m-Isometric Operators with Null Symbol and Elementary Operator Entries

Abstract

A pair $(A,B)$ of Banach space operators is strict $(m,X)$-isometric for a Banach space operator $X\in B(\mathcal{X})$ and a positive integer m if ${▵}_{A,B}^m(X)=\left({\displaystyle \sum _{j=0}^m}\left(\begin{array}{c}m\\ j\end{array}\right)L_A^j{R}_B^j\right)(X)=0$ and ${▵}_{A,B}^{m-1}(X)\ne 0$, where $L_A$ and $R_B\in B(B(\mathcal{X}))$ are, respectively, the operators of left multiplication by A and right multiplication by B. Define operators $E_{A,B}$ and ${\mathcal{E}}_{A,B}(X)$ by $E_{A,B}=L_A{R}_B$ and ${\left({\mathcal{E}}_{A,B}(X)\right)}^n=E_{A,B}^n(X)$ for all non-negative integers n. Using little more than an algebraic argument, the following generalised version of a result of relating $(m,X)$-isometric properties of pairs $(A_1,A_2)$ and $(B_1,B_2)$ to pairs $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ and $(E_{A_1,A_2},E_{B_1,B_2})$ is proved: if $A_i,B_i,S_i,X$ are operators in $B(\mathcal{X})$, $1\le i\le 2$ and X a quasi-affinity, then the pair $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ (resp., the pair $(E_{A_1,A_2},E_{B_1,B_2})$) is strict $(m,X)$-isometric for all $X\in B(\mathcal{X})$ if and only if there exist positive integers $m_i\le m$, $1\le i\le 2$ and $m=m_1+m_2-1$, and a non-zero scalar $\beta$ such that $I-{\mathcal{E}}_{\beta A_1,A_2}(S_1)$ is (strict) $m_1$-nilpotent and $I-{\mathcal{E}}_{{\displaystyle \frac{1}{\beta }}{B}_1,B_2}(S_2)$ is (strict) $m_2$-nilpotent (resp., $(\beta A_1,B_1)$ is strict $(m_1,I)$-isometric and $({\displaystyle \frac{1}{\beta }}{B}_2,A_2)$ is strict $(m_2,I)$-isometric).

1. Introduction

Let $B(\mathcal{X})$ (resp., $B(\mathcal{H})$) denote the algebra of operators, i.e., bounded linear transformations, on a complex infinite dimensional Banach space $\mathcal{X}$ (resp., Hilbert space $\mathcal{H}$) into itself. An operator pair $(A,B)\in B(\mathcal{X})\times B(\mathcal{X})$ is $(m,X)$-isometric for some positive integer m, $m\in {\mathtt{N}}_{+}$, and operator $X\in B(\mathcal{X})$, if

$$\begin{array}{ccc}\hfill {▵}_{A,B}^m(X)& =& {(I-L_A{R}_B)}^m(X)=\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_A^j{R}_B^j\right)(X)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{j}X{B}^j\hfill \\ & =& 0,\hfill \end{array}$$

where I is the identity of $B(\mathcal{X})$, $L_A$ and $R_B\in B(B(\mathcal{X}))$ are the operators

$$L_A(Z)=AZ\mathrm{and}{R}_B(Z)=ZB,Z\in B(\mathcal{X}),$$

of left multiplication by A and right multiplication by B, respectively. $(m,X)$-isometric operators arise naturally in classical Function Theory and a study of the structure of such operators has been carried out by a number of authors in the recent past (see [1,2,3,4,5,6,7,8] and some of the references cited there). A number of the properties of $(m,X)$-isometric operators lie on the surface and are readily obtained. Thus,

$$\begin{array}{ccc}& & (A,B)\in (m,X)-isometric⟹{▵}_{A,B}^{m+t}(X)={▵}_{A,B}^t({▵}_{A,B}^m(X))=0\mathrm{for}\mathrm{all}t\in \mathtt{N};\hfill \\ & & {▵}_{A^n,B^n}^m(X)=0⟺{(I-L_A^n{R}_B^n)}^m(X)={\left(\sum _{j=0}^{n-1}{(L_A{R}_B)}^j\right)}^m({▵}_{A,B}^m(X))=0\mathrm{for}\mathrm{all}n\in \mathtt{N};\hfill \\ & & {▵}_{A,B}^m(X)=0⟺{▵}_{A.B}^{m-1}(X)=L_A{R}_B({▵}_{A,B}^{m-1}(X))⟺{▵}_{A.B}^{m-1}(X)=L_A^n{R}_B^n({▵}_{A,B}^{m-1}(X))\mathrm{for}\mathrm{all}n\in \mathtt{N}.\hfill \end{array}$$

Some other properties of $(m,X)$-isometric operators lie deeper and their proof requires some argument. For example, if ${▵}_{A,B}^m(X)=0$, $N_1\in B(\mathcal{X})$ is an $n_1$-nilpotent operator for some positive integer $n_1\in \mathtt{N}$, $[A,N_1]=A{N}_1-N_{1}A=0$ (i.e., A commutes with $N_1$), then

$$\begin{array}{ccc}\hfill {▵}_{A+N_1,B}^{m+n_1-1}(X)& =& \sum _{j=0}^{m+n_1-1}{(-1)}^j\left(\begin{array}{c}m+n_1-1\\ j\end{array}\right){(L_{N_1}{R}_B)}^j{▵}_{A,B}^{m+n_1-1-j}(X)\hfill \\ & =& 0,\hfill \end{array}$$

since ${▵}_{A,B}^{m+n_1-1-j}(X)=0$ for all $m+n_1-1-j\ge m$, equivalently, $j\le n_1-1$, and $L_{N_1}^j=0$ for all $j\ge n_1$. Furthermore, if $N_2\in B(\mathcal{X})$ is also an $n_2$-nilpotent operator which commutes with B, then

$${▵}_{A+N_1,B+N_2}^{m+n_1+n_2-2}(X)=\sum _{j=0}^{m+n_1+n_2-2}{(-1)}^j\left(\begin{array}{c}m+n_1+n_2-2\\ j\end{array}\right){(L_{A+N_1}{R}_{N_2})}^j{▵}_{A+N_1,B}^{m+n_1+n_2-2-j}(X)=0,$$

since ${▵}_{A+N_1,B}^{m+n_1+n_2-2-j}(X)=0$ if $m+n_1+n_2-2-j\ge m+n_1-1$, equivalently, $j\le n_2-1$, and $R_{N_2}^j=0$ if $j\ge n_2$. Conclusion: if $(A,B)$ is $(m,X)$-isometric, $N_i$ are $n_i$-nilpotent operators, and $[A,N_1]=[B,N_2]=0$, then $(A+N_1,B+N_2)$ is $(m+n_1+n_2-2,X)$-isometric.

Let $A,B,S,A_i,B_i,S_i\in B(\mathcal{X})$, $1\le i\le 2$, and let $E_{A,B}\in B(B(\mathcal{X}))$ and ${\mathcal{E}}_{A,B}(S)\in B(\mathcal{X})$ be the operators defined by

$$\begin{array}{ccc}& & E_{A,B}=L_A{R}_B\mathrm{so}\mathrm{that}{E}_{A,B}^n(S)={(L_A{R}_B)}^n(S)=A^{n}S{B}^n\mathrm{for}\mathrm{all}n\in \mathtt{N},\hfill \\ & & {{\mathcal{E}}_{A,B}(S)}^n=E_{A,B}^n(S)=A^{n}S{B}^n\mathrm{for}\mathrm{all}n\in \mathtt{N},\mathrm{in}\mathrm{particular}{\mathcal{E}}_{A,B}{(S)}^0=S.\hfill \end{array}$$

This paper considers $(m,X)$-isometric pairs $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ such that ${▵}_{A_1,A_2}^{m_1}(S_1)=0={▵}_{B_1,B_2}^{m_2}(S_2)$ for some integers $m_1,m_2\in {\mathtt{N}}_{+}$, and pairs $(E_{A_1,A_2},E_{B_1,B_2})$. ($(m,X)$-isometric operators with entries $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ have been called $(m,X)$-isometric operators with null symbol entries [2,7]; m-isometric operators with entries of type $(E_{A_1,A_2},E_{B_1,B_2})$ have been considered by Gu [1], and Duggal and Kim [9,10].) Let X be a quasi-affinity. (Thus, X is injective and has a dense range.) Using little more than linear algebra, we generalise [2] (Theorems 1 and 2(iii)) to prove that “any two of the conditions (i) $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is $(m,X)$-isometric; (ii) there exists a positive integer $m_1\le m$ such that $I-{\mathcal{E}}_{A_1,A_2}(S_1)$ is $m_1$-nilpotent; (iii) there exists a positive integer $m_2\le m$ such that $I-{\mathcal{E}}_{B_1,B_2}(S_2)$ is $m_2$-nilpotent implies the third.” Recall that an $(m,X)$-isometric pair is strict$(m,X)$-isometric if ${▵}_{A,B}^m(X)=0$ and ${▵}_{A,B}^{m-1}(X)\ne 0$. In a similar vein, we say that an operator A is strict m-nilpotent if $A^m=0$ and $A^{m-1}\ne 0$. Answering an open problem raised in [2] (Section 4), we give an elementary proof that the pair $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ (resp., the pair $(E_{A_1,A_2},E_{B_1,B_2})$) is strict $(m,X)$-isometric for all $X\in B(\mathcal{X})$ if and only if there exist positive integers $m_i\le m$, $1\le i\le 2$ and $m=m_1+m_2-1$, and a non-zero scalar $\beta$ such that $I-{\mathcal{E}}_{\beta A_1,A_2}(S_1)$ is strict $m_1$-nilpotent and $I-{\mathcal{E}}_{{\displaystyle \frac{1}{\beta }}{B}_1,B_2}(S_2)$ is strict $m_2$-nilpotent (resp., $(\beta A_1,B_1)$ is strict $(m_1,I)$-isometric and $({\displaystyle \frac{1}{\beta }}{B}_2,A_2)$ is strict $(m_2,I)$-isometric).

Most of our notation is standard (and any non-standard notation will be explained at the point of its introduction). We write $A-\lambda$ for $A-\lambda I$, and ${\sigma }_a(A)$ for the approximate point spectrum of the operator A. We say that the pair $(A,B)$ of operators in $B(\mathcal{X})\times B(\mathcal{X})$ is m-isometric if it is $(m,X)$-isometric for all $X\in B(\mathcal{X})$.

The plan of the paper is as follows. We consider null symbol operator pairs $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ in Section 2, Section 3 considers pairs $(E_{A_1,A_2},E_{B_1,B_2})$, Section 4 consists of a concluding remark.

2. Null Symbol Entries: Pairs $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$

If $(A,B)\in B(\mathcal{X})\times B(\mathcal{X})$ is $(m,X)$-isometric for some $X\in B(\mathcal{X})$, then

$$\begin{array}{ccc}\hfill 0& =& {▵}_{A,B}^m(X)=\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(L_A{R}_B)}^j\right)(X)\hfill \\ & =& \left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)E_{A,B}^j\right)(X)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\mathcal{E}}_{A,B}{(X)}^j\hfill \\ & =& {(I-{\mathcal{E}}_{A,B}(X))}^m={\nabla }_{A,B}^m(X)\hfill \end{array}$$

(by the definition of the operator ${\mathcal{E}}_{A,B}{(X)}^n$, $n\in {\mathtt{N}}_{+}$). Hence, the pair $(A,B)$ is (strictly) $(m,X)$-isometric if and only if the operator ${\nabla }_{A,B}(X)=(I-{\mathcal{E}}_{A,B}(X))$ is (strictly) m-nilpotent.

Recalling that $(A,B)$ is strictly $(m,X)$-isometric if and only if ${▵}_{A,B}^m(X)=0$ and ${▵}_{A,B}^{m-1}(X)\ne 0$, it follows that if ${\nabla }_{A,B}(X)$ is strictly m-nilpotent, then the sequence of operators ${\left\{{\nabla }_{A,B}^j(X)\right\}}_{j=0}^{m-1}$ is linearly independent. Again, if $(A,B)$ is $(m,X)$-isometric, then

$$\begin{array}{ccc}& & {\nabla }_{A,B}^m(X)=(I-{\mathcal{E}}_{A,B}(X)){\nabla }_{A,B}^{m-1}(X)=0\hfill \\ & ⟺& {\nabla }_{A,B}^{m-1}(X)={\mathcal{E}}_{A,B}(X){\nabla }_{A,B}^{m-1}(X)\hfill \\ & ⟺& {\nabla }_{A,B}^{m-1}(X)={\mathcal{E}}_{A,B}^n(X){\nabla }_{A,B}^{m-1}(X)\hfill \end{array}$$

for all $n\in \mathtt{N}$. Let $A_i,B_i$ and $S_i$, $1\le i\le 2$, be operators in $B(\mathcal{X})$ such that

$${\nabla }_{A_1,A_2}^{m_1}(S_1)=0={\nabla }_{B_1,B_2}^{m_2}(S_2),m_1\mathrm{and}{m}_2\in {\mathtt{N}}_{+}.$$

The following proposition relates $(m_1,S_1)$-isometric pairs $(A_1,A_2)$ and $(m_2,S_2)$-isometric pairs $(B_1,B_2)$ to $(m,X)$-isometric pairs $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$.

Proposition 1. 

If $X\in B(\mathcal{X})$ is a quasi-affinity, then any two of the following conditions implies the third:

(i) there exists $m\in N_{+}$ such that $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is $(m,X)$-isometric;

(ii) there exists $m_1\in N_{+}$, $m_1\le m$, such that ${\nabla }_{A_1,A_2}(S_1)$ is $m_1$-nilpotent;

(iii) there exists $m_2\in N_{+}$, $m_2\le m$, such that ${\nabla }_{B_1,B_2}(S_2)$ is $m_2$-nilpotent.

(Here, if (ii) and (iii) hold, then $m=m_1+m_2-1$.)

Proof. 

Let, for convenience, ${\mathcal{E}}_{A_1,A_2}(S_1)=A$ and ${\mathcal{E}}_{B_1,B_2}(S_2)=B$.

$(ii)+(iii)⟹(i)$. Let $m_1+m_2-1=m$. By definition

$$\begin{array}{ccc}\hfill {▵}_{A,B}^m(X)& =& {(I-L_A{R}_B)}^m(X)={\left((I-L_A)R_B+(I-R_B)\right)}^m(X)\hfill \\ & =& \left(\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){((I-L_A)R_B)}^{m-j}{(I-R_B)}^j\right)(X)\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){(I-A)}^{m-j}X{B}^{m-j}{(I-B)}^j\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){\nabla }_{A_1,A_2}^{m-j}(S_1)X{B}^{m-j}{\nabla }_{B_1,B_2}^j(S_2)\hfill \end{array}$$

(since $[I-L_A,R_B]=[R_B,I-R_B]=0$). Since ${\nabla }_{A_1,A_2}^{m-j}(S_1)=0$ for all $m-j\ge m_1$, equivalently, $j\le m-m_1=m_2-1$, and since ${\nabla }_{B_1,B_2}^j(S_2)=0$ for all $j\ge m_2$, ${▵}_{A,B}^{m_0}(X)=0$ for all $m_0\ge m$.

$(i)+(iii)⟹(ii)$. Considering next $(i)$ and $(iii)$, we may assume without loss of generality that $(A,B)$ is strict $(m,X)$-isometric and ${\nabla }_{B_1,B_2}(S_2)$ is strict $m_2$-nilpotent. Since

$${\nabla }_{B_1,B_2}^{m_2}(S_2)=0⟺{\nabla }_{B_1,B_2}^{m_2-1}(S_2)=B^n{\nabla }_{B_1,B_2}^{m_2-1}(S_2)$$

for all $n\in \mathtt{N}$, the strictness implies that the sequence

$$\left\{B^n{\nabla }_{B_1,B_2}^j(S_2)\right\},0\le j\le m_2-1$$

is linearly independent for all $n\in \mathtt{N}$. If $(i)$ holds, then (see above)

$$\begin{array}{ccc}\hfill 0& =& {▵}_{A,B}^m(X)=\left(\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)L_{{\nabla }_{A_1,A_2}(S_1)}^{m-j}{R}_B^{m-j}{R}_{{\nabla }_{B_1,B_2}(S_2)}^j\right)(X)\hfill \\ & =& \sum _{j=0}^{m_2-1}\left(\begin{array}{c}m\\ j\end{array}\right){\nabla }_{A_1,A_2}^{m-j}(S_1)X{B}^{m-j}{\nabla }_{B_1,B_2}^j(S_2)\hfill \end{array}$$

(since ${\nabla }_{B_1,B_2}^t(S_2)=0$ for all $t\ge m_2$). The linear independence of the sequence ${\left\{B^n{\nabla }_{B_1,B_2}^j(S_2)\right\}}_{j=0}^{m_2-1}$, taken along with the fact that X has a dense range, implies that

$${\nabla }_{A_1,A_2}^{m-j}(S_1)X=0⟺{\nabla }_{A_1,A_2}^{m-j}(S_1)=0,0\le j\le m_2-1.$$

In particular, letting $j=m_2-1$,

$${\nabla }_{A_1,A_2}^{m-m_2+1}(S_1)={\nabla }_{A_1,A_2}^{m_1}(S_1)=0.$$

$(i)+(ii)⟹(iii)$. The proof here is similar to that of the previous case except for the fact that we now consider the adjoint operators

$${({▵}_{A,B}^m(X))}^{*}={▵}_{B^{*},A^{*}}^m(X^{*})={▵}_{{\mathcal{E}}_{B_2^{*},B_1^{*}}(S_2^{*}),{\mathcal{E}}_{A_2^{*},A_1^{*}}(S_1^{*})}^m(X^{*})$$

and

$${({\nabla }_{A_1,A_2}^{m_1}(S_1))}^{*}={\nabla }_{A_2^{*},A_1^{*}}^{m_1}(S_1^{*}).$$

Assuming strictness, it is seen that the sequence $\left\{A^{*n}{\nabla }_{A_2^{*},A_1^{*}}^j(S_1^{*})\right\}$ is linearly independent for all $0\le j\le m_1-1$ and $n\in \mathtt{N}$; hence, since X is a quasi-affinity,

$${\nabla }_{B_2^{*},B_1^{*}}^{m-j}(S_2^{*})X^{*}=0⟺{\nabla }_{B_2^{*},B_1^{*}}^{m-j}(S_2^{*})=0⟺{\nabla }_{B_1,B_2}^{m-j}(S_2)=0$$

for all $0\le j\le m_1-1$. In particular, ${\nabla }_{B_1,B_2}^{m-m_1+1}(S_2)={\nabla }_{B_1,B_2}^{m_2}(S_2)=0$. □

If $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is strict $(m,X)$-isometric and ${\nabla }_{B_1,B_2}(S_2)$ is strict $m_2$-nilpotent, then the argument of the proof of the proposition shows that ${\nabla }_{A_1,A_2}(S_1)$ is strict $m_1=m-m_2+1$ nilpotent. (Reason: if ${\nabla }_{A_1,A_2}(S_1)$ is t-nilpotent for some $t<m_1$, then $(ii)$ and $(iii)$ taken together imply $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is $(t+m_1-1(<m),X)$-isometric—a contradiction.) Indeed, if $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is strict $(m,X)$-isometric, ${\nabla }_{A_1,A_2}(S_1)$ is $m_1$-nilpotent, and ${\nabla }_{B_1,B_2}(S_2)$ is $m_2$ nilpotent, then

$$\begin{array}{ccc}& & {▵}_{A,B}^{m-1}(X)=\sum _{j=0}^{m-1}\left(\begin{array}{c}m-1\\ j\end{array}\right){\nabla }_{A_1,A_2}^{m-j-1}(S_1)X{B}^{m-j-1}{\nabla }_{B_1,B_2}^j(S_2)\hfill \\ & =& \sum _{j=0}^{m_2-1}\left(\begin{array}{c}m-1\\ j\end{array}\right){\nabla }_{A_1,A_2}^{m-j-1}(S_1)X{B}^{m-j-1}{\nabla }_{B_1,B_2}^j(S_2)\hfill \\ & & ({\nabla }_{B_1,B_2}^j(S_2)=0\mathrm{for}\mathrm{all}j\ge m_2)\hfill \\ & =& \left(\begin{array}{c}m-1\\ m_2-1\end{array}\right){\nabla }_{A_1,A_2}^{m_1-1}(S_1)X{B}^{m-j-1}{\nabla }_{B_1,B_2}^{m_2-1}(S_2)\hfill \\ & & ({\nabla }_{A_1,A_2}^{m-j-1}(S_1)=0\mathrm{for}\mathrm{all}m-j-1\ge m_1)\hfill \\ & =& \left(\begin{array}{c}m-1\\ m_2-1\end{array}\right){\nabla }_{A_1,A_2}^{m_1-1}(S_1)X{\nabla }_{B_1,B_2}^{m_2-1}(S_2)\hfill \\ & & ({\nabla }_{B_1,B_2}^{m_2}(S_2)=0⟹B^{m-j-1}{\nabla }_{B_1,B_2}^j(S_2)={\nabla }_{B_1,B_2}^j(S_2)\mathrm{for}\mathrm{all}j\ge m_2)\hfill \\ & \ne & 0.\hfill \end{array}$$

Thus

$${(I-A)}^{m_1-1}X{(I-B)}^{m_2-1}\ne 0,$$

and the conditions ${(I-A)}^{m_1-1}\ne 0$ and ${(I-B)}^{m_2-1}\ne 0$ are necessary for $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ to be strict $(m,X)$-isometric. These conditions are, however, not sufficient. For example, if we choose X to be such that it maps the range of $(I-B)$ into the null space of $(I-A)$, then ${(I-A)}^{m_1-1}X{(I-B)}^{m_2-1}=0$ even though neither of ${(I-A)}^{m_1-1}$ and ${(I-B)}^{m_2-1}$ is the 0 operator. If, however, $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is strict m-isometric (i.e., strict $(m,X)$-isometric for all X), then, necessarily, ${(I-A)}^{m_1-1}X{(I-B)}^{m_2-1}=0$ if and only if one ${(I-A)}^{m_1-1}$ or ${(I-B)}^{m_2-1}$ is 0. The proof of the following proposition uses little more than linear algebra to prove a necessary and sufficient condition for the operator pair $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ to be strict $(m,X)$-isometric for a given quasi-affinity X.

Proposition 2. 

Given operators $A_i,B_i,S_i\in B(\mathcal{X})$, $1\le i\le 2$, and a quasi-affinity $X\in B(\mathcal{X})$, the pair of operators $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is a strict $(m,X)$-isometry if and only if there exist integers $m_i\in N_{+}$ ($1\le i\le 2$) and a non-trivial scalar β such that $m=m_1+m_2-1$,

(i) ${\nabla }_{\beta A_1,A_2}(S_1)$ is a strict $m_1$ nilpotent, ${\nabla }_{{\displaystyle \frac{1}{\beta }}{B}_1,B_2}(S_2)$ is a strict $m_2$-nilpotent;

(ii) ${\nabla }_{A_1,A_2}^{m_1-1}(S_1)X{\nabla }_{B_1,B_2}^{m_2-1}(S_2)\ne 0$.

Proof. 

As before, for convenience, we let

$$\begin{array}{ccc}& & {\mathcal{E}}_{A_1,A_2}(S_1)=A,{\mathcal{E}}_{B_1,B_2}(S_2)=B,I-\beta A(=I-{\mathcal{E}}_{\beta A_1,A_2}(S_1))=\nabla (\beta A),\mathrm{and}\hfill \\ & & I-{\displaystyle \frac{1}{\beta }}B=I-{\displaystyle \frac{1}{\beta }}{\mathcal{E}}_{B_1,B_2}(S_2)=1-{\mathcal{E}}_{{\displaystyle \frac{1}{\beta }}{B}_1,B_2}(S_2)=\nabla (\beta B).\hfill \end{array}$$

To prove the “if part” of the theorem, we start by observing that

$${▵}_{\beta A,{\displaystyle \frac{1}{\beta }}B}^m(X)=\left(\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)L_{\beta A}^j{R}_{{\displaystyle \frac{1}{\beta }}B}^j\right)(X)={▵}_{A,B}^m(X).$$

If $\nabla (\beta A)$ is $m_1$-nilpotent and $\nabla (\beta B)$ is $m_2$-nilpotent, then

$${▵}_{\beta A,{\displaystyle \frac{1}{\beta }}B}^{m_1+m_2-1}(X)={▵}_{A,B}^{m_1+m_2-1}(X)={▵}_{A,B}^m(X)=0.$$

Also, if condition $(ii)$ is satisfied, then ${▵}_{A,B}^{m-1}(X)\ne 0$, i.e., $(A,B)$ is strict $(m,X)$-isometric.

For the “only if part” of the proof, we start by recalling that

$$\begin{array}{ccc}\hfill {▵}_{A,B}^m(X)& =& \left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_A^j{R}_B^j\right)(X)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\mathcal{E}}_{A_1,A_2}^j(S_1)X{\mathcal{E}}_{B_1,B_2}^j(S_2).\hfill \end{array}$$

We claim that there exists a non-trivial $\beta \in {\sigma }_a(B)$. For this it will suffice to prove that $0\notin {\sigma }_a(B)$. If $0\in {\sigma }_a(B)$, then there exist sequences $\left\{x_n\right\}$ and $\left\{x_n^{\prime }\right\}$, in $\mathcal{X}$ and (the dual space) $X^{*}$, respectively, such that

$$x_n^{\prime }(x_n)=\langle x_n,x_n^{\prime }\rangle =1\mathrm{for}\mathrm{all}n\in {\mathtt{N}}_{+}\mathrm{and}\underset{n\to \infty }{lim}\langle B{x}_n,x_n^{\prime }\rangle =0.$$

We have

$$\begin{array}{ccc}\hfill {▵}_{A,B}^m(X)& =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{j}X\underset{n\to \infty }{lim}\langle B^j{x}_n,x_n^{\prime }\rangle \hfill \\ & =& 0\hfill \end{array}$$

for all j except for $j=0$ when we have ${▵}_{A,B}^m(X)=X=0$. The operator X being a quasi-affinity, this is a contradiction and our claim is proved.

Let $(0\ne )\beta \in {\sigma }_a(B)$. Then, by an argument similar to the one used above to prove $0\notin {\sigma }_a(B)$,

$$\begin{array}{ccc}\hfill 0& =& {▵}_{A,B}^m(X)=\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\beta }^j{A}^j\right)(X)\hfill \\ & ⟺& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\beta }^j{A}^j={(I-\beta A)}^m=0.\hfill \end{array}$$

Let $m_1$ be the least positive integer such that

$${(I-\beta A)}^{m_1}={(I-\beta {\mathcal{E}}_{A_1,A_2}(S_1))}^{m_1}={(I-{\mathcal{E}}_{\beta A_1,A_2}(S_1))}^{m_1}=0.$$

Then $I-\beta A$ is strictly $m_1$-nilpotent and the sequence ${\left\{{\nabla }^j(\beta A)\right\}}_{j=0}^{m_1-1}$ is linearly independent. Since

$$\begin{array}{ccc}\hfill 0& =& {▵}_{A,B}^m(X)={▵}_{\beta A,{\displaystyle \frac{1}{\beta }}B}^m(X)\hfill \\ & =& {\left((I-R_{{\displaystyle \frac{1}{\beta }}B})+(I-L_{\beta A})R_{{\displaystyle \frac{1}{\beta }}B}\right)}^m(X)\hfill \\ & =& \left(\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){(I-R_{{\displaystyle \frac{1}{\beta }}B})}^{m-j}{R}_{{\displaystyle \frac{1}{\beta }}B}^j{(I-L_{\beta A})}^j\right)(X)\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){\nabla }^j(\beta A)X{({\displaystyle \frac{1}{\beta }}B)}^j{\nabla }^{m-j}(\beta B)\hfill \\ & =& \sum _{j=0}^{m_1-1}\left(\begin{array}{c}m\\ j\end{array}\right){\nabla }^j(\beta A)X{({\displaystyle \frac{1}{\beta }}B)}^j{\nabla }^{m-j}(\beta B)\hfill \\ & & (\mathrm{since}{\nabla }^t(\beta A)=0\mathrm{for}\mathrm{all}t\ge m_1)\hfill \\ & ⟹& X{({\displaystyle \frac{1}{\beta }}B)}^j{\nabla }^{m-j}(\beta B)=0\mathrm{for}\mathrm{all}0\le j\le m_1-1.\hfill \end{array}$$

Since both X and B are injective, recall that X is a quasi-affinity and B is left invertible, ${\nabla }^{m-j}(\beta B)=0$ for all $0\le j\le m_1-1$. In particular,

$${\nabla }^{m-m_1+1}(\beta B)={\nabla }^{m_2}(\beta B)=0.$$

The strict $(m,X)$-isometric property of the pair $(A,B)$, taken in conjunction with the fact that $m_2=m-m_1+1$, implies that $\nabla (\beta B)$ is strict $m_2$-nilpotent. The necessity of condition $(ii)$ having already been seen (above), the proof is complete. □

The Hilbert space case In the case in which $A_1^{*}=A_2=A$ and $B_1^{*}=B_2=B$ are Hilbert space operators,

$$\begin{array}{ccc}& & {\mathcal{E}}_{A^{*},A}{(I)}^n=E_{A^{*},A}^n(I)=A^{*n}{A}^n,{\mathcal{E}}_{B^{*},B}{(I)}^n=E_{B^{*},B}^n(I)=B^{*n}{B}^n,\hfill \\ & & {▵}_{{\mathcal{E}}_{A^{*},A}(I),{\mathcal{E}}_{B^{*},B}(I)}^n(X)=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)A^{*j}{A}^{j}X{B}^{*j}{B}^j\hfill \end{array}$$

for all $n\in {\mathtt{N}}_{+}$. Proposition 2 in such a case takes the following form.

Corollary 1. 

Given operators $A,B,X\in B(\mathcal{H})$, X a quasi-affinity, the pair $({\mathcal{E}}_{A^{*},A}(I),{\mathcal{E}}_{B^{*},B}(I))$ is strict $(m,X)$-isometric if and only if there exist positive integers $m_i\le m$, $m=m_1+m_2-1$, and a (non-trivial) positive scalar β such that $(\beta A^{*},A)$ is strict $(m_1,I)$-isometric, $({\displaystyle \frac{1}{\beta }}{B}^{*},B)$ is strict $(m_2,I)$-isometric, and ${▵}_{\beta A^{*},A}^{m_1-1}(I)X{▵}_{{\displaystyle \frac{1}{\beta }}{B}^{*},B}^{m_2-1}(I)\ne 0$.

In the absence of the property ${\mathcal{E}}_{A,B}{(I)}^n=E_{A,B}^n(I)$ for all $n\in \mathtt{N}$, the strict $(m,X)$-isometric property of the pair $(E_{A^{*},A}(I),E_{B^{*},B}(I))$ for a quasi-affinty X implies

$${▵}_{E_{A^{*},A}(I),E_{B^{*},B}(I)}^n(X)=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right){\left|A\right|}^{2j}X{\left|B\right|}^{2j};$$

there exist positive integers $m_i\le m$, $m=m_1+m_2-1$, and a (non-trivial) positive scalar $\beta$ such that $(\beta |A|,|A|)$ is strict $(m_1,I)$-isometric, $({\displaystyle \frac{1}{\beta }}|B|,|B|)$ is strict $(m_2,I)$-isometric and ${▵}_{\beta \left|A\right|,\left|A\right|}^{m_1-1}(I)X{▵}_{{\displaystyle \frac{1}{\beta }}\left|B\right|,\left|B\right|}^{m_2-1}(I)\ne 0$.

3. Pairs $(E_{A_1,A_2},E_{B_1,B_2})$

Proposition 1 fails in the absence of the (fairly restrictive) hypothesis ${\mathcal{E}}_{A,B}{(S)}^n=E_{A,B}^n(S)$. This follows from the following elementary example.

Example 1. 

Trivially, the pair $(I,I)$ satisfies ${▵}_{I,I}(S)=0$ for all $S\in B(\mathcal{X})$. Considering the pair $({\mathcal{E}}_{I,I}(S_1),{\mathcal{E}}_{I,I}(S_2))$, the validity of $(ii)+(iii)$ implies $(i)$ in the proof of the proposition implies ${▵}_{{\mathcal{E}}_{I,I}(S_1),{\mathcal{E}}_{I,I}(S_2)}(X)=0$ for all $X\in B(\mathcal{X})$, i.e., $X=S_{1}X{S}_2$ for all $S_1,S_2$ and $X\in B(\mathcal{X})$. This is absurd.

Observe that $E_{I,I}^n(S)=S\ne S^n={({\mathcal{E}}_{I,I}(S))}^n$ and $E_{A,B}^n(I)=A^n{B}^n\ne {(AB)}^n={\mathcal{E}}_{A,B}{(I)}^n$. An immediate consequence of examples of the above type is that the proposition cannot be used, contrary to the claim made in [2] (Corollary 1), to deduce results of the the type “$(A_1,A_2)$ is $(m_1,I)$-isometric and $(B_1,B_2)$ is $(m_2,I)$-isometric, then $(E_{A_1,A_2},E_{B_1,B_2})$ is $(m_1+m_2-1,X)$-isometric for all $X\in B(\mathcal{X})$”. Indeed, if we let the pair $(E_{A_1,A_2},E_{B_1,B_2})$ be such that $(A_1,A_2)=(I,I)$ and choose the pair $(B_1,B_2)$ to be $(m,I)$-isometric for some $m\in {\mathtt{N}}_{+}$, then for all $X\in B(\mathcal{X})$,

$$\begin{array}{ccc}\hfill {▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)& =& \left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{E_{A_1,A_2}}^j{R}_{E_{B_1,B_2}}^j\right)(X)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{E_{A_1,A_2}}^j\left(L_{B_1}^{j}X{R}_{B_2}^j\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{E_{A_1,A_2}}^j\left(B_1^{j}X{B}_2^j\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{A_1}^j\left(B_1^{j}X{B}_2^j\right)R_{A_2}^j\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A_1^j{B}_1^{j}X{B}_2^j{A}_2^j\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)B_1^{j}X{B}_2^j={▵}_{B_1,B_2}^m(X).\hfill \end{array}$$

Evidently, ${▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)$ cannot be 0 for all X (i.e., $(ii)$ and $(iii)$ imply $(i)$ of Proposition 1 fails for pairs $(E_{A_1,A_2},E_{B_1,B_2})$). We remark here that even though [2] (Theorem 1) makes no explicit mention of the hypothesis ${\mathcal{E}}_{A,B}{(X)}^n=E_{A,B}^n(X)$, its use is implicit in the proof of the theorem.

A pair $(\mathcal{X},\tilde{\mathcal{X}})$ of Banach spaces is a dual pairing if either $\mathcal{X}={\mathcal{X}}^{*}$ or $\mathcal{X}={\tilde{\mathcal{X}}}^{*}$. If we let $x\otimes y^{\prime }$, $x\in \mathcal{X}$ and $y^{\prime }\in {\mathcal{Y}}^{*}$, $\mathcal{Y}$ a Banach space, denote the rank one operator $\mathcal{Y}\to \mathcal{X},y\to \langle y,y^{\prime }\rangle x$, then the operator ideal $\mathcal{J}$ between $\mathcal{Y}$ and $\mathcal{X}$ is a linear subspace $B(\mathcal{Y},\mathcal{X})$ equipped with a Banach norm $\nu$ such that (i) $x\otimes y^{\prime }\in \mathcal{J}$; $\nu (x\otimes y^{\prime })=\parallel x\parallel \parallel y^{\prime }\parallel$ and (ii) ${\mathcal{E}}_{A,B}(X)=L_A{R}_B(X)=AXB$, $\nu (AXB)\le \parallel A\parallel \nu (\parallel X\parallel )\parallel B\parallel$ for all $x\in \mathcal{X}$, $y^{\prime }\in {\mathcal{Y}}^{*}$, $X\in \mathcal{J}$, and $A\in B(\mathcal{X})$ and $B\in B(\mathcal{Y})$. Thus defined, each $\mathcal{J}$ is a tensor product relative to the dual pairings $(\mathcal{X},{\mathcal{X}}^{*})$ and $(\mathcal{Y},{\mathcal{Y}}^{*})$ and the bilinear mapping

$$\begin{array}{ccc}& & \mathcal{X}\times {\mathcal{Y}}^{*}\to \mathcal{J},\langle x,y^{\prime }\rangle \to x\otimes y^{\prime },\hfill \\ & & B(\mathcal{X})\times B({\mathcal{Y}}^{*})\to B(\mathcal{J}),(A,B^{*})\to A\otimes B^{*},\hfill \end{array}$$

where $A\otimes B^{*}(X)=AXB=E_{A,B}(X)$ [11] (page 51). It is known, see [9] (Corollary 2) (see also [1,7]), that for $A_i,B_i\in B(\mathcal{X})$, $1\le i\le 2$, $(E_{A_1,A_2},E_{B_1,B_2})$ is strict $(m,X)$-isometric if and only if there exist positive integers $m_i\le m$ and a non-zero scalar $\beta$ such that $m=m_1+m_2-1$, $(\beta A_1,A_2)$ is strict $(m_1,I)$-isometric, $({\displaystyle \frac{1}{\beta }}{B}_2,A_2)$ is strict $(m_2,I)$-isometric, and ${▵}_{\beta A_1,B_1}^{m_1-1}\left({▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(X)\right)\ne 0$ for $X\in \mathcal{J}$. This result does not follow from Proposition 2, even for the case in which $\mathcal{X}$ is a Hilbert space and the pair $(E_{A_1,A_2},E_{B_1,B_2})$ is the pair $(E_{A^{*},A},E_{B^{*},B})$. The following theorem, our main result, uses an algebraic argument to prove this result for the case in which the operator pair $(E_{A_1,A_2},E_{B_1,B_2})$ is strict $(m,X)$-isometric for a quasi-affinity $X\in B(\mathcal{X})$.

Theorem 1. 

Given operators $A_i,B_i\in B(\mathcal{X})$, $1\le i\le 2$, the pair $(E_{A_1,A_2},E_{B_1,B_2})$ is strict $(m,X)$-isometric for a quasi-affinity $X\in B(\mathcal{X})$ if and only if there exist positive integers $m_i\le m$ and a non-trivial scalar β such that $m=m_1+m_2-1$, $(\beta A_1,B_1)$ is strict $(m_1,I)$-isometric, $({\displaystyle \frac{1}{\beta }}{B}_2,A_2)$ is strict $(m_2,I)$-isometric, and ${▵}_{\beta A_1,B_1}^{m_1-1}(I)X({▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I))\ne 0$.

Proof. 

We start by proving the “only if” part: the proof depends upon a judicious use of the properties of the operator $L_{E_{A_1,A_2}}^j{R}_{E_{B_1,B_2}}^j$. If the pair $(E_{A_1,A_2},E_{B_1,B_2})$ is $(m,X)$-isometric, then

$${▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)=\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)L_{E_{A_1,A_2}}^j{R}_{E_{B_1,B_2}}^j\right)(X)=0.$$

By definition

$$\begin{array}{cc}& L_{E_{A_1,A_2}}^j{R}_{E_{B_1,B_2}}^j(X)=L_{L_{A_1}{R}_{A_2}}^j{R}_{L_{B_1}{R}_{B_2}}^j(X)\hfill \\ =& \left(L_{L_{A_1}}^j{L}_{R_{A_2}}^j{R}_{L_{B_1}}^j{R}_{R_{B_2}}^j\right)(X)\hfill \\ =& \left(L_{L_{A_1}}^j{R}_{L_{B_1}}^j{L}_{R_{A_2}}^j{R}_{R_{B_2}}^j\right)(X)([L_C,R_D]=0\mathrm{for}\mathrm{all}\mathrm{operators}C,D)\hfill \\ =& {\left(L_{L_{A_1}}{R}_{L_{B_1}}\right)}^{j}X{\left(L_{R_{A_2}}{R}_{R_{B_2}}\right)}^j\hfill \\ =& L_{L_{A_1}}^j{R}_{L_{B_1}}^j\left(L_{R_{A_2}}^{j}X{R}_{R_{B_2}}^j\right)\hfill \\ =& L_{L_{A_1}}^j{R}_{L_{B_1}}^j\left(R_{A_2}^{j}X{R}_{B_2}^j\right)=L_{L_{A_1}}^j{R}_{L_{B_1}}^j\left(X{B}_2^j{A}_2^j\right)\hfill \end{array}$$

(1)

$$\begin{array}{cc}=& L_{A_1}^j\left(X{B}_2^j{A}_2^j\right)L_{B_1}^j=L_{A_1}^j{B}_1^j\left(X{B}_2^j{A}_2^j\right)=A_1^j{B}_1^{j}X{B}_2^j{A}_2^j.\hfill \end{array}$$

(2)

For convenience, set $L_{L_{A_1}}{R}_{L_{B_1}}=C$ and $L_{R_{A_2}}{R}_{R_{B_2}}=D$. We claim that $\sigma (D)\ne \left\{0\right\}$. Let us suppose to the contrary that $\sigma (D)=\left\{0\right\}$. Then there exist sequences $\left\{Z_n\right\}$ and $\left\{Z_n^{\prime }\right\}$ of unit vectors (in $B(B(\mathcal{X}))$ and its dual space, respectively) such that

$$Z_n^{\prime }(Z_n)=\langle Z_n,Z_n^{\prime }\rangle =1\mathrm{for}\mathrm{all}n\in {\mathtt{N}}_{+}\mathrm{and}\underset{n\to \infty }{lim}D{Z}_n=0.$$

We have

$${▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)C^{j}X\underset{n\to \infty }{lim}\langle D^j{Z}_n,Z_n^{\prime }\rangle =0$$

for all j except $j=0$ when we obtain $X=0$. The operator X being a quasi-affinity, we have a contradiction. Hence ($0\notin {\sigma }_a(D)$ and) there exists a non-trivial scalar $\beta \in {\sigma }_a(D)$. Assuming $\left\{Z_n\right\}$ and $\left\{Z_n^{\prime }\right\}$ to be sequences of unit vectors such that $Z_n^{\prime }(Z_n)=1$ for all $n\in {\mathtt{N}}_{+}$ and ${lim}_{n\to \infty }\langle D{Z}_n,Z_n^{\prime }\rangle =\beta$, we have

$$\begin{array}{ccc}\hfill 0& =& {▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)C^{j}X\underset{n\to \infty }{lim}\langle D^j{Z}_n,Z_n^{\prime }\rangle \hfill \\ & =& \left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(\beta C)}^j\right)X\hfill \\ & =& \left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(\beta A_1)}^j{B}_1^j\right)X(\mathrm{see}(2))\hfill \\ & =& \left({▵}_{\beta A_1,B_1}^m(I)\right)X.\hfill \end{array}$$

The operator X being a quasi-affinity, we conclude $(\beta A_1,B_1)$ is $(m,I)$-isometric. Consequently there exists a positive integer $m_1\le m$ such that $(\beta A_1,B_1)$ is strict $(m_1,I)$-isometric, and hence the set

$${\left\{{▵}_{\beta A_1,B_1}^j(I)\right\}}_{j=0}^{m_1-1},\mathrm{equivalently}{\left\{{▵}_{\beta A_1,B_1}^{m-j}(I)\right\}}_{j=m-m_1+1}^m,\mathrm{is}\mathrm{linearly}\mathrm{independent}.$$

Once again, for convenience, set $L_{L_{\beta A_1}}{R}_{L_{B_1}}=C_{\beta }$ and $L_{R_{A_2}}{R}_{R_{{\displaystyle \frac{1}{\beta }}{B}_2}}=D_{\beta }$. Then

$$\begin{array}{ccc}& & {▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)={▵}_{E_{\beta A_1,A_2},E_{B_1,{\displaystyle \frac{1}{\beta }}{B}_2}}^m(X)\hfill \\ & =& {\left(I-L_{L_{\beta A_1}}{R}_{L_{B_1}}{L}_{R_{A_2}}{R}_{R_{{\displaystyle \frac{1}{\beta }}{B}_2}}\right)}^m(X)\hfill \\ & =& {\left(I-C_{\beta }{D}_{\beta }\right)}^m(X)\hfill \\ & =& {\left((I-C_{\beta })D_{\beta }+(I-D_{\beta })\right)}^m(X)\hfill \\ & =& \left(\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){(I-C_{\beta })}^{m-j}{D}_{\beta }^{m-j}{(I-D_{\beta })}^j\right)(X)\hfill \\ & =& \left(\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right)C_{\beta }^p\right)D_{\beta }^{m-j}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)D_{\beta }^k\right)\right)(X).\hfill \end{array}$$

Since

$$\begin{array}{ccc}& & D_{\beta }^{m-j}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)D_{\beta }^k\right)(X)\hfill \\ & =& X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-j}{A}_2^{m-j}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){({\displaystyle \frac{1}{\beta }}{B}_2)}^k{A}_2^k\right)\hfill \\ & =& X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-j}{A}_2^{m-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right)C_{\beta }^p\right)X\hfill \\ & =& \left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){(\beta A_1)}^p{B}_1^p\right)X\hfill \\ & =& {▵}_{\beta A_1,B_1}^{m-j}(I)X,\hfill \end{array}$$

we have

$$\begin{array}{ccc}\hfill {▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)& =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){▵}_{\beta A_1,B_1}^{m-j}(I)X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-j}{A}_2^{m-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)\hfill \\ & =& 0.\hfill \end{array}$$

By the linear independence of the set $\left\{{▵}_{\beta A_1,B_1}^{m-j}(I)\right\}$ for all $m-m_1+1\le j\le m$,

$$X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-j}{A}_2^{m-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)=0$$

for all $0\le j\le m_1-1$. Since X is a quasi-affinity and $0\notin {\sigma }_a(D_{\beta })$ (implies $0\notin {\sigma }_a(B_2^{m-j}{A}_2^{m-j}$)),

$${▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)=0\mathrm{for}\mathrm{all}0\le j\le m_1-1.$$

In particular, upon letting $m-m_1+1=m_2$,

$${▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2}(I)=0.$$

The strict $(m,X)$-isometric property of the pair $(E_{A_1,A_2},E_{B_1,B_2})$ implies also that

$$\begin{array}{ccc}\hfill 0& \ne & {▵}_{E_{A_1,A_2},E_{B_1,B_2}}^{m-1}(X)\hfill \\ \hfill & =& {\sum }_{j=0}^{m-1}\left(\begin{array}{c}m-1\\ j\end{array}\right){▵}_{\beta A_1,B_1}^{m-1-j}(I)X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-1-j}{A}_2^{m-1-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)\hfill \\ \hfill & =& {\sum }_{j=0}^{m_2-1}\left(\begin{array}{c}m-1\\ j\end{array}\right){▵}_{\beta A_1,B_1}^{m-1-j}(I)X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-1-j}{A}_2^{m-1-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)\hfill \\ \hfill & & ({▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)=0\mathrm{for}\mathrm{all}j\ge m_2)\hfill \\ \hfill & =& \left(\begin{array}{c}m-1\\ m_2-1\end{array}\right){▵}_{\beta A_1,B_1}^{m_1-1}(I)X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-1-j}{A}_2^{m-1-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I)\hfill \\ \hfill & & ({▵}_{\beta A_1,B_1}^{m-1-j}(I)=0\mathrm{for}\mathrm{all}m-1-j\ge m_1,\mathrm{equivalently},j\le m_2-2)\hfill \\ \hfill & =& \left(\begin{array}{c}m-1\\ m_2-1\end{array}\right){▵}_{\beta A_1,B_1}^{m_1-1}(I)X{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I)\hfill \\ \hfill & & \left(\mathrm{recall}:{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2}(I)=0⟹{({\displaystyle \frac{1}{\beta }}{B}_2)}^t{A}_2^t{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I)={▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I)\mathrm{for}\mathrm{all}t\in \mathtt{N}\right).\hfill \end{array}$$

(3)

Since ${▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I)=0$ would contradict this condition, the pair $({\displaystyle \frac{1}{\beta }}{B}_2,A_2)$ is strict $(m_2,I)$-isometric.

The proof of the reverse implication is straightforward. Thus, if $(\beta A_1,B_1)$ is strict $(m_1,I)$-isometric, $({\displaystyle \frac{1}{\beta }}{B}_2,A_2)$ is strict $(m_2,I)$-isometric, and $m=m_1+m_2-1$, then for every quasi-affinity $X\in B(\mathcal{X})$,

$$\begin{array}{ccc}\hfill {▵}_{E_{A_1,A_2},E_{B_1,B_2}}^m(X)& =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){▵}_{\beta A_1,B_1}^{m-j}(I)X{({\displaystyle \frac{1}{\beta }}{B}_2)}^{m-j}{A}_2^{m-j}{▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)\hfill \\ & =& 0,\hfill \end{array}$$

since ${▵}_{\beta A_1,B_1}^{m-j}(I)=0$ for all $m-j\ge m_1$, equivalently, $j\le m-m_1=m_2-1$, and ${▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^j(I)=0$ for all $j\ge m_2$. The strictness implies

$${▵}_{\beta A_1,B_1}^{m_1-1}(I)\ne 0\ne {▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I)$$

and this in turn implies ${▵}_{\beta A_1,B_1}^{m_1-1}(I)X({▵}_{{\displaystyle \frac{1}{\beta }}{B}_2,A_2}^{m_2-1}(I))\ne 0$. □

The proof of Theorem 1 in the case in which $A_1^{*}=B_1=A$ and $A_2^{*}=B_2=B$, A and B Hilbert space operators, is more straightforward. Thus, if the pair $(E_{A^{*},B^{*}},E_{A,B})$ is $(m,X)$-isometric for some quasi-affinity $X\in B(\mathcal{H})$, then

$${▵}_{E_{A^{*},B^{*}},E_{A,B}}^m(X)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{*j}{A}^{j}X{B}^j{B}^{*j},$$

$0\notin {\sigma }_a(B^{*})$, and $(0\ne ){\beta }_0\in {\sigma }_a(B^{*})$ implies

$$\begin{array}{ccc}& & \left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\beta }^j{A}^{*j}{A}^j\right)X=0(\beta =|{\beta }_0{|}^2)\hfill \\ & ⟺& {▵}_{\beta A^{*},A}^m(I)=0.\hfill \end{array}$$

Corollary 2. 

Given operators $A,B\in B(\mathcal{H})$, the pair $(E_{A^{*},B^{*}},E_{A,B}))$ is strict $(m,X)$-isometric for a quasi-affinity $X\in B(\mathcal{H})$ if and only if there exist positive integers $m_i\le m$ and a non-trivial scalar β such that $m=m_1+m_2-1$, $(\beta A^{*},A)$ is strict $(m_1,I)$-isometric, $({\displaystyle \frac{1}{\beta }}B,B^{*})$ is strict $(m_2,I)$-isometric, and ${▵}_{\beta A^{*},A}^{m_1-1}(I)X({▵}_{{\displaystyle \frac{1}{\beta }}B,B^{*}}^{m_2-1}(I))\ne 0$.

Since ${▵}_{\beta A^{*},A}^m(I)=0$ if and only if ${▵}_{\overline{{\beta }_0}{A}^{*},{\beta }_{0}A}^m(I)=0$, $\overline{{\beta }_0}{\beta }_0=\beta$, the pair of operators $(\overline{{\beta }_0}{A}^{*},{\beta }_{0}A)$ is $(m,I)$-isometric; hence, ${\sigma }_a({\beta }_{0}A)$ lies in the boundary $\partial \mathtt{D}$ of the unit disc $\mathtt{D}$ in $\mathtt{C}$ and $1-(\overline{{\beta }_0}{\sigma }_a(A^{*})({\beta }_0{\sigma }_a(A))=0$. There exists a non-trivial scalar $\lambda$, $\left|\lambda \right|=1$, such that ${\beta }_0\alpha =\lambda$ and $\overline{{\beta }_0\alpha }={\displaystyle \frac{1}{\lambda }}=\overline{\lambda }$ for all $\alpha \in {\sigma }_a(A)$. We assert that ${\sigma }_a(B^{*})$ consists, at most, of two points. For if there exist non-trivial $\overline{\mu },\overline{\nu }\in {\sigma }_a(B^{*})$, $\mu \ne {\beta }_0\ne \nu$, then ${\sigma }_a(\mu A)={\sigma }_a({\beta }_{0}A)+{\sigma }_a((\mu -{\beta }_0)A)$ and ${\sigma }_a(\nu A)={\sigma }_a({\beta }_{0}A)+{\sigma }_a((\nu -{\beta }_0)A)$: since $0\notin {\sigma }_a(A)$, not both of these translations of ${\sigma }_a({\beta }_{0}A)$ are in $\partial \mathtt{D}$. This argument applies equally to ${\sigma }_a(A)$; hence $\sigma (A)$ and $\sigma (B)$ consist at most of two points. A similar statement holds for operators $A,B\in B(\mathcal{H})$ such that the pair $({\mathcal{E}}_{A^{*},A}(I),{\mathcal{E}}_{B^{*},B}(I))$ is a strict $(m,X)$- isometry for some quasi-affinity $X\in B(\mathcal{H})$. For in this case for every $\beta \in {\sigma }_a(B)$, $\beta$ being necessarily non-trivial, the pair $(\overline{\beta }{A}^{*},\beta A)$ is $(m_1,I)$-isometric for some positive integer $m_1\le m$ (see Corollary 1). Thus ${\sigma }_a(\beta A)\subset \partial \mathtt{D}$. Let $\lambda \in {\sigma }_a(A)$, and suppose that there exist distinct scalars $\mu ,\nu \in {\sigma }_a(B)$; $\mu ,\nu \ne \beta$. Then ${\sigma }_a(\mu A)={\sigma }_a(\beta A)+{\sigma }_a((\mu -\beta )A)$, ${\sigma }_a(\nu A)={\sigma }_a(\beta A)+{\sigma }_a((\nu -\beta )A)$, and not both these translations of ${\sigma }_a(\beta A)$ can be in $\partial (\mathtt{D})$. We remark that both of these classes of operators belong to the class of $(m,I)$-isometric operators with a finite spectrum [1,9,10,12,13].

4. A Concluding Remark

Let ${\mathcal{C}}_2(\mathcal{H})\subseteq B(\mathcal{H})$ denote the operator ideal of Hilbert–Schmidt operators (equipped with the Hilbert–Schmidt operator norm). Given $A,B\in B(\mathcal{H})$, Gu [1] (Theorem 7) proves that $E_{A,B}\in {\mathcal{C}}_2(\mathcal{H})$ is a strict m-isometry, i.e., the pair $(E_{A,B}^{*},E_{A,B})$ is strict $(m,T)$-isometric for all $T\in {\mathcal{C}}_2(H)$, if and only if there exists a non-trivial scalar $\beta$ and integers $m_1,m_2\in \mathtt{N}$ such that $m=m_1+m_2-1$, $\beta A$ is strict $(m_1,I)$-isometric and ${\displaystyle \frac{1}{\beta }}B$ is strict $(m_2,I)$-isometric. Observe that if $A,B\in B(\mathcal{H})$ and $(A,B)$ is $(m,X)$-isometric for a quasi-affinity $X\in B(\mathcal{H})$, then (our purely algebraic argument from Section 2 shows that) there exist $\mu \in {\sigma }_a(B)$ and $\overline{\nu }\in {\sigma }_a(A^{*})$ such that ${(I-\mu A)}^m=0={(I-\overline{\nu }{B}^{*})}^m$. Since $\sigma ({▵}_{A,B})=1-\sigma (A)\sigma (B)$, $\mu \nu =1$, ${(I-\mu A)}^m=0={(I-{\displaystyle \frac{1}{\mu }}B)}^m$. The operators ${\mathcal{E}}_{A_1,A_2}(S_1)$ and ${\mathcal{E}}_{B_1,B_2}(S_2)$ are operators in $B(\mathcal{H})$; if $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ is $(m,X)$-isometric, $X\in B(\mathcal{H})$ a quasi-affinity, then there exists a non-zero scalar $\beta$ such that $I-\beta {\mathcal{E}}_{A_1,A_2}(S_1)$ and $I-{\displaystyle \frac{1}{\beta }}{\mathcal{E}}_{B_1,B_2}(S_2)$ are m-nilpotent. Thus, if ${\mathcal{E}}_{A_1,A_2}^n(X)=E_{A_1,A_2}^n(X)$ for all $n\in \mathtt{N}$, then there exists a non-zero scalar $\beta$ such that $(\beta A_1,A_2)$ is $(m,S_1)$-isometric and $({\displaystyle \frac{1}{\beta }}{B}_1,B_2)$ is $(m,S_2)$-isometric. Proposition 2 is a Banach space generalisation of this result. The extension of this algebraic argument to the pair $(E_{A_1,A_2},E_{B_1,B_2})\in B{(B(\mathcal{X}))}^2$ requires a judicious use of the algebraic, especially the commutative, properties of the left/right regularisation operators (in the terminology of Taylor and Lay [14] (Page 392)) $L_A$ and $R_B$ of the algebra $B(\mathcal{X})$. We remark in closing that a proof of the Hilbert space version of this result for the m-null symbols pair $({\mathcal{E}}_{A_1,A_2}(S_1),{\mathcal{E}}_{B_1,B_2}(S_2))$ using arithmetic progressions and a combinatorial argument, thus avoiding analytic arguments, has been given by Marrero [15].