New extension of quasi-$M$-hypnormal operators

1.   Introduction

Let $\mathcal{B}[\mathcal{K}]$ be the the algebra of all bounded linear operators acting on a complex Hilbert space $\mathcal{K}$ with inner product $\langle.\, |.\; \rangle$ and

The elements of $\mathcal{B}(\mathcal{K})$ are called positive operators. For every $\mathcal{U} \in \mathcal{B}[\mathcal{K}]$, $\ker(\mathcal{U}), \mathcal{R}(\mathcal{U})$, and $\overline{\mathcal{R}(\mathcal{U})}$ represent, respectively, the null space, the range, and the closure of the range of $\mathcal{U}$. Its adjoint operator is denoted by $\mathcal{U}^*$. In addition, if $\mathcal{U}_1, \mathcal{U}_2 \in \mathcal{B}[\mathcal{K}]$, then $\mathcal{U}_1 \geq \mathcal{U}_2$ means that $\mathcal{U}_1- \mathcal{U}_2 \in \mathcal{B}[\mathcal{K}]^ +.$

For $\mathcal{U} \in \mathcal{B}[\mathcal{K}]$, let $\sigma_p(\mathcal{U})$, $\sigma(\mathcal{U})$, $\sigma_s(\mathcal{U})$, and $\sigma_a(\mathcal{U})$ denote the point spectrum, the spectrum, the surjective spectrum, and, approximate point spectrum of $\mathcal{U}$. If $\mu \in \sigma_p(\mathcal{U})$ and $\overline{\mu} \in \sigma_p(\mathcal{U}^*)$, then $\mu$ is in the joint point spectrum, $\sigma_{jp}(\mathcal{U}).$ If $\mu \in \sigma_a(\mathcal{U})$ and $\overline{\mu} \in \sigma_a(\mathcal{U}^*)$, then we say that $\mu$ is in the joint approximate point spectrum, $\sigma_{ja}(\mathcal{U}).$ The following classes of operators have been studied by many authors. An operator $\mathcal{U}\in \mathcal{B}[\mathcal{K}]$ is said to be

(1) normal if $\mathcal{U}^*\mathcal{U} = \mathcal{U}\mathcal{U}^*$ ^[1,2,3],

(2) hyponormal operator if $\mathcal{U}^*\mathcal{U}\geq\mathcal{ U}\mathcal{U}^*$ ^[4,5],

(3) $M$-hyponormal operator ^[6] if there exists a real positive number $M$ such that

or, equivalently, if

(4) quasi-$M$-hyponormal ^[7,8] if there exits a real positive number $M$ such that

(5) $k$-quasi-$M$-hyponormal operator if there exists a real positive number $M$ such that

where $k$ is a natural number ^[9].

In the papers ^[10,11], the authors have introduced the class of polynomially normal as follows: An operator $\mathcal{U}$ is said to be polynomially normal if there exists a nontrivial polynomial $P = \sum\limits_{0\leq k \leq n}b_kz^k \in C[z]$ with

In the following, we introduce a new class of operators called the class of polynomially quasi-$M$-hyponormal operators denoted by $[\mathcal{PQK}]_M$.

Definition 1.1. An operator ${\mathcal{U}}\in \mathcal{B}[\mathcal{K}]$ is said to be a polynomially quasi-$M$-hyponormal operator if there exists a nonconstant polynomial $P\in \mathbb{C}[z]$ and a positive constant $M$ such that,

for all $\varrho \in {\mathbb{C}}$ or

Remark 1.1. (1) Every $M$-hyponormal operator is in $[\mathcal{PQK}]_M$.

(2) Every quasi-$M$-hyponormal operator is in $[\mathcal{PQK}]_M$ with $P(z) = z.$

(3) Every $k$-quasi-$M$-hyponormal operator is in $[\mathcal{PQK}]_M.$ with $P(z) = z^k$.

2.   Main results

In this section, we will show several properties of the class $[\mathcal{PQK}]_M.$

Theorem 2.1. Let $\mathcal{U} \in \mathcal{B}[\mathcal{K}]$. Then $\mathcal{U} \in [\mathcal{PQK}]_M$ if and only if there exists a real number $M$ that is positive, such that

Proof. Assume that $\mathcal{U} \in [\mathcal{PQK}]_M$, then there exists $P\in \mathbb{C}[z]$ and $M > 0$ for which

Conversely, assume that $\mathcal{U}$ satisfies

for each $\omega \in \mathcal{K}$, so one can obtain that

So one can obtain that

Therefore

Hence, one can obtain

Therefore, $\mathcal{U}$ is polynomially quasi-$M$-hyponormal operator. □

Corollary 2.1. Let $\mathcal{U}\in [\mathcal{PQK}]_M$ such that $\overline{\mathcal{R}(P({\mathcal{U}}))} = \mathcal{K}$, then $\mathcal{U}$ is an $M$-hyponormal operator.

Proof. Supposing $\overline{\mathcal{R}(P(\mathcal{U}))} = \mathcal{K}$, let $\omega \in \mathcal{K}$. Then there is a sequence $\omega_{n} \in \mathcal{K}$ such that $P(\mathcal{U})\omega_{n}\longrightarrow \omega$ as $n\longrightarrow\infty$. Now, from the hypothesis of this corollary and Theorem 2.1, we have

This implies

By taking the limit $n\to \infty$ we obtain

Therefore, $\mathcal{U}$ is $M$-hyponormal operator. □

A characterization of some members of $[\mathcal{PQK}]_M$ will be given in the following theorem.

Theorem 2.2. Let $\mathcal{U}\in {\mathcal B}[{\mathcal K}]$ such that $\overline{\mathcal{R}(P(\mathcal{U}))}\not = \mathcal{K}$ for some $P\in \mathbb{C}[z]$, then the following are equivalent.

(1) $\mathcal{U} \in [\mathcal{PQK}]_M$.

(2) $\mathcal{U} = \left(\begin{array}{cc} \mathcal{U}_1 & \mathcal{U}_2\\ 0 & \mathcal{U}_3 \\ \end{array} \right)$ on ${\mathcal K} = \overline{{\mathcal R}(P(\mathcal{U}))}\oplus \ker(P(\mathcal{U})^{*})$, where $\mathcal{U}_1 = \mathcal{U}_{\big| \overline{{\mathcal R}(P(\mathcal{U}))}}$ satisfies

and $P(\mathcal{U}_3) = 0$. Furthermore, $\sigma(\mathcal{U}) = \sigma(\mathcal{U}_1)\cup \sigma(\mathcal{U}_3).$

Proof. (1) $\Rightarrow$ (2). By taking into account the matrix representation of ${\mathcal{U}}$ with respect to the decomposition ${\mathcal K} = \overline{{\mathcal R}(P(\mathcal{U}))} \oplus \ker(P(\mathcal{U})^{\ast})$ : $\mathcal{U} = \left(\begin{array}{cc} \mathcal{U}_1 & \mathcal{U}_2 \\ 0 & \mathcal{U}_3 \\ \end{array} \right)$. Let $P_{\big|\overline{{\mathcal R}(P(\mathcal{U})}}$ be the projection onto $\overline{{\mathcal R}(P(\mathcal{U})}$. Then $\left(\begin{array}{cc} \mathcal{U}_1 & 0 \\ 0 & 0\\ \end{array} \right) = \mathcal{U}P_{\big|\overline{{\mathcal R}(P(\mathcal{U}))}} = P_{\big|\overline{{\mathcal R}(P(\mathcal{U}))}}\mathcal{U}P_{\big|\overline{{\mathcal R}(P(\mathcal{U}))}}$. Since $\mathcal{U}\in [\mathcal{PQK}]_M$, from Definition 2.1, we have

That is

for all $\varrho \in \mathbb{C}.$

On the other hand, let $\omega = \omega_1+\omega_2\in {\mathcal K} = \overline{{\mathcal R}(P(\mathcal{U}))}\oplus \ker(P(\mathcal{U})^*).$ A simple computation shows that

So, $P(\mathcal{U}_3) = 0.$ The proof of the identity $\sigma(\mathcal{U}) = \sigma(\mathcal{U}_1)\cup \sigma(\mathcal{U}_3)$ is deduced by an argument similar to the one given in [12, Corollaries 7 and 8].

(2) $\Rightarrow$ (1) Suppose that $\mathcal{U} = \left(\begin{array}{cc} \mathcal{U}_1 & \mathcal{U}_2\\ \\ 0 & \mathcal{U}_3\\ \end{array} \right)$ onto ${\mathcal K} = \overline{{\mathcal R}(P(\mathcal{U}))}\oplus \ker(P(\mathcal{U})^{*})$, with

for all $\varrho \in \mathbb{C}$ and $P(\mathcal{U}_3) = 0.$

Since $\mathcal{U}^m = \left(\begin{array}{cc} \mathcal{U}_1^m & \sum\limits_{ j = 0}^{ m-1}\mathcal{U}_1^j\mathcal{U}_2\mathcal{U}_3^{k-1-j} \\ \\ 0 & \mathcal{U}_3^m \\ \end{array} \right)$, $P(\mathcal{U}) = \left(\begin{array}{cc} P(\mathcal{U}_1) & X \\ \\ 0 & 0 \\ \end{array} \right),$

and

Further

where $D = P(\mathcal{U}_1)P(\mathcal{U}_1)^{*} +XX^* = D^*.$

Hence, for all $\varrho \in \mathbb{C}$, we have

It follows that

This means that

on ${\mathcal K} = \overline{{\mathcal R}(P(\mathcal{U})^{*})}\oplus \ker(P(\mathcal{U}))$. Consequently, $\mathcal{U}\in [\mathcal{PQK}]_M.$

□

In the following theorem we prove that part of $[\mathcal{PQK}]_M$ on a closed subspace is again $[\mathcal{PQK}]_M$.

Theorem 2.3. Let $\mathcal{U}\in [\mathcal{PQK}]_M$. If ${\mathcal M}\subset {\mathcal K}$ is a closed invariant subspace for $\mathcal{U}$, then the restriction $\mathcal{U}_{\big| {\mathcal M}}$ is in $[\mathcal{PQK}]_M$.

Proof. With respect to the decomposition ${\mathcal K} = {\mathcal M} \oplus {\mathcal M}^{\bot },$ $\mathcal{U}$ can be written

Hence, for all integer $k,$ $k\geq 1,$ we get

for some $X\in \mathcal{B}[\mathcal{K}]$ and

and

Since $\mathcal{\mathcal{U}} \in [\mathcal{PQK}]_M$, there exists $P\in \mathbb{C}[z]$ and $M\geq0$ such that for all $\varrho \in \mathbb{C}$

Hence, we obtain

where

and some operator $Z\in {\mathcal B}[{\mathcal K}].$ By [13, Theorem 6], $\left(\begin{array}{cc} \Phi & \Psi \\ \Psi^{\ast } & Z \end{array} \right) \geq 0$ if and only if $\Phi, Z\geq 0$ and $\Psi = \Phi^{\frac{1}{2}}WZ^{\frac{1 }{2}}$ for some contraction $W.$ Thus,

According to $\mathcal{U}_2\mathcal{U}_2^{\ast }\geq 0,$ it follows that

Consequently, the restriction $\mathcal{U}_1 = \mathcal{U}_{\big| \mathcal{M}}\in [\mathcal{PQK}]_M$.

□

Theorem 2.4. Let $P \in \mathbb{C}[z]$ be a polynomial and $\mathcal{U} = \left(\begin{array}{cc} \mathcal{U}_1 & \mathcal{U}_2 \\ 0 & \mathcal{U}_3 \\ \end{array} \right) \in{\mathcal B}({\mathcal K}\oplus {\mathcal K})$. If $\mathcal{U}_1 \in [\mathcal{PQK}]_M$, $P(\mathcal{U}_3) = 0$ and $\sigma_s(\mathcal{U}_1)\cap\sigma_a(\mathcal{U}_3) = \emptyset$, then $\mathcal{U}$ is similar to a direct sum of a member of $[\mathcal{PQK}]_M$ and an algebraic operator.

Proof. According to the condition $\sigma_s(\mathcal{U}_1)\cap\sigma_a(\mathcal{U}_3) = \emptyset$, it follows from [14, Theorem 3.5.1,(c)] that there exists $B \in {\mathcal B}({\mathcal K})$ such that $\mathcal{U}_1B - B\mathcal{U}_3 = \mathcal{U}_2.$ In view of the equality,

It is clear that $\mathcal{U}$ is similar to $\Psi = \left(\begin{array}{cc} \mathcal{U}_1 & 0 \\ 0 & \mathcal{U}_3 \\ \end{array} \right).$ From the assumption that $\mathcal{U}_1\in [\mathcal{PQK}]_M$ and $P(\mathcal{U}_3) = 0$, we get by easy calculation that

Following this, $\mathcal{U}$ is similar to a member of $[\mathcal{PQK}]_M$ and an algebraic operator.

□

Theorem 2.5. Let $N\in {\mathcal B}\big[\mathcal{K}\big]$ be an invertible operator and $\mathcal{U} \in {\mathcal B}\big[\mathcal{K}\big]$ be an operator such that $\mathcal{U}$ commutes with ${N}^*{N}$. Then $\mathcal{U} \in [\mathcal{PQK}]_M$ if and only if ${N}\mathcal{U}{N}^{-1}\in [\mathcal{PQK}]_M$.

Proof. Assume that $\mathcal{U}\in [\mathcal{PQK}]_M$. Then there exists $P\in \mathbb{C}[z]$ and $M > 0$ such that

From this, we have that

A computation shows that

This shows that the operator ${N}{N}^*$ commutes with the operator

Hence, the operator $({N}{N}^*)^{-1}$ also commutes with the operator

Using the fact that the operators $(N{N}^*)^{-1}$ and

are positive, and since they commute with each other. We get that their product is also a positive operator

This implies that

From the fact that $\mathcal{U}{N}^*N = {N}^*{N}{\mathcal{U}}$, it follows that

Hence,

On the other hand,

Now we show that ${N}{\mathcal{U}}{N}^{-1} \in [\mathcal{PQK}]_M$. Indeed

Based on these calculations, we conclude that ${N}{\mathcal{U}}{N}^{-1} \in [\mathcal{PQK}]_M.$

Conversely, assume that ${N}{\mathcal{\mathcal{U}}}{N}^{-1} \in [\mathcal{PQK}]_M$. Then

Similar to before, we get

Hence,

or equivalently

By using that, ${N}^*{N}$ commutes with operator ${\mathcal{U}}$, and hence commutes with operators

It follows that $\big({N}^*{N}\big)^{-1}$ commute with

By observing that $\big({N}^*{N}\big)^{-1}$ and

are positive, and since they commutes with each other, we have

Therefore,

Whit does it mean that ${\mathcal{U}} \in [\mathcal{PQK}]_M$. □

Theorem 2.6. Let ${\mathcal{U}}\in [\mathcal{PQK}]_M$ for $P\in \mathbb{C}[z]$. Then

for all $\mu\in \mathbb{C}$ such that $P(\mu)\not = 0.$

Proof. Let $\omega \in \ker({\mathcal{U}}-\mu)$. Since ${\mathcal{U}} \in[\mathcal{PQK}]_M$ for $P\in \mathbb{C}[z]$, it follows in view of Theorem 2.1,

Since ${\mathcal{U}}\omega = \mu \omega$, we get $P(\mathcal{U})\omega = P(\mu)\omega$, and therefore

According to $({\mathcal{U}}-\mu)\omega = 0$ we obtain $\|\big({\mathcal{U}}-\mu\big)^*P(\mu)\omega\| = 0$ or $|P(\mu)|\|\big(\mathcal{U}-\mu\big)^*\omega\| = 0$. Since $P(\mu)\not = 0$ we get $\big({\mathcal{U}}-\mu\big)^*\omega = 0.$ Therefore, the proof is complete. □

Remark 2.1. When $P(z) = z$, Theorem 2.6 coincides with [8, Proposition 1.9].

Corollary 2.2. Let ${\mathcal{U}} \in [\mathcal{PQK}]_M$ for some $P\in \mathcal{C}[z]$. If $\alpha, \beta \in \sigma_p({\mathcal{U}})-\{0\}$ with $\alpha\not = \beta$ and $P(\beta)\not = 0.$ Then

Proof. Let $\omega_1 \in \ker({\mathcal{U}}-\alpha)$ and $\omega_2 \in \ker({\mathcal{U}}-\beta)$, then ${\mathcal{U}}\omega_1 = \alpha \omega_1$ and ${\mathcal{U}}\omega_2 = \beta \omega_2.$ Therefore

We deduce that $(\alpha-\beta)\langle \omega_1\mid\omega_2\rangle = 0$ and so that $\langle \omega_1\mid\omega_2\rangle = 0$ $(\alpha\not = \beta)$. Thus, $\ker({\mathcal{U}}-\alpha)\bot\ker({\mathcal{U}}-\beta).$ □

Remark 2.2. When $P(z) = z$, Corollary 2 coincides with [8, Corollary 1.10].

Theorem 2.7. ^[15] Let $\mathcal{H}$ be a complex Hilbert space. Then there exists a Hilbert space $\mathcal{K} \supset \mathcal{H}$ and $\psi: \mathcal{B}(\mathcal{H}) \longrightarrow \mathcal{B}(\mathcal{K})$ satisfying the following properties for every $T, S \in \mathcal{B}(\mathcal{H})$ and $\varrho, \mu\in \mathbb{C}$.

(1) $\psi(T^*) = \psi(T)^*, \; \psi(I_{\mathcal{H}}) = I_\mathcal{K}, \psi(\varrho T+\mu S) = \varrho \psi(T)+\mu \psi(S)$,

(2) $\psi(TS) = \psi(T)\psi(S), \; \|\psi(T)\| = \|T\|, \psi(T)\geq \psi(S)$, for $T\geq S$,

(3) $\psi(T) \geq0$ if $T\geq0$,

(4) $\sigma_a(T) = \sigma_a(\psi(T)) = \sigma_p(\psi(T))$,

(5) $\sigma_{ja}(T) = \sigma_{jp}(\psi(T)).$

Theorem 2.8. Let $\mathcal{U} \in [\mathcal{PUK}]_M$ for some $P\in \mathcal{C}[z]$ such that ${P}(\mu)\not = 0$ for all $\mu\in\sigma_a(\mathcal{U})$. Then $\sigma_a(\mathcal{U}) = \sigma_{ja}(\mathcal{U}).$

Proof. Since $\mathcal{U}\in [\mathcal{PQK}]_M$, then there exists $P\in \mathbb{C}[z]$ and constant $M > 0$ such that

In view of Theorem 2.7, we have

Hence $\psi(\mathcal{U}) \in [\mathcal{PQK}]_M$.

From Theorem 2.7, we have $\sigma_{a}(\mathcal{U}) = \sigma_{p}(\psi(\mathcal{U})).$ Since $\psi(\mathcal{U}) \in [\mathcal{PQK}]_M$, we have $\ker\big(\psi(\mathcal{U})-\mu\big)\subset \ker\big(\psi(\mathcal{U})-\mu\big)^*$ (from Theorem 2.6). Hence $\sigma_p(\psi(\mathcal{U})) = \sigma_{jp}(\psi(\mathcal{U}))$. According to Theorem 2.7, we have $\sigma_{jp}(\psi(\mathcal{U})) = \sigma_{ja}(\mathcal{U})$. Hence, $\sigma_{a}(\mathcal{U}) = \sigma_{ja}(\mathcal{U}).$ □

In the following theorem, we will prove, under suitable conditions, the stability of the class $[\mathcal{PQK}]_M$ under the sum of operators.

Theorem 2.9. Let $\mathcal{U}_{k}\in [\mathcal{PQK}]_M$ for $k = 1, 2$. If $\mathcal{U}_1$ and $\mathcal{\mathcal{U}}_2$ satisfy the following conditions for some $P\in \mathbb{C}[z]$:

Then $\mathcal{U}_{1}+{\mathcal{U}}_{2} \in [\mathcal{PQK}]_M$.

Proof. Set $P(z) = \sum\limits_{0\leq k\leq n}a_kz^k.$ Since ${\mathcal{U}}_1{\mathcal{U}}_2 = {\mathcal{U}}_2{\mathcal{U}}_1 = 0$, we obtain

From the hypothesis that $\mathcal{U}_{1}$ and $\mathcal{U}_{2}$ are in $[\mathcal{PQK}]_M$, then both of them satisfy (2.1), and by our hypothesis

and

To show that $\mathcal{U}_{1}+\mathcal{U}_{2} \in [\mathcal{PQK}]_M$, we have

Therefore, $\mathcal{U}_1+\mathcal{U}_2 \in [\mathcal{PQK}]_M$. □

3.   Conclusions

In this paper, we have presented a study of new class of operators which considered as an extension of previous work in this field. This study will contribute to further studies in the field of operator theory.

Author contributions

O. B. Almutairi and S. A. O. A. Mahmoud: Conceptualization, Validation, Formal analysis, Supervision, Writing-review and Editing. All authors contributed equally to the writing of this article. All authors have read and approved the final version of the manuscript for publication.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors would like to express their gratitude to the four anonymous reviewers for their useful comments and editorial suggestions, which improved the comprehension of the manuscript.

Conflict of interest

The authors declare that they have no conflict of interest.