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On n-quasi-\([m,C]\)-isometric operators

Abstract

For positive integers m and n, an operator \(T \in B ( H )\) is said to be an n-quasi-\([m,C]\)-isometric operator if there exists some conjugation C such that Tn(j=0m(1)j(mj)CTmjC.Tmj)Tn=0. In this paper, some basic structural properties of n-quasi-\([m,C]\)-isometric operators are established with the help of operator matrix representation. As an application, we obtain that a power of an n-quasi-\([m,C]\)-isometric operator is again an n-quasi-\([m,C]\)-isometric operator. Moreover, we show that the class of n-quasi-\([m,C]\)-isometric operators is norm closed. Finally, we examine the stability of n-quasi-\([m,C]\)-isometric operator under perturbation by nilpotent operators commuting with T.

Introduction

Let \(\mathbb{N}\) and \(\mathbb{C}\) be the sets of natural numbers and complex numbers, respectively, and let \(B(H)\) denote the algebra of all bounded linear operators on a separable complex Hilbert space H. If \(T\in B(H)\), we shall write \(N(T)\), \(R(T)\), and \(\sigma (T)\) for the null space, range space, and the spectrum of T, respectively. The closure of a set M will be denoted by .

An antilinear operator C on H is said to be conjugation if C satisfies \(C^{2} = I\) and \((Cx,Cy)=(y,x)\) for all \(x,y\in H\). In 1990s, Agler and Stankus [1] studied the theory of m-isometric operators which are connected to Toeplitz operators, ordinary differential equations, classical function theory, classical conjugate point theory, distributions, Fejer–Riesz factorization, stochastic processes, and other topics. For fixed \(m \in \mathbb{N}\), an operator \(T\in B(H)\) is said to be an m-isometric operator if it satisfies the identity

$$ \sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}T^{*m-j}T^{m-j}=0, $$

where (mj) is the binomial coefficient. Several authors have studied the m-isometric operator. We refer the reader to [26, 9, 11] for further details.

In [7], Chō, Ko, and Lee introduced (m, C)-isometric operators with conjugation C as follows: For an operator \(T\in B(H)\) and an integer \(m\geq 1\), T is said to be an (m, C)-isometric operator if there exists some conjugation C such that

$$ \sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}T^{*m-j}.CT^{m-j}C=0. $$

In [8], Chō, Lee, and Motoyoshi introduced [m, C]-isometric operators with conjugation C as follows: For an operator \(T\in B(H)\) and an integer \(m\geq 1\), T is said to be an [m, C]-isometric operator if there exists some conjugation C such that

$$ \sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j}=0. $$

For an operator \(T\in B(H)\) and a conjugation C, define the operator \(\lambda _{m}(T)\) by

$$ \lambda _{m}(T)=\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j}. $$

Then T is an [m, C]-isometric operator if and only if

$$ \lambda _{m}(T)=0. $$

Moreover,

$$ CTC.\lambda _{m}(T).T-\lambda _{m}(T)= \lambda _{m+1}(T ) $$

holds. Hence, an [m, C]-isometric operator is an [n, C]-isometric operator for every \(n\geq m\).

In [13], Mahmoud Sid Ahmed, Chō, and Lee introduced n-quasi-\((m,C)\)-isometric operators, which generalize \((m,C)\)-isometric operators. For positive integers m and n, an operator \(T\in B(H)\) is said to be an n-quasi-\((m,C)\)-isometric operator if there exists some conjugation C such that

$$ T^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}T^{*m-j}.CT^{m-j}C \Biggr)T^{n}=0. $$

In [10], Duggal studied n-quasi-\([m,C]\)-isometric operators and gave some properties of them. For positive integers m and n, an operator \(T \in B ( H )\) is said to be an n-quasi-\([m,C]\)-isometric operator if there exists some conjugation C such that

$$ T^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T^{n}=0. $$

It is clear that every \([m,C]\)-isometric operator is an n-quasi-\([m,C]\)-isometric operator.

The following example provides an operator which is an n-quasi-\([2,C]\)-isometric operator, but not a \([2,C]\)-isometric operator.

Example 1.1

Let \(H=\mathbb{C}^{2}\) and let C be a conjugation on H given by \(C(x, y) = (\overline{y}, \overline{x})\). If T=(0100) on H, then CTC=(0010). Hence

$$\begin{aligned} &(CTC)^{2}.T^{2}-2CTC.T+I \\ &\quad = \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix}^{2} \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}^{2}-2 \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix} \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}+ \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} \\ &\quad = \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}, \end{aligned}$$

i.e., T is not a \([2,C]\)-isometric operator.

On the other hand, since

$$\begin{aligned} &T^{*2}\bigl((CTC)^{2}.T^{2}-2CTC.T+I \bigr)T^{2} \\ &\quad = \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix}^{2} \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix} \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}^{2} \\ &\quad =0. \end{aligned}$$

Hence T is a 2-quasi-\([2,C]\)-isometric operator.

Remark 1.1

Let \(T \in B ( H )\) and let C be a conjugation on H.

Note that

$$\begin{aligned} &T^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T^{n} \\ &\quad =C\bigl(CT^{*n}C\bigr) \Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}T^{m-j}.CT^{m-j}C \Biggr)CT ^{n}C)C. \end{aligned}$$

It is clear that T is an n-quasi-\([m,C]\)-isometric operator if and only if \(CTC\) is an n-quasi-\([m,C]\)-isometric operator.

Remark 1.2

It is clear that every quasi-\([m,C]\)-isometric operator is an n-quasi-\([m,C]\)-isometric operator for \(n \geq 2\). The converse is not true in general as shown in the following example.

Example 1.2

Let T=(000000100)B(C3), and let \(C: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}\) satisfy \(C(x_{1}, x_{2}, x_{3}) = (-\overline{x_{3}}, \overline{x_{2}}, -\overline{x_{1}})\). We have CTC=(001000000) and

$$\begin{aligned} T^{*}(CTC.T-I)T &= \begin{pmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0 \end{pmatrix} \begin{pmatrix} 0&\quad 0&\quad 0 \\ 0&-1&\quad 0 \\ 0&\quad 0&-1 \end{pmatrix} \begin{pmatrix} 0&0&0 \\ 0&0&0 \\ 1&0&0 \end{pmatrix} \\ &=- \begin{pmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{pmatrix}. \end{aligned}$$

Hence T is not a quasi-\([1,C]\)-isometric operator.

On the other hand, since

$$\begin{aligned} &T^{*2}(CTCT-I)T^{2} \\ &\quad = \begin{pmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0 \end{pmatrix}^{2} \begin{pmatrix} 0&\quad 0&\quad 0 \\ 0&-1&\quad 0 \\ 0&\quad 0&-1 \end{pmatrix} \begin{pmatrix} 0&0&0 \\ 0&0&0 \\ 1&0&0 \end{pmatrix}^{2} \\ &\quad =0, \end{aligned}$$

it follows that T is a 2-quasi-\([1,C]\)-isometric operator.

n-quasi-\([m,C]\)-isometric operators

In this section we give some basic properties of n-quasi-\([m,C]\)-isometric operators. We begin with the following theorem, which is a structural theorem for n-quasi-\([m,C]\)-isometric operators.

Theorem 2.1

Let \(C = C _{1}\oplus C _{2}\)be a conjugation onHwhere \(C_{1}\)and \(C_{2}\)are conjugations on \(\overline{R(T^{n})}\)and \(N(T^{*n})\), respectively. If \(T^{n}\neq 0\)does not have a dense range, then the following statements are equivalent:

  1. (1)

    Tis ann-quasi-\([m,C]\)-isometric operator;

  2. (2)

    T=(T1T20T3)on \(H=\overline{R(T^{n})}\oplus N(T^{*n})\), where \(T_{1}\)is an \([m,C_{1}]\)-isometric operator and \(T_{3}^{n}=0\). Furthermore, \(\sigma (T)=\sigma (T_{1})\cup \{0\}\).

Proof

\((1)\Rightarrow (2)\) Consider the matrix representation of T with respect to the decomposition \(H=\overline{R(T^{n})}\oplus N(T^{n*})\):

$$ T= \begin{pmatrix} T_{1} &T_{2} \\ 0&T_{3} \end{pmatrix}. $$

Let P be the projection onto \(\overline{R(T^{n})}\). Since T is an n-quasi-\([m,C]\)-isometric operator, it follows that

$$ P\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)P= 0. $$

This means that

$$ \sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}C_{1}T_{1}^{m-j}C_{1}.T_{1}^{m-j}= 0. $$

Hence \(T_{1}\) is an \([m,C_{1}]\)-isometric operator on \(\overline{R(T ^{n})}\). On the other hand, for any \(x=(x_{1}, x_{2})\in \overline{R(T ^{n})}\oplus N(T^{*n})=H\), we have

$$ \bigl(T_{3}^{n}x_{2},x_{2} \bigr)=\bigl(T^{n}(I-P)x, (I-P)x\bigr)=\bigl((I-P)x, T^{*n}(I-P)x\bigr)=0, $$

which implies \(T_{3}^{n}=0\). Since \(\sigma (T)\cup M=\sigma (T_{1}) \cup \sigma (T_{3})\), where M is the union of the holes in \(\sigma (T)\), which happens to be a subset of \(\sigma (T_{1})\cap \sigma (T_{3})\) by Corollary 7 of [12], and \(\sigma (T _{1})\cap \sigma (T_{3})\) has no interior point since \(T_{3}\) is nilpotent, we have \(\sigma (T)=\sigma (T_{1})\cup \{0\}\).

\((2)\Rightarrow (1)\) Suppose that T=(T1T20T3) on \(H=\overline{R(T^{n})}\oplus N(T^{*n})\), where j=0m(1)j(mj)C1×T1mjC1.T1mj=0 and \(T_{3}^{n}=0\). Since

$$ T^{n}= \begin{pmatrix} T_{1}^{n} &\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j} \\ 0& 0 \end{pmatrix}, $$

we have

$$\begin{aligned} &T^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T^{n} \\ &\quad = \begin{pmatrix} T_{1} & T_{2} \\ 0 & T_{3} \end{pmatrix} ^{*n} \\ &\qquad {}\times \left(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix} \begin{pmatrix} C_{1} &0 \\ 0 & C_{2} \end{pmatrix} \begin{pmatrix} T_{1} & T_{2} \\ 0 & T_{3} \end{pmatrix}^{m-j} \begin{pmatrix} C_{1} &0 \\ 0 & C_{2} \end{pmatrix} \begin{pmatrix} T_{1} & T_{2} \\ 0 & T_{3} \end{pmatrix}^{m-j} \right) \\ &\qquad {}\times \begin{pmatrix} T_{1} & T_{2} \\ 0 & T_{3} \end{pmatrix}^{n} \\ &\quad = \begin{pmatrix} T_{1}^{*n}FT_{1}^{n}& T_{1}^{*n}F\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j} \\ (\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j})^{*}FT_{1}^{n}& (\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j})^{*}F\sum_{j=0}^{n-1}T_{1}^{j}T_{2}T_{3}^{n-1-j} \end{pmatrix}, \end{aligned}$$

where F=j=0m(1)j(mj)C1T1mjC1.T1mj. Hence

$$ T^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T^{n}=0, $$

i.e., T is an n-quasi-\([m,C]\)-isometric operator. □

As a consequence, we obtain the following corollaries.

Corollary 2.1

Let \(T\in B(H)\)and letCbe a conjugation onH. IfTis ann-quasi-\([m,C]\)-isometric operator and \(T^{n}\)has a dense range, thenTis an \([m,C]\)-isometric operator.

Corollary 2.2

Let \(T\in B(H)\)and letCbe a conjugation onH. IfTis an invertiblen-quasi-\([m,C]\)-isometric operator, then \(T^{-1}\)is ann-quasi-\([m,C]\)-isometric operator.

Proof

Suppose that T is an invertible n-quasi-\([m,C]\)-isometric operator. Then T is an \([m,C]\)-isometric operator, and so is \(T^{-1}\) by [8]. Hence \(T^{-1}\) is an n-quasi-\([m,C]\)-isometric operator. □

Corollary 2.3

Let T=(T1T20T3)on \(H=\overline{R(T^{n})}\oplus N(T^{*n})\), and let \(C = C _{1}\oplus C _{2}\)be a conjugation onHwhere \(C_{1}\)and \(C_{2}\)are conjugations on \(\overline{R(T^{n})}\)and \(N(T^{*n})\), respectively. IfTis ann-quasi-\([m,C]\)-isometric operator and \(T_{1}\)is invertible, thenTis similar to a direct sum of an \([m,C_{1}]\)-isometric operator and a nilpotent operator.

Proof

Since \(T_{1}\) is invertible, we have \(\sigma (T_{1})\cap \sigma (T _{3})=\phi \). Then there exists an operator S such that \(T_{1}S-ST _{3}=T_{2}\) [14]. Since

$$ \begin{pmatrix} I &S \\ 0&I \end{pmatrix}^{-1}= \begin{pmatrix} I &-S \\ 0& I \end{pmatrix}, $$

it follows that

$$ T= \begin{pmatrix} T_{1} &T_{2} \\ 0&T_{3} \end{pmatrix}= \begin{pmatrix} I &S \\ 0&I \end{pmatrix}^{-1} \begin{pmatrix} T_{1} &0 \\ 0&T_{3} \end{pmatrix} \begin{pmatrix} I &S \\ 0&I \end{pmatrix}. $$

 □

Corollary 2.4

Let \(T\in B(H)\)and let \(C = C _{1}\oplus C _{2}\)be a conjugation onHwhere \(C_{1}\)and \(C_{2}\)are conjugations on \(\overline{R(T^{n})}\)and \(N(T^{*n})\), respectively. IfTis ann-quasi-\([m,C]\)-isometric operator, then \(T^{k}\)is also ann-quasi-\([m,C]\)-isometric operator for any \(k\in \mathbb{N}\).

Proof

If \(T^{n}\) has a dense range, then T is an \([m,C]\)-isometric operator, and so is \(T^{k}\) for any \(k\in \mathbb{N}\) by [8, Theorem 3.4]. If \(T^{n}\) does not have a dense range, we decompose T as T=(T1T20T3) on \(H=\overline{R(T^{n})}\oplus N(T^{*n})\), where \(T_{1}\) is an \([m,C_{1}]\)-isometric operator, and so is \(T_{1}^{k}\). Since

$$ T^{k}= \begin{pmatrix} T_{1}^{k} &\sum_{j=0}^{k-1}T_{1}^{j}T_{2}T_{3}^{k-1-j} \\ 0&T_{3}^{k} \end{pmatrix}\quad \text{on } H=\overline{R \bigl(T^{k}\bigr)}\oplus N\bigl(T^{*k}\bigr), $$

it follows from Theorem 2.1 that \(T^{k}\) is an n-quasi-\([m,C]\)-isometric operator for any \(k\in \mathbb{N}\). □

Remark 2.1

The converse of Corollary 2.4 is not true in general as shown in the following example.

Example 2.1

Let T=(011000000)B(C3), and let \(C: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}\) satisfy \(C(x_{1}, x_{2}, x_{3}) = (-\overline{x_{3}}, \overline{x_{2}}, -\overline{x_{1}})\). A simple calculation shows that \(T^{*2}(CT^{2}C.T^{2}-I)T^{2}=0\) and \(T^{*}(CTC.T-I)T\neq 0\). So, we obtain that \(T^{2}\) is a quasi-\([1,C]\)-isometric operator, but T is not a quasi-\([1,C]\)-isometric operator.

Theorem 2.2

Let \(T\in B(H)\)and let \(C = C _{1}\oplus C _{2}\)be a conjugation onHwhere \(C_{1}\)and \(C_{2}\)are conjugations on \(\overline{R(T^{n})}\)and \(N(T^{*n})\), respectively. IfTis ann-quasi-\([m,C]\)-isometric operator, thenTis ann-quasi-\([k,C]\)-isometric operator for every positive integer \(k\geq m\).

Proof

If \(T^{n}\) has a dense range, then T is an \([m,C]\)-isometric operator, and hence T is a \([k,C]\)-isometric operator for every positive integer \(k\geq m\). If \(T^{n}\) does not have a dense range, we decompose T as T=(T1T20T3) on \(H=\overline{R(T^{n})}\oplus N(T^{*n})\), where \(T_{1}\) is an \([m,C_{1}]\)-isometric operator and \(T_{3}^{n}=0\). Hence \(T_{1}\) is a \([k,C_{1}]\)-isometric operator for every positive integer \(k\geq m\). It follows from Theorem 2.1 that T is an n-quasi-\([k,C]\)-isometric operator. □

Theorem 2.3

Let \(T\in B(H)\)and letCbe a conjugation onH. If \(\{T_{k}\}\)is a sequence ofn-quasi-\([m,C]\)-isometric operators such that \(\lim_{k\rightarrow \infty }\|T_{k}-T\|=0\), thenTis ann-quasi-\([m,C]\)-isometric operator.

Proof

Suppose that \(\{T_{k}\}\) is a sequence of n-quasi-\([m,C]\)-isometric operators such that \(\lim_{n\rightarrow \infty }\|T_{k}-T\|=0\). Then

$$\begin{aligned} &\Biggl\| T_{k}^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT_{k}^{m-j}C.T_{k} ^{m-j}\Biggr)T_{k}^{n} -T^{*n} \Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T ^{m-j} \Biggr)T^{n}\Biggr\| \\ &\quad \leq \Biggl\| T_{k}^{*n}\Biggl(\sum _{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT_{k}^{m-j}C.T _{k}^{m-j}\Biggr)T_{k}^{n}\\ &\qquad {}-T_{k}^{*n} \Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT ^{m-j}C.T^{m-j} \Biggr)T^{n} \Biggr\| \\ &\qquad {}+ \Biggl\| T_{k}^{*n}\Biggl(\sum _{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T ^{n}\\ &\qquad {}-T^{*n}\Biggl(\sum _{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T^{n} \Biggr\| \\ & \quad \leq \bigl\| T_{k}^{*n} \bigr\| \sum_{j=0}^{m} \begin{pmatrix}m\\j\end{pmatrix} \bigr\| T_{k}^{m-j}C.T_{k}^{m-j+n}-T ^{m-j}C.T^{m-j+n} \bigr\| \\ &\qquad {}+ \bigl\| T_{k}^{n}-T^{n} \bigr\| \sum_{j=0}^{m} \begin{pmatrix}m\\j\end{pmatrix} \bigl\| T^{m-j}C.T^{m-j+n} \bigr\| \rightarrow 0. \end{aligned}$$

Since \(\{T_{k}\}\) is an n-quasi-\([m,C]\)-isometric operator,

$$ T_{k}^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT_{k}^{m-j}C.T_{k}^{m-j} \Biggr)T _{k}^{n}=0, $$

we have

$$ T^{*n}\Biggl(\sum_{j=0}^{m}(-1)^{j} \begin{pmatrix}m\\j\end{pmatrix}CT^{m-j}C.T^{m-j} \Biggr)T^{n}=0, $$

i.e., T is an n-quasi-\([m,C]\)-isometric operator. □

Theorem 2.4

Let \(T, N\in B(H)\)and letCbe a conjugation onH. Assume that \(T^{*}CNC=CNCT^{*}\)and \(T^{*}CTC=CTCT^{*}\). IfTis ann-quasi-\([m,C]\)-isometric operator andNis a nilpotent operator of orderpsuch that \(TN=NT\), then \(T+N\)is an \(n+p\)-quasi-\([m+2p-2,C]\)-isometric operator.

Proof

Since

$$ \lambda _{m}(T+N)=\sum_{i+j+k=m} \begin{pmatrix} m\\ i,j,k \end{pmatrix} C(T+N)^{i}C.CN^{j}C. \lambda _{k}(T).T^{j}.N^{i}, $$

where (mi,j,k)=m!i!.j!,k! and \(\lambda _{0}(*)=I\) by [8].

We have

$$\begin{aligned} &(T+N)^{*n+p}\lambda _{m+2p-2}(T+N) (T+N)^{n+p} \\ &\quad =\Biggl(\sum_{s=0}^{n+p} \begin{pmatrix}n+p\\ s \end{pmatrix} T^{*n+p-s}N^{*s} \Biggr) \\ &\qquad {}\times \biggl(\sum_{i+j+k=m+2p-2}\begin{pmatrix}m+2p-2\\ i,j,k \end{pmatrix} C(T+N)^{i}C.CN^{j}C.\lambda _{k}(T).T^{j}.N^{i}\biggr) \\ &\qquad {}\times \Biggl(\sum_{t=0}^{n+p} \begin{pmatrix}n+p \\t \end{pmatrix} T^{n+p-t}N^{t} \Biggr). \end{aligned}$$
  1. (i)

    If \(\operatorname{max} \{i,j\}\geq p\), then \(CN^{j}C=0\) or \(N^{i}=0\).

  2. (ii)

    If \(\operatorname{max} \{i,j\}\leq p-1\), then \(k\geq m\). Since T is an n-quasi-\([m,C]\)-isometric operator, \(T^{*}CNC=CNCT^{*}\) and \(T^{*}CTC=CTCT^{*}\), we obtain

    $$ T^{*n+p-s}\lambda _{k}(T)T^{n+p-t}=0 \quad \text{for } 0\leq s,t\leq p $$

    and

    $$ N^{*s}=0 \quad \text{or}\quad N^{t}=0 \quad \text{for } p+1 \leq s\leq n+p \text{ or } p+1\leq t\leq n+p. $$

    By (i) and (ii), \((T+N)^{*n+p}\lambda _{m+2p-2}(T+N)(T+N)^{n+p}=0\). Therefore \(T+N\) is an \(n+p\)-quasi-\([m+2p-2,C]\)-isometric operator.

 □

Example 2.2

Let C be a conjugation given by \(C(z_{1}, z_{2}, z_{3}) = (\overline{z _{3}}, \overline{z_{2}}, \overline{z_{1}})\) on \(\mathbb{C}^{3}\).

If T=(1mm010001) on \(\mathbb{C}^{3}\), we have CTC=(100010mm1), then

$$\begin{aligned} &I-3CTC.T+3(CTC)^{2}.T^{2}-(CTC)^{3}T^{3} \\ &\quad = \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix}-3 \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ \overline{m}&\overline{m}&1 \end{pmatrix} \begin{pmatrix} 1&m&m \\ 0&1&0 \\ 0&0&1 \end{pmatrix} \\ &\qquad {}+3 \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ \overline{m}&\overline{m}&1 \end{pmatrix}^{2} \begin{pmatrix} 1&m&m \\ 0&1&0 \\ 0&0&1 \end{pmatrix}^{2} - \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ \overline{m}&\overline{m}&1 \end{pmatrix}^{3} \begin{pmatrix} 1&m&m \\ 0&1&0 \\ 0&0&1 \end{pmatrix}^{3} \\ &\quad = \begin{pmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{pmatrix}. \end{aligned}$$

Hence \(T^{*2}(I-3CTC.T+3(CTC)^{2}.T^{2}-(CTC)^{3}T^{3})T^{2}=0\), i.e., T is a 2-quasi-\([3,C]\)-isometric operator with conjugation C.

On the other hand, since \(T = I + N\), where N=(0mm000000), \(N^{2} = 0\), it follows from Theorem 2.4 that T is a 2-quasi-\([3,C]\)-isometric operator with conjugation C.

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Funding

This research is supported by the National Natural Science Foundation of China (No. 11601130).

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Correspondence to Junli Shen.

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Shen, J. On n-quasi-\([m,C]\)-isometric operators. J Inequal Appl 2019, 313 (2019). https://doi.org/10.1186/s13660-019-2268-3

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MSC

  • 47B20
  • 47A05

Keywords

  • n-quasi-\([m,C]\)-isometric operator
  • Perturbation
  • Nilpotent operator