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 $T^{\ast n}({\sum }_{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)C{T}^{m-j}C.T^{m-j})T^n=0$. 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.

1 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 M̅.

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 $\left(\begin{array}{c}m\\ j\end{array}\right)$ is the binomial coefficient. Several authors have studied the m-isometric operator. We refer the reader to [2–6, 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 $T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ on H, then $CTC=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. 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 $T=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 1& 0& 0\end{array}\right)\in B({\mathbb{C}}^3)$, 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 $CTC=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$ 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.

2 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) Tis ann-quasi-$[m,C]$-isometric operator;

• (2) $T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)$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 $T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)$ on $H=\overline{R(T^{n})}\oplus N(T^{*n})$, where ${\sum }_{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)C_1\times T_1^{m-j}{C}_1.T_1^{m-j}=0$ 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 $F={\sum }_{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)C_1{T}_1^{m-j}{C}_1.T_1^{m-j}$. 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 $T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)$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 $T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)$ 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 $T=\left(\begin{array}{ccc}0& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\in B({\mathbb{C}}^3)$, 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 $T=\left(\begin{array}{cc}{T}_1& T_2\\ 0& T_3\end{array}\right)$ 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 $\left(\begin{array}{c}m\\ i,j,k\end{array}\right)=\frac{m!}{i!.j!,k!}$ and $\lambda _{0}(*)=I$ by [8].

We have

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

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

• (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 $T=\left(\begin{array}{ccc}1& m& m\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$ on $\mathbb{C}^{3}$, we have $CTC=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ \bar{m}& \bar{m}& 1\end{array}\right)$, 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 $N=\left(\begin{array}{ccc}0& m& m\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$, $N^{2} = 0$, it follows from Theorem 2.4 that T is a 2-quasi-$[3,C]$-isometric operator with conjugation C.