On the Symbols of Strictly m-Null Elementary Operators

Abstract

This paper extends the previous work by the author on m-null pairs of operators in Hilbert space. If an elementary operator L has elementary symbols A and B that are p-null and q-null, respectively, then L is $(p+q-1)$-null. Here, we prove the converse under strictness conditions, modulo some nonzero multiplicative constant—if L is strictly $(p+q-1)$-null, then a scalar $\lambda \ne 0$ exists such that $\lambda A$ is strictly p-null and ${\lambda }^{-1}B$ is strictly q-null. Our constructive argument relies essentially on algebraic and combinatorial methods. Thus, the result obtained by Gu on m-isometries is recovered without resorting to spectral analysis. For several operator classes that generalize m-isometries and are subsumed by m-null operators, the result is new.

1. Introduction

The concept of m-isometry was introduced by Agler and Stankus in a series of seminal papers [1,2,3]. Since then, m-isometries have been the subject of active investigation. In fact, the case $m=1$, corresponding to standard isometries, had already been extensively studied due to its central role in the development of operator theory and its applications, particularly in modeling contractive operators through their isometric dilations [4], its function-theoretic ties to the classical Hardy space [5,6], and its applications in $H^{\infty }$-control problems [7]. For $m=2$, the theory connects to discrete-time nonstationary processes with stationary increments, a class that generalizes Brownian motion [2,3]. More broadly, m-isometries with $m>1$ appear naturally in the disconjugacy theory for Toeplitz operators with smooth symbols, as introduced by Boutet de Monvel and Guillemin [8], as well as in the analysis of shift operators on Dirichlet-type spaces. For instance, multiplication by z on the Dirichlet space over the unit disk is not an isometry but a 2-isometry ([9], Proposition 9.3.1). Richter [10,11] established that every cyclic 2-isometry arises from multiplication by z on certain Dirichlet-type spaces, creating a link with the study of shift-invariant subspaces; for recent progress on this topic, see [12]. Other relevant contributions are [13,14,15,16,17]. All these applications emphasize the significance of m-isometries across analysis, probability, and mathematical physics, motivating the search for generalizations.

The original notion of m-isometry has indeed been extended in numerous directions. Bayart [18] introduced $(m,p)$-isometries in Banach spaces. Hoffman et al. [19] explored the role of the second parameter, p, in this definition, allowing $p\in (0,\infty ]$. Subsequently, other classes of operators encompassing or related to m-isometries have been proposed, usually first in Hilbert spaces and later in the Banach or metric spaces setting. Without aiming for exhaustiveness, some of them are as follows: Chō et al. [20,21] introduced the so-called $(m,C)$- and $[m,C]$-isometric operators, where C is some conjugation of the Hilbert space. Motivated by m-isometries, Sid Ahmed generalized the notions of left and right inverses to m-left and m-right inverses, respectively, on Banach spaces [22]. The notions of m-left and m-right generalized inverse operators—which extend both m-left and m-right inverses and generalized inverses—have been studied on Banach spaces by Ezzahraoui [23]. Saddi et al. defined m-partial isometries on Hilbert spaces [24] and $(A,m)$-partial isometries in semi-Hilbertian spaces [25], as an extension of the former. Aouichaoui [26] introduced $(m,{\mathcal{N}}_A)$-isometries. Further extensions include (but are not limited to) $(B,n,p)$-isometric mappings on Banach spaces, for $p\in (0,\infty ]$ (see [27] and references therein), along with $\psi (m,q)$-isometric mappings on metric spaces [28]; n-quasi-$(m,C)$-isometric operators [29]; $(m,n)$-isosymmetric operators [30]; exponentially m-isometric operators [31]; and q-partial-$(m,A)$-isometries [32]. Another natural generalization of m-isometries are (spherical) m-isometric d-tuples, as introduced by Gleason and Richter [33]. Spherical m-isometries have attracted additional interest due to their relation to a moment problem [34]. The properties of this class have also been investigated in [35], while interesting examples can be found in [36]. In the last years, other single-variable operator families of m-isometries have been extended to d-tuples; see, e.g., [37,38,39,40].

Botelho et al. [41,42] characterized the elementary operators of length 1 acting on the Hilbert–Schmidt class of a separable Hilbert space that are 2-isometries and 3-isometries, and they conjectured the following: Given $p,q\in \mathbb{N}$, if there exists a nonzero scalar $\lambda$ such that $\lambda A$ is a p-isometry and ${\lambda }^{-1}{B}^{*}$ is a q-isometry, then the elementary operator $L\left(T\right)=ATB$ is a $(p+q-1)$-isometry. Duggal [43] and others subsequently confirmed this conjecture, cf. [44] and references therein. Later on, Gu [45] proved its converse—the elementary operator with symbols $A,B\in \mathcal{B}\left(H\right)$, acting on the Hilbert–Schmidt class of a separable Hilbert space, is a strict m-isometry if and only if there exists a constant $\lambda \ne 0$ such that $\lambda A$ is a strict p-isometry and ${\lambda }^{-1}{B}^{*}$ is a strict q-isometry, where $p+q-1=m$. Unlike the if part, which can be established through algebraic and combinatorial methods, the proof for the only if part uses analytical arguments relying on properties of the spectral radius and the approximate point spectrum of an operator, which can be traced back to Magajna [46]. This fact might explain why, as compared to Duggal’s —which has been versioned in almost all generalizations of the notion of m-isometry—Gu’s theorem has received less attention in the literature. Actually, to our knowledge, no analog of Gu’s necessary condition exists for the generalizations listed above.

The initial approach of Botelho and Jamison [41], based on a result by Fong and Sourour [47] (of wider applicability), does not seem to have been continued in later research on this topic. Aiming to provide a proof of the above results along the lines of [41], in [48], we introduced the concept of m-null pairs of operators and established some properties and characterizations of m-null elementary operators, which led to a generalization of Duggal’s theorem. Also, we raised the question of whether Gu’s necessity proof inherently required analytical techniques. In the present paper, we answer it in the negative by relying essentially on algebra and combinatorics to extend Gu’s theorem to m-null elementary operators. Our procedure, which involves explicitly solving a difference equation in terms of binomial coefficients, is quite constructive, and we hope that it provides deeper insight into the role and scope of Fong and Sourour’s theorem.

Investigations on the aforementioned extensions of m-isometries have employed a wide range of techniques, comprising combinatorial arguments, arithmetic progressions, Lagrange interpolation, properties of operator roots of polynomials, tensor products, factorization into left and right multiplication operators, and the hereditary functional calculus. Despite this methodological diversity, a number of such extensions are special cases of $(m,T)$-null pairs of operators (see Examples 1 to 10 below). Our approach provides a common framework for those generalizations, thereby enabling a unified treatment. In particular, a version of Gu’s theorem is immediately obtained for all those generalizations of m-isometries that can be expressed in terms of $(m,T)$-null pairs. Furthermore, the use of combinatorial and algebraic arguments, as opposed to analytic ones, is particularly well suited to more general contexts where conventional tools from spectral theory may be unavailable, such as general metric spaces or Banach spaces with a limited operator structure, a fact that suggests new research directions. Finally, the constructive nature of our approach not only strengthens the understanding of the mechanisms governing m-null elementary operators but also opens new pathways for computational applications, where explicit representations are essential.

The paper is structured as follows: In Section 2, we recall some previous material from [48], along with the theorem from [47] on which this research is based. The main result is achieved in Section 3 (Theorem 3 and Corollary 4). The last Section 4 briefly elaborates on the applications, versatility, limitations, and scope of our findings.

2. Preliminaries

Let H be a Hilbert space, and let $\mathcal{B}\left(H\right)$ denote the set of all bounded linear operators on H. Given $(A,B)\in \mathcal{B}\left(H\right)\times \mathcal{B}\left(H\right)$, the elementary operator of length 1 with symbols A and B is the operator $Q_{A,B}:\mathcal{B}\left(H\right)\to \mathcal{B}\left(H\right)$, defined by

$$Q_{A,B}\left(T\right):=ATB(T\in \mathcal{B}\left(H\right)).$$

It can be easily checked that the adjoint $Q_{A,B}^{*}$ of $Q_{A,B}$ is $Q_{A^{*},B^{*}}$. For simplicity, the elementary operator $Q_{I,I}$ will be represented by I, as this notation is unambiguous in context.

Definition 1.

Let m be a positive integer and $T\in \mathcal{B}\left(H\right)$. A pair of symbols $(A,B)\in \mathcal{B}\left(H\right)\times \mathcal{B}\left(H\right)$ is said to be $(m,T)$-null, provided that

$$\begin{array}{cc}\hfill P(A,B,m)\left(T\right)& :={(I-Q_{A,B})}^m\left(T\right)\hfill \\ \hfill & =\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)Q_{A,B}^j\left(T\right)\hfill \\ \hfill & =\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A^{j}T{B}^j=0.\hfill \end{array}$$

If this is true for all $T\in \mathcal{B}\left(H\right)$, then we shall just say that $(A,B)$ is m-null. The elementary operator $Q_{A,B}$ is called $(m,T)$-null, respectively m-null, when so is $(A,B)$. Either one of these conditions will be termed strict if it holds for m, but not for $m-1$.

The flat condition in Definition 1 can be lifted from $\mathcal{B}\left(H\right)$ to $\mathcal{B}\left(\mathcal{B}\right(H\left)\right)$, as follows:

Definition 2.

Let m be a positive integer and $T\in \mathcal{B}\left(H\right)$. A pair of superoperators $(A,B)\in \mathcal{B}\left(\mathcal{B}\right(H\left)\right)\times \mathcal{B}\left(\mathcal{B}\right(H\left)\right)$ is said to be $(m,T)$-null at $(R,S)\in \mathcal{B}\left(H\right)\times \mathcal{B}\left(H\right)$, provided that

$$\begin{array}{c}\hfill \sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A^j\left(R\right)T{B}^j\left(S\right)=0,\end{array}$$

(1)

where $A^j$ and $B^j$, respectively, denote the j-th iterates of A and B $(0\le j\le m)$. Alternatively, we will abbreviate this condition by saying that $\left(A\right(R),B(S\left)\right)$ is $(m,T)$-null. The pair $(A,B)$ is m-null at $(R,S)$ if (1) holds for all $T\in \mathcal{B}\left(H\right)$. Again, either condition will be termed strict if it holds for m, but not for $m-1$. When $R=S$, we shall just refer to nullity at S, for short.

Note that the m-nullity of $\left(A\right(R),B(S\left)\right)$ does not mean ${\sum }_{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A{\left(R\right)}^{j}TB{\left(S\right)}^j=0$, as $A\left(R\right),B\left(S\right)\in \mathcal{B}\left(H\right)$, and their iterates differ from those of $A,B\in \mathcal{B}\left(\mathcal{B}\right(H\left)\right)$. In particular, if $A\left(X\right)=Q_{A_1,A_2}\left(X\right)=A_{1}X{A}_2$ and $B\left(X\right)=Q_{B_2,B_1}\left(X\right)=B_{2}X{B}_1$$(X\in \mathcal{B}(H\left)\right)$ are elementary operators with symbols $A_1,A_2,B_1,B_2\in \mathcal{B}\left(H\right)$, we have $A^j\left(R\right)=A_1^{j}R{A}_2^j$ and $B^j\left(S\right)=B_2^{j}S{B}_1^j$$(0\le j\le m)$, so that (1) becomes the following:

$$\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A_1^{j}R{A}_2^{j}T{B}_2^{j}S{B}_1^j=0.$$

The flat condition in Definition 1 can be recovered from the lifted one in Definition 2 by choosing the superoperators $A\left(X\right)=Q_{A_1,I}\left(X\right)=A_{1}X$ and $B\left(X\right)=Q_{I,B_1}\left(X\right)=X{B}_1$$(X\in \mathcal{B}(H\left)\right)$ as left and right multiplications, respectively, and setting $R=S=I$.

Some examples illustrating the above are listed below.

Example 1

(Flat sense). Recall from [1,2,3] that an operator $S\in \mathcal{B}\left(H\right)$ is said to be an m-isometry if

$$\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)S^{*j}{S}^j=0$$

for some $m\in \mathbb{N}$. Thus, in the notation above, $S\in \mathcal{B}\left(H\right)$ is an m-isometry, provided that $P_{S^{*},S,m}\left(I\right)=0$. Also, the operator S is an m-isometry if the pair $(S^{*},S)$ is $(m,I)$-null, and a strict m-isometry if $(S^{*},S)$ is $(m,I)$-null, but not $(m-1,I)$-null.

Example 2

(Flat sense). An operator $S\in \mathcal{B}\left(H\right)$ is said to be $(m,C)$-isometric [20] if there exists some conjugation C such that

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

for some $m\in \mathbb{N}$. If C is a conjugation and $S\in \mathcal{B}\left(H\right)$, then $C{S}^{j}C={\left(CSC\right)}^j$ for every $j\in \mathbb{N}$. Thus, S is $(m,C)$-isometric if the pair $(S^{*},CSC)$ is $(m,I)$-null.

Example 3

(Flat sense). An operator $S\in \mathcal{B}\left(H\right)$ is said to be $[m,C]$-isometric [21] if there exists some conjugation C such that

$$\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)\left(C{S}^{j}C\right)S^j=0$$

for some $m\in \mathbb{N}$. Thus, S is $[m,C]$-isometric if the pair $(CSC,S)$ is $(m,I)$-null.

Example 4

(Flat sense). The notions of m-left and m-right invertible operators were introduced in [22]. Let $m\in \mathbb{N}$. An operator $B\in \mathcal{B}\left(H\right)$ is m-left invertible if there exists $A\in \mathcal{B}\left(H\right)$ such that

$$\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A^j{B}^j=0.$$

(2)

Similarly, $A\in \mathcal{B}\left(H\right)$ is m-right invertible if there exists $B\in \mathcal{B}\left(H\right)$ such that (2) holds. Thus, B is m-left invertible (and A is m-right invertible) if the pair $(A,B)$ is $(m,I)$-null.

Example 5

(Lifted sense). The notions of m-left and m-right generalized inverses were introduced in [23]. Let $m\in \mathbb{N}$ and $S,R\in \mathcal{B}\left(H\right)$. We say that S is an m-left generalized inverse of R if

$$R\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)S^j{R}^j=0.$$

Similarly, S is an m-right generalized inverse of R if

$$S\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)R^j{S}^j=0.$$

Introducing the elementary operators $M\left(X\right)=SXR$ and $N\left(X\right)=RXS$$(X\in \mathcal{B}(H\left)\right)$, we obtain that S is an m-left generalized inverse of R if the pair $(I,M)$ is $(m,R)$-null at I, and an m-right generalized inverse of R if the pair $(I,N)$ is $(m,S)$-null at I.

Example 6

(Lifted sense). An operator $S\in \mathcal{B}\left(H\right)$ is an m-partial isometry [24] if

$$S\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)S^{*j}{S}^j=0$$

for some $m\in \mathbb{N}$. Introducing the elementary operator $M\left(X\right)=S^{*}XS$$(X\in \mathcal{B}(H\left)\right)$, we obtain that S is an m-partial isometry if the pair $(I,M)$ is $(m,S)$-null at I.

Example 7

(Lifted sense). Let $A\in \mathcal{B}\left(H\right)$ be a positive operator and $m\in \mathbb{N}$. An operator $S\in \mathcal{B}\left(H\right)$ is an $(A,m)$-partial isometry [25] if

$$S\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)S^{*j}A{S}^j=0.$$

Introducing the elementary operator $M\left(X\right)=S^{*}XS$$(X\in \mathcal{B}(H\left)\right)$, we obtain that S is an $(A,m)$-partial isometry if the pair $(I,M)$ is $(m,S)$-null at $(I,A)$.

Example 8

(Lifted sense). Let $A\in \mathcal{B}\left(H\right)$ and $m\in \mathbb{N}$. An operator $S\in \mathcal{B}\left(H\right)$ is said to be an $\left(m,{\mathcal{N}}_A\right)$-isometry [26] if

$$A\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)S^{*j}{S}^j=0.$$

Introducing the elementary operator $M\left(X\right)=S^{*}XS$$(X\in \mathcal{B}(H\left)\right)$, we obtain that S is an $\left(m,{\mathcal{N}}_A\right)$-isometry if the pair $(I,M)$ is $(m,A)$-null at I.

Example 9

(Flat sense). Let $m,n\in \mathbb{N}$. An operator $S\in \mathcal{B}\left(H\right)$ is said to be $(m,n)$-isosymmetric [30] if

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

Thus, S is $(m,n)$-isosymmetric if the pair $(S^{*},S)$ is $(m,T)$-null, with

$$T=\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)S^{*(n-k)}{S}^k.$$

Example 10

(Flat sense). An operator $S\in \mathcal{B}\left(H\right)$ is exponentially m-isometric [31] if

$$\sum _{j=0}^m{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)e^{j{S}^{*}}{e}^{jS}=0$$

for some $m\in \mathbb{N}$. Thus, S is exponentially m-isometric if the pair $(e^{S^{*}},e^S)$ is $(m,I)$-null.

The next result is ([48], Proposition 3). We include a (partially different) proof for the sake of completeness.

Proposition 1.

Let $m\in \mathbb{N}$ and $A,B,T\in \mathcal{B}\left(H\right)$. The pair $(A,B)$ is $(m,T)$-null if and only if

$$Q_{A,B}^k\left(T\right)=\sum _{i=0}^{m-1}{(-1)}^{m-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k-i-1}{m-i-1}}\right)Q_{A,B}^i\left(T\right)(k\in {\mathbb{N}}_0).$$

(3)

Proof. 

Fix $k\in {\mathbb{N}}_0$ and assume that $(A,B)$ is $(m,T)$-null. Since ${(Q_{A,B}-I)}^j\left(T\right)=0$ for $j\ge m$, we obtain:

$$\begin{array}{cc}\hfill Q_{A,B}^k\left(T\right)& ={\left[\left(Q_{A,B}-I\right)+I\right]}^k\left(T\right)\hfill \\ \hfill & =\sum _{j=0}^k\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){(Q_{A,B}-I)}^j\left(T\right)\hfill \\ \hfill & =\sum _{j=0}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\sum _{i=0}^j{(-1)}^{j-i}\left({\displaystyle \genfrac{}{}{0pt}{}{j}{i}}\right)Q_{A,B}^i\left(T\right)\hfill \\ \hfill & =\sum _{i=0}^{m-1}\sum _{j=i}^{m-1}{(-1)}^{j-i}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{j}{i}}\right)Q_{A,B}^i\left(T\right)\hfill \\ \hfill & =\sum _{i=0}^{m-1}{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)\sum _{j=i}^{m-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{j-i}}\right)Q_{A,B}^i\left(T\right).\hfill \end{array}$$

By ([49], Corollary 2.3):

$$\begin{array}{cc}\hfill \sum _{j=i}^{m-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{j-i}}\right)& =\sum _{j=0}^{m-i-1}{(-1)}^{j-i}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{j}}\right)\hfill \\ \hfill & ={(-1)}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i-1}{m-i-1}}\right)(0\le i\le m-1).\hfill \end{array}$$

Thus, (3) follows.

For the converse, it suffices to particularize $k=m$ in (3). □

Proposition 1 shows that the sequence of powers ${\left\{Q_{A,B}^k\right\}}_{k=0}^{\infty }$ of an $(m,T)$-null operator $Q_{A,B}$ with symbols $A,B\in \mathcal{B}\left(H\right)$ belongs to the class of operator arithmetic progressions of order $m-1$, consisting of all those operator sequences ${\left\{V_k\right\}}_{k=0}^{\infty }$ whose $(m-1)$-th differences are constant (and whose m-th differences are therefore null). In fact, for $x\in \mathbb{R}$, the expression

$$F\left(x\right):=\sum _{i=0}^{m-1}\sum _{j=i}^{m-1}{(-1)}^{j-i}\left({\displaystyle \genfrac{}{}{0pt}{}{x}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{j}{i}}\right)V_i,$$

where

$$\left({\displaystyle \genfrac{}{}{0pt}{}{x}{j}}\right):={\displaystyle \frac{x(x-1)\cdots (x-j+1)}{j!}},$$

defines a polynomial in x of degree at most $m-1$ with coefficients in $\mathcal{B}\left(H\right)$, such that $V_k=F\left(k\right)$ for all $k\in {\mathbb{N}}_0$ (see, e.g., [50]).

The following two results were also established in [48]. The second one is essentially ([43], Corollary 2.11), while the first one extends this result to $(m,T)$-null operators.

Theorem 1

([48], Theorem 1). Let $p,q\in \mathbb{N}$, let $S\in \mathcal{B}\left(H\right)$, and assume that the pairs $(A_1,A_2)$ and $(B_2,B_1)$ of operators in $\mathcal{B}\left(H\right)$ are $(p,S)$-null and $(q,S)$-null, respectively. Define the elementary operators

$$A\left(S\right):=Q_{A_1,A_2}\left(S\right)=A_{1}S{A}_2,B\left(S\right):=Q_{B_2,B_1}\left(S\right)=B_{2}S{B}_1.$$

Then, the pair $\left(A\left(S\right),B\left(S\right)\right)$ is $(p+q-1)$-null, that is,

$$\sum _{k=0}^{p+q-1}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p+q-1}{k}}\right)A_1^{k}S{A}_2^{k}T{B}_2^{k}S{B}_1^k=0(T\in \mathcal{B}\left(H\right)).$$

Corollary 1

([48], Corollary 1). Let $p,q\in \mathbb{N}$ and $M,N\in \mathcal{B}\left(H\right)$. Assume that M is a p-isometry and $N^{*}$ is a q-isometry. Then, the operator $L\left(T\right)=MTN$$(T\in \mathcal{B}(H\left)\right)$ is a $(p+q-1)$-isometry.

The theorem of Fong and Sourour on which our main results are based is stated next.

Lemma 1

([47], Theorem 1). Let ${\left\{M_k\right\}}_{k=1}^m$ and ${\left\{N_k\right\}}_{k=1}^m$ be bounded operators on the Hilbert space H, and let Φ be the operator given by

$$\Phi \left(T\right)=\sum _{k=1}^m{M}_{k}T{N}_k(T\in \mathcal{B}\left(H\right)),$$

where not all the $M_k$’s equal 0. If $\Phi \left(T\right)=0$ for all $T\in \mathcal{B}\left(H\right)$, then $\left\{N_1,N_2,\dots ,N_m\right\}$ is linearly dependent. Furthermore, if $\left\{N_1,N_2,\dots ,N_q\right\}$$(q\le m)$ is a maximal linearly independent subset of $\left\{N_1,N_2,\dots ,N_m\right\}$, and $c_{ij}$$(j,i\in \mathbb{N},1\le j\le q,q+1\le i\le m)$ denote constants for which

$$N_i=\sum _{j=1}^q{c}_{ij}{N}_j(q+1\le i\le m),$$

(4)

then $\Phi \left(T\right)=0$ for all $T\in \mathcal{B}\left(H\right)$ if and only if

$$M_j=-\sum _{i=q+1}^m{c}_{ij}{M}_i(1\le j\le q).$$

(5)

(In case $m=q$, identity (4) becomes vacuous and condition (5) should be interpreted as $M_1=M_2=\dots =M_m=0$).

The following derivation of ([48], Lemma 3) will be useful in the sequel.

Lemma 2.

Let $m\in \mathbb{N}$, and let $A,B,T\in \mathcal{B}\left(H\right)$. If $Q_{A,B}$ is strictly $(m,T)$-null, then the set of powers ${\left\{Q_{A,B}^k\left(T\right)\right\}}_{k=0}^{m-1}$ is linearly independent.

We end this section by recalling one of the main results from [48]. Although it was fully proved there, we find it instructive to provide a new, explicit proof of the converse statement in Part $\left(iii\right)$ and highlight some if its consequences. The proof of the converse to Part $\left(iii\right)$ will be given as Proposition 2 below.

Theorem 2

([48], Theorem 2). Notation is as in Theorem 1. Fix $m\in \mathbb{N}$ and $S\in \mathcal{B}\left(H\right)$. Let the elementary operator $Q_S$ be given by the following:

$$\begin{array}{cc}\hfill Q_S\left(T\right)& :=Q_{A\left(S\right),B\left(S\right)}\left(T\right)\hfill \\ \hfill & =A\left(S\right)TB\left(S\right)\hfill \\ \hfill & =Q_{A_1,A_2}\left(S\right)T{Q}_{B_2,B_1}\left(S\right)(T\in \mathcal{B}\left(H\right)).\hfill \end{array}$$

(i) 

Further, let $q\in \mathbb{N}$ be such that ${\left\{B_2^{j}S{B}_1^j\right\}}_{j=0}^{q-1}$ is a maximal linearly independent subset of the set ${\left\{B_2^{k}S{B}_1^k\right\}}_{k=0}^m$, so that

$$B_2^{i}S{B}_1^i=\sum _{j=0}^{q-1}{c}_{ij}{B}_2^{j}S{B}_1^j(q\le i\le m)$$

(6)

for some constants $c_{ij}$$(j,i\in {\mathbb{N}}_0,0\le j\le q-1,q\le i\le m)$. Then, $Q_S$ is m-null if and only if

$$\begin{array}{cc}\hfill & {(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A_1^{j}S{A}_2^j\hfill \\ \hfill & =-\sum _{i=q}^m{(-1)}^i{c}_{ij}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A_1^{i}S{A}_2^i(0\le j\le q-1).\hfill \end{array}$$

(7)

(ii) 

If q is as in Part $\left(i\right)$, then $(B_2,B_1)$ is $(q,S)$-null if and only if

$$\begin{array}{cc}\hfill c_{ij}& ={(-1)}^{q-j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j-1}{q-j-1}}\right)\hfill \\ \hfill & ={(-1)}^{q-j-1}{\displaystyle \frac{q-j}{i-j}}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{j}}\right)(0\le j\le q-1,q\le i\le m).\hfill \end{array}$$

(iii) 

Let q be as in Part $\left(i\right)$, let $p=m-q+1$, and assume $Q_S$ is m-null. If $(B_2,B_1)$ is (necessarily strictly) $(q,S)$-null, then $(A_1,A_2)$ is $(p,S)$-null. Conversely, if $(A_1,A_2)$ is strictly $(p,S)$-null, then $(B_2,B_1)$ is $(q,S)$-null.

It is worth noting that Theorem 2 can be particularized to m-isometries.

Corollary 2.

Fix $m\in \mathbb{N}$ and $M,N\in \mathcal{B}\left(H\right)$.

(i) 

Let $q\in \mathbb{N}$ be such that ${\left\{N^j{N}^{*j}\right\}}_{j=0}^{q-1}$ is a maximal linearly independent subset of the set ${\left\{N^k{N}^{*k}\right\}}_{k=0}^m$, so that

$$N^i{N}^{*i}=\sum _{j=0}^{q-1}{c}_{ij}{N}^j{N}^{*j}(q\le i\le m)$$

for some constants $c_{ij}$$(j,i\in {\mathbb{N}}_0,0\le j\le q-1,q\le i\le m)$. Then, $L\left(T\right)=MTN$$(T\in \mathcal{B}(H\left)\right)$ is an m-isometry if and only if

$$\begin{array}{cc}\hfill & {(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)M^{*j}{M}^j\hfill \\ \hfill & =-\sum _{i=q}^m{(-1)}^i{c}_{ij}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)M^{*i}{M}^i(0\le j\le q-1).\hfill \end{array}$$

(ii) 

If q is as in Part $\left(i\right)$, then $N^{*}$ is a q-isometry if and only if

$$\begin{array}{cc}\hfill c_{ij}& ={(-1)}^{q-j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j-1}{q-j-1}}\right)\hfill \\ \hfill & ={(-1)}^{q-j-1}{\displaystyle \frac{q-j}{i-j}}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{j}}\right)(0\le j\le q-1,q\le i\le m).\hfill \end{array}$$

(iii) 

Let q be as in Part $\left(i\right)$, let $p=m-q+1$, and assume $L\left(T\right)=MTN$$(T\in \mathcal{B}(H\left)\right)$ is an m-isometry. If $N^{*}$ is a (necessarily strict) q-isometry, then M is a p-isometry. Conversely, if M is a strict p-isometry, then $N^{*}$ is a q-isometry.

Proof. 

It is enough to apply Theorem 2, with $S=I$, $\left(A_1,A_2\right)=\left(M^{*},M\right)$, and $\left(B_2,B_1\right)=\left(N,N^{*}\right)$. □

The next corollary clarifies the role of the strictness conditions.

Corollary 3.

Notation is as in Theorems 1 and 2. If $(B_2,B_1)$ is (strictly) $(q,S)$-null, then $Q_S$ is strictly m-null if and only if $(A_1,A_2)$ is strictly $(p,S)$-null. Similarly, if $(A_1,A_2)$ is strictly $(p,S)$-null, then $Q_S$ is strictly m-null if and only if $(B_2,B_1)$ is (strictly) $(q,S)$-null.

Proof. 

The property of $(B_2,B_1)$ being $(q,S)$-null must be strict; otherwise, Proposition 1 would render $B_2^{q-1}S{B}_1^{q-1}$ as a linear combination of ${\left\{B_2^{j}S{B}_1^j\right\}}_{j=0}^{q-2}$, contradicting the linear independence of ${\left\{B_2^{j}S{B}_1^j\right\}}_{j=0}^{q-1}$. Since the proof of the second part can be reduced to that of the first part, we will concentrate on the latter. Thus, assume $(B_2,B_1)$ is $(q,S)$-null. From Part $\left(iii\right)$ of Theorem 2, we find that $(A_1,A_2)$ is $(p,S)$-null.

If $(A_1,A_2)$ were $(r,S)$-null for some $r<p$ then, by Theorem 1, $Q_S$ would be n-null with $n=r+q-1<p+q-1=m$, thereby preventing $Q_S$ from being strictly m-null.

Conversely, let $Q_S$ be n-null for some $n<m$. As ${\left\{B_2^{j}S{B}_1^j\right\}}_{j=0}^{q-1}$ is a maximal linearly independent subset of ${\left\{B_2^{k}S{B}_1^k\right\}}_{k=0}^m$, Lemma 1 yields $r\in \mathbb{N}$, with $n=r+q-1$, such that Part $\left(i\right)$ of Theorem 2 holds for n instead of m. But then, by Part $\left(iii\right)$ of the same theorem, $(A_1,A_2)$ is $(r,S)$-null with $r=n-q+1<m-q+1=p$, so that $(A_1,A_2)$ is not strictly p-null. □

3. Main Results

In order to provide an explicit proof of the converse to Part $\left(iii\right)$ of Theorem 2, in the following two lemmas, we analyze the recurrence relation satisfied by the coefficients appearing in Part $\left(i\right)$ of that theorem. For notational simplicity, the analysis will be performed in a more general context.

Lemma 3.

Assume $S\in \mathcal{B}\left(H\right)$ is such that

$$S^i=\sum _{j=0}^{q-1}{c}_{ij}{S}^j(q\le i\le p+q-1,0\le j\le q-1).$$

Then, the double sequence of coefficients $\{c_{ij}:q\le i\le p+q-1,0\le j\le q-1\}$ satisfies the recurrence relation

$$c_{i+1,j}=c_{i,j-1}+c_{i,q-1}{c}_{qj}(q\le i\le p+q-2,0\le j\le q-1),$$

(8)

provided that $c_{i,-1}=0$$(q\le i\le p+q-1)$.

Proof. 

Fix $q\le i\le p+q-2$. On the one hand,

$$S^{i+1}=\sum _{j=0}^{q-1}{c}_{i+1,j}{S}^j.$$

(9)

On the other hand,

$$\begin{array}{cc}\hfill S^{i+1}& =S{S}^i=S\left(\sum _{j=0}^{q-1}{c}_{ij}{S}^j\right)=\sum _{j=0}^{q-1}{c}_{ij}{S}^{j+1}=\sum _{j=1}^q{c}_{i,j-1}{S}^j\hfill \\ \hfill & =\sum _{j=1}^{q-1}{c}_{i,j-1}{S}^j+c_{i,q-1}{S}^q=\sum _{j=1}^{q-1}{c}_{i,j-1}{S}^j+c_{i,q-1}\sum _{j=0}^{q-1}{c}_{qj}{S}^j\hfill \\ \hfill & =c_{i,q-1}{c}_{q0}I+\sum _{j=1}^{q-1}(c_{i,j-1}+c_{i,q-1}{c}_{qj})S^j.\hfill \end{array}$$

(10)

By linear independence, a comparison between (9) and (10) yields the following:

$$\begin{array}{cc}\hfill & c_{i+1,0}=c_{i,q-1}{c}_{q0},\hfill \\ \hfill & c_{i+1,j}=c_{i,j-1}+c_{i,q-1}{c}_{qj}.\hfill \end{array}$$

Setting $c_{i,-1}=0$, we obtain (8). □

Lemma 4.

The solution to the recurrence relation (8) with initial conditions

$$c_{i,-1}=0,c_{i,q-1}=\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)(q\le i\le p+q-1)$$

(11)

is the sequence

$$b_{ij}={(-1)}^{q-j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j-1}{q-j-1}}\right)(q\le i\le p+q-1,0\le j\le q-1).$$

(12)

Proof. 

Write (8) in the form:

$$c_{i,j}=c_{i+1,j+1}-c_{i,q-1}{c}_{q,j+1}(q\le i\le p+q-2,0\le j\le q-2).$$

(13)

Since

$$\left({\displaystyle \genfrac{}{}{0pt}{}{i}{-1}}\right)=0,\left({\displaystyle \genfrac{}{}{0pt}{}{i-q}{0}}\right)=1(q\le i\le p+q-1),$$

it is apparent that the proposed solution (12) satisfies the initial conditions (11) for all $q\le i\le p+q-1$ and $j=q-1$. Let us verify by induction that (12) also satisfies (13) for any $q\le i\le p+q-2$ and $j=q-2$. The base case $j=q-1$, along with the Pascal formula, leads to:

$$\begin{array}{cc}\hfill c_{i,q-2}& =c_{i+1,q-1}-c_{i,q-1}{c}_{q,q-1}\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{i+1}{q-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{i+1}{q-1}}\right)-q\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)-q\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)(1-q)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)\hfill \\ \hfill & =-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)(i-q+2)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)=-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)(i-q-1).\hfill \end{array}$$

On the other hand, for the proposed solution, we obtain:

$$b_{i,q-2}=-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+1}{1}}\right)=-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-2}}\right)(i-q+1).$$

This disposes of this induction step. Assuming now that (12) satisfies (8) for all $q\le i\le p+q-2$ and $j=q-k$, with $0\le k\le q-1$, let us prove the same for all $q\le i\le p+q-2$ and $j=q-k-1$. Again, by the inductive hypothesis and the Pascal identity,

$$\begin{array}{cc}\hfill c_{i,q-k-1}& =c_{i+1,q-k}-c_{i,q-1}{c}_{q,q-k}\hfill \\ \hfill & ={(-1)}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i+1}{q-k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)-{(-1)}^{k-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{k}}\right)\hfill \\ \hfill & ={(-1)}^{k-1}\left\{\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\right]\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{k}}\right)\right\}\hfill \\ \hfill & ={(-1)}^{k-1}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)\right.\hfill \\ \hfill & +\left.\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{k}}\right)\right]\hfill \\ \hfill & ={(-1)}^{k-1}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q-1}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{k}}\right)\right]\hfill \\ \hfill & ={(-1)}^{k-1}\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left[\left({\displaystyle \genfrac{}{}{0pt}{}{q-1}{k-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{q}{k}}\right)\right]+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)\right\}\hfill \\ \hfill & ={(-1)}^k\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q-1}{k}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k-1}}\right)\right]\hfill \\ \hfill & ={(-1)}^k\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q-1}{k}}\right)\right.\hfill \\ \hfill & -\left.\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k+1}{k}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k}}\right)\right]\right\}\hfill \\ \hfill & ={(-1)}^k\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q-1}{k}}\right)\right.\hfill \\ \hfill & -\left.\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k+1}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k}}\right)\right]\hfill \\ \hfill & ={(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k}}\right).\hfill \end{array}$$

On the other hand, for the proposed solution, we obtain:

$$b_{i,q-k-1}={(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-k-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-q+k}{k}}\right).$$

At this point, the induction process shows that $c_{ij}=b_{ij}$ for all $q\le i\le p+q-2$ and $0\le j\le q-1$. But then, by (8), equality holds for $i=p+q-1$ as well, as shown by the following:

$$\begin{array}{cc}\hfill c_{p+q-1,j}& =c_{p+q-2,j-1}+c_{p+q-2,q-1}{c}_{q,j}\hfill \\ \hfill & =b_{p+q-2,j-1}+b_{p+q-2,q-1}{b}_{q,j}\hfill \\ \hfill & =b_{p+q-1,j}(0\le j\le q-1).\hfill \end{array}$$

The proof is thus complete. □

We are now in a position to provide the announced explicit proof of the converse statement in Part $\left(iii\right)$ of Theorem 2. The notation introduced will be retained.

Proposition 2.

Assume $Q_S$ is m-null, and let $p,q$ be as in Part $\left(iii\right)$ of Theorem 2. If $(A_1,A_2)$ is strictly $(p,S)$-null, then $(B_2,B_1)$ is $(q,S)$-null.

Proof. 

Since the pair $(A_1,A_2)$ is strictly $(p,S)$-null, the set of powers ${\left\{A^i\left(S\right)\right\}}_{i=0}^{p-1}$ is linearly independent (Lemma 2). Now, $Q_S$ being m-null, Part $\left(i\right)$ of Theorem 2 yields

$$A^j\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{q+i-j-1}{c}_{q+i,j}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)}^{-1}{A}^{q+i}\left(S\right)$$

(14)

for all $0\le j\le q-1$. Particularize $j=0$ in (14) to obtain

$$\begin{array}{c}\hfill S=R{A}^q\left(S\right)=A^{q}R\left(S\right),\end{array}$$

where

$$R:=\sum _{i=0}^{p-1}{(-1)}^{q+i-1}{c}_{q+i,0}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i}}\right)A^i.$$

Specializing $j=q-1$ in (14) and shifting indices, we also obtain

$$\begin{array}{cc}\hfill A^q\left(S\right)& =A{A}^{q-1}\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^i{c}_{q+i,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)}^{-1}{A}^{q+i+1}\left(S\right)\hfill \\ \hfill & =\sum _{i=1}^p{(-1)}^{i-1}{c}_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)}^{-1}{A}^{q+i}\left(S\right),\hfill \end{array}$$

or

$$\left[I+\sum _{i=1}^p{(-1)}^i{c}_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)}^{-1}{A}^i\right]A^q\left(S\right)=0.$$

As R commutes with the operator in brackets, it follows that

$$\begin{array}{cc}\hfill 0& =R\left[I+\sum _{i=1}^p{(-1)}^i{c}_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)}^{-1}{A}^i\right]A^q\left(S\right)\hfill \\ \hfill & =\left[I+\sum _{i=1}^p{(-1)}^i{c}_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)}^{-1}{A}^i\right]R{A}^q\left(S\right)\hfill \\ \hfill & =S+\sum _{i=1}^p{(-1)}^i{c}_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)}^{-1}{A}^i\left(S\right).\hfill \end{array}$$

On the other hand, $(A_1,A_2)$ being $(p,S)$-null,

$$A^p\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)A^i\left(S\right).$$

Therefore, writing $m=p+q-1$,

$$\begin{array}{cc}\hfill 0& =\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)S+\sum _{i=1}^{p-1}{(-1)}^i{c}_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)A^i\left(S\right)\hfill \\ \hfill & +{(-1)}^p{c}_{m,q-1}\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)A^i\left(S\right)\hfill \\ \hfill & =\left[\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)-c_{m,q-1}\right]A^0\left(S\right)\hfill \\ \hfill & +\sum _{i=1}^{p-1}{(-1)}^i\left[c_{q+i-1,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)-c_{m,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)\right]A^i\left(S\right).\hfill \end{array}$$

All the brackets in this expression must vanish, because the set ${\left\{A^i\left(S\right)\right\}}_{i=0}^{p-1}$ is linearly independent. Consequently,

$$c_{m,q-1}=\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right),$$

and, for $1\le i\le p-1$,

$$\begin{array}{cc}\hfill c_{q+i-1,q-1}& =c_{m,q-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)}^{-1}=\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)}^{-1}\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{m}{p}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{m}{p-i}}\right)}^{-1}=\left({\displaystyle \genfrac{}{}{0pt}{}{m-p+i}{i}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{q+i-1}{q-1}}\right).\hfill \end{array}$$

To complete the proof, it suffices to invoke Lemma 4, along with Part $\left(ii\right)$ of Theorem 2. □

In ([48], Section 4), the question was raised whether a converse of Theorem 1—hence, the m-isometric version of this converse ([45], Theorem 7)—could be proven without resorting to analytic tools. Theorem 3 answers this question in the affirmative. A previous result is required.

Lemma 5.

Notation is as in Theorem 2. If $Q_S$ is strictly m-null, then $c_{i,q-1}\ne 0$ and $c_{i0}\ne 0$$(q\le i\le m)$. Moreover, the set of powers ${\left\{A^k\left(S\right)\right\}}_{k=0}^{p-1}$ is linearly independent.

Proof. 

To prove the first assertion, assume that $c_{r,q-1}=0$ for some $r\in \mathbb{N}$ with $q\le r\le m$. Let $T\in \mathcal{B}\left(H\right)$. The m-nullity of $Q_S$ yields:

$$\begin{array}{cc}\hfill \sum _{j=0}^{q-1}& {(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A^j\left(S\right)T{B}^j\left(S\right)+\sum _{i=q}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T{B}^i\left(S\right)\hfill \\ \hfill & =\sum _{k=0}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)A^k\left(S\right)T{B}^k\left(S\right)=0.\hfill \end{array}$$

(15)

Substitute the relation (7),

$${(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A^j\left(S\right)=-\sum _{i=q}^m{(-1)}^i{c}_{ij}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)(0\le j\le q-1),$$

into the first sum on the right-hand side of (15), to obtain:

$$\begin{array}{cc}\hfill \sum _{j=0}^{q-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)A^j\left(S\right)T{B}^j\left(S\right)& =\sum _{j=0}^{q-1}\left[-\sum _{i=q}^m{(-1)}^i{c}_{ij}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)\right]T{B}^j\left(S\right)\hfill \\ \hfill & =-\sum _{i=q}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T\sum _{j=0}^{q-1}{c}_{ij}{B}^j\left(S\right).\hfill \end{array}$$

Inserting this one back into (15) results in the following equation:

$$-\sum _{i=q}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T\sum _{j=0}^{q-1}{c}_{ij}{B}^j\left(S\right)+\sum _{i=q}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T{B}^i\left(S\right)=0.$$

Since $c_{r,q-1}=0$ (where $q\le r\le m$), and, by (6),

$$B^i\left(S\right)=\sum _{j=0}^{q-1}{c}_{ij}{B}^j\left(S\right)(q\le i\le m),$$

we infer that

$$\begin{array}{cc}\hfill & -\sum _{\begin{array}{c}i=q\\ i\ne r\end{array}}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T\sum _{j=0}^{q-1}{c}_{ij}{B}^j\left(S\right)\hfill \\ \hfill & +\sum _{\begin{array}{c}i=q\\ i\ne r\end{array}}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T{B}^i\left(S\right)+{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)A^r\left(S\right)T{B}^r\left(S\right)\hfill \\ \hfill & =\sum _{\begin{array}{c}i=q\\ i\ne r\end{array}}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)A^i\left(S\right)T\left[B^i\left(S\right)-\sum _{j=0}^{q-1}{c}_{ij}{B}^j\left(S\right)\right]\hfill \\ \hfill & +{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)A^r\left(S\right)T{B}^r\left(S\right)\hfill \\ \hfill & ={(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)A^r\left(S\right)T{B}^r\left(S\right)=0,\hfill \end{array}$$

where $q\le r\le m$. If $q\le r\le m-1$, this conclusion prevents ${\left\{A^k\left(S\right)T{B}^k\left(S\right)\right\}}_{k=0}^{m-1}$ from being linearly independent, which, in view of Lemma 2, contradicts our hypothesis that $Q_S$ is strictly m-null. If $r=m$, then (15) provides a zero linear combination of ${\left\{A^k\left(S\right)T{B}^k\left(S\right)\right\}}_{k=0}^{m-1}$ with nonzero coefficients, violating again the linear independence of this set of powers. Consequently, $c_{i,q-1}\ne 0$ for all $q\le i\le m$. That $c_{i0}\ne 0$$(q\le i\le m)$ now follows from (8) and the fact that $c_{q0}$ must be nonzero, to avoid a contradiction with the case $j=0$ of Equation (14).

To prove the second statement, assume that some $r\in \mathbb{N}$ is such that

$$A^r\left(S\right)=\sum _{\begin{array}{c}k=q\\ k\ne r\end{array}}^m{\lambda }_k{A}^k\left(S\right)$$

for suitable scalars ${\lambda }_k$$(q\le k\le m,k\ne r)$. Substituting this identity and (6) in (15) yields the following:

$$\begin{array}{cc}\hfill 0& =\sum _{k=0}^{q-1}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)A^k\left(S\right)T{B}^k\left(S\right)+\sum _{\begin{array}{c}k=q\\ k\ne r\end{array}}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)A^k\left(S\right)T{B}^k\left(S\right)\hfill \\ \hfill & +{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)A^r\left(S\right)T{B}^r\left(S\right)\hfill \\ \hfill & =\sum _{k=0}^{q-1}{A}^k\left(S\right)T\left[{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B^k\left(S\right)\right]+\sum _{\begin{array}{c}k=q\\ k\ne r\end{array}}^m{A}^k\left(S\right)T\left[{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B^k\left(S\right)\right]\hfill \\ \hfill & +{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)\sum _{\begin{array}{c}k=q\\ k\ne r\end{array}}^m{\lambda }_k{A}^k\left(S\right)T{B}^r\left(S\right)\hfill \\ \hfill & =\sum _{k=0}^{q-1}{A}^k\left(S\right)T\left[{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B^k\left(S\right)\right]\hfill \\ \hfill & +\sum _{\begin{array}{c}k=q\\ k\ne r\end{array}}^m{A}^k\left(S\right)T\left[{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B^k\left(S\right)+{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right){\lambda }_k{B}^r\left(S\right)\right]\hfill \\ \hfill & =\sum _{\begin{array}{c}k=0\\ k\ne r\end{array}}^m{A}^k\left(S\right)T{V}_k\left(S\right),\hfill \end{array}$$

where

$$V_k\left(S\right):={(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B^k\left(S\right)$$

if $0\le k\le q-1$, and

$$\begin{array}{cc}\hfill V_k\left(S\right)& :={(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B^k\left(S\right)+{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right){\lambda }_k{B}^r\left(S\right)\hfill \\ \hfill & =\sum _{j=0}^{q-1}\left[{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)c_{kj}+{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right){\lambda }_k{c}_{rj}\right]B^j\left(S\right)\hfill \end{array}$$

if $q\le k\le m$, $k\ne r$. Since ${\left\{V_j\left(S\right)\right\}}_{j=0}^{q-1}$ is a maximal linearly independent subset of ${\left\{V_k\left(S\right)\right\}}_{k=0,k\ne r}^m$, the Fong–Sourour theorem (Lemma 1 above) entails that

$$A^j\left(S\right)=-\sum _{\begin{array}{c}i=q\\ i\ne r\end{array}}^m\left[{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)c_{ij}+{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right){\lambda }_i{c}_{rj}\right]A^i\left(S\right)(0\le j\le q-1).$$

For $0\le j\le q-1$, combining this with (7) allows us to obtain:

$$\begin{array}{cc}\hfill A^j\left(S\right)& =-\sum _{\begin{array}{c}i=q\\ i\ne r\end{array}}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)c_{ij}{A}^i\left(S\right)-{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)c_{rj}{A}^r\left(S\right)\hfill \\ \hfill & =-{(-1)}^j{\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)}^{-1}\sum _{\begin{array}{c}i=q\\ i\ne r\end{array}}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)c_{ij}{A}^i\left(S\right)\hfill \\ \hfill & -{(-1)}^j{\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)}^{-1}{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r}}\right)c_{rj}{A}^r\left(S\right),\hfill \end{array}$$

whence

$$\left[{(-1)}^j{\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)}^{-1}-1\right]\sum _{\begin{array}{c}i=q\end{array}}^m{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{i}}\right)c_{ij}{A}^i\left(S\right)=0.$$

A comparison with (7) shows that either $A^j\left(S\right)=0$, or

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right)={(-1)}^j.$$

The second possibility can hold only if $j=0$ or $j=m$ and m is even. This means that, unless $q=1$ or $q=m$, we must have $A^j\left(S\right)=0$ whenever $0\le j\le q-1$. But this contradicts the strict m-nullity of $Q_S$ and proves that ${\left\{A^k\left(S\right)\right\}}_{k=0}^{p-1}$ is linearly independent, as asserted.

If $q=1$, so that $m=p$, Equation (6) yields the following:

$$B^k\left(S\right)=\sum _{j=0}^{q-1}{c}_{kj}{B}^j\left(S\right)=c_{k0}S(1=q\le k\le m=p).$$

In particular, $B\left(S\right)=c_{10}S$ implies $B^k\left(S\right)=c_{10}^{k}S$ for all $0\le k\le p$. As $Q_S$ is strictly p-null,

$$\begin{array}{cc}\hfill 0& =\sum _{k=0}^p{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)A^k\left(S\right)T{B}^k\left(S\right)=\sum _{k=0}^p{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)A^k\left(S\right)T{c}_{10}^{k}S\hfill \\ \hfill & =\sum _{k=0}^p{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right){\left(c_{10}A\right)}^k\left(S\right)TS.\hfill \end{array}$$

Lemma 2 then renders ${\left\{{\left(c_{10}A\right)}^k\left(S\right)TS\right\}}_{k=0}^{p-1}$ linearly independent, along with ${\left\{A^k\left(S\right)\right\}}_{k=0}^{p-1}$. Indeed, assume that ${\sum }_{k=0}^{p-1}{\mu }_k{\left(c_{10}A\right)}^k\left(S\right)=0$ for some scalars ${\mu }_0,\dots ,{\mu }_{p-1}$, and, given any $x\in H$, consider the rank-one operator $Tz=⟨z,Sx⟩x$$(z\in H)$. Then,

$$\sum _{k=0}^{p-1}{\mu }_k{\left(c_{10}A\right)}^k\left(S\right)TSx=⟨Sx,Sx⟩\sum _{k=0}^{p-1}{\mu }_k{\left(c_{10}A\right)}^k\left(S\right)x=0.$$

The linear independence of ${\left\{{\left(c_{10}A\right)}^k\left(S\right)TS\right\}}_{k=0}^{p-1}$ forces ${\mu }_0=\dots ={\mu }_{p-1}=0$. Since $c_{10}\ne 0$, the desired conclusion follows.

If $q=m$, so that $p=1$, then ${\left\{A^k\left(S\right)\right\}}_{k=0}^{p-1}=\left\{S\right\}$ is linearly independent. This completes the proof. □

Definition 3.

Always in the notation of Theorem 2, we shall say that $\mu A$ is (strictly) $(r,S)$-null for some scalar μ and $r\in \mathbb{N}$ whenever $(\mu A_1,A_2)$ is (strictly) $(r,S)$-null (which implies that $(A_1,\mu A_2)$ is (strictly) $(r,S)$-null). An analogous convention applies for $\mu B$.

Theorem 3.

Assume $Q_S$ is strictly m-null, and let $p,q$ be as in Theorem 2. Then, there exists a constant $\lambda \ne 0$ such that $\lambda A$ is $(p,S)$-null and ${\lambda }^{-1}B$ is $(q,S)$-null.

Proof. 

The argument in the proof of Proposition 2 leads us to the following:

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)S+\sum _{i=1}^{p-1}{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)c_{q+i-1,q-1}{A}^i\left(S\right)+{(-1)}^p{c}_{m,q-1}{A}^p\left(S\right)=0.$$

Multiplying both sides of this identity by $c_{q0}\ne 0$, we find from (8) that

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right)c_{q0}S+\sum _{i=1}^{p-1}{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)c_{q+i,0}{A}^i\left(S\right)+{(-1)}^p{c}_{m+1,0}{A}^p\left(S\right)=0,$$

(16)

where $c_{m+1,0}:=c_{m,q-1}{c}_{q0}\ne 0$, by Lemma 5. Now, from Equation (16), we obtain:

$$A^p\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\displaystyle \frac{c_{q+i,0}}{c_{m+1,0}}}{A}^i\left(S\right).$$

(17)

On the other hand, rewriting (16) as

$$\begin{array}{cc}\hfill & \sum _{i=0}^p{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)c_{q+i,0}{A}^i\left(S\right)\hfill \\ \hfill & =\sum _{i=0}^p{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right)}^{-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)c_{q+i,0}{A}^i\left(S\right)\hfill \\ \hfill & =0,\hfill \end{array}$$

or equivalently, after simplifying the binomial coefficients,

$$\sum _{i=0}^p{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\displaystyle \frac{i!}{(q+i-1)!}}{c}_{q+i,0}{A}^i\left(S\right)=0,$$

we see that the sequence given by

$$T_i\left(S\right)={\displaystyle \frac{i!}{(q+i-1)!}}{c}_{q+i,0}{A}^i\left(S\right)(i\in {\mathbb{N}}_0)$$

is an operator arithmetic progression of order $p-1$, and therefore

$$T_k\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k-i-1}{p-i-1}}\right)T_i\left(S\right)(k\in {\mathbb{N}}_0);$$

that is,

$$\begin{array}{cc}\hfill & {\displaystyle \frac{k!}{(q+k-1)!}}{c}_{q+k,0}{A}^k\left(S\right)\hfill \\ \hfill & =\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{k-i-1}{p-i-1}}\right){\displaystyle \frac{i!}{(q+i-1)!}}{c}_{q+i,0}{A}^i\left(S\right)(k\in {\mathbb{N}}_0).\hfill \end{array}$$

(18)

Again, Lemma 5 ensures that $c_{q+k,0}\ne 0$, so that (18) reduces to

$$A^k\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i-1}{p-i-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{q+i-1}}\right){\displaystyle \frac{c_{q+i,0}}{c_{q+k,0}}}{A}^i\left(S\right)(k\in {\mathbb{N}}_0).$$

In particular,

$$A^{k+1}\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{p-i-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k}{q+i-1}}\right){\displaystyle \frac{c_{q+i,0}}{c_{q+k+1,0}}}{A}^i\left(S\right)(k\in {\mathbb{N}}_0).$$

Also, writing $m=p+q-1$, we have

$$\begin{array}{cc}\hfill A^{k+1}\left(S\right)& =A{A}^k\left(S\right)\hfill \\ \hfill & =\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i-1}{p-i-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{q+i-1}}\right){\displaystyle \frac{c_{q+i,0}}{c_{q+k,0}}}{A}^{i+1}\left(S\right)\hfill \\ \hfill & =\sum _{i=1}^p{(-1)}^{p-i}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{p-i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{q+i-2}}\right){\displaystyle \frac{c_{q+i-1,0}}{c_{q+k,0}}}{A}^i\left(S\right)\hfill \\ \hfill & =\sum _{i=1}^{p-1}{(-1)}^{p-i}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{p-i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{q+i-2}}\right){\displaystyle \frac{c_{q+i-1,0}}{c_{q+k,0}}}{A}^i\left(S\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{m-1}}\right){\displaystyle \frac{c_{m0}}{c_{q+k,0}}}{A}^p\left(S\right)\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{m-1}}\right){\displaystyle \frac{c_{m0}}{c_{q+k,0}}}\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\displaystyle \frac{c_{q+i,0}}{c_{m+1,0}}}{A}^i\left(S\right)\hfill \\ \hfill & +\sum _{i=1}^{p-1}{(-1)}^{p-i}\left({\displaystyle \genfrac{}{}{0pt}{}{k-i}{p-i}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{q+i-2}}\right){\displaystyle \frac{c_{q+i-1,0}}{c_{q+k,0}}}{A}^i\left(S\right)(k\in {\mathbb{N}}_0).\hfill \end{array}$$

The linear independence of the set of powers ${\left\{A^i\left(S\right)\right\}}_{i=0}^{p-1}$, guaranteed by Lemma 5, allows us to equate the coefficients for $i=0$ in the two representations of $A^{k+1}$, namely

$$\begin{array}{cc}\hfill & {(-1)}^{p-1}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q+k}{q-1}}\right){\displaystyle \frac{c_{q0}}{c_{q+k+1,0}}}\hfill \\ \hfill & ={(-1)}^{p-1}\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{m-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right){\displaystyle \frac{c_{m0}}{c_{q+k,0}}}{\displaystyle \frac{c_{q0}}{c_{m+1,0}}}(k\in {\mathbb{N}}_0),\hfill \end{array}$$

whence

$${\displaystyle \frac{c_{q+k,0}}{c_{q+k+1,0}}}=\left({\displaystyle \genfrac{}{}{0pt}{}{q+k-1}{m-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q-1}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{q+k}{q-1}}\right)}^{-1}{\left({\displaystyle \genfrac{}{}{0pt}{}{k}{p-1}}\right)}^{-1}{\displaystyle \frac{c_{m0}}{c_{m+1,0}}}(k\in {\mathbb{N}}_0).$$

This simplifies to

$${\displaystyle \frac{c_{q+k,0}}{c_{q+k+1,0}}}={\displaystyle \frac{(k+1)}{(k+q)}}{\displaystyle \frac{m{c}_{m0}}{p{c}_{m+1,0}}}(k\in {\mathbb{N}}_0).$$

It follows that

$$\begin{array}{cc}\hfill {\displaystyle \frac{c_{q+i,0}}{c_{q+k,0}}}& ={\displaystyle \frac{c_{q+i,0}}{c_{q+i+1,0}}}{\displaystyle \frac{c_{q+i+1,0}}{c_{q+i+2,0}}}\cdots {\displaystyle \frac{c_{q+k-1,0}}{c_{q+k,0}}}\hfill \\ \hfill & =\prod _{j=i}^{k-1}{\displaystyle \frac{(j+1)}{(j+q)}}{\left({\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}\right)}^{i-k}(k,i\in {\mathbb{N}}_0,k\ge p>i).\hfill \end{array}$$

At this point, we carry the previous identity, with $k=p$, over to (17),

$$\begin{array}{cc}\hfill A^p\left(S\right)& =\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right){\displaystyle \frac{c_{q+i,0}}{c_{m+1,0}}}{A}^i\left(S\right)\hfill \\ \hfill & =\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)\prod _{j=i}^{p-1}{\displaystyle \frac{(j+1)}{(j+q)}}{\left({\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}\right)}^{i-p}{A}^i\left(S\right),\hfill \end{array}$$

and obtain:

$${\left({\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}A\right)}^p\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)\prod _{j=i}^{p-1}{\displaystyle \frac{(j+1)}{(j+q)}}{\left({\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}A\right)}^i\left(S\right).$$

Now,

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{q+i-1}}\right)\prod _{j=i}^{p-1}{\displaystyle \frac{(j+1)}{(j+q)}}={\displaystyle \frac{p\cdots (i+1)}{m\cdots (q+i)}}{\displaystyle \frac{i!m!}{(q+i-1)!i!(p-i)!}}=\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right).$$

Therefore,

$${\left({\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}A\right)}^p\left(S\right)=\sum _{i=0}^{p-1}{(-1)}^{p-i-1}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{i}}\right){\left({\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}A\right)}^i\left(S\right),$$

which shows that $\lambda A$ is $(p,S)$-null, with

$$\lambda ={\displaystyle \frac{p{c}_{m+1,0}}{m{c}_{m0}}}={\displaystyle \frac{p{c}_{m,q-1}}{m{c}_{m-1,q-1}}}\ne 0.$$

From the hypothesis, we may write:

$$\sum _{k=0}^{p+q-1}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p+q-1}{k}}\right){\left(\lambda A\right)}^k\left(S\right)T{\left({\lambda }^{-1}B\right)}^k\left(S\right)=0.$$

Since $\lambda A$ is (strictly) $(p,S)$-null, Proposition 2 shows that ${\lambda }^{-1}B$ is $(q,S)$-null and completes the proof. □

The announced generalization of ([45], Theorem 7) can now be stated as follows.

Corollary 4.

The operator $Q_S$ is strictly m-null if and only if there exists a nonzero constant λ such that $\lambda A$ is strictly $(p,S)$-null and ${\lambda }^{-1}B$ is strictly $(q,S)$-null, with $p+q-1=m$.

Proof. 

The result is a straightforward consequence of Theorems 1 and 3, together with Corollary 3. □

We finish by deriving some consequences about the coefficients in Part $\left(i\right)$ of Theorem 2 and their recurrence relation (8).

Corollary 5.

Assume $Q_S$ is m-null, and let q and $c_{ij}$$(0\le j\le q-1,q\le i\le m)$ be as in Part $\left(i\right)$ of Theorem 2. Then, some constant $\lambda \ne 0$ is such that

$$c_{ij}={(-1)}^{q-j-1}{\lambda }^{i-j}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j-1}{q-j-1}}\right)(q\le i\le m,0\le j\le q-1).$$

(19)

Proof. 

Corollary 4 provides a scalar $\lambda \ne 0$ such that ${\lambda }^{-1}B$ is $(q,S)$-null. Hence (Proposition 1),

$${\left({\lambda }^{-1}B\right)}^i\left(S\right)=\sum _{j=0}^{q-1}{(-1)}^{q-j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j-1}{q-j-1}}\right){\left({\lambda }^{-1}B\right)}^j\left(S\right),$$

or

$$\begin{array}{cc}\hfill B^i\left(S\right)& =\sum _{j=0}^{q-1}{(-1)}^{q-j-1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j-1}{q-j-1}}\right){\lambda }^{i-j}{B}^j\left(S\right)\hfill \\ \hfill & =\sum _{j=0}^{q-1}{c}_{ij}{B}^j\left(S\right).\hfill \end{array}$$

By linear independence, (19) follows. □

Our last result is immediate from the preceding one, but a direct proof is provided for completeness.

Proposition 3.

For any scalar λ, the sequence (19) is a solution to the recurrence relation (8), with initial conditions $c_{i,-1}=0$$(q\le i\le m)$.

Proof. 

We want to check that

$$c_{i+1,j}=c_{i,j-1}+c_{i,q-1}{c}_{qj}(q\le i\le p+q-2,0\le j\le q-1).$$

Fix $q\le i\le p+q-2$. For $j=0$, this becomes

$$c_{i+1,0}=c_{i,q-1}{c}_{q0}.$$

Now, for the proposed solution, we have

$$c_{i+1,0}={(-1)}^{q-1}{\lambda }^{i+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right),c_{i,q-1}={\lambda }^{i-q+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right),c_{q0}={(-1)}^{q-1}{\lambda }^q,$$

so that

$$c_{i+1,0}={(-1)}^{q-1}{\lambda }^{i+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)=c_{i,q-1}{c}_{q0}.$$

Next, we consider $1\le j\le q-1$. On the one hand,

$$\begin{array}{cc}\hfill c_{i+1,j}& ={(-1)}^{q-j-1}{\lambda }^{i-j+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i+1}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right),\hfill \\ \hfill c_{i,j-1}& ={(-1)}^{q-j}{\lambda }^{i-j+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j}}\right).\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill & c_{i+1,j}-c_{i,j-1}\hfill \\ \hfill & ={(-1)}^{q-j-1}{\lambda }^{i-j+1}\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i+1}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j}}\right)\right].\hfill \end{array}$$

On the other hand,

$$c_{i,q-1}={\lambda }^{i-q+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right),c_{qj}={(-1)}^{q-j-1}{\lambda }^{q-j}\left({\displaystyle \genfrac{}{}{0pt}{}{q}{j}}\right).$$

Thus,

$$c_{i,q-1}{c}_{qj}={(-1)}^{q-j-1}{\lambda }^{i-j+1}\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{j}}\right).$$

So, we must prove that

$$\left({\displaystyle \genfrac{}{}{0pt}{}{i+1}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{j}}\right).$$

In fact, two applications of the Pascal recurrence for binomial coefficients yield

$$\begin{array}{cc}\hfill \left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)& \left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j}}\right)\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)\left[\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j}}\right)\right]+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j+1}{q-j}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{i}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{i-j}{q-j-1}}\right)\hfill \\ \hfill & ={\displaystyle \frac{i!}{(j-1)!(q-j)!(i-q+1)!}}+{\displaystyle \frac{i!}{j!(q-j-1)!(i-q+1)!}}\hfill \\ \hfill & ={\displaystyle \frac{i!}{(j-1)!(q-j-1)!(i-q+1)!}}\left({\displaystyle \frac{1}{q-j}}+{\displaystyle \frac{1}{j}}\right)\hfill \\ \hfill & ={\displaystyle \frac{i!}{(j-1)!(q-j-1)!(i-q+1)!}}{\displaystyle \frac{q}{j(q-j)}}\hfill \\ \hfill & ={\displaystyle \frac{i!q}{j!(q-j)!(i-q+1)!}}={\displaystyle \frac{i!}{(q-1)!(i-q+1)!}}{\displaystyle \frac{q!}{j!(q-j)!}}\hfill \\ \hfill & =\left({\displaystyle \genfrac{}{}{0pt}{}{i}{q-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{q}{j}}\right),\hfill \end{array}$$

as required. □

4. Discussion

This paper was motivated by an open question regarding the need for analytical tools (spectral radius and approximate point spectrum) in proving Gu’s result on the symbols of elementary operators that are strict m-isometries [45] (Theorem 7). Here, the question has been answered in the negative within the framework of m-null operators, a generalization of m-isometries introduced and studied by the author in [48]. Specifically, we have demonstrated that algebraic and combinatorial methods suffice, thereby eliminating dependencies on spectral theory.

Our constructive approach, which involves explicitly solving a difference equation in terms of binomial coefficients, not only resolves this question but also revitalizes a line of research initiated by Botelho et al. [41,42], on the basis of a prior theorem by Fong and Sourour [47] with broader implications. The findings are placed in the context of operator theory and functional analysis, where m-isometries and related concepts have been intensively generalized and investigated through diverse methodologies. We show that a number of such generalizations can be expressed as m-null operators, a unification that enables direct extensions of Gu’s result. Furthermore, by removing its apparent reliance on spectral theory, we pave the way for new extensions beyond the m-null and Hilbert space settings.

The primary motivation for considering m-null and strictly m-null pairs of operators is twofold. First, as just recalled, they encompass m-isometries and strict m-isometries along with some of their generalizations and related classes, enriching operator theory and its applications by offering a framework that captures a wider range of operator behaviors. Second, they offer new insights into the algebraic and combinatorial nature of m-isometries, suggesting novel methods to handle similar polynomially defined families.

The algebraic character of our methods supports computational applications, such as numerical algorithms for operator decomposition or simulation of discrete-time processes in signal processing and control theory. Moreover, the m-null framework’s versatility suggests applicability to operator semigroups, noncommutative algebras, or graph-theoretic structures.

A primary limitation of the present work is its restriction to Hilbert space operators, raising the question of whether our methods can be effectively adapted to Banach spaces or general metric spaces with minimal structural assumptions, as could be expected. Additionally, due to the recursive nature of m-null conditions, the computational complexity of our algorithms may grow exponentially with the parameter m, potentially limiting scalability in numerical applications.

Thus, future research could focus on extending our results to non-Hilbert settings and enhancing the computational efficiency of our algorithms. A compelling question involves characterizing the structure of specific operator classes, such as weighted shifts, composition operators, or Toeplitz operators, within the m-null framework. Examining links with fields like combinatorics, representation theory, or quantum information theory, where operator-theoretic methods apply to quantum channel analysis, might uncover deeper structural properties and novel applications.

5. Final Remarks

While this paper was under review, B.P. Duggal kindly shared a preprint proposing a methodologically distinct answer to the question in ([48], Section 4) in the Banach space setting. Nevertheless, our results remain fully significant due to the constructive nature of their proofs, their independence from spectral theory, and the explicit combinatorial solution to a difference equation, as discussed above.