n-Quasi-m-Complex Symmetric Transformations

Abstract

Our aim in this study is to consider a generalization of the concept of m-complex symmetric transformations to n-quasi-m-complex symmetric transformations. A map $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is said to be an n-quasi-m-complex symmetric transformation if there exists a conjugation C on $\mathcal{Y}$ such that S satisfies the condition $S^{*n}\left({\displaystyle \sum _{0\le k\le m}}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S^n=0$, for some positive integers n and m. This class of transformation contains the class of m-complex symmetric transformations as a proper subset. Some basic structural properties of n-quasi-m-complex symmetric linear transformations are established with the help of transformation matrix representation. In particular, we obtain that a power of an n-quasi-m-complex symmetric is again an n-quasi-m-complex symmetric operator. Moreover, if T and S are such that T is an $n_1$-quasi-$m_1$-complex symmetric and S is an $n_2$-quasi-$m_2$-complex symmetric, their product $TS$ is an $max\{n_1,n_2\}$-quasi-$(m_1+m_2-1)$-complex symmetric under suitable conditions. We examine the stability of n-quasi-m-complex symmetric operators under perturbation by nilpotent operators.

1. Introduction

In this paper, $\mathcal{Y}$ stands for an finite or infinite complex Hilbert space with inner product $\langle ·|·\rangle$. From $\mathcal{B}\left(\mathcal{Y}\right)$, we denote the Banach algebra of all bounded linear transformations on $\mathcal{Y}$. Letting $m\in \mathbb{N}$ be the set of positive integers and $S\in \mathcal{B}\left(\mathcal{Y}\right)$, set

$${\mathbf{S}}_m\left(S\right):=\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}{S}^{m-k},$$

(1)

and

$${\mathbf{Q}}_m\left(S\right):=\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}{S}^k.$$

(2)

An immediate consequence of (2) is that

$${\mathbf{Q}}_{m+1}\left(S\right)=S^{*}{\mathbf{Q}}_m\left(S\right)S-{\mathbf{Q}}_m\left(S\right).$$

(3)

A transformation $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is said to be m symmetric [1] if

$${\mathbf{S}}_m\left(S\right)=0,$$

(4)

and it is said to be m isometric [2] if S satisfies

$${\mathbf{Q}}_m\left(S\right)=0.$$

(5)

From (2), any such transformation is also an $(m+1)$ isometry. For more details on this class of transformations, the reader is referred to [2,3,4,5,6,7].

A conjugation is a conjugate-linear transformation $C:\mathcal{Y}⟶\mathcal{Y}$, which is both involutive (i.e., $C^2=I)$ and isometric (i.e., $\langle Cx|Cy\rangle =\langle y|x\rangle (\forall x,y\in \mathcal{Y})$).

Recall that if C is a conjugation on $\mathcal{Y}$, then $\parallel C\parallel =1$, ${\left(CSC\right)}^k=C{S}^{k}C$ and ${\left(CSC\right)}^{*}=C{S}^{*}C$ for every positive integer k (see [8,9] for more details).

Using the identity (4) and a conjugation C, Chō et al. [10] defined $(m,C)$-isometric operators as follows: an operator $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is said to be an $(m,C)$-isometric operator if there exists some conjugation C for which

$$\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{k}C=0.$$

(6)

for some $m\in \mathbb{N}$. Put ${\displaystyle {\Lambda }_m(S,C):=\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{k}C}$. Then, S is an $(m,C)$-isometric transformation if and only if ${\Lambda }_m(S,C)=0.$ Note that

$$S.{\Lambda }_m(S,C)\left(CSC\right)-{\Lambda }_m(S,C)={\Lambda }_{m+1}(S,C).$$

Hence, if ${\Lambda }_m(S,C)=0,$ then ${\Lambda }_n(S,C)=0$ for all $n\ge m.$ Moreover, S is an ($m,C)$ isometry if and only if $CSC$ is an $(m,C)$ isometry (see [10]).

A transformation $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is said to be an m-complex symmetric if there exists a conjugation operator C on $\mathcal{Y}$ such that S satisfies

$${\Phi }_m(S,C):=\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C=0,$$

(7)

for some positive integer m. In this case, we say that S is an m-complex symmetric transformation with conjugation C. When $m=1$, 1-complex symmetric transformation is a complex symmetric transformation, i.e., $CSC-S^{*}=0$ and when $m=2$, the 2-complex symmetric transformation satisfies

$$C{S}^{2}C-2S^{*}CSC+S^{*2}=0.$$

It is well known that the class of m-isometric transformations has been extended to the so-called n-quasi-m-isometric transformation. For $S\in \mathcal{B}\left(\mathcal{Y}\right)$, we say that S is

(i)

Quasi-isometry if $S^{*2}{S}^2-S^{*}S=0 {\displaystyle (}{S}^{*}(S^{*}S-I)S=0{\displaystyle )}$ [11,12],

(ii)

n-quasi-isometry if $S^{*(n+1)}{S}^{n+1}-S^{*n}{S}^n=0 {\displaystyle (}{S}^{*n}(S^{*}S-I)S^n=0{\displaystyle )}$ [13],

(iii)

n-quasi-m-isometric transformation if $S^{*n}{\mathbf{Q}}_m\left(S\right)S^n=0$ for some integers n and m [14,15]. Similarly, the class of $(m,C)$-isometric transformations has been extended to the so-called n-quasi-$(m,C)$-isometric transformations as $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is called n-quasi-$(m,C)$-isometric transformation if $S^{*n}{\Lambda }_m(S,C)S^n=0$ for some integers n and m (see [16,17]).

Recently, Zuo et al. [18] introduced the concept of n-quasi-m-symmetric transformations as follows: $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is called n-quasi-m-symmetric if

$$S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}{S}^{m-k}\right)S^n=0,$$

(8)

for some positive integers n and m.

The theory of complex symmetric operators played a pivotal role in the development of operators theory. Several applications of this approach deal with Schrodinger operators with spectral gaps and scaled Hamiltonians appearing in the complex scaling theory of resonances [19,20]. In the numerous studies considering the concepts of m-complex symmetric transformations [21,22,23], $\left[m\right]$-complex symmetric transformations, skew m-complex symmetric transformations [24], and skew $\left[m\right]$-complex symmetric transformations [25], it was natural to introduce the concept of n-quasi-m-complex symmetric transformations. This was our goal in this study.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$. We stated that $S$ is an n-quasi-m-complex symmetric transformation if there exists a conjugation C on $\mathcal{Y}$ such that

$$S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S^n=0,$$

(9)

for some positive integers n and m. In another way,

$$S^{*n}{\Phi }_m(S,C)S^n=0.$$

Remark 1.

$S$ is 1-quasi-1-complex symmetric with conjugation C if

$$S^{*}\left(CSC-S^{*}\right)S=0.$$

Remark 2.

$S$ is n-quasi-2-complex symmetric with conjugation C if

$$S^{*n}\left(C{S}^{2}C-2S^{*}CSC+S^{*2}\right)S^n=0.$$

Remark 3.

$S$ is n-quasi-3-complex symmetric with conjugation C if

$$S^{*n}\left(S^{*3}-3S^{*2}CSC+3S^{*}C{S}^{2}C-C{S}^{3}C\right)S^n=0.$$

The paper is organized as follows: In Section 2, we give a matrix characterization of an n-quasi-m-complex symmetric transformation using the decomposition $\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left({S^{*}}^n\right)$. Several properties of this class are obtained by exploiting the special kind of transformation matrix representation associated with it. In the course of our investigation, we find some properties of m-complex symmetric transformations, which are retained by n-quasi-m-complex symmetric transformations. In particular, we show that if $S\in \mathcal{B}\left(\mathcal{Y}\right)$ is an n-quasi-m-complex symmetric with conjugation C, then its power is n-quasi-m-complex symmetric with the same conjugation C. Moreover, if T and S are such that T is $n_1$-quasi-$m_1$-complex symmetric and S is $n_2$-quasi-$m_2$-complex symmetric, then their product $TS$ is $max\{n_1,n_2\}$-quasi-$(m_1+m_2-1)$-complex symmetric under suitable conditions. We also prove the sum of an n-quasi-m-complex symmetric transformation with conjugation C and a commuting nilpotent transformation of degree q is a $(n+q)$-quasi-$(m+2q-2)$-complex symmetric with same conjugation.

2. n-Quasi-m-Complex Symmetric Transformations

We start this section with several remarks and examples, which try to clarify the context of the concept of n-quasi-m-complex symmetric transformations.

Remark 4.

(1) A 1-Quasi-m-complex symmetric transformation is called a quasi-m-complex symmetric transformation.

(2) It is obvious that every m-complex symmetric transformation is an n-quasi-m-complex symmetric transformation and a quasi-m-complex symmetric transformation is an n-quasi-m-complex symmetric transformation.

Remark 5.

(1) Note that if $S\in \mathcal{B}\left(\mathcal{Y}\right)$ and if C is a conjugation, we have

$$\begin{array}{ccc}& & S^{*n}\left(\sum _{0\le k\le m}{(-1)}^k\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S^n\hfill \\ & =& C{\left(C{S}^{*}C\right)}^n\left(\sum _{0\le k\le m}{(-1)}^k\left(\genfrac{}{}{0pt}{}{m}{k}\right){\left(C{S}^{*}C\right)}^{k}C{\left(CSC\right)}^{m-k}C\right){\left(CSC\right)}^{n}C.\hfill \end{array}$$

It is clear that S is an n-quasi-m-complex symmetric transformation with conjugation C if and only if $CSC$ is an n-quasi-m-complex symmetric transformation with conjugation C.

(2) If $CS=SC$, then S is an n-quasi-m-complex symmetric transformation if and only if S is an n-quasi-m-symmetric transformation [18].

Example 1.

In the following, we give an example of a transformation that is n-quasi-m-complex symmetric but not is m-complex symmetric for some positive integers n and m. Let C be a conjugation defined on ${\mathbb{C}}^3$ by $C(u_1,u_2,u_3)=(\overline{u_1},\overline{u_2},\overline{u_3})$. Let S in $\mathcal{B}\left({\mathbb{C}}^3\right)$ be given by $S=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& \sqrt{2}\\ 0& 0& 0\\ & \end{array}\right)$. It is seen that $S^2=\left(\begin{array}{ccc}0& 0& \sqrt{2}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) and{S}^3=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),$ and from this, it easily follows that

$$\sum _{0\le k\le 4}{(-1)}^k\left(\genfrac{}{}{0pt}{}{4}{k}\right)S^{*k}C{S}^{4-k}C=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 12\end{array}\right)$$

and

$$S^{*}\left(\sum _{0\le k\le 4}{(-1)}^k\left(\genfrac{}{}{0pt}{}{4}{k}\right)S^{*k}C{S}^{4-k}C\right)S=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right).$$

Therefore, S is a quasi-4-complex symmetric transformation with conjugation C, while S is not 4-complex symmetric.

Remark 6.

It is proven in Example 1 that there is a transformation that is n-quasi-m-complex symmetric but not m-complex symmetric for some positive integers n and m. Thus, our proposed new class of transformations contains the class of m-complex symmetric transformations as a proper subset.

Remark 7.

It is clear that every quasi-m-complex symmetric transformation is an n-quasi-m-complex symmetric transformation for $n\ge 2$. The converse is not true in general, as shown in the following example.

Example 2.

Let $S=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\\ & \end{array}\right)\in \mathcal{B}\left({\mathbb{C}}^3\right)$ and $C:{\mathbb{C}}^3⟶{\mathbb{C}}^3$ such that $C(x_1,x_2,x_3)=(\overline{x_3},\overline{x_2},\overline{x_1})$. Then, $S^{*}CSCS\ne 0$ and $S^2=0$, so that $S^{*}(CSC-S^{*})S\ne 0$. Therefore, S is not a quasi-complex symmetric transformation, while S is a 2-quasi-complex symmetric transformation.

In the following theorem, we show that under suitable conditions on an n-quasi-m-complex symmetric transformation, $n\ge 2$, it becomes a quasi-m-complex symmetric transformation.

Theorem 1.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$ be an n-quasi-m-complex symmetric transformation with conjugation C for $n\ge 2$. If $ker\left(S^{*}\right)=ker\left(S^{*2}\right)$. Then, S is a quasi-m-complex symmetric transformation with conjugation C.

Proof. 

Under the assumption that $ker\left(S^{*}\right)=ker\left(S^{*2}\right)$, it follows that $ker\left(S^{*}\right)=ker\left(S^{*n}\right)$ for $n=1,2,\cdots .$ From the identity,

$$\begin{array}{c}\hfill S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S^n=0,\end{array}$$

we obtain

$$\begin{array}{c}\hfill S^{*}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S^n=0.\end{array}$$

Thus,

$$\begin{array}{c}\hfill S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right){\left(S^{*k}C{S}^{m-k}C\right)}^{*}\right)S=0.\end{array}$$

Using again the condition $ker\left(S^{*}\right)=ker\left(S^{*n}\right)$, we obtain

$$\begin{array}{c}\hfill S^{*}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right){\left(S^{*k}C{S}^{m-k}C\right)}^{*}\right)S=0.\end{array}$$

Thus, we have

$$\begin{array}{c}\hfill S^{*}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S=0.\end{array}$$

This means that S is a quasi-m-complex symmetric transformation, and the proof is completed. □

Proposition 1.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$, and let $\mathcal{M}$ be a closed subspace of $\mathcal{Y}$, which reduces S. If S is an n-quasi-m-complex symmetric transformation with conjugation $C=C_1\oplus C_2$, where $C_1$ and $C_2$ are conjugations on $\mathcal{M}$ and ${\mathcal{M}}^{\perp }$, respectively, then $S{|}_{\mathcal{M}}$ is an n-quasi-m-complex symmetric transformation with conjugation $C_1$.

Proof. 

Since $\mathcal{M}$ is a reducing subspace of S, we can write

$$S=\left(\begin{array}{ccc}{S}_1& 0& \\ 0& S_2\end{array}\right)\mathrm{on}\mathcal{Y}=\mathcal{M}\oplus {\mathcal{M}}^{\perp }.$$

A simple computation shows that

$$\begin{array}{ccc}& & S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\begin{array}{c}m\\ k\end{array}\right)S^{*m-k}C{S}^{m-k}C\right)S^n\hfill \\ & =& {\left(\begin{array}{cc}{S}_1& 0\\ 0& S_2\end{array}\right)}^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\begin{array}{c}m\\ k\end{array}\right){\left(\begin{array}{cc}{S}_1& 0\\ 0& S_2\end{array}\right)}^{*k}\left(\begin{array}{cc}{C}_1& 0\\ 0& C_2\\ \end{array}\right){\left(\begin{array}{cc}{S}_1& 0\\ 0& S_2\end{array}\right)}^{m-k}\left(\begin{array}{cc}{C}_1& 0\\ 0& C_2\\ \end{array}\right)\right)\hfill \\ & & \times {\left(\begin{array}{cc}{S}_1& 0\\ 0& S_2\end{array}\right)}^n\hfill \\ & =& \left(\begin{array}{cc}{S}_1^{*n}F{S}_1^n& X\\ W& Z\end{array}\right),\hfill \end{array}$$

where

$$F=\sum _{0\le k\le m}{(-1)}^{m-k}\left(\begin{array}{c}m\\ k\end{array}\right)S_1^{*k}{C}_1{S}_1^{m-k}{C}_1.$$

From the condition that S is an n-quasi-m-complex symmetric with conjugation C, we have

$$S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}{S}^{*k}C{S}^{m-k}C\right)S^n=0.$$

This means that

$${\displaystyle S_1^{*n}\left(\sum _{0\le k\le m}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)S_1^{*k}{C}_1{S}_1^{m-k}{C}_1\right)S_1^n=0.}$$

Therefore, $S_1=S{|}_{\mathcal{M}}$ is an n-quasi-m-complex symmetric transformation with conjugation $C_1$. □

In the following theorem, we give a structural theorem for n-quasi-m-complex symmetric transformations.

Theorem 2.

Let $C=C_1\oplus C_2$ be a conjugation on $\mathcal{Y}$, where $C_1$ and $C_2$ are conjugations on $\overline{Im\left(S^n\right)}$ and $ker\left({S^{*}}^n\right)$, respectively. Assuming that $\overline{Im\left(S^n\right)}\ne Y$, the following statements are equivalent:

(1) S is an n-quasi-m-complex symmetric transformation with conjugation C.

(2) $S=\left(\begin{array}{ccc}T& R& \\ 0& V\end{array}\right)$ on $\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left({S^{*}}^n\right)$, where $T=S{|}_{\overline{Im\left(S^n\right)}}$ is an m-complex symmetric transformation on $\overline{Im\left(S^n\right)}$ with conjugations $C_1$, $V^n=0$, and $\sigma \left(S\right)=\sigma \left(T\right)\cup \left\{0\right\}$, where $\sigma \left(S\right)$ is the spectrum of S.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$. Considering the matrix representation of $S=\left(\begin{array}{ccc}T& R& \\ 0& V\end{array}\right)$ on the decomposition $\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left({S^{*}}^n\right).$

Let P be the projection of $\mathcal{Y}$ onto $\overline{Im\left(S^n\right)}.$ With S being an n-quasi-m-complex symmetric transformation, it follows that

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

This implies that

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

Therefore, T is an m-complex symmetric transformation on $\overline{Im\left(S^n\right)}$.

Let $x=u\oplus v\in \overline{Im\left(S^n\right)}\oplus ker\left({S^{*}}^n\right)=\mathcal{Y}$. If $x\in ker\left({S^{*}}^n\right)$, then

$$\begin{array}{ccc}\hfill \langle V^{n}v,v\rangle & =& \langle S^n(I-P)x,(I-P)x\rangle \hfill \\ & =& \langle (I-P)x,{S^{*}}^n(I-P)x\rangle =0.\hfill \end{array}$$

Hence, $V^n=0$. Using similar technics as in [26] (Corollary 7 and Corollary 8), we can show that $\sigma \left(S\right)=\sigma \left(T\right)\cup \sigma \left(V\right)=\sigma \left(T\right)\cup \left\{0\right\}.$

$\left(2\right)\Rightarrow \left(1\right).$ Suppose that

$$S=\left(\begin{array}{cc}T& R\\ 0& V\end{array}\right)\mathrm{on}\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left(S^{*n}\right)$$

where $\overline{Im\left(S^n\right)}$ is the closure of $Im\left(S^n\right)$, T is an m-complex symmetric transformation on $\overline{Im\left(S^n\right)}$, and $V^n=0$. Since

$$S^n=\left(\begin{array}{cc}{T}^n& {\displaystyle \sum _{0\le j\le n-1}{T}^{j}R{V}^{n-1-j}}\\ 0& 0\end{array}\right),$$

we have

$$\begin{array}{ccc}& & S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\begin{array}{c}m\\ k\end{array}\right)S^{*m-k}C{S}^{m-k}C\right)S^n\hfill \\ & =& {\left(\begin{array}{cc}T& R\\ 0& V\end{array}\right)}^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\begin{array}{c}m\\ k\end{array}\right){\left(\begin{array}{cc}T& R\\ 0& V\end{array}\right)}^{*k}\left(\begin{array}{cc}{C}_1& 0\\ 0& C_2\\ \end{array}\right){\left(\begin{array}{cc}T& R\\ 0& V\end{array}\right)}^{m-k}\left(\begin{array}{cc}{C}_1& 0\\ 0& C_2\\ \end{array}\right)\right)\hfill \\ & & \times {\left(\begin{array}{cc}T& R\\ 0& V\end{array}\right)}^n\hfill \\ & =& \left(\begin{array}{cc}{T}^{*n}D{T}^n& {\displaystyle T^{*n}D\sum _{0\le j\le n-1}{T}^{j}R{V}^{n-1-j}}\\ {\displaystyle {\left(\sum _{0\le j\le n-1}{T}^{j}R{V}^{n-1-j}\right)}^{*}D{T}^n}& {\displaystyle {\left(\sum _{0\le j\le n-1}{T}^{j}R{V}^{n-1-j}\right)}^{*}D\sum _{0\le j\le n-1}{T}^{j}R{V}^{n-1-j}}\end{array}\right),\hfill \end{array}$$

where

$$D=\sum _{0\le k\le m}{(-1)}^{m-k}\left(\begin{array}{c}m\\ k\end{array}\right)T^{*k}{C}_1{T}^{m-k}{C}_1.$$

Hence,

$${\displaystyle S^{*n}\left(\sum _{0\le k\le m}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)S^{*k}C{S}^{m-k}C\right)S^n=0\mathrm{o}\mathrm{n}\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left(S^{*n}\right).}$$

Therefore, S is n-quasi-m-complex symmetric. □

Corollary 1.

If S is an n-quasi-m-complex symmetric transformation with conjugation C and $\overline{Im\left(S^n\right)}=\mathcal{Y}$, then S is an m-complex symmetric transformation with conjugation C.

Proof. 

$$\begin{array}{ccc}& & {\displaystyle S^{*n}\left(\sum _{0\le k\le m}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)S^{*k}C{S}^{m-k}C\right)S^n=0}\hfill \\ & \Rightarrow & {\displaystyle \sum _{0\le k\le m}{(-1)}^k\left(\begin{array}{c}m\\ k\end{array}\right)S^{*k}C{S}^{m-k}C=0\mathrm{on}\overline{Im\left(S^n\right)}=\mathcal{Y}.}\hfill \end{array}$$

□

Corollary 2.

Under the same hypothesis as in Theorem 2, if $S=\left(\begin{array}{ccc}T& R& \\ 0& V\end{array}\right)$ on $\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left({S^{*}}^n\right)$ is such that T is invertible, then S is similar to a direct sum of an m-complex symmetric transformation and a nilpotent transformation.

Proof. 

From the fact that T is an invertible transformation, we have $\sigma \left(T\right)\cap \sigma \left(V\right)=\varnothing$. Then, there exists a transformation W such that $TW-WV=R$ by [27]. Therefore,

$$S=\left(\begin{array}{ccc}T& R& \\ 0& V\end{array}\right)={\left(\begin{array}{ccc}I& W& \\ 0& I\end{array}\right)}^{-1}\left(\begin{array}{ccc}T& 0& \\ 0& V\end{array}\right)\left(\begin{array}{ccc}I& W& \\ 0& I\end{array}\right).$$

From Theorem 2, we have that T is m-complex symmetric and V is nilpotent; from this, we deduce the required consequence. □

In [10], the authors showed that a power of an m-complex symmetric transformation is again an m-complex symmetric transformation. In the following theorem, we show that this remains true for an n-quasi-m-complex symmetric transformation.

Theorem 3.

Let $C=C_1\oplus C_2$ be a conjugation on $\mathcal{Y}$, where $C_1$ and $C_2$ are conjugations on $\overline{Im\left(S^n\right)}$ and $ker\left({S^{*}}^n\right)$, respectively. If S is an n-quasi-m-complex symmetric transformation with conjugation C, then $S^p$ is an n-quasi-m-complex symmetric transformation with conjugation C for every natural number p.

Proof. 

If $Im\left(S^n\right)$ is dense, then S is an m-complex symmetric transformation and so is $S^p$ for every positive integer p according to ([10], Theorem 2.1).

If $Im\left(S^n\right)$ is not dense, from Theorem 2, we can use the matrix representation of S on $\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left(S^{*n}\right)$ given by

$$S=\left(\begin{array}{cc}T& R\\ 0& V\end{array}\right),$$

where T is an m-complex symmetric transformation on $\overline{Im\left(S^n\right)}$. From [10] (Theorem 2.1), it is known that $T^p$ is an m-complex symmetric transformation with conjugation $C_1$. By observing

$$S^p=\left(\begin{array}{cc}{T}^p& {\displaystyle \sum _{0\le j\le p-1}{T}^{j}R{V}^{p-1-j}}\\ 0& V^p\end{array}\right)\mathrm{on}\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left(S^{*n}\right),$$

we deduce that $S^p$ is an n-quasi-m-complex symmetric transformation through the application of Theorem 2. □

Proposition 2.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$, and let $C=C_1\oplus C_2$ be a conjugation on $\mathcal{Y}$, where $C_1$ and $C_2$ are conjugations on $\overline{Im\left(S^n\right)}$ and $ker\left({S^{*}}^n\right)$, respectively. If S is an n-quasi-m-complex symmetric transformation with conjugation C, then S is an q-quasi-k-complex symmetric transformation with conjugation C for every positive integer $k\ge m$ and $q\ge n$.

Proof. 

The proof is two-fold: (i) If $\overline{Im\left(S^n\right)}=\mathcal{Y}$, then S is an m-complex symmetric transformation and, by taking into account [23], S is an k-complex symmetric transformation with conjugation C for every positive integer $k\ge m$.

If $\overline{Im\left(S^n\right)}\ne \mathcal{Y}$, from Theorem 2, we can use the matrix representation of S associated to the decomposition $\mathcal{Y}=\overline{Im\left(S^n\right)}\oplus ker\left(S^{*n}\right)$ as $S=\left(\begin{array}{ccc}T& R& \\ 0& V\end{array}\right)$ where $T=S{|}_{\overline{Im\left(S^n\right)}}$ is an m-complex symmetric transformation with conjugation $C_1$ and $V^n=0$. We have T as an k-complex symmetric transformation with conjugation $C_1$ for every integer $k\ge m$. From statement $\left(2\right)$ of Theorem 2, we obtain that S is an n-quasi-k-complex symmetric transformation for $k\ge m$; therefore, S is an q-quasi-k-complex symmetric transformation for $q\ge n.$ □

The following lemma gives another criteria for which an n-quasi-m-complex symmetric transformation with conjugation C becomes an n-quasi-k-complex symmetric transformation with conjugation C for each $k\ge m$.

Lemma 1.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$ be n-quasi-m-complex symmetric with conjugation C on $\mathcal{Y}$. If $[S,CSC]=0$; then, S is an n-quasi-k-complex symmetric transformation with conjugation C for every positive integer $k\ge m$.

Proof. 

As observed in [22], we have

$${\Phi }_{m+1}(S,C)=S^{*}{\Phi }_m(S,C)S-{\Phi }_m(S,C)\left(CSC\right).$$

In view of the assumption that S is an n-quasi-m-complex symmetric transformation with conjugation C satisfying $[S,CSC]=0$, it follows that

$$S^{*n}{\Phi }_{m+1}(S,C)S^n=S^{*n+1}{\Phi }_m(S,C)S^{n+1}-S^{*n}{\Phi }_m\left(S\right)S^n\left(CSC\right)=0.$$

Therefore, S is n-quasi-$(m+1)$-complex symmetric with the same conjugation C. □

The perturbations of an n-quasi-m-complex symmetric transformation by a nilpotent transformation are investigated in the following theorem.

Theorem 4.

Let S and $N\in \mathcal{B}\left(\mathcal{Y}\right)$ and C be a conjugation on $\mathcal{Y}$ satisfying the following conditions:

$$[S,N]=[S,N^{*}]=[S,CNC]=0.$$

If S is an n-quasi-m-complex symmetric transformation, and N is a nilpotent transformation of order q, then $S+N$ is an $(n+q)$-quasi-$(m+2q-2)$-complex symmetric transformation with conjugation C.

Proof. 

We need to prove that

$${\left(S+N\right)}^{\beta }{\Phi }_{m+2q-2}\left(S+N\right){\left(S+N\right)}^{\beta }=0\mathrm{where}\beta =n+q.$$

By taking into account [23] (Lemma 3.7), we obtain

$${\Phi }_{m+2q-2}\left(S+N,C\right)=\sum _{0\le k\le m+2q-2}\left(\genfrac{}{}{0pt}{}{m+2q-2}{k}\right){\Phi }_k(S,C){\Phi }_{m+2q-2-k}(N,C)$$

and, moreover,

$$\begin{array}{ccc}& & {\left(S+N\right)}^{*\beta }{\Phi }_{m+2q-2}\left(S+N,C\right){\left(S+N\right)}^{\beta }\hfill \\ & =& \left(\sum _{0\le r\le \beta }\left(\genfrac{}{}{0pt}{}{\beta }{r}\right)S^{*(\beta -r)}{N}^{*r}\right)\left(\sum _{0\le k\le m+2q-2}\left(\genfrac{}{}{0pt}{}{m+2q-2}{,k}\right){\Phi }_k(S,C){\Phi }_{m+2q-2-k}(N,C)\right)\hfill \\ & & \times \left(\sum _{0\le s\le \beta }\left(\genfrac{}{}{0pt}{}{\beta }{s}\right)S^{\beta -s}{N}^s\right).\hfill \end{array}$$

Since $N^j=0$ for $j\ge q$, we have

$$\begin{array}{ccc}& & {\left(S+N\right)}^{*\beta }{\Phi }_{m+2q-2}\left(S+N,C\right){\left(S+N\right)}^{\beta }\hfill \\ & =& \left(\sum _{0\le r\le q-1}\left(\genfrac{}{}{0pt}{}{\beta }{r}\right)S^{*(\beta -r)}{N}^{*r}\right)\left(\sum _{0\le k\le m+2q-2}\left(\genfrac{}{}{0pt}{}{m+2q-2}{k}\right){\Phi }_k(S,C){\Phi }_{m+2q-2-k}(N,C)\right)\hfill \\ & & \times \left(\sum _{0\le s\le q-1}\left(\genfrac{}{}{0pt}{}{\beta }{s}\right)S^{\beta -s}{N}^s\right).\hfill \end{array}$$

For $0\le r\le q-1$ and $0\le s\le q-1$, we obtain $n+1\le \beta -r\le \beta$ and $n+1\le \beta -s\le \beta$. By using Proposition 2, $N^{*}S=S{N}^{*}$, and $S\left(CNC\right)=\left(CNC\right)S$, we obtain that

$$S^{*\beta -r}{\Phi }_k(S,C)S^{\beta -s}=0fork\ge m,r,s=0,\cdots ,q-1.$$

However, if $0\le k\le m-1$, then $2q-1\le m+2q-2-k$; hence, ${\Phi }_{m+2q-2-k}(N,C)=0$. Combining the above cases, we can deduce that

$${\left(S+N\right)}^{n+q}{\Lambda }_{m+2q-2}\left(S+N\right){\left(S+N\right)}^{n+q}=0.$$

So, $S+N$ is an $(n+q)$-quasi-$(m+2q-2)$-complex symmetric transformation with conjugation C. Therefore, the theorem is proven. □

Proposition 3.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$ and $T\in \mathcal{B}\left(\mathcal{Y}\right)$, and let $V\in \mathcal{B}\left(\mathcal{Y}\right)$ be a unitary transformation such that $VT=SV$ if C is a conjugation transformation on $\mathcal{Y}.$ Then, we have

$${\Phi }_k\left(T,C\right)=V^{*}{\Phi }_k\left(S,VC{V}^{*}\right)V.$$

(10)

In particular, $T$ is an n-quasi-m-complex symmetric transformation with conjugation C if and only if $S$ is n-quasi-m-complex symmetric with conjugation $VC{V}^{*}$.

Proof. 

Note that if C is a conjugation operator, then $VC{V}^{*}$ has the same property. Moreover, for all $k\in \{0,1,\cdots ,m\}$, we have

$$T^{*k}C{T}^{m-k}C=V^{*}{S}^{*k}VC{V}^{*}{S}^{m-k}VC=V^{*}{S}^{*k}\left(VC{V}^{*}\right)S^{m-k}\left(VC{V}^{*}\right)V.$$

This means that

$$\begin{array}{ccc}\hfill {\Phi }_m\left(T,C\right)& =& \sum _{0\le k\le m}{(-1)}^k\left(\genfrac{}{}{0pt}{}{m}{k}\right)T^{*k}C{T}^{m-k}C\hfill \\ & =& V^{*}\left(\sum _{0\le k\le m}{(-1)}^k\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}\left(VC{V}^{*}\right)S^{m-k}\left(VC{V}^{*}\right)\right)V\hfill \\ & =& V^{*}{\Phi }_m\left(S,VC{V}^{*}\right)V.\hfill \end{array}$$

Using (10), we can assert that

$$\begin{array}{ccc}\hfill T^{*n}{\Phi }_m\left(T,C\right)T^n& =& V^{*}{S}^{*n}V{V}^{*}{\Phi }_m\left(S,VC{V}^{*}\right)V.V^{*}{S}^{n}V\hfill \\ & =& V^{*}{S}^{*n}{\Phi }_m\left(S,VC{V}^{*}\right)S^{n}V.\hfill \end{array}$$

Consequently, $T$ is an n-quasi-m-complex symmetric transformation with conjugation C if and only if $S$ is n-quasi-m-complex symmetric with conjugation $VC{V}^{*}$. □

Theorem 5.

Let S and T be $\mathcal{B}\left(\mathcal{Y}\right)$ and let $C=C_1\oplus C_2$ be a conjugation on $\mathcal{Y}$, where $C_1$ and $C_2$ are conjugations on $\overline{Im\left(S^n\right)}$ and $ker\left({S^{*}}^n\right)$, respectively. Assume that $[T,S]=[T^{*},S]=0$ and $[T,\left(CTC\right)]=[T,\left(CSC\right)]=[S^{*},CTC]=0.$ If T is an $n_1$-quasi-k-complex symmetric transformation, and S is an $n_2$-quasi-m-complex symmetric transformation with conjugation C, then $TS$ is an $n^{\prime }=max\{n_1,n_2\}$-quasi-$(k+m-1)$-complex symmetric transformation.

Proof. 

Firstly, since $[T,S]=[S^{*},CTC]=0$, we have, by taking into account [22] (Lemma 4.7),

$${\Phi }_{m+k-1}(TS,C)=\sum _{0\le j\le k+m-1}{\Phi }_{k+m-1-j}(T,C)S^{*m+k-1-j}C{T}^{j}C{\Phi }_j(S,C)$$

Furthermore, as $[T,S^{*}]=[T,CSC]=[CTC,S^{*}]=[T,CTC]=0$, we have

$$\begin{array}{ccc}& & {\left(TS\right)}^{*n^{\prime }}{\Phi }_{k+m-1}\left(TS\right){\left(TS\right)}^{n^{\prime }}\hfill \\ & =& \sum _{0\le j\le k+m-1}\left(\genfrac{}{}{0pt}{}{k+m-1}{j}\right)T^{*n^{\prime }}{\Phi }_{k+m-1-j}(T,C)T^{n^{\prime }}{S}^{*m+k-1}C{T}^{j}C{S}^{*n^{\prime }}{\Phi }_j(S,C)S^{n^{\prime }}.\hfill \end{array}$$

Since S is an $n_2$-quasi-m-complex symmetric transformation, we obtain from Proposition 2 that $S^{*n^{\prime }}{\Phi }_j(S,C)S^{n^{\prime }}=0$ for $j\ge m$ and $n^{\prime }\ge n_2$. On the other hand, if $j\le m-1$, then $k+m-1-j\ge k+m-1-m+1=k$ and so $T^{{*}^{n^{\prime }}}{\Phi }_{k+m-1-j}(T,C)T^{n^{\prime }}=0$ from Lemma 1. Hence, $TS$ is an $n^{\prime }$-quasi-$(k+m-1)$-complex symmetric transformation, as required. □

Corollary 3.

Let S and $T\in \mathcal{B}\left(\mathcal{Y}\right)$. Let $C=C_1\oplus C_2$ be a conjugation on $\mathcal{H}$, where $C_1$ and $C_2$ are conjugations on $\overline{Im\left(S^n\right)}$ and $ker\left({S^{*}}^n\right)$, respectively. Assume that $[T,S]=[T,S^{*}]=[T,CSC]=[T,CTC]=[S^{*},CTC]=0$. If T is an $n_1$-quasi-k-complex symmetric with conjugation C, and S is an $n_2$-quasi-m-complex symmetric with conjugation C, then $T{S}^q$ is an $n^{\prime }=max\{n_1,n_2\}$-quasi-$(k+m-1)$-complex symmetric transformation with conjugation C for some positive integer q.

Proof. 

From Theorem 3, we have that $S^q$ is an $n_2$-quasi-m-complex symmetric transformation with conjugation C. However, $[T,C{S}^{q}C]=[T,CTC]=[S^{*q},CTC]=0$. Clearly, T and $S^q$ satisfy the conditions of Theorem 5. Hence, $T{S}^q$ is an $n^{\prime }$-quasi-$(k+m-1)$-complex symmetric transformation with conjugation C. □

Lemma 2.

Let $S\in \mathcal{B}\left(\mathcal{Y}\right)$ and let C and D be conjugations on $\mathcal{Y}$. Then, the following statements are true:

(1) S is an n-quasi-m-complex symmetric transformation with conjugation C if and only if the tensor product $S\otimes I$ is an n-quasi-m-complex symmetric transformation with conjugation $C\otimes D$.

(2) S is an n-quasi-m-complex symmetric transformation with conjugation C if and only if the tensor product $I\otimes S$ is an n-quasi-m-complex symmetric transformation with conjugation $D\otimes C.$

Proof. 

It was observed in ([28] (Lemma 4.5) that if C and D are conjugations on $\mathcal{Y}$, then $C\otimes D$ is a conjugation on $\mathcal{Y}\overline{\otimes }\mathcal{Y}$. A direct computation shows that

$$\begin{array}{ccc}& & {\left(S\otimes I\right)}^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right){\left(S\otimes I\right)}^{*k}\left(C\otimes D\right){\left(S\otimes I\right)}^{m-k}\left(C\otimes D\right)\right){\left(S\otimes I\right)}^n\hfill \\ & =& S^{*n}\left(\sum _{0\le k\le m}{(-1)}^{m-k}\left(\genfrac{}{}{0pt}{}{m}{k}\right)S^{*k}C{S}^{m-k}C\right)S^n\otimes I\hfill \\ & =& S^{*n}{\Phi }_m(S,C)S^n\otimes I.\hfill \end{array}$$

From this, it follows S is an n-quasi-m-complex symmetric transformation if and only if $S\otimes I$ is an n-quasi-m-complex symmetric transformation with conjugation $C\otimes D$. □

Theorem 6.

Let $T,S\in \mathcal{B}\left(\mathcal{Y}\right)$ and $C,D$ are conjugations on $\mathcal{Y}$ such that $C\otimes D=A\oplus B$, where A and B are conjugations on $\overline{Im(I\otimes S^n)}$ and $ker\left({I\otimes S^{*}}^n\right)$, respectively. If T is an $n_1$-quasi-$m_1$-complex symmetric transformation with conjugation C, and S is an $n_2$-quasi-$m_2$-complex symmetric transformation with conjugation D such that $[T,CTC]=0$, then $T\otimes S$ is an $n^{\prime }=max\{n_1,n_2\}$-quasi-$(m_1+m_2-1)$-complex symmetric transformation with conjugation $C\otimes D$.

Proof. 

$T\otimes S=\left(T\otimes I\right)\left(I\otimes S\right)=\left(I\otimes S\right)\left(T\otimes I\right)$ and, by taking into account, Lemma 2, we obtain that $T\otimes I$ is an $n_1$-quasi-$m_1$-complex symmetric transformation with conjugation $C\otimes D)$ and, similarly, $I\otimes S$ is an $n_2$-quasi-$m_1$-complex symmetric transformation with conjugation $C\otimes D$. On the other hand, note that $T\otimes I$ and $I\otimes S$ satisfy all conditions required in Theorem 5 as

$$[T\otimes I,I\otimes S]=[{(T\otimes I)}^{*},I\otimes S]=0,$$

$$[T\otimes I,(C\otimes D\left)\right(T\otimes I\left)\right(C\otimes D\left)\right]=[T\otimes I,(C\otimes D\left)\right(I\otimes S\left)\right(C\otimes D\left)\right]=0$$

and

$$[{(I\otimes S)}^{*},(C\otimes D)(T\otimes I)(C\otimes D)]=0.$$

Consequently, $(T\otimes I)(I\otimes S)$ is an $n^{\prime }$-quasi-$(m_1+m_2-1)$-complex symmetric transformation with conjugation $C\otimes D$. □

Proposition 4.

Let S and $N\in \mathcal{B}\left(\mathcal{Y}\right)$ and C be a conjugation on $\mathcal{Y}$. If S is an n-quasi-m-complex symmetric transformation, and N is a nilpotent transformation of order q, then $S\otimes I+I\otimes N$ and $I\otimes S+N\otimes I$ are $(n+q)$-quasi-$(m+2q-2)$-complex symmetric transformations with conjugation $C\otimes C$.

Proof. 

By applying Lemma 2, we can see that $S\otimes I$ is n-quasi-m-complex symmetric with conjugation $C\otimes C$. Moreover, since N is nilpotent of order q, it follows that $I\otimes N$ is nilpotent of the same order. Using the properties of the tensor product, it can be checked that

$$[S\otimes I,I\otimes N]=[S\otimes I,I\otimes N^{*}]=[S\otimes I,(C\otimes C)(I\otimes N)(C\otimes C)]=0.$$

Accordingly, the conditions of Theorem 4 are fulfilled by $S\otimes I$ and $I\otimes N$. Therefore, $S\otimes I+I\otimes N$ is an $(n+q)$-quasi-$(m+2q-2)$-complex symmetric transformation with conjugation $C\otimes C$. □

3. Conclusions

The paper was devoted to some class of operators on the Hilbert space, which is a generalization of m-complex symmetric operators. More precisely, we introduced a new class of operators, which is called the class of n-quasi-m-complex symmetric operators. It was proved with an example that there is an operator that is n-quasi-m-complex symmetric but not m-complex symmetric. Thus, the proposed new class of operators contains the class of m-complex symmetric operators as a proper subset. Some basic structural properties of this class of operators were established with the help of a special kind of operator matrix representation associated with such class of operators.