# Structure ofk-Quasi-m,n-Isosymmetric Operators

Authors: Sid Ahmed Ould Ahmed Mahmoud; El Moctar Ould Beiba; Sidi Hamidou Jah; Maawiya Ould Sidi
Journal: Journal of Mathematics
 (2022)

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## Abstract

The investigation of new operators belonging to some specific classes has been quite fashionable since the beginning of the century, and sometimes it is indeed relevant. In this study, we introduce and study a new class of operators calledk-quasi-m,n-isosymmetric operators on Hilbert spaces. This new class of operators emerges as a generalization of the m,n-isosymmetric operators. We give a characterization for any operator to be k-quasi-m,n-isosymmetric operator. Using this characterization, we prove that any power of an k-quasi-m,n-isosymmetric operator is also an k-quasi-m,n-isosymmetric operator. Furthermore, we study the perturbation of an k-quasi-m,n-isosymmetric operator with a nilpotent operator. The product and tensor products of two k-quasi-m,n-isosymmetries are investigated.


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## Body

## 1. Introduction

Letℬℋ be the C∗-algebra of bounded linear operators on a complex Hilbert space ℋ and let (1)ℂu,v=Pu,v=∑0≤k≤m0≤l≤nck,lukvl,ck,l∈ℂ.ForR∈ℬℋ, we will write ranR, kerR, and R∗ the range, the kernel (or null space), and the adjoint of R, respectively. Also, σpR,σapR,σR, and σsR denote the point spectrum, the approximate spectrum, the spectrum, and the surjective spectrum R.The hereditary functional calculus defines(2)PR=∑0≤k≤m0≤l≤nck,lR∗lRk,forP∈ℂu,v.It is easy to check that forP∈ℂu,v and Q∈ℂu,v, we have(3)PQR=∑k,lck,lR∗lQRRk=∑k,ldk,lR∗lPRRk.Recall that an operatorR∈ℬℋ is called a hereditary root or simply root of P∈ℂu,v if PR=0. For more details on the hereditary functional calculus, we refer the reader to [4, 5].In recent years, the concepts ofm-isometric operators and the related classes of operators, namely, n-quasi-m-isometries, m,C-isometries, and n-quasi-m,C-isometries have received substantial attention. Several authors have been introduced, and these classes of operators are studied intensively in the papers [12–18], [20–23, 28, 32]. It has been proved that some products of m-isometries are again m-isometry [8, 11], the powers of an m-isometry are m- isometries, and the perturbation of m-isometries by nilpotent operators has been studied in [6, 9, 10]. The dynamics of m-isometries has been explored in [7]. Almost all of these properties have been extended to n-quasi-m-isometric operators, m,C-isometries, and n-quasi-m,C-isometries. The reader can refer to the papers [12–14, 17, 18, 20–23, 32] for more details.LetR∈ℬℋ and m, n, and k be positive integers.(1)
R is called m-isometry [1–3] if it is a root of Pu,v=vu−1m, that is,(4)∑0≤k≤m−1kmkR∗kRk=0.(2)
R is called m-symmetry [15, 27] if it is a root of Pu,v=v−um, that is,(5)∑0≤k≤m−1kmkR∗m−kRk=0.(3)
R is called k-quasi-m-isometry [19, 20, 31] if it is a root of Pu,v=vkvu−1muk, that is,(6)R∗k∑0≤j≤m−1jmjR∗jRjRk=0.(4)
R is called k-quasi-n-symmetric [33] if it is a root of Pu,v=vkv−umuk, that is,(7)R∗k∑0≤j≤n−1jnjR∗n−jRjRk=0.(5)
R is called m,n-isosymmetric [29, 30] if it is a root of(8)Pu,v=vu−1mv−un,that is,(9)∑0≤j≤m−1jmjR∗m−j∑0≤k≤n−1knkR∗n−kRkRm−j.=∑0≤k≤n−1knkR∗n−k∑0≤j≤m−1jmjR∗m−jRm−jRk=0.Forn,m∈ℕ, set αnu,v=v−un, βmu,v=vu−1m, and(10)γm,nu,v=βmu,vαnu,v=αnu,vβmu,v.ForR∈ℬℋ, we have(11)αnR=∑0≤k≤n−1knkR∗n−kRk,βmR=∑0≤k≤m−1kmkR∗m−kRm−k,γm,nR=∑0≤k≤n−1knkR∗n−kβmRRk,or∑0≤k≤m−1kmkR∗m−kαnRRm−k..It is easy to see that(12)γm+1,nR=R∗γm,nRR−γm,nR,γm,n+1R=R∗γm,nR−γm,nRR..It is well known that a common way to prepare a scientific study is to introduce some new mathematical objects and then state several results related to them. The investigation of new operators belonging to some specific classes has been quite fashionable since the beginning of the century, and sometimes it is indeed relevant. The motivation of this study is to introduce and study the concept ofk-Quasi-m,n-isosymmetric operators on Hilbert spaces. This new class of operators emerges as a generalization of the m,n-isosymmetric operators. It is proved that there is an operator which is a k-quasi-n;m-isosymmetric operator, but not n,m-isosymmetric, and thus, the proposed new class is larger than the class of n,m-isosymmetric operators. In section two, we give a matrix characterization of k-quasi-m,n-isosymmetric operators in terms of m,n-isosymmetric operators. We give some results related to this class by using this matrix representation. In section three, we investigate some spectral properties of k-quasi-m,n-isosymmetric operators; in particular, we explore conditions for k-quasi-n,m-isosymmetric operators to be k-quasi-m-isometric operators or k-quasi-n-symmetric operators. Finally, in section forth, we study the sum of an k-quasi-m,n-isosymmetric operator with a nilpotent operator. We also study the product and tensor product of two k-quasi-m,n-isosymmetric operators.


## 2. Structure ofk-Quasi-m,n-Isosymmetric Operators

In the present section, we give the definition and basic properties ofk-quasi-m,n-isosymmetric operators. The obtained results improve and generalize some works on m-isometries, n-quasi-m-isometric, n-symmetries, and k-quasi-n-symmetric operators.Definition 1.
An operatorR∈ℬℋ is said to be k-quasi-m,n-isosymmetric operator if R is a root of the polynomial(13)Pm,n,ku,v=vkγm,nu,vuk,for some positive integers m, n, and k, or equivalently,(14)R∗k∑0≤j≤m−1jmjR∗m−j∑0≤r≤n−1rnrR∗n−rRrRm−jRk=R∗k⋅∑0≤r≤n−1rnrR∗n−r∑0≤j≤m−1jmjR∗m−jRm−jRrRk=0.Example 1.
(i)
Everym-isometric operator is an k-quasi-m,n-isosymmetric, and every n-symmetric operator is an k-quasi-m,n-isosymmetric operator(ii)
Everym,n-isosymmetric operator is an k-quasi-m,n-isosymmetric operator(iii)
Everyk-quasi-m-isometric operator is an k-quasi-m,n-isosymmetric operator(iv)
Everyk-quasi-n-symmetric operator is an k-quasi-m,n-isosymmetric operatorRemark 1.
Sinceq≥k, r≥m, and s≥n, the polynomial vkvu−1mv−unuk divides vqvu−1rv−usuq; it follows that if r is a k-quasi-m,n-isosymmetric operator, then it is a q-quasi-r,s-isosymmetric operator.Remark 2.
(1)
Ifn=m=k=1, 1-quasi-1,1-isosymmetric operator is a quasi-isosymmetric, i.e., an operator R is quasi-isosymmetric if and only if(15)R∗R∗2R−R∗R2−R∗+RR=R∗3R2−R∗2R3−R∗2R+R∗R2=0.(2)
An operatorR is quasi-2,1-isosymmetric if and only if(16)R∗R∗3R2−R∗2R3−2R∗2R+2R∗R2+R∗−RR=0.(3)
An operatorT is quasi-1,2-isosymmetric if and only if(17)R∗R∗3R−2R∗2R2+R∗R3−R∗2+2R∗R−R2R=0.Remark 3.
In the following example, we show that there is an operator which isk-quasi-m,n-isosymmetric, but not m,n-isosymmetric for some positive integers n,m,k≥1, and therefore, the proposed new class is large than the class of n,m-isosymmetric operators.Example 2.
LetR=001000000∈ℬℂ3. The direct calculation shows that(18)R∗R∗2R−R∗R2−R∗+RR=0andR∗2R−R∗R2−R∗+R≠0.
Thus,R is a quasi-isosymmetric but not isosymmetric operator.Example 3.
(1)
LetR=II00∈ℬℋ⊕ℋ. A simple calculation shows that(19)αnR≠0,βmR≠0,andγm,n,kR=0.Thus,R is a k-quasi-m,n-isosymmetric operator; however, R is neither m-isometry nor n-symmetry.(2)
LetR=0I00∈ℬℋ⊕ℋ. A simple calculation shows that(20)R∗R∗R−IR≠0,R∗R∗−RR≠0andR∗R∗2R−R∗R2−R∗+RR=0.Thus,R is a quasi-isosymmetric operator; however, R is neither quasi-isometry nor quasi-symmetry.Remark 4.
The following inclusions hold:(21)m−isometry⊈k−quasi​−m−​isometry⊈k−quasi−m,n−isosymmetry,n−symmetry⊈k−quasi−n−symmetry⊈k−quasi−m,n−isosymmetry,m,n−isosymmetry⊈k−quasi−m,n−isosymmetry.Proposition 1.
LetR∈ℬℋ, then the following statements are equivalent:(1)
R is k-quasi-m,n-isosymmetric operator(2)
γm,nRy|y=0,∀y∈ranRk¯Proof.
(22)Ris ak−quasi−m,n−isosymmetric⇔R∗kγm,nRRk=0,⇔R∗kγm,nRRkx|x=0,∀x∈ℋ⇔γm,nRRkx|Rkx=0,∀x∈ℋ⇔γm,nRy|y=0,∀y∈ranRk¯.Corollary 1.
LetR∈ℬℋ with k and q are two nonnegative integers such thatranRk¯=ranRq¯, then R is k-quasi-m,n-isosymmetric operator if and only if R is q-quasi-m,n-isosymmetric operator.Proof.
Straightforward from Proposition 1.Theorem 1.
LetR∈ℬℋbek-quasi-m,n-isosymmetric operator. IfkerR∗q=kerR∗q+1for some1≤q≤k−1, thenRisq-quasi-m,n-isosymmetric.Proof.
From the assumptionskerR∗q=kerR∗q+1 and q≤k−1, it follows that kerR∗q=kerR∗k. Thus, ranRk¯=ranRq¯. Applying Corollary 1, we get the desired conclusion.Proposition 2.
Letℳbe a closed subspace ofℋwhich reducesR. IfRisk-quasi-(m,n)-isosymmetric, thenR|ℳisk-quasi-m,n-isosymmetric.Proof.
LetS=R|ℳ be the restriction of R to ℳ. On the one hand, we have(23)γm,nS=γm,nR|ℳ.
On the other hand, we haveranSk¯ℳ⊂ranRk¯, where ranSk¯ℳ is the closer of ranSk in ℳ. Thus, γm,nSy|y=0,∀y∈ranSk¯ since γm,nRy|y=0,∀y∈ranRk¯. Therefore, by statement (2) of Proposition 1, S=R|ℳ is k-quasi-m,n-isosymmetric on ℳ.The following theorem characterizes the members ofk-quasi-m,n-isosymmetric operators.Theorem 2.
LetR∈ℬℋsuch thatℋ≠ranRk¯. Then, the following properties are equivalent:(1)
Risk-quasi-m,n-isosymmetric operator(2)
R=R1R20R3onℋ=ranRk¯⊕kerR∗k, whereR1=R|ranRk¯is anm,n-isosymmetric operator andR3k=0Proof.
1⇒2. By taking into account the matrix representation related to the decomposition ℋ=ranRk¯⊕kerR∗k as R=R1R20R3, we get(24)R1000=RP|ranRk¯=P|ranRk¯RP|ranRk¯,where P|ranRk¯ is the orthogonal projection onto ranRk¯.
From the condition thatR is k-quasi-m,n-isosymmetric operator, we have(25)P|ranRk¯∑0≤j≤m−1jmkR∗jαnRRjP|ranRk¯=0,or(26)P|ranRk¯∑0≤r≤n−1rnrR∗n−rβmRRrP|ranRk¯=0.
From this, we deduce that(27)∑0≤j≤m−1jmjR1∗jαnR1R1j=0,or(28)∑0≤r≤n−1rnrR1∗n−rβmR1R1r=0.
Therefore,R1 is m,n-an isosymmetric operator.
On the other hand, letz=z1+z2∈ℋ=ranRk¯⊕kerR∗k. The direct calculation shows that(29)R3kz2|z2=RkI−P|ranRk¯z,I−P|ranRk¯z=I−P|ranRk¯z,R∗kI−P|ranRk¯z=0.
So that,R3k=0.
2⇒1 Assume that R=R1R20R3 onto ℋ=ranRk¯⊕kerR∗k, with(30)γm,nR1=∑0≤r≤n−1rnrR1∗n−rβmR1R1r=∑0≤j≤m−1jmkR1∗m−jαnR1R1m−j=0,and R3k=0.
Direct calculation shows thatRk=R1k∑0≤j≤k−1R1jR2R3k−1−j0R3k=R1kwk00 and therefore, R∗k=R1∗k0wk∗0. Moreover,(31)RkR∗k=R1kR1∗k+wkwk∗000=Ck000,where Ck=R1kR1∗k+wkwk∗=Ck∗.(32)RkR∗k∑0≤j≤m−1jmjR∗jαnRRjRkR∗k=Ck000αm,nR1ABDCk000=Ckαm,nR1Ck000=0.
This means that(33)RkR∗k∑0≤j≤m−1jmjR∗jαnRRjRkR∗kx|x=0⇒R∗k∑0≤j≤m−1jmjR∗jαnRRjRkR∗kx|R∗kx=0⇒R∗k∑0≤j≤m−1jmjR∗jαnRRjRk=0onranR∗k¯=kerRk⊥.
Obviously,R∗k∑0≤j≤m−1jmjR∗jαnRRjRk=0 on kerRk, and consequently,(34)R∗k∑0≤j≤m−1jmjR∗jαnRRjRk=0,on ℋ=kerRk⊕kerRk⊥.
Therefore,R is a k-quasi-m,n-isosymmetric operator.Corollary 2.
IfR∈ℬℋis ank-quasi-m,n-isosymmetric such thatranRkis dense, thenRis anm,n-isosymmetric.Proof.
This is a direct consequence of Theorem2.Corollary 3.
IfR∈ℬℋis an invertiblek-quasi-m,n-isosymmetric operator, thenR−1is ak-quasi-m,n-isosymmetric operator.Proof.
Under the assumption thatR is an invertible k-quasi-m,n-isosymmetric operator, it follows that R is an m,n-isosymmetric operator, and so is R−1 by Theorem 2.4 in [24]. Therefore, R−1 is a k-quasi-m,n-isosymmetric operator.Corollary 4.
LetR∈ℬℋbe ak-quasi-m,n-isosymmetric operator such thatranRk¯≠ℋ. If the restrictionR1=R|ranRk¯is invertible, thenRis similar to a direct sum of anm,n-isosymmetric operator and a nilpotent operator with an index of nilpotence less than or equalk.Proof.
Consider the matrix decomposition ofR as(35)T=R1R20R3onℋ=ranRk¯⊕kerR∗k.
Then,R1 is an m,n-isosymmetric operator by Theorem 2. Since R1 is invertible, we have 0∉σR1, which implies σR1∩σR3=∅. By Rosenblum’s corollary [26], it follows that there exists A∈ℬℋ for which R1A−AR3=R2. Therefore, we have(36)R=I−A0IR100R3IA0I=IA0I−1R100R3IA0I=IA0I−1R1⊕R3IA0I.It was proved that power ofm-isometric (resp m-symmetric) operator is again m-isometric (resp m-symmetric) operator. The following corollary shows that the same property holds for k-quasi-m,n-isosymmetric operators.Corollary 5.
IfR∈ℬℋis ak-quasi-m,n-isosymmetric operator, thenRqis also ak-quasi-m,n-isosymmetric operator for any positive integerq.Proof.
LetranRk¯=ℋ, then R is an m,n-isosymmetric and so is Rq for any q∈ℕ by Theorem 2.4 in [24]. If ranRk¯≠ℋ, we can use the decomposition of R=R1R20R3 on ℋ=ranRk¯⊕kerR∗k, where R1 is an m,n-isosymmetric operator and so is R1q. On the other hand, observing that(37)Rq=R1q∑0≤j≤q−1R1jR2R3q−1−j0R3q,it follows that Rq is k-quasi-m,n-isosymmetric operator by applying Theorem 2.Theorem 3.
LetR=R1R20R3∈ℬℋ⊕ℋ. IfR1is a surjectivem,n-isosymmetric operator andR3k=0, thenRis similar to ank-quasi-m,n-isosymmetric operator.Proof.
Under the assumptions onR1 and R3, we have σsR1∩σapR3=∅. From the axiom c in Theorem 3.5.1 in [16], it follows that there exists some operator A∈ℬℋ such that R1A−AR3=R2. Hence,(38)IA0IR1R20R3=R100R3IA0I.
From this, we deduce thatR is similar to W=R100R3.
By using the facts thatR1 is a m,n-isosymmetric and R3k=0, we can obtain(39)W∗k∑0≤j≤m−1jmjW∗jαnWWjWk=R1∗k∑0≤j≤m−1jmjR1∗jαnR1R1jR1k000=0.
Consequently,R is similar to a k-quasi-m,n-isosymmetric operator.


## 3. Spectral Properties

In this section, we investigate some spectral properties ofk-quasi-m,n-isosymmetric operators.LetP∈ℂu,v and R∈ℬℋ. Let u,v∈ℋ and λ,μ∈ℂ be such that Ru=λu and Rv=μv, we get by a little calculation that(40)PRu|v=Pλ,μ¯u|v.It follows from (40) that if R is a root of p and λ is a spectrum point of R, then pλ,λ¯=0.We have the following theorem.Theorem 4.
LetP∈ℂu,vandR∈ℬℋ. Ifλ,μ∈σapRandunn∈ℕ,vnn∈ℕbe two sequences of unit vectors such thatR−λun⟶0andR−μvn⟶0asn⟶∞, then(41)PRun,vn−pλ,μ¯un|vn⟶0asn⟶∞.Proof.
Foru∈ℋ, α∈ℂ, and r∈ℕ, we have(42)Rr−αru=∑0≤j≤r−1Rjαr−j−1R−αU.
Thus, we get(43)Rr−αru≤∑0≤j≤r−1Rjαr−j−1R−αu.
Letm,n∈ℕ. Set(44)wk,n=Rnuk−λnuk,zk,m=Rmvk−μmvk.
Applying (43) to λ and μ, respectively, we obtain(45)wk,n≤∑0≤j≤n−1Rjλn−j−1Ruk−λuk,zk,m≤∑0≤j≤m−1Rjμm−j−1Rvk−μvk.
Thus,wk,n⟶0 and zk,m⟶0 as k⟶0.
On the other hand, it is easy to verify(46)R∗mRnuk,vk−μ¯mλnuk,vk=wk,n,zk,m+wk,n,μmyk+λnuk,zk,m.
Since the right side of the previous equality tends to0, we obtain(47)R∗mRnuk,vk−μ¯mλnuk,vk⟶0.
Taking linear combinations, we get(48)PRuk,vk−Pλ,μ¯uk,vk⟶0.Corollary 6.
LetP∈ℂu,vandR∈ℬℋ. Letλ,μ∈σapRandukk∈ℕ⊂ℋandvkk∈ℕ⊂ℋsuch thatuk=vk=1,Ruk−λuk⟶0, andRvk−μvk⟶0. IfRis a root ofP, then one of the following statements holds:(1)
uk|vk⟶0ask⟶∞(2)
Pλ,μ¯=0Proof.
SinceR is a root of P, we have PRuk,vk=0,∀k. This yields, from (41), the desired conclusion.Corollary 7.
LetP∈ℂu,vandR∈ℬℋ. Letu,v∈ℋbe nonzero vectors andλ,μ∈ℂbe such thatRu=λuandRv=μv. IfRis a root ofP, then one of the following two statements holds:(1)
u,v=0(2)
Pλ,μ¯=0Proof.
Takeuk=u and vk=v in Corollary 6.As a consequence of Theorem4, we have the following two results due to Stankus [30].Corollary 8.
(Proposition 21 [30]). LetR∈ℬℋandP∈ℂu,v. Then,(49)σapR⊂λ∈ℂ,Pλ,λ¯∈WPR¯.Corollary 9.
(Corollary 22 in [30]). LetR∈ℬℋandP∈ℂu,v. IfRis a root ofP, then(50)σapR⊂λ∈ℂ,Pλ,λ¯=0.Letℋ and K be two Hilbert spaces, R∈ℬℋ and S∈ℬK and(51)Pu,v=∑k,l≥0ck,lvluk∈ℂu,v.Consider the bounded linear transformation(52)ΦR,S,P:ℬℋ,K⟶ℬℋ,K,defined by(53)ΦR,S,PX=∑k,l≥0ck,lSlXRkforX∈ℬℋ,K.We also define the mapsAR,LS:ℬℋ,K⟶ℬℋ,K by(54)ARX=XR,LSX=SXforX∈ℬℋ,K.AR and LS commute. Indeed, for X∈ℬℋ,K, we have(55)ARLSx=ARSX=SXR=LSXA=LSARX=LSARX,Lemma 1.
Letℋ,K,S,R,AT,LS, andP be as above. Then,(56)ΦR,S,P=PAR,LS.Proof.
It is easy to check that form∈ℕ, we have(57)ARm=ARmandLSm=LSm.
Thus, forn,m∈ℕ, we have(58)ARmLSnX=SnXRm,X∈ℬℋ,K.
This yields(59)ΦR,S,PX=∑m,n≥0cm,nARmLSnX=PAR,LSXforX∈ℬℋ.
Therefore,ΦR,S,P=PAR,LS.We need the following lemmas.Lemma 2.
(Lemma 0.11 in [25]). IfAandBare commuting operators on the Banach spaceX, thenσPA,B⊂PσA,σB=Pλ,μ,λ∈σA,μ∈σBfor every polynomialP∈ℂx,y.Lemma 3.
(Lemma 27 in [30]). LetR∈ℬℋ, Pu,v=∑m,n≥0cm,nvnum∈ℂu,v, Q∈ℂu,v, thenPQR=∑m,n≥0cm,nR∗nQRRm.Remark 5.
From (57), Lemma 3 is equivalent to(60)ΦR,R∗,PQR=PAR,LR∗QR=PQR.Remark 6.
It is easy to verify that ifR is an isomorphism, then LR and AR are also isomorphisms and we have LR−1=LR−1 and AR−1=AR−1. Since, for λ∈ℂ, LR−λ=LR−λ, and AR−λ=AR−λ, we get σLR⊂σR and σAR⊂σR.Proposition 3.
LetS,R∈ℬℋ, andP∈ℂu,v, then(61)σΦR,S,P⊂PσR,σS=Pλ,μ,λ∈σR,μ∈σS.Proof.
From (57), we have σΦR,S,P=σPAR,LS. Thus, by Lemma 2, we obtain σΦR,S,P⊂PσAR,σLS. This yields, from Remark 6,(62)σΦR,S,P⊂PσR,σS.Corollary 10.
LetR∈ℬℋandP∈ℂu,v, then(63)σΦR,R∗,P⊂Pλ,μ¯,λ,μ∈σR.Proof.
Apply Proposition3 by taking S=R∗.Proposition 4.
LetR∈ℬℋandP,Q∈ℂu,v. IfPQR=0, then either0∈σΦR,R∗,PorQR=0.Proof.
Suppose that0∉σΦR,R∗,p. This means that ΦR,R∗,P is invertible. Thus, from (60), we obtain QR=0.From the following, we give a sufficient condition for ak-quasi-m,n-isosymmetric operator to be k-quasi m-isometric operator or k-quasi-n-symmetric operator.Theorem 5.
LetR∈ℬℋbe ak-quasi-m,n-isosymmetric operator. The following statements hold:(1)
IfσR∩σR∗=∅, thenRis ak-quasi-m-isometry(2)
If0∉λμ¯−1,λ,μ∈σR, thenRis ak-quasi-n-symmetric operatorProof.
(1)
SetPm,n,k=PQ, where Pu,v=v−un and Qu,v=vkvu−1muk. Since σR∩σR∗=∅, we have 0∉Pλ,μ¯,λ,μ∈σR. Thus, from (63), we get 0∉σΦR,R∗,P. This yields, by Proposition 4, QR=0. Therefore, R is a k-quasi--misometry.(2)
SetPm,n,k=PQ, where Pu,v=vu−1m and Qu,v=vkv−unuk. Since 0∉λμ¯−1,λμ∈σR, similarly, as in (1), we obtain that 0∉σΦR,R∗,P. Applying again Proposition 4, we obtain QR=0. Thus, R is a k-quasi-n-symmetric operator.Theorem 6.
LetR∈ℬℋbek-quasi-m,n-isosymmetric operator, thenRhas the single-valued extension property (SEVP).Proof.
Ifℋ=ranRk¯, then R is an m,n-isosymmetric operator, and therefore R has SVEP by Theorem 2.20 in [24]. If ℋ≠RRk¯, we use the matrix decomposition of R as(64)R=R1R20R3onℋ=ranRk¯⊕kerR∗k.
From Theorem2, R1 is a m,n-isosymmetric operator and R3 is a nilpotent operator. Hence, R1 and R3 have SVEP; then, by simple calculations, we see that R has SVEP as required.Rachid [24] showed that if R is m,n-isosymmetric operator, then σR⊂∂D∪ℝ, where ∂D is the unit circle. Now, we extend this result to k-quasi-m,n-isosymmetric operators.Theorem 7.
LetRben-quasi-m,n-isosymmetric operator, thenσapR⊂∂D∪ℝ.Proof.
Letγ∈σapR, then there exists a sequence wrr≥1⊂ℋ, with wr=1 such that R−γIwr⟶0 as p⟶∞. We have Rj−γjIwr⟶0 as r⟶∞ for all positive integersj. From the condition that R is an k-quasi-m,n-isosymmetric operator, one has(65)0=R∗k∑0≤j≤m−1jmjR∗m−jαnRRm−jRkwr|wr=∑0≤j≤m−1jmjαnRRm+k−jwr|Rm+k−jwr=∑0≤j≤m−1jmjαnRRm+k−j−γm+k−jwr+γm+k−jwr|Rm+k−j−γm+k−jwr+γm+k−jwr=γ2kγ2−1mαnwr|wrr⟶∞=γ2kγ2−1m∑0≤j≤n−1jmjR∗n−jRjwr|wrr⟶∞=γ2kγ2−1m∑0≤j≤n−1jmjRjwr|Rn−jwrr⟶∞=γ2kγ2−1m∑0≤j≤n−1jmjRj−γj+γjwr|Rn−j−γn−j+γn−jwrr⟶∞=γ2kγ2−1mγ−γ¯nwr2=γ2kγ2−1m2Imγn.
Consequently,γ=0 or γ=1 or γ∈ℝ. This completes the proof.Proposition 5.
LetRbek-quasi-m,n-isosymmetric operator. Ifγ=a+ibwithb≠0is an approximate eigenvalue ofR, thenγ¯is an approximate eigenvalue ofR∗.Proof.
Assume thatγ=a+ib∈σapR with b≠0, then there exists wrr∈ℋ : wr=1 such that R−γwr⟶ as r⟶∞. By taking into account that R is a k-quasi-m,n-isosymmetric operator, it follows that(66)0=R∗k∑0≤j≤m−1jmjR∗m−jαnRRm−jRkwr=R∗k∑0≤j≤m−1jmjR∗m−jαnRRm+k−j−γm+k−j+γm+k−jwr=R∗k∑0≤j≤m−1jmjR∗m−jαnRRm+k−j−γm+k−jwr+R∗k∑0≤j≤m−1jmjR∗m−jαnRγm+k−jwr=R∗k∑0≤j≤m−1jmjR∗m−jαnRSm+k−j−γm+k−jwr+γnR∗kI−γR∗mαnRwr.
By observing that(67)αnRwr=∑0≤j≤n−1jnjRn−jRjwr=∑0≤j≤n−jnjRn−jRj−γj+γjwr=∑0≤j≤n−jnjRn−jRj−γjwr+γ−R∗nwr,and limr⟶∞Rs−γswr=0 for all positive integers s, we get from the above relations that(68)γkR∗kI−γR∗mγ−R∗nwr⟶0,asr⟶∞.
IfI−γR∗ is bounded from below, then so is I−γR∗m and therefore there exists a positive constant K>0 such that(69)I−γR∗mw≥Kw,∀w∈ℋ.
From this, we deduce that(70)‖I−γR∗mR∗kγ−R∗nwr|≥KR∗kγ−R∗nwr.
Consequently,R∗kγ−R∗nwr⟶0,asr⟶∞. So, we have(71)0=limr⟶∞R∗kγ−R∗nwr|wr=limr⟶∞wr|γ¯−RnRkwr=limr⟶∞wr|γ¯−RnRk−γkwr+limr⟶∞wr|γkγ¯−Rnwr=γ¯klimr⟶∞wr|γ¯−Rnwr=γ¯kγ−γ¯n.
Thus,γ¯n=0 or γ−γ¯=0, which are a contradiction. Hence, I−γR∗ is not bounded from below. In view of Theorem 7, we have γ=1 and so I−γR∗=γγ¯−R∗, which implies that γ¯−R∗ is not bounded from below. This proves the statement of the proposition.


## 4. Products and Perturbation ofk-Quasi-(N, m)-Isosymmetric Operators

In this section, we study the perturbation of ank-quasi-m,n-isosymmetric operator by a nilpotent operator and we study the product and tensor product of two k-quasi-m,n-isosymmetric operators.Lemma 4.
Forp,m,n∈ℕ, the following identity holds:(72)vpup−1mvp−upn=∑0≤r≤mp−1∑0≤j≤np−1λrμjvm+np−1−r+jvu−1mv−unuj+mp−1−r,whereλrandμjare some constants.Proof.
See the proof of statement (ii) in Theorem 3.3 in [14].The following theorem shows that the power of ak-quasi-m,n-isosymmetric operator is again a k-quasi-m,n-isosymmetric which is similar to that of Corollary 5 but with another proof.Theorem 8.
IfRisk-quasi-m,n-isosymmetric, thenRpisk-quasi -m,n-isosymmetric for anyp∈ℕ0.Proof.
Two different proofs of this statement will be given.
First Proof. We need to prove that Rp∗kγm,nRpRpk=0for anyp∈ℕ.
In fact, from Lemma4, we obtain that(73)Rp∗kγm,nRpRpk=Rp∗k∑0≤r≤mp−1∑0≤j≤np−1λrμjR∗pm+np−1−r+jγmnRRpj+mp−1−rRpk=∑0≤r≤mp−1∑0≤j≤np−1λrμjR∗pm+np−1−r+jRp∗kγmnRRpkRpj+mp−1−r.
SinceR is k-quasi-m,n-isosymmetric, we have R∗kγm,nRRk=0 and therefore(74)Rp∗kγm,nRpRpk=0.
This means thatRp is k-quasi-m,n-isosymmetric for any positive integer p.
Second Proof. Let Pu,v=Pm,n,kup,vp. It is easy to verify that Pm,n,kRp=PR. A little calculation shows that there exists a polynomial Q such that P=QPm,n,k. Thus, by (60), we get(75)Pm,n,kRp=PR=ΦR,R∗,QPm,n,kR,which yields Pm,n,kRp=0 since Pm,n,kR=0. Therefore, Rp is k-quasi-m,n-isosymmetric operator.Lemma 5.
LetR,S∈ℬℋsuch thatR,S=R∗,S=0. Then, the following identity holds:(76)γm,nRS=∑0≤k≤m∑0≤j≤nmknjR∗j+kγm−k,n−jRγk,jSRkSn−j.Proof.
In view of the following identity (see [14]),(77)cdab−1mcd−abn=∑0≤k≤m∑0≤j≤nmknjcj+kca−1m−kc−an−jad−1kd−bjakbn−k,it follows by taking a=R,c=R∗,b=S, and d=S∗ with RS=SR and RS∗=S∗R that(78)γm,nRS=∑0≤k≤m∑0≤j≤nmknjR∗j+kγm−k,n−jRγk,jSRkSn−j.Theorem 9.
LetR∈ℬℋandS∈ℬℋsuch thatR,S=R,S∗=0. IfRis ank-quasi-m,n-isosymmetric andSis ank′-quasi-m′-isometric andn′-symmetric, thenRSis ank0=maxk,k′-quasi-m+m′−1,n+n′−1-isosymmetric.Proof.
We prove thatRS∗k0γm+m′−1,n+n′−1RSRSk0=0.
In fact, by taking into account Lemma5, we get(79)RS∗k0γm+m′−1,n+n′−1RSRSk0=RS∗k0∑0≤k≤m+m′−1∑0≤j≤n+n′−1m+m′−1kn+n′−1jR∗j+kγm+m′−1−k,n+n′−1−jRγk,jSRkSn+n′−1−jRSk0=∑0≤k≤m+m′−1∑0≤j≤n+n′−1m+m′−1kn+n′−1jR∗j+kR∗k0γm+m′−1−k,n+n′−1−jRRk0×S∗k0γk,jSRkSn+n′−1−jSk0.
We have the following observations:(i)
Ifk≥m′ or j≥n′, then S∗k0γk,jSSk0=0.(ii)
Ifk≤m′−1 and j≤n′−1, then m+m′−1−k≥m and n+n′−1−j≥n. So that(80)R∗k0γm+m′−1−k,n+n′−1−jRRk0=0.Therefore,(81)RS∗k0γm+m′−1,n+n′−1RSRSk0=0.Hence,RS is an k0=maxk,k′-quasi-m+m′−1,n+n′−1-isosymmetric.Corollary 11.
LetR∈ℬℋandS∈ℬℋsuch thatR,S=R,S∗=0. IfRis ank-quasi-m,n-isosymmetric andSis ank′-quasi-m′-isometric andn′-symmetric, thenRpSqis ank0=maxk,k′-quasi-m+m′−1,n+n′−1-isosymmetric for any positive integerspandq.Proof.
In view of Theorem8, we have Rp as k-quasi-m,n-isosymmetric for any positive integer p. Similarly, from Theorem 12 in [18] and corollary 3.1 in [27], we have Sq is k′-quasi-m′-isometric and n′-symmetric for any positive integer q.
Applying Theorem9, we get RpSq is an k0=maxk,k′-quasi-m+m′−1,n+n′−1-isosymmetric for any positive integers p and q.Corollary 12.
LetR∈ℬℋandS∈ℬℋ. IfRis ank1-quasi-m1,n1-isosymmetric andSis ank2-quasi-m2-isometric andn2-symmetric, thenR⊗Sisk0=maxk1,k2-quasi-m+m1−1,n+n1−1-isosymmetric.Proof.
It is well known thatR⊗S=R⊗II⊗S and moreover(82)R⊗I,I⊗S=R⊗I∗,I⊗S=0.
On the other hand,R is an k1-quasi-m1,n1-isosymmetric if and only if R⊗I is an k1-quasi-m1,n1-isosymmetric and S is an k2-quasi-m2-isometric and n2-symmetric if and only if I⊗S is an k2-quasi-m2-isometric and n2-symmetric. From Theorem 9, it follows that R⊗S is a k0-quasi-m+m1−1,n+n1−1-isosymmetric.Lemma 6.
LetR,S∈ℬℋsuch thatR,S=R∗,S=0, then the following identity holds:(83)γm,nR+Q=∑0≤j≤n∑i+l+k=mmi,l,knj×R∗+Q∗iQ∗lγk,n−jRαjQRlQi.Proof.
By the equation (see [14]),(84)c+da+b−1mc+d−a+dn=,∑0≤j≤n∑i+l+k=mnjmi,l,kc+didlca−1hc−an−jd−bjalbi.
Ifa=R, C=R∗, B=S, and d=S∗ with R,S=R∗,S=0, we get(85)γm,nR+Q=∑0≤j≤n∑i+l+k=mmi,l,knj×R∗+Q∗iQ∗lγk,n−jRαjQRlQi.Theorem 10.
LetR,Q∈ℬℋbe doubly commuting. IfRis ank-quasi-m,n-isosymmetric andQis a nilpotent operator of orderq, thenR+Qis ak+q-quasi-m+2q−2,n+2q−1-isosymmetric operator.Proof.
We need to show thatR+Q∗k+qγm+2q−2,n+2q−2R+QR+Qk+q=0.
Note that by Lemma6, we have(86)γm+2q−2,n+2q−1R+Q=∑i+l+k=m+2q−2∑0≤j≤n+2q−1m+2q−2i,l,kn+2q−1j×R∗+Q∗iQ∗lγk,n+2q−1−jRαjQRlQi.
However,(87)R+Q∗k+qγm+2q−2,n+2q−2R+QR+Qk+q=∑0≤r≤k+qk+qrR∗k+q−rQ∗r∑i+l+k=m+2q−2∑0≤j≤n+2q−1m+2q−2i,l,kn+2q−1j×R∗+Q∗iQ∗lγk,n+2q−1−jRαjQRlQi×∑0≤r≤k+qk+qrRk+q−rQr.
Now, observe that ifj≥2q or i≥q or l≥q, then αjQ=0 or Qi=0 or Q∗l=0 and hence(88)R∗+Q∗iQ∗lγk,n+2q−1−jRαjQRlQi=0.
However, ifj≤2q−1, i≤q−1, and l≤q−1, we obtain(89)k=m+2q−1−i−l≥m−2q−1−q+1−q+1≥mandn+2q−1−j≥n.
From the fact thatR is a k-quasi-n-m,n-isosymmetric, we get(90)R∗k+q−rγk,n+2q−1RRk+q−r=0forr=0,…,q,R∗k+q−rQ∗rγk,n+2q−1−jRRk+q−rQr=0forr=q+1,…,k+q.
Consequently, we obtainR+Q∗k+qγm+2q−2,n+2q−2R+QR+Qk+q=0.
Therefore,R+Q is k+q-quasi-m+2q−2,n+2q−1-isosymmetric operator.Corollary 13.
LetM,N∈ℬℋsuch thatM,N=M∗,N=0. IfMis ank-quasi-m,n-isosymmetric, then the operatorS=MN0M∈ℬℋ⊕ℋisk+2-quasi-m+2,n+3-isosymmetric.Proof.
ConsiderR=M00M and Q=0N00. Clearly, R is k-quasi-m,n-isosymmetric and Q2=0 (i.e., Q is 2-nilpotent). On the other hand, since(91)M,N=M∗,N=0,it follows that R,Q=R∗,Q=0. In view of Theorem 10, we deduce that S=R+Q is k+2-quasi-m+2,n+3-isosymmetric.Corollary 14.
LetR∈ℬℋbek-quasi-m,n-isosymmetric andQ∈ℬℋbeq-nilpotent. Then, R⊗I+I⊗Qisk+q-quasi-m+2q−2,n+2q−1-isosymmetric.Proof.
Observe thatR⊗I∈ℬℋ⊗¯ℋ is k-quasi-m,n-isosymmetric and I⊗Q∈ℬℋ⊗¯ℋ is q-nilpotent. Moreover,(92)R⊗I,I⊗Q=R⊗I∗,I⊗Q=0.
Applying Theorem10, we deduce that R⊗I+I⊗Q is k+q-quasi-m+2q−2,n+2q−1-isosymmetric.



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