On m-Isometric and m-Symmetric Operators of Elementary Operators

1. Introduction

Given a complex infinite dimensional Hilbert space $(\mathcal{H};\langle .,.\rangle )$, let $B\left(\mathcal{H}\right)$ denote the algebra of operators, equivalently bounded linear transformations, on $\mathcal{H}$ into itself. For an operator $A\in B\left(\mathcal{H}\right)$, let $L_A$ and $R_A\in B\left(B\left(\mathcal{H}\right)\right)$ denote, respectively, the operators

$$L_A\left(X\right)=AX\mathrm{and}{R}_A\left(X\right)=XA$$

of left multiplication by A and right multiplication by A. For $A,B\in B\left(\mathcal{H}\right)$, the elementary operator ${▵}_{A,B}$ of length two and the generalised derivation ${\delta }_{A,B}\in B\left(B\left(\mathcal{H}\right)\right)$ are defined by

$${▵}_{A,B}\left(X\right)=(I-L_A{R}_B)\left(X\right)=X-AXB\mathrm{and}{\delta }_{A,B}\left(X\right)=(L_A-R_B)\left(X\right)=AX-XB,$$

respectively. We say that the pair $(A,B)$ of operators in $B\left(\mathcal{H}\right)\times B\left(\mathcal{H}\right)$ is m-isometric (resp. , m-symmetric) for some positive integer m if

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

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

(In the case in which the pair $(A,B)$ is the pair $(T^{*},T)$ for an operator $T\in B\left(\mathcal{H}\right)$, we shorten $(T^{*},T)$ is m-isometric (resp., m-symmetric) to T is m-isometric (resp., m-symmetric).) Structural properties of m-isometric and m-symmetric pairs $(A,B)$, and their connections with classical function theory, non-stationary stochastic processes, Toeplitz operators and nilpotent perturbations of Hermitian operators (etc.), have been studied by a number of authors in the recent past: see [2–4,8,18,25,37,39] for further references.

A characterisation of m-isometric operators with a finite spectrum has been carried out in [29,34,37]. An example of such an operator, first considered by Botelho and Jamison [12,13] for the cases $m=2$ and $m=3$, is the operator

$${▵}_{L_{A^{*}}{R}_{B^{*}},L_A{R}_B}^m\left(I\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{*j}{A}^j{B}^j{B}^{*j}=0;A,B\in B\left(\mathcal{H}\right).$$

It is straightforward to see that if $\overline{\lambda }$ is in the approximate point spectrum ${\sigma }_a\left(B^{*}\right)$ of $B^{*}$, then ${▵}_{\overline{\lambda }{A}^{*},\lambda A}^m\left(I\right)=0$, $\sigma \left(\lambda A\right)$ is a subset of the boundary $\partial \mathtt{D}$ of the unit disc $\mathtt{D}\subset \mathtt{C}$ and $1-\left(\overline{\lambda }{\sigma }_a\left(A^{*}\right)\right)\left(\lambda {\sigma }_a\left(A\right)\right)=0$. There exists a non-trivial scalar $\beta$, $\left|\beta \right|=1$, such that $\lambda \alpha =\beta$ and $\overline{\lambda \alpha }={\displaystyle \frac{1}{\beta }}=\overline{\beta }$ for all $\alpha \in {\sigma }_a\left(A\right)$. We assert that ${\sigma }_a\left(B^{*}\right)$ consists, at most, of two points. For if there exist non-trivial $\overline{\mu },\overline{\nu }\in {\sigma }_a\left(B^{*}\right)$, $\mu \ne \lambda \ne \nu$, then ${\sigma }_a\left(\mu A\right)={\sigma }_a\left(\lambda A\right)+{\sigma }_a\left((\mu -\lambda )A\right)$ and ${\sigma }_a\left(\nu A\right)={\sigma }_a\left(\lambda A\right)+{\sigma }_a\left((\nu -\lambda )A\right)$: since $0\notin {\sigma }_a\left(A\right)$, not both of these translates of ${\sigma }_a\left(\lambda A\right)$ are in $\partial \mathtt{D}$. This argument applies equally to ${\sigma }_a\left(A\right)$; hence $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points. A version of the preceding argument applies to m-isometric operators ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$. Indeed Gu, [24] proves that if ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ for some operators $A,B\in B\left(\mathcal{H}\right)$, then $A,B$ have spectrum consisting at most of two points. An extension of this result to operators ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ has recently been considered by Duggal and Kim [22].

For operators $A,B\in B\left(\mathcal{H}\right)$, let $D_{d_1,d_2}^m\left(I\right)$ denote the operator defined by the choices

$$D=▵\mathrm{or}\delta ,d_1={▵}_{A^{*},B^{*}}\mathrm{or}{\delta }_{A^{*},B^{*}}\mathrm{and}{d}_2={▵}_{A,B}\mathrm{or}{\delta }_{A,B}.$$

This paper considers the structure of the resulting eight operators, two of which have already been discussed in [22,24], to prove that the spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at best of two points. It is fairly easily seen that $D_{d_1,d_2}^m\left(I\right)=0$ implies $D_{d_1,d_2}^n\left(I\right)=0$ for all integers $n\ge m$. The reverse problem of does $D_{d_1,d_2}^m\left(I\right)=0$ imply $D_{d_1,d_2}^n\left(I\right)=0$ for some positive integer $n<m$ does not, in general, have a positive answer. We prove that a necessary and sufficient condition for $D_{d_1,d_2}^m\left(I\right)=0$ to imply $D_{d_1,d_2}\left(I\right)=0$ is that the pairs of operators $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)$, ${\alpha }_i$ and ${\beta }_i$ ($i=1,2$) scalars, satisfy a Putnam-Fuglede commutativity property [28,32,35] The paper considers also the case in which the operator $D_{d_1,d_2}^m$ is compact. Here it is seen that the operators A and B (are algebraic operators, and as such) have a finite spectrum consisting of poles (of the resolvent) of the operators.

2. Complementary Results: The Adjoint Operator ${\left(D_{d_1,d_2}^m\left(I\right)\right)}^{*}$

In the following, $A,B$ will denote Hilbert space operators such that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B\right)$, and m will denote a positive integer. Given complex scalars ${\alpha }_i$ and ${\beta }_i$, $1\le i\le 2$, we say that the pair of operators $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)$ satisfies the Putnam-Fuglede commutativity property, shortened to $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)\in \left(PF\right)$, if

$$D_{{\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2}^{-1}\left(0\right)\subseteq D_{\overline{{\alpha }_1}{A}^{*}+\overline{{\beta }_1},\overline{{\alpha }_2}{B}^{*}+\overline{{\beta }_2}}^{-1}\left(0\right);D=▵\mathrm{or}\delta .$$

It is well known, see [28,36], that if $A,B$ are normal operators, then the pair $({\alpha }_{1}A+{\beta }_1,{\alpha }_{2}B+{\beta }_2)\in \left(PF\right)$.

The ascent (resp., descent) of A, denoted $\mathrm{asc}\left(A\right)$ (resp., $\mathrm{dsc}\left(A\right)$), is the least positive integer n such that

$$A^{-(n+1)}\left(0\right)\subseteq A^{-n}\left(0\right)(\mathrm{resp}.A^n\left(\mathcal{H}\right)\subseteq A^{n+1}\left(\mathcal{H}\right)).$$

Finite ascent and finite descent implies their equality [5,38]; we say that a point $\lambda$ of the spectrum $\sigma \left(A\right)$ of A is a pole (of the resolvent) of A if $\mathrm{asc}(A-\lambda )=\mathrm{dsc}(A-\lambda )$ [5,38]. (Here, and in the sequel, we write $A-\lambda$ for $A-\lambda I$.) We observe here that if $\lambda \in \sigma \left(A\right)$ is a pole of A, then necessarily $\lambda$ is an isolated point of $\sigma \left(A\right)$ (i.e., $\lambda \in \mathrm{iso}\sigma \left(A\right)$). We say that $D_{A,B}^{-1}\left(0\right)\perp D_{A,B}\left(\mathcal{H}\right)$, i.e. the kernel of the operator $D_{A,B}$ is orthogonal to the range of $D_{A,B}$ in the sense of James [14,31], if

$$\left|\right|X\left|\right|\le \left|\right|X+Y\left|\right|,\mathrm{all}X\in D_{A,B}^{-1}\left(0\right)\mathrm{and}Y\in D_{A,B}\left(B\left(\mathcal{H}\right)\right).$$

The following lemma, linking “range-kernel orthogonality" to the “ascent" of operators satisfying the $\left(PF\right)$-property, will be useful in our considerations below.

Lemma 2.1.

[16, Proposition 2.26]. If $D_{A,B}\in \left(PF\right)$, then ($D_{A,B}^{-1}\left(0\right)\perp D_{A,B}\left(\mathcal{H}\right)$, hence) $\mathrm{asc}\left(D_{A,B}\right)\le 1$.

The quasi-nilpotent part $H_0(A-\lambda )$ of an operator $A\in B\left(\mathcal{H}\right)$ at $\lambda \in \mathtt{C}$ is the set

$$H_0(A-\lambda )=\{x\in \mathcal{H}:\underset{n\to \infty }{lim}{\left|\right|(A-\lambda )}^n{\left|\right|}^{{\displaystyle \frac{1}{n}}}\left|\right|=0\}$$

[5, Page 119]. A has SVEP, the single-valued extension property, at ${\lambda }_0$ if for every open disc ${\mathtt{D}}_{{\lambda }_0}$ centered at ${\lambda }_0$ the only analytic function $f:{\mathtt{D}}_{{\lambda }_0}\to \mathcal{H}$ satsifying $(A-\lambda )f\left(\lambda \right)=0$ for all $\lambda \in {\mathtt{D}}_{{\lambda }_0}$ is the function $f\equiv 0$ [5,23]; A has SVEP if it has SVEP at all $\lambda$. Operators A with a countable spectrum have SVEP, and if $\sigma \left(A\right)=\{{\alpha }_1,\dots ,{\alpha }_p\}$ for some natural number p, then $A={\oplus }_{i=0}^n{A|}_{H_0(A-{\alpha }_i)}$ [5, Page 303, Lemma 4.17]. A necessary and sufficient condition for the points ${\alpha }_i$ to be poles of (the resolvent of) A of order $t_i$, for some positive integers $t_i$, $1\le i\le p$, is that $H_0(A-{\alpha }_i)={(A-{\alpha }_i)}^{-t_i}\left(0\right)$ [5]. Observe that if $\sigma \left(A\right)=\{{\alpha }_1,\dots ,{\alpha }_p\}$ and the points ${\alpha }_i$ are poles of A of order $t_i$, then $A={\oplus }_{i=1}^p{A|}_{{(A-{\alpha }_i)}^{-t_i}\left(0\right)}+N$ for some nilpotent operator N.

The operator A is algebraic if there exists a non-trivial polynomial $p(.)$ such that $p\left(A\right)=0$. It is well known, see for example [5, Theorem 3.83], that A is algebraic if and only if $\sigma \left(A\right)$ is a finite set consisting of the poles of A (i.e., if and only if $\sigma \left(A\right)$ is finite and A is polaroid [5]). Recall from Boasso [10] that A is polaroid, i.e. points $\lambda \in \mathrm{iso}\sigma \left(A\right)$ are poles of A, if and only if $L_A$ and $R_B$ are polaroid. The (Banach space) operator ${\delta }_{A,B}$ is algebraic if and only if $A,B$ are algebraic. $L_A{R}_B$ algebraic does not imply A and B algebraic, only that at least one of A and B is algebraic [16, Proposition 2.6].

A is nilpotent, more precisely A is n-nilpotent for some positive integer n, if $A^n=0$. Either of A and B nilpotent implies $L_A{R}_B$ nilpotent; conversely, $L_A{R}_B$ nilpotent implies, at best, that at least one of A and B is nilpotent. We consider next the adjoint of the operator $D_{d_1,d_2}^m\left(I\right)$.

Start by observing that for all $X\in B\left(\mathcal{H}\right)$,

$${\left({\delta }_{A,B}\left(X\right)\right)}^{*}={\left((L_A-R_B)\left(X\right)\right)}^{*}=(R_{A^{*}}-L_{B^{*}})\left(X^{*}\right)=(-1){\delta }_{B^{*},A^{*}}\left(X^{*}\right)$$

and

$${\left({▵}_{A,B}\left(X\right)\right)}^{*}={\left((I-L_A{R}_B)\left(X\right)\right)}^{*}=(I-L_{B^{*}}{R}_{A^{*}})\left(X^{*}\right)={▵}_{B^{*},A^{*}}\left(X^{*}\right).$$

Hence:

$$\begin{array}{ccc}& & {\left({\delta }_{A,B}^n\left(X\right)\right)}^{*}={(-1)}^n{\delta }_{B^{*},A^{*}}^n\left(X^{*}\right);\hfill \\ & & {\left({▵}_{A,B}^n\left(X\right)\right)}^{*}={▵}_{B^{*},A^{*}}^n\left(X^{*}\right);\hfill \\ & & {\left({▵}_{A^{*},B^{*}}\left({▵}_{A,B}\left(X\right)\right)\right)}^{*}={▵}_{B,A}\left({\left({▵}_{A,B}\left(X\right)\right)}^{*}\right)={▵}_{B,A}\left({▵}_{B^{*},A^{*}}\left(X^{*}\right)\right);\hfill \\ & & {\left({\delta }_{A^{*},B^{*}}\left({\delta }_{A,B}\left(X\right)\right)\right)}^{*}=(-1){\delta }_{B,A}{\left({\delta }_{A,B}\left(X\right)\right)}^{*}{)=(-1)}^2{\delta }_{B,A}\left({\delta }_{B^{*},A^{*}}\left(X^{*}\right)\right);\hfill \\ & & {\left({▵}_{A^{*},B^{*}}\left({\delta }_{A,B}\right)\left(X\right))\right)}^{*}={▵}_{B,A}\left({\left({\delta }_{A,B}\left(X\right))\right)}^{*}\right)=(-1){▵}_{B,A}\left({\delta }_{B^{*},A^{*}}\left(X^{*}\right)\right)\mathrm{and}\hfill \\ & & {\left({\delta }_{A^{*},B^{*}}\left({▵}_{A,B}\left(X\right)\right)\right)}^{*}=(-1){\delta }_{B,A}\left({\left({▵}_{A,B}\left(X\right)\right)}^{*}\right)=(-1){\delta }_{B,A}\left({▵}_{B^{*},A^{*}}\left(X^{*}\right)\right).\hfill \end{array}$$

Letting

$${\nabla }_{E,F}\left(X\right)=(I+L_E{R}_F)\left(X\right)\mathrm{and}{♢}_{E,F}\left(X\right)=(L_E+R_F)\left(X\right)$$

for Banach space operators $E,F$ and X, we have:

$$\begin{array}{ccc}\hfill \left(1\right)& & {\left({▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}\hfill \\ & =& {\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{A^{*},B^{*}}^j\left({▵}_{A,B}^j\left(I\right)\right)\right)}^{*}\hfill \\ & =& \sum _{j=0}^m{(-1)}^{2j}\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{B,A}^j{\left({▵}_{A,B}^j\left(I\right)\right)}^{*}\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{B,A}^j\left({▵}_{B^{*},A^{*}}^j\left(I\right)\right)\hfill \\ & =& {\nabla }_{{\delta }_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \left(2\right)& & {\left({\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}\hfill \\ & =& {\left(\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{A^{*},B^{*}}^{m-j}\left({▵}_{A,B}^j\left(I\right)\right)\right)}^{*}\hfill \\ & =& \sum _{j=0}^m{(-1)}^m\left(\begin{array}{c}m\\ j\end{array}\right){\delta }_{B,A}^{m-j}\left({▵}_{B^{*},A^{*}}^j\left(I\right)\right)\hfill \\ & =& {(-1)}^m{♢}_{{\delta }_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);\hfill \end{array}$$

$$\left(3\right){\left({▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={\nabla }_{{▵}_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right);$$

$$\left(4\right){\left({\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={♢}_{{▵}_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right);$$

$$\left(5\right){\left({▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}={▵}_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);$$

$$\left(6\right){\left({\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={(-1)}^m{\delta }_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)$$

$$\left(7\right){\left({\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\right)}^{*}={\delta }_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right);$$

$$\left(8\right){\left({▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)\right)}^{*}={▵}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right).$$

3. The Spectrum

The approximate point spectrum of an operator being a non-empty set, if the pair $(A,B)$ satisfies the operator equation $D_{d_1,d_2}^m\left(I\right)=0$, $D=▵$ or $\delta$, $d_1={▵}_{A^{*},B^{*}}$ or ${\delta }_{A^{*},B^{*}}$ and $d_2={▵}_{A,B}$ or ${\delta }_{A,B}$, then there exist non-trivial scalars $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. (We assume in the following that $\sigma \left(A\right)\ne \left\{0\right\}\ne \sigma \left(B^{*}\right)$; see [30] for the spectra of left and right multiplication operators.) By definition

$$\begin{array}{ccc}& & {▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){\left(L_{{\delta }_{A^{*},B^{*}}}{R}_{{\delta }_{A,B}}\right)}^j\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)L_{A^{*}}^{j-p}{R}_{B^{*}}^j\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)L_A^{j-k}{R}_B^k\right)\right]\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right]A^{*(j-p)}{A}^{j-k}{B}^k{B}^{*p}\hfill \\ & =& C_{\sum }{\mathcal{E}}_{A^{*(j-p)}{A}^{j-k},B^k{B}^{*p}},\hfill \end{array}$$

where $C_{\sum }=C(m,j,p,k)$ denotes

$$C_{\sum }=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right]$$

and ${\mathcal{E}}_{E,F}$ is the operator

$${\mathcal{E}}_{E,F}\left(X\right)=EXF.$$

A similar argument shows that

$${▵}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)=C_{\sum }{\mathcal{E}}_{B^{j-p}{B}^{*(j-p)},A^{*k}{A}^p}.$$

If $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, and $\left\{x_n\right\},\left\{y_n\right\}\subseteq \mathcal{H}$ are sequences of unit vectors such that ${lim}_{n\to \infty }(B^{*}-\overline{\lambda })x_n=0={lim}_{n\to \infty }(A-\alpha )y_n$, then for all $x,y\in \mathcal{H}$

$$\begin{array}{ccc}& & \left[{\mathcal{E}}_{A^{*(j-p)}{A}^{j-k},B^k{B}^{*p}}(x\otimes x_n)\right]x_n\hfill \\ & =& A^{*(j-p)}{A}^{(j-k)}x\langle x_n,B^p{B}^{*k}{x}_n\rangle \hfill \end{array}$$

and

$$\begin{array}{ccc}& & \left[{\mathcal{E}}_{B^{j-p}{B}^{*(j-p)},A^{*k}{A}^p}(y\otimes y_n)\right]y_n\hfill \\ & =& B^{j-p}{B}^{*(j-k)}y\langle y_n,A^{*p}{A}^k{y}_n\rangle .\hfill \end{array}$$

(Here, and in the sequel, $\langle .,.\rangle$ denotes the inner product on $\mathcal{H}$.) Thus, if ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then (for all $x\in \mathcal{H}$)

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}{C}_{\sum }{A}^{*(j-p)}{A}^{(j-k)}x\langle x_n,B^p{B}^{*k}{x}_n\rangle \hfill \\ & =& C_{\sum }\left({\overline{\lambda }}^p{A}^{*(j-p)}\right)\left({\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right){\overline{\lambda }}^p{A}^{*(j-p)}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)\right]x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(A^{*}-\overline{\lambda })}^j{(A-\lambda )}^{j}x,\hfill \end{array}$$

i.e.,

$${▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0.\left(9a\right)$$

A similar argument shows that if ${▵}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)=0$, then

$${▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)=0.\left(9b\right)$$

If ${▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)$, then ${\sigma }_a(A-\lambda )$ and ${\sigma }_a(B^{*}-\overline{\alpha })$ are subsets of the boundary $\partial \mathtt{D}$ of the unit disc in the complex plane $\mathtt{C}$. Considering ${\sigma }_a(A-\lambda )$, if $\overline{\mu }\in {\sigma }_a(B^{*}-\overline{\lambda })$, then ${\sigma }_a(A-\lambda -\mu )\in \partial \mathtt{D}$ and, for a $\beta \in {\sigma }_a(A-\lambda )$, $1=|{\sigma }_a(A-\lambda -\mu )|=|{\sigma }_a(A-\lambda )-\mu |=|\beta -\mu |$. Let $\mu =e^{it}\left|\mu \right|$, then $\left|\mu \right|\le 2$ and

$$\begin{array}{ccc}& & |\beta -\mu |=1⟹{\beta }^2-e^{it}\left|\mu \right|\beta +e^{2it}=0\hfill \\ & ⟺& \beta =e^{it}\left({\displaystyle \frac{\left|\mu \right|\pm i\sqrt{4-{\left|\mu \right|}^2}}{2}}\right).\hfill \end{array}$$

A similar argument shows if $\nu =e^{i\theta }\left|\nu \right|\in {\sigma }_a(A-\alpha )$, then $\left|\nu \right|\le 2$ and, for every $\overline{\tau }\in {\sigma }_a(B^{*}-\overline{\alpha })$,

$$\begin{array}{ccc}& & |\tau -\nu |=1⟹{\tau }^2-e^{i\theta }\left|\nu \right|\tau +e^{2i\theta }=0\hfill \\ & ⟺& \tau =e^{i\theta }\left({\displaystyle \frac{\left|\nu \right|\pm i\sqrt{4-{\left|\nu \right|}^2}}{2}}\right).\hfill \end{array}$$

Spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points: for if $\overline{\mu },\overline{\eta }\in {\sigma }_a(B^{*}-\overline{\lambda })$, then ${\sigma }_a(A-\lambda -\eta )={\sigma }_a(A-\lambda -\mu )+(\mu -\eta )$, and both ${\sigma }_a(A-\lambda -\eta )$ and its non-trivial translates can not be in $\partial \mathtt{D}$. The following proposition encompasses this known result.

Proposition 3.1.

Given $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, if

(i) ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then

$${▵}_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={▵}_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right),$$

$$\sigma (A-\lambda )={\sigma }_a(A-\lambda )=\{e^{it}\left({\displaystyle \frac{\left|\mu \right|\pm i\sqrt{4-{\left|\mu \right|}^2}}{2}}\right):\overline{\mu }=e^{-it}\left|\mu \right|\in {\sigma }_a(B^{*}-\overline{\lambda }),0\le t<2\pi ,\left|\mu \right|\le 2\}$$

and

$$\sigma (B-\alpha )={\sigma }_a(B-\alpha )=\left\{e^{it}\left({\displaystyle \frac{\left|\mu \right|\pm i\sqrt{4-{\left|\mu \right|}^2}}{2}}\right):\mu =e^{it}\left|\mu \right|\in {\sigma }_a(A-\alpha ),0\le t<2\pi ,\left|\mu \right|\le 2\right\}.$$

(ii) ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then

(a) ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$;

(b) there exist real scalars ${\beta }_1$ and ${\beta }_2$, ${\beta }_1\le {\beta }_2$,such that

$$\left\{\lambda \sigma \left(A\right)\right\}\wedge \left\{\alpha \sigma \left(B\right)\right\}\subseteq [1-{\beta }_2,1-{\beta }_1];$$

(c) $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points.

Proof. Part (i) having already been proved, we start proving part (ii) by proving ${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0={\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$. By definition

$$\begin{array}{ccc}& & {\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=C_{{\sum }_1}{\mathcal{E}}_{A^{*p}{A}^k,B^k{B}^{*p}}\mathrm{and}\hfill \\ & & {\delta }_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right)=C_{{\sum }_1}{\mathcal{E}}_{B^p{B}^{*k},A^{*k}{A}^p},\hfill \end{array}$$

where

$$C_{{\sum }_1}=C(m,j,p,k)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right].$$

Arguing as above, it is seen that

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}{\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\hfill \\ & =& \underset{n\to \infty }{lim}{C}_{{\sum }_1}{A}^{*p}{A}^{k}x\langle x_n,B^k{B}^{*p}{x}_n\rangle \hfill \\ & =& C_{{\sum }_1}\left({\overline{\lambda }}^p{A}^{*p}\right)\left({\lambda }^k{A}^k\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\overline{\lambda }}^p{A}^{*p}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^k\right)\right]x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^{m-j}{(I-\lambda A)}^{j}x,\hfill \end{array}$$

for all $x\in \mathcal{H}$ and

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}{\delta }_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right)\hfill \\ & =& \underset{n\to \infty }{lim}{C}_{{\sum }_1}{B}^p{B}^{*k}y\langle y_n,A^{*p}{A}^k{y}_n\rangle \hfill \\ & =& C_{{\sum }_1}\left({\alpha }^p{B}^p\right)\left({\overline{\alpha }}^k{B}^{*k}\right)y\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\alpha B)}^{m-j}{(I-\overline{\alpha }{A}^{*})}^{j}y\hfill \end{array}$$

for all $y\in \mathcal{H}$. Hence

$${\delta }_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0\left(10a\right)$$

and

$${\delta }_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)=0\left(10b\right).$$

This proves $\left(a\right)$. To prove $\left(b\right)$, we start by observing from the spectral mapping theorem that

$$\begin{array}{ccc}& & {\sigma }_a(I-\overline{\lambda }{A}^{*})-{\sigma }_a(I-\lambda A)=(I-\overline{\lambda }{\sigma }_a\left(A^{*}\right))-(1-\lambda {\sigma }_a\left(A\right))\hfill \\ & =0& \\ & =& {\sigma }_a(I-\alpha B)-{\sigma }_a(I-\overline{\alpha }{B}^{*})=(1-\alpha {\sigma }_a\left(B\right))-(1-\overline{\alpha }{\sigma }_a\left(B^{*}\right)).\hfill \end{array}$$

Hence there exist real numbers ${\beta }_i$, $1\le i\le 2$, such that

$$1-\lambda \sigma \left(A\right)\subseteq [{\beta }_1,{\beta }_2],\mathrm{equivalently}\lambda \sigma \left(A\right)\subseteq [1-{\beta }_2,1-{\beta }_1].$$

(This inclusion being true for all $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$, we have also that $\alpha \sigma \left(B\right)\subseteq [1-{\beta }_2,1-{\beta }_1]$.) To prove $\left(c\right)$, assume that there exist distinct scalars $\mu ,\nu$ such that $\overline{\mu },\overline{\nu }\in {\sigma }_a(B^{*}-\overline{\lambda })$. Then $\overline{\lambda +\mu }$ and $\overline{\lambda +\nu }\in {\sigma }_a\left(B^{*}\right)$, and hence

$$\left\{(\lambda +\mu )\sigma \left(A\right)\right\}\wedge \left\{(\lambda +\nu )\sigma \left(A\right)\right\}\subseteq [1-{\beta }_2,1-{\beta }_1].$$

Recalling the hypothesis that ${\sigma }_a\left(A\right)\ne \left\{0\right\}$, we have

$$\begin{array}{ccc}& & {\sigma }_a(I-(\lambda +\nu )A)={\sigma }_a(I-(\lambda +\mu )A)+(\mu -\nu ){\sigma }_a\left(A\right)\hfill \\ & ⟹& (\lambda +\nu ){\sigma }_a\left(A\right)\subseteq [1-{\beta }_2,1-{\beta }_1]+(\mu -\nu ){\sigma }_a\left(A\right)\neg \subseteq [1-{\beta }_2,1-{\beta }_1],\hfill \end{array}$$

since $1-{\beta }_1+(\mu -\nu ){\sigma }_a\left(A\right)>1-{\beta }_1$ if $(\mu -\nu ){\sigma }_a\left(A\right)>0$ and $1-{\beta }_2+(\mu -\nu ){\sigma }_a\left(A\right)<1-{\beta }_2$ if $(\mu -\nu ){\sigma }_a\left(A\right)<0$. A similar argument proves that ${\sigma }_a\left(A\right)$ consists at most of two points.  □

The point 0 can not be in both ${\sigma }_a\left(B^{*}\right)$ and ${\sigma }_a\left(A\right)$ in the case in which ${▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$. This follows from the following argument. If $0\in {\sigma }_a\left(B^{*}\right)$, then (see above) for all $x\in \mathcal{H}$,

$$\begin{array}{ccc}& & {▵}_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0⟹C_{\sum }{A}^{*(j-p)}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{x}_n,A^{*k}{x}_n\rangle \hfill \\ & =& 0\mathrm{for}\mathrm{all}p,k>0\mathrm{and}\mathrm{if}p=k=0,\mathrm{then}\hfill \\ & & \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)A^{*j}{A}^{j}x={▵}_{A^{*},A}^m\left(I\right)x=0\hfill \\ & ⟹& {▵}_{A^{*},A}^m\left(I\right)=0.\hfill \end{array}$$

Thus, A is m-isometric, and hence ${\sigma }_a\left(A\right)\subseteq \partial \mathtt{D}$. A similar argument shows that if $0\in {\sigma }_a\left(A\right)$, then $B^{*}$ is m-isometric and ${\sigma }_a\left(B^{*}\right)\subseteq \partial \mathtt{D}$. This fails for operators $A,B$ satisfying ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$. Consider, for example, the operators $A,B$ such that $\sigma \left(A\right)=\{a{e}^{i\theta },0\}$, $\sigma \left(B\right)=\{b{e}^{-i\theta },0\}$ for some real numbers $a,b$ and $0\le \theta <2\pi$, $A=a{e}^{i\theta }{I}_1\oplus 0I_2$ and $B=b{e}^{-i\theta }{I}_1\oplus 0I_2$; $I=I_1\oplus I_2$. Then ${\delta }_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=C_{{\sum }_1}(I_1\oplus 0I_2)=0$.

The following theorem proves that the conclusion $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points holds for the remaining six choices of the operators $D,d_1$ and $d_2$ (in $D_{d_1,d_2}^m\left(I\right)=0$). For continuity, we keep the numbering above – thus our theorem starts with case $\left(iii\right)$ and ends with case $\left(viii\right)$. We assume in the proof of the theorem that $\left\{x_n\right\}$ and $\left\{y_n\right\}$ are sequences of unit vectors in $\mathcal{H}$ such that ${lim}_{n\to \infty }(A-\alpha )x_n=0={lim}_{n\to \infty }(B^{*}-\overline{\lambda })y_n=0$. (Thus, $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$.)

Theorem 3.2.

(iii). If ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then ${▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m=0={▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$ and for each $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$ there exists a θ, $0\le \theta <2\pi$, such that $\left|\alpha \lambda \right|=2(1-cos\theta )$.

(iv) . If ${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then ${▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m=0={\nabla }_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$ and there exists a non-zero scalar β such that

$$\lambda ={\displaystyle \frac{-{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\overline{\beta }}}\mathrm{and}\alpha ={\displaystyle \frac{{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\overline{\beta }}}.$$

(v). If ${▵}_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$, then ${▵}_{A^{*}-\overline{\lambda },I-\lambda A}^m=0={\nabla }_{B-\alpha ,I-\overline{\alpha }{B}^{*}}^m\left(I\right)$ and there exists a non-zero scalar β such that

$$\lambda ={\displaystyle \frac{-{\left|\beta \right|}^2\pm i\sqrt{4\beta (1-\beta )-{\left|\beta \right|}^4}}{2\beta }}\mathrm{and}\alpha ={\displaystyle \frac{{\left|\beta \right|}^2\pm i\sqrt{4\beta (1-\beta )-{\left|\beta \right|}^4}}{2\beta }}.$$

(vi). If ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then

$${\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right),$$

$0\in {\sigma }_a\left(A\right)$ (resp., $0\in {\sigma }_a\left(B^{*}\right)$) implies $B^{*}$ (resp., A) m-symmetric, and ${\sigma }_a\left(A\right)\subseteq \{r+\lambda :\overline{\lambda }\in {\sigma }_a\left(B^{*}\right),r\mathrm{real}\}$.

(vii). If ${\delta }_{{\delta }_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$ and $A-I,B+I$ are non-nilpotent, then

$${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={♢}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$$

and there exists a scalar β such that

$$\lambda ={\displaystyle \frac{-\overline{\beta }\pm i\sqrt{4(\beta -1)-{\overline{\beta }}^2}}{2}}\mathrm{and}\alpha ={\displaystyle \frac{\overline{\beta }\pm i{\sqrt{4(\beta -1)-\overline{\beta }}}^2}{2}}.$$

(viii). If ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$ and $A-I,B+I$ are non-nilpotent, then

$${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={♢}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$$

and there exists a scalar β such that

$$\lambda ={\displaystyle \frac{-\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}\mathrm{and}\alpha ={\displaystyle \frac{\beta \pm i\sqrt{4(\overline{\beta }-1)}-{\beta }^2}{2}}.$$

Furthermore, the spectra $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist at most of two points in each of the (six) cases.

Proof. The proof of $\left(iv\right)$ and $\left(v\right)$, and $\left(vii\right)$ and $\left(viii\right)$, is similar: we prove $\left(iii\right),\left(iv\right),\left(vi\right)$ and $\left(vii\right)$.

Case(iii). We start by proving that $I-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ are m-isometric. By definition

$$\begin{array}{ccc}& & {▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)A^{*p}\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)A^k\right)B^k{B}^{*p}\right].\hfill \end{array}$$

Hence, for all $x\in \mathcal{H}$,

$$\begin{array}{ccc}\hfill 0& =& \left[\underset{n\to \infty }{lim}{▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)(x\otimes x_n)\right]x_n\hfill \\ & =& C_{\sum }{A}^{*p}{A}^{k}x\underset{n\to \infty }{lim}\langle B^{*p}{x}_n,B^{*k}{x}_n\rangle \hfill \\ & =& C_{\sum }{\left(\overline{\lambda }{A}^{*}\right)}^p{\left(\lambda A\right)}^{k}x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^j{(I-\lambda A)}^{j}x\hfill \\ & =& {▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)x.\hfill \end{array}$$

Hence ${▵}_{I-\overline{\lambda }{A}^{*},I-\lambda A}^m\left(I\right)=0$. A similar argument, working this time with

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{▵}_{{▵}_{B,A},{▵}_{B^{*},A^{*}}}^m\left(I\right)(x\otimes y_n)\right]y_n\hfill \\ & =& C_{\sum }{B}^p{B}^{*k}x\underset{n\to \infty }{lim}\langle A^p{y}_n,A^k{y}_n\rangle ,\hfill \end{array}$$

proves ${▵}_{I-\alpha B,I-\overline{\alpha }{B}^{*}}^m\left(I\right)=0$.

We claim that ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$, hence also $\sigma \left(A\right)$ and $\sigma \left(B\right)$, consist at most of two points. Assume to the contrary that (alongwith $\overline{\mu }$) $\overline{\nu }\in {\sigma }_a(B^{*}-\overline{\lambda })$; $\nu \ne \mu$. Then ${\sigma }_a(I-(\lambda +\nu )A)\subseteq \partial \mathtt{D}$ and

$${\sigma }_a(I-(\lambda +\nu )A)={\sigma }_a(I-(\lambda +\mu )A)+(\mu -\nu ){\sigma }_a\left(A\right).$$

Since ${\sigma }_a\left(A\right)\ne \left\{0\right\}$ and ${\sigma }_a(I-(\lambda +\mu )A)\subseteq \partial \mathtt{D}$, ${\sigma }_a((I-(\lambda +\nu )A)$ is a translate by a non-zero scalar of points in $\partial \mathtt{D}$, hence not a point in the boundary of $\mathtt{D}$. This being a contradiction, our claim is proved.

The operators $I-\lambda A$ and $I-\overline{\alpha }{B}^{*}$ being m-isometric, the spectra ${\sigma }_a(I-\lambda A)$, ${\sigma }_a(I-\overline{\alpha }{B}^{*})$ are subsets of $\partial \mathtt{D}$ and the operators $I-\lambda A$, $I-\overline{\alpha }{B}^{*}$ are (left invertible, hence) invertible.The spectral mapping theorem implies that

$$\sigma \left({▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}\right)=0⟹\alpha \lambda +\overline{\alpha \lambda }={\left|\alpha \lambda \right|}^2$$

for $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$. Since $\sigma (I-\lambda A)\subseteq \left\{e^{i\theta }\right\}$,equivalently, $\lambda \alpha \subseteq \{1-e^{i\theta }\}$, for each $\alpha \in \sigma \left(A\right)$ and $\lambda \in \sigma \left(B\right)$ there exists a $0\le \theta <2\pi$ such that $\left|\alpha \lambda \right|=2(1-cos\theta )$.

Case $\left(iv\right)$. We start by proving ${▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m=0={\nabla }_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$ for all $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. Arguing as above,

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)(x\otimes x_n\right]x_n,x\in \mathcal{H}\hfill \\ & =& C_{\sum }{A}^{*p}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{x}_n,B^{*k}{x}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right){\left(\overline{\lambda }{A}^{*}\right)}^p\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^j{(A-\lambda )}^{j}x\hfill \\ & ⟹& {▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\nabla }_{{▵}_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)(x\otimes y_n)\right]y_n,x\in \mathcal{H}\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right)B^p\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)B^{*(j-k)}\right)x\underset{n\to \infty }{lim}\langle A^p{y}_n,A^k{y}_n\rangle \right)\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}j\\ p\end{array}\right){\left(\alpha B\right)}^p\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\overline{\alpha }}^k{B}^{*(j-k)}\right)x\hfill \\ & ⟹& {\nabla }_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)=0.\hfill \end{array}$$

If ${▵}_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0$, then (by the spectral mapping theorem)

$$1-(1-\overline{\lambda }\overline{\alpha })(\alpha -\lambda )=0;\mathrm{all}\alpha \in {\sigma }_a\left(A\right)\mathrm{and}\overline{\lambda }\in {\sigma }_a\left(B^{*}\right).$$

Consequently, there exists a non-zero scalar $\beta$ such that

$$1-\overline{\lambda }\overline{\alpha }={\displaystyle \frac{1}{\overline{\beta }}},\alpha -\lambda =\beta ⟹\beta {\lambda }^2+{\left|\beta \right|}^2\lambda +(1-\beta )=0$$

and this (solving for $\lambda$ and $\alpha$) implies

$$\lambda ={\displaystyle \frac{-{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\beta }}\mathrm{and}\alpha ={\displaystyle \frac{{\left|\beta \right|}^2\pm i\sqrt{4\overline{\beta }(1-\overline{\beta })-{\left|\beta \right|}^4}}{2\beta }},$$

$\sigma \left(A\right)={\sigma }_a\left(A\right)$ and $\sigma \left(B\right)={\sigma }_a\left(B\right)$ consist at most of two points. (In particular, if $0\in \sigma \left(B\right)$, then $\beta =1$, $\sigma \left(A\right)=\{0,1\}$ and $\sigma \left(B\right)=\{-1,0\}$.)

Case $\left(vi\right)$. We have

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)(x\otimes x_n)\right]x_n,x\in \mathcal{H}\hfill \\ & =& C_{{\sum }_1}{A}^{*(m-j-p)}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{y}_n,B^{*k}{y}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\overline{\lambda }}^p{A}^{*(m-j-p)}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(A^{*}-\overline{\lambda })}^{m-j}{(A-\lambda )}^{j}x\hfill \\ & ⟹& {\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0.\hfill \end{array}$$

Again,

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\delta }_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)(x\otimes y_n)\right]y_n\hfill \\ & =& C_{{\sum }_1}{B}^{m-j-p}{B}^{*(j-k)}x\underset{n\to \infty }{lim}\langle A^p{y}_n,A^k{y}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^j{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\alpha }^p{B}^{m-j-p}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)({\overline{\alpha }}^k{B}^{*(j-k)}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(B-\alpha )}^{m-j}{(B^{*}-\overline{\alpha })}^{j}x\hfill \\ & ⟹& {\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)=0.\hfill \end{array}$$

Thus

$${\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0={\delta }_{B-\alpha ,B^{*}-\overline{\alpha }}^m\left(I\right)$$

for all $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$.

Evidently, $A-\lambda$ and $B^{*}-\overline{\alpha }$ are m-symmetric. Furthermore, since ${\delta }_{A^{*}-\overline{\lambda },A-\lambda }^m\left(I\right)=0$ implies $\overline{(\alpha -\lambda )}-(\alpha -\lambda )=0$, $\Im \alpha =\Im \lambda$ for all $\alpha$ and $\lambda$. Hence

$${\sigma }_a\left(A\right)\subseteq \{r+\lambda :\overline{\lambda }\in {\sigma }_a\left(B^{*}\right),r\mathrm{real}\}.$$

Trivially, if $0\in {\sigma }_a\left(A\right)$, then $B^{*}$ is m-symmetric, and if $0\in {\sigma }_a\left(B^{*}\right)$, then A is m-symmetric. The set $\left\{\sigma \right(A)-\lambda \}$ being a compact set, there exist scalars ${\beta }_i$, $1\le i\le 2$, such that (${\beta }_1\le {\beta }_2$ and) $\sigma \left(A\right)-\lambda \subseteq [{\beta }_1,{\beta }_2]$. Suppose that there exist scalars $\overline{\mu },\overline{\nu }\in {\sigma }_a(B^{*}-\overline{\lambda })$, $\mu \ne \nu$. Then ($\overline{\lambda +\mu }$ and $\overline{\lambda +\nu }\in {\sigma }_a\left(B^{*}\right)$, hence) ${\sigma }_a(A-\lambda -\mu )=\sigma (A-\lambda -\mu )$ and ${\sigma }_a((A-\lambda -\nu )=\sigma (A-\lambda -\nu )$ are subsets of $[{\beta }_1,{\beta }_2]$. Since

$$\begin{array}{ccc}& & \sigma (A-\lambda -\nu )=\sigma (A-\lambda -\mu )+(\mu -\nu ),\mathrm{where}0\ne \mu -\nu \mathrm{is}\mathrm{real},\hfill \\ & & {\beta }_1-(\mu -\nu )<{\beta }_1\mathrm{if}(\mu -\nu )>0\mathrm{and}{\beta }_2-(\mu -\nu )>{\beta }_2\mathrm{if}(\mu -\nu )>0,\hfill \end{array}$$

we have a contradiction. Conclusion: $\sigma \left(A\right)$ and $\sigma \left(B\right)$ consist, at most, of two points.

Case(vii). In this case

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)(x\otimes x_n)\right]x_n,x\in \mathcal{H}\hfill \\ & =& C_{{\sum }_1}{A}^{*p}{A}^{j-k}x\underset{n\to \infty }{lim}\langle B^{*p}{x}_n,B^{*k}{x}_n\rangle \hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\left(\overline{\lambda }\right)}^p{A}^{*p}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right){\lambda }^k{A}^{j-k}\right)x\hfill \\ & =& \sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{A}^{*})}^j{(A-\lambda )}^{j}x\hfill \\ & ⟹& {\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0.\hfill \end{array}$$

Let

$$C_{{\sum }_0}=C(m,j,p,k)=\sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left[\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)\right)\right].$$

Then, for all $x\in \mathcal{H}$,

$$\begin{array}{ccc}\hfill 0& =& \underset{n\to \infty }{lim}\left[{♢}_{{\delta }_{B,A},{\delta }_{B^{*},A^{*}}}^m\left(I\right)(x\otimes y_n)\right]y_n\hfill \\ & =& C_{{\sum }_0}{B}^p{B}^{*(j-k)}x\underset{n\to \infty }{lim}\langle A^p{y}_n,A^k{y}_n\rangle \hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right)\left(\sum _{p=0}^{m-j}{(-1)}^p\left(\begin{array}{c}m-j\\ p\end{array}\right){\alpha }^p{B}^p\right)\left(\sum _{k=0}^j{(-1)}^k\left(\begin{array}{c}j\\ k\end{array}\right)({\overline{\alpha }}^k{B}^{*(j-k)}\right)x\hfill \\ & =& \sum _{j=0}^m\left(\begin{array}{c}m\\ j\end{array}\right){(I-\overline{\lambda }{B}^{*})}^{m-j}{(B-\lambda )}^{j}x\hfill \\ & ⟹& {♢}_{I-\lambda B,B^{*}-\overline{\lambda }}^m\left(I\right)=0.\hfill \end{array}$$

This proves ${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0={♢}_{I-\alpha B,B^{*}-\overline{\alpha }}^m\left(I\right)$.

The conclusion ${\delta }_{I-\overline{\lambda }{A}^{*},A-\lambda }^m\left(I\right)=0$ implies (by the spectral mapping theorem) that $(1-\overline{\alpha }\overline{\lambda })-(\alpha -\lambda )=0$ for all non-zero $\alpha \in {\sigma }_a\left(A\right)$ and $\overline{\lambda }\in {\sigma }_a\left(B^{*}\right)$. Hence there exists a non-zero scalar $\beta$ such that

$$\begin{array}{ccc}& & \alpha -\lambda =\beta =1-\overline{\alpha }\overline{\lambda }⟹{\lambda }^2+\beta \lambda +(\overline{\beta }-1)=0\hfill \\ & ⟹& \lambda ={\displaystyle \frac{-\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}\mathrm{and}\hfill \\ & & \alpha ={\displaystyle \frac{\beta \pm i\sqrt{4(\overline{\beta }-1)-{\beta }^2}}{2}}.\hfill \end{array}$$

Evidently, ${\sigma }_a\left(A\right)=\sigma \left(A\right)$ and ${\sigma }_a\left(B\right)=\sigma \left(B\right)$ consist, at most, of two points (which, if $0\in \sigma \left(B\right)$, so that $\beta =1$, are respectively the sets $\{0,1\}$ and $\{-1,0\}$; observe that $I-A$ is m-nilpotent if $0\in \sigma \left(B\right)$ and $I+B$ is m-nilpotent if $0\in \sigma \left(A\right)$ ). □

The conclusions of Theorem 3.2 help in building a picture of the operators $A,B$ satisfying $D_{d_1,d_2}^m\left(I\right)=0$. The following examples consider cases $\left(iii\right)$, $\left(iv\right)$, $\left(vi\right)$ and $\left(vii\right)$ of the theorem.

Example 3.3.

0 may be in both ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$ in the case in which ${▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)=0$: consider, for example, the operators $A,B$ such that ${\sigma }_a\left(A\right)=\{0,{\displaystyle \frac{1-e^{-i\theta }}{a}}\}$, ${\sigma }_a\left(B^{*}\right)=\{0,a\}$, $A=0I_1\oplus {\displaystyle \frac{1-e^{-i\theta }}{a}}{I}_2$ and $B=0I_1\oplus a{I}_2$, $0\ne a$ real and $0<\theta <2\pi$. It is seen that

$$\begin{array}{ccc}& & {▵}_{{▵}_{A^{*},B^{*}},{▵}_{A,B}}^m\left(I\right)\hfill \\ & =& C_{\sum }{\left(0I_1\oplus {\displaystyle \frac{1-e^{i\theta }}{a}}{I}_2\right)}^p{\left(0I_1\oplus {\displaystyle \frac{1-e^{-i\theta }}{a}}{I}_2\right)}^k\left(0I_1\oplus a^{p+k}{I}_2\right)\hfill \\ & =& C_{\sum }\left(0I_1\oplus {(1-e^{i\theta })}^p{(1-e^{-i\theta })}^k{I}_2\right)=\sum _{j=0}^m{(-1)}^j\left(\begin{array}{c}m\\ j\end{array}\right){(I_1\oplus e^{i\theta }{I}_2)}^j{(I_1\oplus e^{-i\theta }{I}_2)}^j\hfill \\ & =& 0.\hfill \end{array}$$

Example 3.4.

If ${▵}_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, then 0 may belong to both ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$ (consider, for example the operators $A=I_1\oplus 0I_2$ and $B=0I_1\oplus -I_2$), or, 0 may just be in one of ${\sigma }_a\left(A\right)$ and ${\sigma }_a\left(B^{*}\right)$ (consider the operators $A=I_1\oplus (1-\sqrt{{\displaystyle \frac{3}{2}}})I_2$ and $B=0I_1\oplus (-1-\sqrt{{\displaystyle \frac{3}{2}}})I_2$).

Example 3.5.

In the case in which ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$, $0\in {\sigma }_a\left(B^{*}\right)$ implies A is m-symmetric (thus $\sigma \left(A\right)$ is real) and $0\in {\sigma }_a\left(A\right)$ implies $B^{*}$ is m-symmetric (thus $\sigma \left(B\right)$ is real). If $a,b,r_1,r_2$ are real numbers, $\sigma \left(A\right)=\{a,r_1+ib\}$, and $\sigma \left(B\right)=\{0,r_2+ib\}$, then $A=a{I}_1\oplus (r_1+ib)I_2$ and $B=0I_1\oplus (r_2+ib)I_2$ is an example of a pair of operators satisfying ${\delta }_{{\delta }_{A^{*},B^{*}},{\delta }_{A,B}}^m\left(I\right)=0$. Again, if we let $\beta =i$ and $A=I_1\oplus (-1-i)I_2$, $B=(1+i)I_1\oplus -I_2$ (in the proof of case $\left(vii\right)$), then ${\delta }_{A,B}\left(I\right)=-iI$, ${▵}_{A^{*},B^{*}}\left(I\right)=iI$ and ${\delta }_{{▵}_{A^{*},B^{*}},{\delta }_{A,B}}\left(I\right)=(I-L_{iI}{R}_{-iI})\left(I\right)=0$.