\end{array}\right),$$

we have

$$\begin{array}{cc}& {\mathcal{Q}}^{\ast n}\left({\displaystyle \sum _{0\le l\le m}}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)C{\mathcal{Q}}^{m-k}C{\mathcal{Q}}^k\right){\mathcal{Q}}^n\hfill \\ =& {\left(\begin{array}{cc}\mathcal{Q}1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^{\ast n}\left({\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)\left(\begin{array}{cc}{\mathbf{C}}_1& 0\\ 0& {\mathcal{C}}_2\end{array}\right){\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^{m-k}\left(\begin{array}{cc}{\mathcal{C}}_1& 0\\ 0& {\mathcal{C}}_2\end{array}\right){\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^k\right)\hfill \\ & \times {\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)}^n\hfill \\ =& \left(\begin{array}{cc}{\mathcal{Q}}_1^{\ast n}\mathbf{F}{\mathcal{Q}}_1^n& {\displaystyle {\mathcal{Q}}_1^{\ast n}\mathbf{F}{\displaystyle \sum _{0\le j\le n-1}}{\mathcal{Q}}_1^j{\mathcal{Q}}_2{\mathcal{Q}}_3^{n-1-j}}\\ {\displaystyle {\left({\displaystyle \sum _{0\le j\le n-1}}{\mathcal{Q}}_1^j{\mathcal{Q}}_2{\mathcal{Q}}_3^{n-1-j}\right)}^{\ast }\mathbf{F}{\mathcal{Q}}_1^n}& {\displaystyle {\left({\displaystyle \sum _{0\le j\le n-1}}{\mathcal{Q}}_1^j{\mathcal{Q}}_2{\mathcal{Q}}_3^{n-1-j}\right)}^{\ast }\mathbf{F}{\displaystyle \sum _{0\le j\le n-1}}{\mathcal{Q}}_1^j{\mathcal{Q}}_2{\mathcal{Q}}_3^{n-1-j}}\end{array}\right),\hfill \end{array}$$

where

$$\mathbf{F}={\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right){\mathcal{C}}_1{\mathcal{Q}}_1^{m-k}{C}_1{\mathcal{Q}}_1^k.$$

Hence,

$${\displaystyle {\mathcal{Q}}^{\ast n}\left({\displaystyle \sum _{0\le k\le m}}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)C{\mathcal{Q}}^{m-k}C{\mathcal{Q}}^k\right){\mathcal{Q}}^n=0\mathrm{on}\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({\mathcal{Q}}^{\ast n}\right).}$$

Thus, $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right].$ □

Remark 4.

If $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ is such that $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}=\mathcal{K}$, then $\mathcal{Q}$ is an $[m,\mathcal{C}]$-symmetric operator.

The authors of [4,5] demonstrated that the power of an $(m,\mathcal{C})$-symmetric operator remains an $(m,\mathcal{C})$-symmetric operator. In [19], it was also proved that the power of an $[m,\mathcal{C}]$-symmetric operator is also an $[m,\mathcal{C}]$-symmetric operator. The following result indicates that the same result is valid, under the conditions of Theorem 1, for members of ${\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$.

Corollary 1.

Let $\mathcal{C}={\mathcal{C}}_1\oplus {\mathcal{C}}_2\in \mathcal{C}\left[\mathcal{K}\right]$, where ${\mathcal{C}}_1\in \mathcal{C}\left[\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\right]$, and ${\mathcal{C}}_2\in \mathcal{C}[ker\left({{\mathcal{Q}}^{\ast }}^n\right)]$. Assume that $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$; then, so too is ${\mathcal{Q}}^k$ for $k\ge 1$.

Proof. 

If $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}=\mathcal{K}$, then $\mathcal{Q}$ is an $[m,\mathcal{C}]$-symmetric operator. Hence, ${\mathcal{Q}}^k$ is an $[m,\mathcal{C}]$-symmetric for every k.

If $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\ne \mathcal{K}$, based on Theorem 1, we have

$$\mathcal{Q}=\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)\mathrm{on}\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({\mathcal{Q}}^{\ast n}\right),$$

where ${\mathcal{Q}}_1$ is an $[m,{\mathcal{C}}_1]$-symmetric operator. By [19] (Theorem 4.1), ${\mathcal{Q}}_1^k$ is an $[m,{\mathcal{C}}_1]$-symmetric operator. We observe that

$${\mathcal{Q}}^k=\left(\begin{array}{cc}{\mathcal{Q}}_1^k& {\displaystyle \sum _{0\le j\le k-1}{\mathcal{Q}}_1^j{\mathcal{Q}}_2{\mathcal{Q}}_3^{k-1-j}}\\ 0& {\mathcal{Q}}_3^k\end{array}\right)\mathrm{on}\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({\mathcal{Q}}^{\ast n}\right).$$

Therefore, ${\mathcal{Q}}^k\in {\left[\mathcal{QCS}\right]}_{(n,m)}\left[\mathcal{K}\right]$ by Theorem 1. □

Remark 5.

The converse of Corollary 1 does not necessarily hold, as can be seen from the following example.

Example 4.

Consider $\mathcal{C}\in \mathcal{C}\left[{\mathbb{C}}^3\right]$ defined by $\mathcal{C}\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}\overline{x_3}\\ -\overline{x_2}\\ \overline{x_1}\end{array}\right)$, and consider the operator matrix $\mathcal{Q}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$ on ${\mathbb{C}}^3$. We show that ${\mathcal{Q}}^{\ast }\left(\mathcal{C}\mathcal{Q}\mathcal{C}-\mathcal{Q}\right)\mathcal{Q}\ne 0$, and ${\mathcal{Q}}^{\ast 2}\left(\mathcal{C}{\mathcal{Q}}^2\mathcal{C}-{\mathcal{Q}}^2\right){\mathcal{Q}}^2=0$. This shows that ${\mathcal{Q}}^2\in {\left[\mathcal{QSC}\right]}_{(1,1)}\left[{\mathbb{C}}^3\right]$, but $\mathcal{Q}\notin {\left[\mathcal{QSC}\right]}_{(1,1)}\left[{\mathbb{C}}^3\right]$.

Proposition 3.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\mathcal{C}={\mathcal{C}}_1\oplus {\mathcal{C}}_2\in \mathcal{C}\left[\mathcal{K}\right]$, where ${\mathcal{C}}_1\in \mathcal{C}\left[\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\right]$ and ${\mathcal{C}}_2\in \mathcal{C}[ker\left({{\mathcal{Q}}^{\ast }}^n\right)]$. If $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$, then $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,k)}\left[\mathcal{K}\right]$ for $k\ge m$.

Proof. 

If $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}=\mathcal{K}$, then $\mathcal{Q}$ is an $[m,\mathcal{C}]$-symmetric operator and $\mathcal{Q}$ is a $[k,\mathcal{C}]$-symmetric operator for $k\ge m$.

If $\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\ne \mathcal{K}$, by Theorem 1, we can write $\mathcal{Q}$ as

$$\mathcal{Q}=\left(\begin{array}{cc}{\mathcal{Q}}_1& {\mathcal{Q}}_2\\ 0& {\mathcal{Q}}_3\end{array}\right)\mathrm{on}\mathcal{K}=\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\oplus ker\left({\mathcal{Q}}^{\ast n}\right),$$

where ${\mathcal{Q}}_1=\mathcal{Q}{|}_{\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}}$ is an $[m,{\mathcal{C}}_1]$-symmetric operator and ${\mathcal{Q}}_3^n=0$. It is well known that ${\mathcal{Q}}_1$ is a $[k,{\mathcal{C}}_1]$-symmetric operator for $k\ge m$. The statement arises directly as a consequence of Theorem 1. □

The following proposition provides an additional condition under which $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ becomes $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,k)}\left[\mathcal{K}\right]$ for $k\ge m$.

Proposition 4.

Let $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$. If $\left[{\mathcal{Q}}^{\ast },\mathcal{C}\mathcal{Q}\mathcal{C}\right]=0$, then $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,k)}\left[\mathcal{K}\right]$ for every $k\ge m$.

Proof. 

From (4), we have

$${\mathsf{\Psi }}_{m+1}(\mathcal{Q};\mathcal{C})=\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathsf{\Psi }}_m(\mathcal{Q};\mathcal{C})-{\mathsf{\Psi }}_m\left(\mathcal{Q};\mathcal{C}\right)\mathcal{Q}.$$

Since $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ satisfies ${\mathcal{Q}}^{\ast }\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right)=\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathcal{Q}}^{\ast }$, it follows that

$${\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m+1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_m\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n-{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_m(\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+1}=0.$$

Thus, $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m+1)}\left[\mathcal{K}\right]$. Hence, by induction, we obtain the desired conclusion. □

Definition 2

([19]). For $\mathcal{Q},\mathbf{T}\in \mathcal{B}\left(\mathcal{K}\right)$, the pair $(\mathcal{Q},\mathbf{T})$ of operators is said to be a $\mathcal{C}$-doubly commuting pair if $\mathcal{Q}\mathbf{T}=\mathbf{T}\mathcal{Q}$ and $\left[\mathcal{C}\mathcal{Q}\mathcal{C},\mathbf{T}\right]=0$ for some $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$.

Lemma 1

([19]). Let $(\mathcal{Q},\mathbf{T})$ be a $\mathcal{C}$-doubly commuting pair, where $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$. Then, it holds that

$${\mathsf{\Psi }}_m\left(\mathcal{Q}+\mathbf{T};\mathcal{C}\right)={\displaystyle \sum _{0\le j\le m}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right){\mathsf{\Psi }}_j\left(\mathcal{Q};\mathcal{C}\right){\mathsf{\Psi }}_{m-j}\left(\mathbf{T},\mathcal{C}\right).$$

(7)

In the subsequent theorem, we consider the perturbations of an element from ${\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$.

Before that, one observes that, if an operator $\mathbf{T}$ is nilpotent with order q, then we have

$$k\ge 2q-1\Rightarrow {\mathsf{\Psi }}_k\left(\mathbf{T};\mathcal{C}\right)=0.$$

(8)

Theorem 2.

Let $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ and let $\mathbf{T}\in \mathcal{B}\left(\mathcal{K}\right)$ be a nilpotent operator of order q. If $\left[{\mathcal{Q}}^{\ast },\mathcal{C}\mathcal{Q}\mathcal{C}\right]=0$, and if $(\mathcal{Q},\mathbf{T})$ and $(\mathcal{Q},{\mathbf{T}}^{\ast })$ are $\mathcal{C}$-doubly commuting pairs, then $\mathcal{Q}+\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(n+q,m+2q-2)}\left[\mathcal{K}\right]$.

Proof. 

We must prove that

$${\left(\mathcal{Q}+\mathbf{T}\right)}^{\ast (n+q)}{\mathsf{\Psi }}_{m+2q-2}\left(\mathcal{Q}+\mathbf{T};\mathcal{C}\right){\left(\mathcal{Q}+\mathbf{T}\right)}^{n+q}=0.$$

From (7), we have

$${\mathsf{\Psi }}_{m+2q-2}\left(\mathcal{Q}+\mathbf{T};\mathcal{C}\right)={\displaystyle \sum _{0\le k\le m+2q-2}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+2q-2}{k}}\right){\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathsf{\Psi }}_{m+2q-2-k}\left(\mathbf{T};\mathcal{C}\right),$$

and further,

$$\begin{array}{cc}& {\left(\mathcal{Q}+\mathbf{T}\right)}^{\ast (n+q)}{\mathsf{\Psi }}_{m+2q-2}\left(\mathcal{Q}+\mathbf{T};\mathcal{C}\right){\left(\mathcal{Q}+\mathbf{T}\right)}^{n+q}\hfill \\ =& \left(\sum _{0\le r\le n+q}\left({\displaystyle \genfrac{}{}{0pt}{}{n+q}{r}}\right){\mathcal{Q}}^{\ast (n+q-r)}{\mathbf{T}}^{\ast r}\right)\left(\sum _{0\le k\le m+2q-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m+2q-2}{k}}\right){\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathsf{\Psi }}_{m+2q-2-k}\left(\mathbf{T};\mathcal{C}\right)\right)\hfill \\ & \times \left(\sum _{0\le s\le n+q}\left({\displaystyle \genfrac{}{}{0pt}{}{n+q}{s}}\right){\mathcal{Q}}^{n+q-s}{\mathbf{T}}^s\right).\hfill \end{array}$$

Since $(\mathcal{Q},\mathbf{T})$ and $(\mathcal{Q},{\mathbf{T}}^{\ast })$ are $\mathcal{C}$-doubly commuting pairs and $\mathbf{T}$ is nilpotent of order q, we may write

$$\begin{array}{cc}& {\left(\mathcal{Q}+\mathbf{T}\right)}^{\ast (n+q)}{\mathsf{\Psi }}_{m+2q-2}\left(\mathcal{Q}+\mathbf{T};\mathcal{C}\right){\left(\mathcal{Q}+\mathbf{T}\right)}^{n+q}=\hfill \\ & \sum _{\begin{array}{c}0\le r\le q-1\\ 0\le s\le q-1\end{array}}\sum _{0\le k\le m+2q-2}\left({\displaystyle \genfrac{}{}{0pt}{}{n+q}{r}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+q}{s}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m+2q-2}{k}}\right)\underbrace{{\mathcal{Q}}^{\ast n+q-r}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+q-s}}\hfill \\ & \times \underbrace{{\mathbf{T}}^{\ast r}{\mathsf{\Psi }}_{m+2q-2-k}\left(\mathbf{T};\mathcal{C}\right){\mathbf{T}}^s}.\hfill \end{array}$$

(i) If $k\ge m$ and $(r,s)\in {\{0,\cdots ,q-1\}}^2$, we obtain, by Proposition 4, that ${\mathcal{Q}}^{\ast n+q-r}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+q-s}=0.$

(ii) If $0\le k\le m-1$, we have $m+2q-2-k\ge 2q-1$, and hence, by (8), we have ${\mathsf{\Psi }}_{m+2q-2-k}\left(\mathbf{T};\mathcal{C}\right)=0$. By combining (i) and (ii), we obtain

$${\left(\mathcal{Q}+\mathbf{T}\right)}^{\ast (n+q)}{\mathsf{\Psi }}_{m+2q-2}\left(\mathcal{Q}+\mathbf{T};\mathcal{C}\right){\left(\mathcal{Q}+\mathbf{T}\right)}^{n+q}=0.$$

□

Theorem 3.

Let $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n_1,m_1)}\left[\mathcal{K}\right]$, and let $\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(n_2,m_2)}\left[\mathcal{K}\right]$. If $(\mathcal{Q},\mathbf{T})$, $(\mathcal{Q},{\mathbf{T}}^{\ast })$ are C-doubly commuting pairs and $\left[{\mathbf{T}}^{\ast },\mathcal{C}\mathbf{T}\mathcal{C}\right]=0$, then $\mathcal{Q}\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(max\{n_1,n_2\},m_1+m_2-1)}\left(\mathcal{K}\right)$.

Proof. 

In view of [19] (Lemma 4.10), we have

$$\begin{array}{cc}& {\mathsf{\Psi }}_{m_1+m_2-1}\left(\mathcal{Q}\mathbf{T};\mathcal{C}\right)\hfill \\ =& \sum _{0\le k\le m_1+m_2-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right){\mathsf{\Psi }}_k\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^{m_1+m_2-1-k}\mathcal{C}{\mathbf{T}}^k\mathcal{C}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left(\mathbf{T};\mathcal{C}\right).\hfill \end{array}$$

This gives

$$\begin{array}{cc}& {\left(\mathcal{Q}\mathbf{T}\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1}\left(\mathcal{Q}\mathbf{T};\mathcal{C}\right){\left(\mathcal{Q}\mathbf{T}\right)}^n=\hfill \\ & {\left(\mathcal{Q}\mathbf{T}\right)}^{\ast n}\left(\sum _{0\le k\le m_1+m_2-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right){\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{m_1+m_2-1-k}\mathcal{C}{\mathbf{T}}^k\mathcal{C}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left(\mathbf{T};\mathcal{C}\right)\right)\hfill \\ & {\left(\mathcal{Q}\mathbf{T}\right)}^n\hfill \\ =& \sum _{0\le k\le m_1+m_2-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right){\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n{\mathcal{Q}}^{m_1+m_2-1-k}\mathcal{C}{\mathbf{T}}^k\mathcal{C}{\mathbf{T}}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left(\mathbf{T};\mathcal{C}\right){\mathbf{T}}^n\hfill \\ =& \sum _{0\le k\le m_1-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right){\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n{\mathcal{Q}}^{m_1+m_2-1-k}\mathcal{C}{\mathbf{T}}^k\mathcal{C}{\mathbf{T}}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left(\mathbf{T};\mathcal{C}\right){\mathbf{T}}^n\hfill \\ & +\sum _{m_1\le k\le m_1+m_2-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right){\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n{\mathcal{Q}}^{m_1+m_2-1-k}\mathcal{C}{\mathbf{T}}^k\mathcal{C}{\mathbf{T}}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left(\mathbf{T};\mathcal{C}\right){\mathbf{T}}^n.\hfill \end{array}$$

Now, consider the following cases.

For $0\le k\le m_1-1$, we have $m_1+m_2-1-k\ge m_2$, and taking into account Lemma 4, we obtain

$${\mathbf{T}}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left(\mathbf{T};\mathcal{C}\right){\mathbf{T}}^n=0.$$

For $k\ge m_1,$ we have ${\mathcal{Q}}^{\ast n}{\mathcal{Q}}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0.$ Taking into account the above observations, we deduce the desired result. □

Corollary 2.

Let $\mathcal{Q}$ and $\mathbf{T}$ be in $\mathcal{B}\left(\mathcal{K}\right)$. Let $\mathcal{C}={\mathcal{C}}_1\oplus {\mathcal{C}}_2\in \mathcal{C}\left[\mathcal{K}\right]$, where ${\mathcal{C}}_1\in \mathcal{C}\left[\overline{\mathcal{R}\left({\mathcal{Q}}^n\right)}\right]$ and ${\mathcal{C}}_2\in \mathcal{C}[ker\left({{\mathcal{Q}}^{\ast }}^n\right)]$. Assume that $(\mathcal{Q},\mathbf{T})$, $(\mathcal{Q},{\mathbf{T}}^{\ast })$ are $\mathcal{C}$-doubly commuting pairs, and $\left[{\mathbf{T}}^{\ast },\mathcal{C}\mathbf{T}\mathcal{C}\right]=0$. If $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m_1)}\left[\mathcal{K}\right]$ and $\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(n^{\prime },m_2)}\left[\mathcal{K}\right]$, then

$${\mathcal{Q}}^q\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(max\{n,n^{\prime }\},m_1-m_2-1)}\left[\mathcal{K}\right]$$

for all positive integers q.

Proof. 

By Corollary 1, we obtain ${\mathcal{Q}}^q\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$. However, $({\mathcal{Q}}^q,\mathbf{T})$, $({\mathcal{Q}}^q,{\mathbf{T}}^{\ast })$ are C-doubly commuting pairs and $\left[{\mathbf{T}}^{\ast },\mathcal{C}\mathbf{T}\mathcal{C}\right]=0$. This shows that $\mathbf{T}$ and ${\mathcal{Q}}^q$ fulfill the conditions set by Theorem 3; therefore, ${\mathcal{Q}}^q\mathbf{T}\in {\left[\mathcal{QSC}\right]}_{(max\{n,n^{\prime }\},m_1-m_2-1)}\left[\mathcal{K}\right]$. □

Lemma 2.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$, and let $\mathcal{C},\mathcal{D}\in \mathcal{C}\left[\mathcal{K}\right]$. The following assertions hold true:

(1)

$\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ if, and only if, ${\mathcal{Q}\otimes \mathbf{I}\in \left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{D}]\right)]}_{(n,m)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right]$.

(2)

$\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$ if, and only if, ${\mathbf{I}\otimes \mathcal{Q}\in \left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{D}]\right)]}_{(n,m)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right].$

Proof. 

In [18] (Lemma 4.5), it has been proved that $\mathcal{C}\otimes \mathcal{D}\in \mathcal{C}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right]$. Now, direct computation gives

$$\begin{array}{cc}& {\left(\mathcal{Q}\otimes \mathbf{I}\right)}^{\ast n}\left(\sum _{0\le k\le m}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\left(\mathcal{C}\otimes \mathcal{D}\right){\left(\mathcal{Q}\otimes \mathbf{I}\right)}^{m-k}\left(\mathcal{C}\otimes \mathcal{D}\right){\left(\mathcal{Q}\otimes \mathbf{I}\right)}^k\right){\left(\mathcal{Q}\otimes \mathbf{I}\right)}^n\hfill \\ =& {\mathcal{Q}}^{\ast n}\left(\sum _{0\le k\le m}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\mathcal{C}{\mathcal{Q}}^{m-k}C{\mathcal{Q}}^k\mathcal{C}\right){\mathcal{Q}}^n\otimes \mathbf{I}\hfill \\ =& {\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_m\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n\otimes \mathbf{I}.\hfill \end{array}$$

Hence, $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$ if, and only if, $\mathcal{Q}\otimes \mathbf{I}\in {[\mathcal{QS}\mathcal{C}\otimes \mathcal{D}]}_{(n,m)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right]$. □

Theorem 4.

Let $\mathcal{Q},\mathbf{T}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\mathcal{C},\mathcal{D}\in \mathcal{C}\left[\mathcal{K}\right]$ such that $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n_1,m_1)}\left[\mathcal{K}\right]$ and $\mathbf{T}\in {\left[\mathcal{QSD}\right]}_{(n_2,m_2)}\left[\mathcal{K}\right]$. If $\left[{\mathbf{T}}^{\ast },\mathcal{C}\mathbf{T}\mathcal{C}\right]=0$, then $\mathcal{Q}\otimes \mathbf{T}\in \in {[\mathcal{QS}\mathcal{C}\otimes \mathcal{D}]}_{(max\{n_1,n_2\},m_1+m_2-1)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right]$.

Proof. 

It is established that $\mathcal{Q}\otimes \mathbf{T}=\left(\mathcal{Q}\otimes \mathbf{I}\right)\left(\mathbf{I}\otimes \mathbf{T}\right)=\left(\mathbf{I}\otimes \mathbf{T}\right)\left(\mathcal{Q}\otimes \mathbf{I}\right)$ and, from Lemma 2, we obtain that $\mathcal{Q}\otimes \mathbf{I}\in {\left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{D})\right]}_{(n_1,m_1)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right]$. Similarly, $\mathbf{I}\otimes \mathbf{T}\in {\left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{D})\right]}_{(n_2,m_2)}\left(\mathcal{K}\overline{\otimes }\mathcal{K}\right)$. On the other hand, note that $\mathcal{Q}\otimes \mathbf{I}$ and $\mathbf{I}\otimes \mathbf{T}$ meet the requirements of Theorem 3. Consequently,

$$(\mathcal{Q}\otimes \mathbf{I})(\mathbf{I}\otimes \mathbf{T})\in {\left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{D})\right]}_{(max\{n_1,n_2\},m_1+m_2-1)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right].$$

□

Proposition 5.

Let $\mathcal{Q}$ and $\mathbf{T}\in \mathcal{B}\left(\mathcal{K}\right)$ and $\mathcal{C}$ be a conjugation on $\mathcal{K}$. If $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left[\mathcal{K}\right]$ and $\mathbf{T}$ is a nilpotent operator of order q, then $\mathcal{Q}\otimes \mathbf{I}+\mathbf{I}\otimes \mathbf{T}$ and $\mathbf{I}\otimes \mathbf{T}+\mathcal{Q}\otimes \mathbf{I}$ are in ${\left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{C})\right]}_{(n+q,m+2q-2)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right].$

Proof. 

In view of Lemma 2, we have $\mathcal{Q}\otimes \mathbf{I}\in {\left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{C})\right]}_{(n,m)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right]$. Moreover, since $\mathbf{T}$ is nilpotent of order q, it follows that $\mathbf{I}\otimes \mathbf{T}$ is nilpotent of the same order. Using the properties of the tensor product, it can be checked that $\mathcal{Q}\otimes \mathbf{I}$ and $\mathbf{I}\otimes \mathbf{T}$ adhere to the conditions outlined in Theorem 2; therefore, $\mathcal{Q}\otimes \mathbf{I}+\mathbf{I}\otimes \mathbf{T}$ is in ${\left[\mathcal{QS}(\mathcal{C}\otimes \mathcal{C})\right]}_{(n+q,m+2q-2)}\left[\mathcal{K}\overline{\otimes }\mathcal{K}\right].$ □

3. $\mathit{n}$-Quasi-Strict $[\mathit{m},\mathcal{C}]$-Symmetric Operators

In this section, we introduce the concept of n-quasi-strict $[m,C]$-symmetric operators, where $n\ge 1$ and $m\ge 2$. Some of the techniques and motivations for this study were inspired by earlier works in operator theory that dealt separately with quasi-isometries and $[m,C]$-isometries (see [25,26,27]). Our contribution here is to establish a framework that merges both views and to study the consequences of such a hybrid structure.

Definition 3.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$. We say that $\mathcal{Q}$ is an n-quasi-strict $[m,\mathcal{C}]$-symmetric operator if $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(\mathcal{K}\right)$, but $\mathcal{Q}\notin {\left[\mathcal{QSC}\right]}_{(n,m-1)}\left(\mathcal{K}\right)$; that is,

$${\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_m\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n=0\mathit{and}{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n\ne 0.$$

Example 5.

Let $\mathcal{Q}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)\mathit{and}\mathcal{C}=\left(\begin{array}{ccc}0& 0& -1\\ 0& 1& 0\\ -1& 0& 0\end{array}\right)$ be operators on the Hilbert space $\mathcal{K}={\mathbb{C}}^3$. Computation shows that

$${\mathcal{Q}}^{\ast }\left(\mathcal{C}\mathcal{Q}\mathcal{C}-\mathcal{Q}\right)\mathcal{Q}\ne 0,\mathit{and}{\mathcal{Q}}^{\ast }\left(\mathcal{C}{\mathcal{Q}}^2\mathcal{C}-2\mathcal{C}\mathcal{Q}\mathcal{C}\mathcal{Q}+{\mathcal{Q}}^2\right)\mathcal{Q}=0.$$

Hence, $\mathcal{Q}$ is a quasi-strict-$[2,\mathcal{C}]$-symmetric operator.

Theorem 5.

Let ${\mathcal{Q}}_1\in \mathcal{B}\left(\mathcal{K}\right)$ be an n-quasi strict-$[m_1,\mathcal{C}]$-symmetric operator and ${\mathcal{Q}}_2$ be an n-quasi strict-$[m_2,\mathcal{C}]$-symmetric operator for some $\mathcal{C}\in \mathcal{C}\left[\mathcal{K}\right]$, where $m_1\ge 2$ and $m_2\ge 2.$ If $({\mathcal{Q}}_1,{\mathcal{Q}}_2)$, $({\mathcal{Q}}_1,{\mathcal{Q}}_2^{\ast })$ are C-doubly commuting pairs and $\left[{\mathcal{Q}}_j^{\ast },\mathcal{C}{\mathcal{Q}}_j\mathcal{C}\right]=0$ for $j=1,2$, then ${\mathcal{Q}}_1{\mathcal{Q}}_2$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}]$-symmetric operator if, and only if,

$${\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}_1^{n+m_2-1}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1}\ne 0.$$

Proof. 

Under the assumptions we have, by Theorem 3,

$${\left({\mathcal{Q}}_1{\mathcal{Q}}_2\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1}\left({\mathcal{Q}}_1{\mathcal{Q}}_2,\mathcal{C}\right){\left({\mathcal{Q}}_1{\mathcal{Q}}_2\right)}^{\ast n}=0.$$

However,

$$\begin{array}{cc}& {\left({\mathcal{Q}}_1{\mathcal{Q}}_2\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2}\left({\mathcal{Q}}_1{\mathcal{Q}}_2;\mathcal{C}\right){\left({\mathcal{Q}}_1{\mathcal{Q}}_2\right)}^n=\hfill \\ & {\left({\mathcal{Q}}_1{\mathcal{Q}}_2\right)}^{\ast n}\left(\sum _{0\le k\le m_1+m_2-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathsf{\Psi }}_k\left({\mathcal{Q}}_1,\mathcal{C}\right){\mathcal{Q}}_1^{m_1+m_2-2-k}\mathcal{C}{\mathcal{Q}}_2^k\mathcal{C}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right)\right)\hfill \\ & {\left({\mathcal{Q}}_1{\mathcal{Q}}_2\right)}^n\hfill \\ =& \sum _{0\le k\le m_1+m_2-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1,\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-1-k}\mathcal{C}{\mathcal{Q}}_2^k\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ =& \sum _{0\le k\le m_1-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n{\mathcal{Q}}_1^{m_1+m_2-2-k}\mathcal{C}{\mathcal{Q}}_2^k\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ & +\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{m_1-1}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left({\mathcal{Q}}_1,\mathcal{C}\right){\mathcal{Q}}_1^{n+m_2-1}\mathcal{C}{\mathcal{Q}}_2^{m_1-1}\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ & +\sum _{m_1\le k\le m_1+m_2-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1,\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-2-k}\mathcal{C}{\mathcal{Q}}_2^k\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right)\hfill \\ & {\mathcal{Q}}_2^n.\hfill \end{array}$$

If $k\in \{0,\cdots ,m_1+m_2-2\}$, then $m_1+m_2-2-k\ge m_2$; therefore,

$${\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2,\mathcal{C}\right){\mathcal{Q}}_2^n=0$$

by Lemma 4. If $k\ge m_1$, then ${\mathcal{Q}}_1^n{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n=0$, also in view of Lemma 4.

Consequently, ${\mathcal{Q}}_1{\mathcal{Q}}_2$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}]$-symmetric operator if, and only if,

$${\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}_1^{n+m_2-1}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1}\ne 0.$$

From (3), we have

$$\mathcal{C}{\mathcal{Q}}_2^{m_1-1}\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n={\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1};$$

therefore, ${\mathcal{Q}}_1{\mathcal{Q}}_2$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}]$-symmetric operator if, and only if,

$${\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left({\mathcal{Q}}_1,\mathcal{C}\right){\mathcal{Q}}_1^{n+m_2-1}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1}\ne 0.$$

Hence, the proof is complete. □

Theorem 6.

Let ${\mathcal{Q}}_1\in \mathcal{B}\left(\mathcal{K}\right)$ and ${\mathcal{Q}}_2\in \mathcal{B}\left(\mathcal{K}\right)$, where ${\mathcal{Q}}_1$ is an n-quasi strict-$[m_1,\mathcal{C}]$-symmetric operator and ${\mathcal{Q}}_2$ is an n-quasi strict-$[m_2,\mathcal{C}]$-symmetric operator for $m_1,m_2\ge 2$. If $\left[{\mathcal{Q}}_j^{\ast },\mathcal{C}{\mathcal{Q}}_j\mathcal{C}\right]=0$ for $j=1,2$, then ${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2$ on $\mathcal{K}\overline{\otimes }\mathcal{K}$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}\otimes \mathcal{C}]$-symmetric operator.

Proof. 

Since ${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2=\left({\mathcal{Q}}_1\otimes \mathbf{I}\right)\left(\mathbf{I}\otimes {\mathcal{Q}}_2\right)$, it is easy to check that ${\mathcal{Q}}_1\otimes \mathbf{I}$ and $\mathbf{I}\otimes {\mathcal{Q}}_2$ fulfill the conditions of Theorem 5. We may write

$$\begin{array}{ccc}\hfill {\mathsf{\Psi }}_l\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2;\mathcal{C}\otimes \mathcal{C}\right)& =& {\mathcal{Q}}_l\left(\left({\mathcal{Q}}_1\otimes \mathbf{I}\right)\left(\mathbf{I}\otimes {\mathcal{Q}}_2\right);\mathcal{C}\otimes \mathcal{C}\right)\hfill \\ & =& \sum _{0\le k\le l}\left({\displaystyle \genfrac{}{}{0pt}{}{q}{k}}\right){\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^{m-j}\otimes \mathcal{C}{\mathcal{Q}}_2^j\mathcal{C}{\mathsf{\Psi }}_{m-k}\left({\mathcal{Q}}_2;\mathcal{C}\right).\hfill \end{array}$$

After the calculation, we have

$$\begin{array}{cc}& {\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2}\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2;\mathcal{C}\otimes \mathcal{C}\right){\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^n\hfill \\ =& {\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^{\ast n}\left(\sum _{0\le k\le m_1+m_2-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^{m-j}\otimes \mathcal{C}{\mathcal{Q}}_2^j\mathcal{C}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right)\right)\hfill \\ & \times {\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^n\hfill \\ =& \sum _{0\le k\le m_1+m_2-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathcal{Q}}_1^{\ast }{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-2-j}\otimes \mathcal{C}{\mathcal{Q}}_2^j\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ & \sum _{0\le k\le m_1-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-2-j}\otimes \mathcal{C}{\mathcal{Q}}_2^j\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ & +\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{m_1-1}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_2-1}\otimes \mathcal{C}{\mathcal{Q}}_2^{m_1-1}\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ & +\sum _{m_1\le k\le m_1+m_2-2}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{k}}\right)\{\hfill \\ & {\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathsf{\Psi }}_1^{m_1+m_2-2-j}\otimes \mathcal{C}{\mathcal{Q}}_2^j\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\}.\hfill \end{array}$$

Following the same discussion as in the proof of Theorem 5, we obtain

$$\begin{array}{cc}& {\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2}\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2;\mathcal{C}\otimes \mathcal{C}\right){\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^n\hfill \\ =& \left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{m_1-1}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_2-1}\otimes \mathcal{C}{\mathcal{Q}}_2^{m_1-1}\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\hfill \\ =& \left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{m_1-1}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^{n+m_2-1}\otimes {\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1}.\hfill \end{array}$$

Since ${\mathcal{Q}}_r$ is n-quasi strict $[m_r,\mathcal{C}]$-isosymmetric, it follows that ${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}\otimes \mathcal{C}]$-symmetric operator.| □

Lemma 3.

Let $\mathcal{Q}\in \mathcal{B}\left(\mathcal{K}\right)$ be an n-quasi strict-$[m,\mathcal{C}]$-symmetric operator, where $\mathcal{C}$ is a conjugation on $\mathcal{K}$ such that $\left[{\mathcal{Q}}^{\ast },\mathcal{C}\mathcal{Q}\mathcal{C}\right]=0$. The following axioms hold.

(i) The family of linear operators

$$\left\{{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n,k=0,1,\cdots ,m-1\right\}$$

is linearly independent.

(ii) If $\mathcal{R}\left(\mathcal{Q}\right)$ is dense, then the family of linear operators

$$\left\{{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m-k},k=0,1,\cdots ,m-1\right\}$$

is linearly independent.

(iii) If $\mathcal{R}\left(\mathcal{Q}\right)$ is dense, then the family of linear operators

$$\left\{{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+k},k=0,1,\cdots ,m-1\right\}$$

is linearly independent.

Proof. 

From Identity (4), we have

$${\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right)=\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathsf{\Psi }}_{k-1}\left(\mathcal{Q};\mathcal{C}\right)-{\mathsf{\Psi }}_{k-1}\left(\mathcal{Q};\mathcal{C}\right)\mathcal{Q}\mathrm{for}\mathrm{all}k=1,2,\cdots m-1,$$

where ${\mathsf{\Psi }}_0\left(\mathcal{Q};\mathcal{C}\right)=I.$

Since $\left[{\mathcal{Q}}^{\ast },\mathcal{C}\mathcal{Q}\mathcal{C}\right]=0$, we may write

$$\begin{array}{ccc}\hfill {\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n& =& {\mathcal{Q}}^{\ast n}\left(\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathsf{\Psi }}_{k-1}\left(\mathcal{Q};\mathcal{C}\right)-{\mathsf{\Psi }}_{k-1}\left(\mathcal{Q};\mathbf{C}\right)\mathcal{Q}\right){\mathcal{Q}}^n\hfill \\ & =& \left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k-1}\left(\mathcal{Q};\mathbf{C}\right){\mathcal{Q}}^n-{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n\mathcal{Q}.\hfill \end{array}$$

Let ${\omega }_k\in \mathbb{C},$ for $k=0,1,\cdots ,m-1$ such that

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0.$$

(9)

By multiplying Equation (9) on the left by $\mathcal{C}\mathcal{Q}\mathcal{C}$ together with the condition $[{\mathcal{Q}}^{\ast },\mathcal{C}\mathcal{Q}\mathcal{C}]=0,$ we can obtain

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}\mathcal{C}\mathcal{Q}\mathcal{C}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0.$$

(10)

Similarly, by multiplying Equation (9) on the right by $\mathcal{Q}$, we obtain

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n\mathcal{Q}=0.$$

(11)

By subtracting the two Equations (10) and (11), we obtain

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}\left(\left(\mathcal{C}\mathcal{Q}\mathcal{C}\right){\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right)-{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right)\mathcal{Q}\right){\mathcal{Q}}^n=0,$$

(12)

or, equivalently,

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0.$$

(13)

Applying the same procedure to Equation (13) yields

$${\displaystyle \sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+2}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0.}$$

(14)

By continuing with this process, we arrive at

$${\displaystyle \sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+q}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0\mathrm{for}\mathrm{all}q\in \mathbb{N}.}$$

(15)

By taking into account Lemma 4, it is widely recognized that, if $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(n,m)}\left(K\right)$, then $\mathcal{Q}\in {\left[\mathcal{QSC}\right]}_{(p,m)}\left(K\right)$ for all $p\ge m$, and this leads to the conclusion that the following implications hold.

For $q=m-1$, we have

$${\displaystyle \sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0\Rightarrow {\omega }_0{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n=0;}$$

so, ${\omega }_0=0$, because ${\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n\ne 0.$

For $q=m-2$, we obtain

$${\displaystyle \sum _{1\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n=0\Rightarrow {\omega }_1{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+m-1}\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^{n}0;}$$

so, ${\gamma }_1=0.$

By repeating this process for $q\in \{m-3,\cdots ,1\}$ and $q=0$, we can deduce that ${\omega }_k=0$ for $k=2,\cdots ,m-1$. Consequently,

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q},\mathcal{C}\right){\mathcal{Q}}^n=0⟹{\omega }_k=0\mathrm{for}k=0.\cdots ,m-1.$$

(ii) Using similar techniques and Statement (i), we obtain

$${\displaystyle \sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+q}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m-k}=0\mathrm{for}\mathrm{all}q\in \mathbb{N};}$$

(16)

therefore,

$$〈{\displaystyle \sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+q}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m-k}u|v}〉=0,\forall u,v\in \mathcal{K}.$$

(17)

For $q=m-1$, by taking into account Lemma 4, we have

$$\begin{array}{ccc}& & 〈{\displaystyle \sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+q}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m-k}u|v}〉=0\hfill \\ & \Rightarrow & 〈{\omega }_0{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m}u|v〉,\forall u,v\in \mathcal{K}\hfill \\ & \Rightarrow & {\omega }_0{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+m}=0\hfill \\ & \Rightarrow & {\omega }_0{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0\left(\sin \mathrm{ce}\overline{\mathcal{R}\left(\mathcal{Q}\right)}=\mathcal{K}\right)\hfill \\ & \Rightarrow & {\omega }_0=0\left({\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n\ne 0\right).\hfill \end{array}$$

Continuing this process, by putting $q=m-2$ into $q=1$ in Equation (17), yields ${\omega }_1=\cdots ={\omega }_{m-1}=0.$ This justifies the desired result.

(iii) Assume that

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_k\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+k}=0.$$

Following the same procedure as in case (i), we obtain

$$\sum _{0\le k\le m-1}{\omega }_k{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{k+q}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+k}=0,\forall q\in \mathbf{N}.$$

(18)

By choosing $q=m-1$ in (18), we determine by Lemma 4 that

$${\omega }_0{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^n=0\Rightarrow {\omega }_0=0.$$

Similarly, for $q=m-2$, by Lemma 4, we obtain

$${\omega }_1{\mathcal{Q}}^{\ast n}{\mathsf{\Psi }}_{m-1}\left(\mathcal{Q};\mathcal{C}\right){\mathcal{Q}}^{n+1}=0\Rightarrow {\omega }_1=0(\sin \mathrm{ce}\overline{\mathcal{R}\left(\mathcal{Q}\right)}=\mathcal{K}).$$

Continuing this process, by putting $q=m-3$ into $q=1$ in Equation (18), we obtain ${\omega }_2=\cdots ={\omega }_{m-1}=0.$ This justifies the desired result. □

Lemma 4

([26]). Let ${\mathcal{Q}}_k\in \mathcal{B}\left(\mathcal{K}\right)$ and ${\mathbf{T}}_k\in \mathcal{B}\left(\mathcal{K}\right)$ for $k=1,\cdots ,d.$ If

$$\sum _{1\le k\le d}{\mathcal{Q}}_k\otimes {\mathbf{T}}_k=0,$$

and $\{{\mathcal{Q}}_1,\cdots ,{\mathcal{Q}}_d\}$ is linearly independent, then ${\mathbf{T}}_k=0$ for $k=1,\cdots ,d.$

Theorem 7.

Let ${\mathcal{Q}}_1$ and ${\mathcal{Q}}_2\in \mathcal{B}\left(\mathcal{K}\right)$ such that $\left[{\mathcal{Q}}_k^{\ast },\mathcal{C}{\mathcal{Q}}_k\mathcal{C}\right]=0$ for $k=1,2$, where $\mathcal{C}$ is a conjugation on $\mathcal{K}$. Assume that $\mathcal{R}\left({\mathcal{Q}}_k\right)$ is dense for $k=1,2$; then, any two of Statements (i)–(ii) imply the third, where

(i)

${\mathcal{Q}}_1$ is an n-quasi strict-$[m_1,\mathcal{C}]$-symmetric operator.

(ii)

${\mathcal{Q}}_2$ is an n-quasi strict-$[m_2,\mathcal{C}]$-symmetric operator.

(iii)

${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2$ is an n-quasi strict-$[m_1+m_2-1,\mathcal{C}\otimes \mathcal{C}]$-symmetric operator.

Proof. 

Assume that (i) and (ii) are true, and prove that (iii) is true. From Theorem 6, we know that ${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\in {[\mathcal{QS}\mathcal{C}\otimes \mathcal{C}]}_{(n,m_1+m_2-1)}\left(\mathcal{K}\right)$. Moreover,

$$\begin{array}{ccc}& & {\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-2}\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2;\mathcal{C}\otimes \mathcal{C}\right){\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^n\hfill \\ & =& \left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-2}{m_1-1}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_{m_1-1}\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^{n+m_2-1}\otimes {\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1}\hfill \\ & \ne & 0.\hfill \end{array}$$

Assume that (i) and (iii) are true, and prove that (ii) is true. In fact, we have

$$\begin{array}{ccc}\hfill 0& =& {\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1}\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2;\mathcal{C}\otimes \mathcal{C}\right){\left({\mathcal{Q}}_1\otimes {\mathcal{Q}}_2\right)}^n\hfill \\ & =& \sum _{0\le k\le m_1+m_2-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right)\hfill \\ & & \left\{{\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-1-k}\otimes \mathcal{C}{\mathcal{Q}}_2^k\mathcal{C}{\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\right\}\hfill \\ & =& \sum _{0\le k\le m_1-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m_1+m_2-1}{k}}\right){\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-1-k}\hfill \\ & & \otimes {\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+k}.\hfill \end{array}$$

Since $\left\{{\mathcal{Q}}_1^{\ast n}{\mathsf{\Psi }}_k\left({\mathcal{Q}}_1;\mathcal{C}\right){\mathcal{Q}}_1^n{\mathcal{Q}}_1^{m_1+m_2-1-k};k=0,\cdots ,m_1-1\right\}$ are linearly independent, it follows from Lemma 4 that

$${\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_1+m_2-1-k}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+k}=0\mathrm{for}k=0,\cdots ,m_1-1.$$

For $k=m_1-1$, we obtain

$${\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^{n+m_1-1}=\left({\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\right){\mathcal{Q}}_2^{m_1-1}=0;$$

hence, ${\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n=0,$ since $\overline{\mathcal{R}\left({\mathcal{Q}}_2\right)}=\mathcal{K}.$ Taking into account that ${\mathcal{Q}}_1\otimes {\mathcal{Q}}_2$ is a quasi-strict $[m_1+m_2-1,\mathcal{C}\otimes \mathcal{C}]$, we deduce that ${\mathcal{Q}}_2^{\ast n}{\mathsf{\Psi }}_{m_2-1}\left({\mathcal{Q}}_2;\mathcal{C}\right){\mathcal{Q}}_2^n\ne 0$. Hence, Statement (ii) is true.

Similarly, we can prove that (ii) and (iii) imply (i). □

4. Conclusions

In this work, we developed the algebraic framework and corresponding operator identities that define a new class of n-quasi-$[m,\mathcal{C}]$-symmetric operators.

In addition, the effects of perturbations by nilpotent operators were examined, providing deeper insight into the robustness of the n-quasi-$[m,C]$-symmetric framework. The results presented here not only enrich the general theory of symmetric-type operators, but also create new avenues for further research. In particular, future investigations could focus on the spectral properties and invariant subspace structure of n-quasi-$[m,\mathcal{C}]$-symmetric operators.