# Jordan(α,β)-Derivations on Operator Algebras

Authors: Quanyuan Chen; Xiaochun Fang; Changjing Li
Journal: Journal of Function Spaces
 (2017)

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## Abstract

LetA be a CSL subalgebra of a von Neumann algebra acting on a Hilbert space H. It is shown that any Jordan (α,β)-derivation on A is an (α,β)-derivation, where α,β are any automorphisms on A. Moreover, the nth power (α,β)-maps on A are investigated.


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## Body

## 1. Introduction and Preliminaries

Throughout the paper, letH be a complex Hilbert space. Denote by B(H) the algebra of all bounded linear operators on H and by I the identity operator on H. The terms “projection" and “subspace" will mean “orthogonal projection" and “norm closed linear manifold." For convenience, a subspace will be identified with the projection onto it. If Mα is a collection of subspaces of H, then the smallest subspace containing each Mα will be denoted by ⋁Mα and the largest subspace contained in each Mα will be denoted by ⋀Mα. A subspace lattice L is a strongly closed lattice of subspaces (or projections) that is closed under the usual lattice operations ⋁ and ⋀, containing 0 and I. If L is a subspace lattice, denote by AlgL the algebra of all bounded operators in B(H) which leave every subspace in L invariant; that is, AlgL={A:A∈B(H)andAP=PAPfor  allP∈L}. Dually, for a subalgebra A of B(H), denote by LatA the lattice of all closed subspaces left invariant under every operator in A; that is, LatA={P:PisasubspaceofHandAP=PAPforallA∈A}. A subspace lattice L is reflexive if L=LatAlgL; an algebra A is reflexive if A=AlgLatA. If all projections in a subspace lattice L commutate pairwise, then the subspace lattice L is called a commutative subspace lattice, or a CSL for short, and AlgL is called a CSL algebra. If L is a CSL, whose projections are contained in a von Neumann algebra N acting on a Hilbert space H, then A=N∩AlgL is called a CSL subalgebra of the von Neumann algebra N.LetA be an algebra. A linear map δ:A→A is called a derivation if δ(AB)=δ(A)B+Aδ(B) for all A,B∈A. Let us review some generalizations of the notion of derivations. Suppose that α and β are two automorphisms on A. Recall that a linear map δ:A→A is an (α,β)-derivation if δ(AB)=δ(A)β(B)+α(A)δ(B) for all A,B∈A; and δ:A→A is a Jordan (α,β)-derivation if δ(A2)=δ(A)β(A)+α(A)δ(A) for all A∈A. For the identity isomorphism idA and an automorphism σ on A, we refer to a (σ,idA)-derivation as a σ-derivation and a Jordan (σ,idA)-derivation as a Jordan σ-derivation. Note that an (idA,idA)-derivation is just a derivation. A linear map δ on A is called a generalized (α,β)-derivation if δ(AB)=δ(A)β(B)+α(A)d(B) for all A,B∈A, where d is an (α,β)-derivation on A. The concept of generalized (α,β)-derivations covers both the concepts of (α,β)-derivations and of generalized derivations. The notions of (α,β)-derivation as a generalization of the concept of derivation have been studied in ([1–5] and references therein).It is straightforward to show that ifδ is an (α,β)-derivation of A and n≥2 is an integer, then δ(An)=∑j=0n-1α(Aj)δ(A)β(An-j-1) for any A in A, where α(A0)B=B=Bβ(A0) for any element B in A. This equation is called the nth power (α,β)-property. When n=2, the nth power (α,β)-property makes δ a Jordan (α,β)-derivation. The natural question is that if δ is a linear map on A and satisfies the nth power (α,β)-property, must δ be an (α,β)-derivation?The purpose of the present article is to answer the above question ifA is a CSL subalgebra of a von Neumann algebra acting on a Hilbert space. It can be proved that δ is a Jordan (α,β)-derivation if δ is a linear map on A and satisfies the nth power (α,β)-property. So we need to prove that a Jordan (α,β)-derivation on A is an (α,β)-derivation.The characterization of Jordan(α,β)-derivations on algebras or rings is a subject in various areas. Clearly, any (α,β)-derivation is a Jordan (α,β)-derivation. Ashraf et al. [6] and Lanski [7] provided us with beautiful counterexamples to illustrate that the converse statements are not true in general. However, a classical result of Herstein ([8], Theorem3.3) shows that a Jordan derivation of a 2-torsion free prime ring is a derivation. This result was extended to 2-torsion free semiprime rings [9]. Brešar and Vukman proved [10] that a Jordan (α,β)-derivation on a 2-torsion free prime ring is an (α,β)-derivation. Then it was proved [7] that a Jordan (α,β)-derivation on a 2-torsion free semiprime ring is an (α,β)-derivation. It was proved [11] that a Jordan derivation on a CSL algebra is a derivation. And it was extended [12] that a Jordan derivation on a CSL subalgebra of von Neumann algebra is a derivation.Motivated by these results, we are concerned with a Jordan(α,β)-derivation and a Jordan σ-derivation on A, a CSL subalgebra of a von Neumann algebra, which is not a semiprime ring.The main results are as follows.Theorem 1.
LetN be a von Neumann algebra acting on a Hilbert space H, let L be a CSL, whose projections are contained in N, and let A=N∩AlgL be the CSL subalgebra of the von Neumann algebra N. Suppose that σ is any automorphism on A; then a Jordan σ-derivation on A is a σ-derivation.Theorem 2.
LetN be a von Neumann algebra acting on a Hilbert space H, let L be a CSL, whose projections are contained in N, and let A=N∩AlgL be the CSL subalgebra of the von Neumann algebra N. Suppose that α,β are any automorphisms on A; then a Jordan (α,β)-derivation on A is an (α,β)-derivation.Then we extend the Jordan(α,β)-derivation case to the nth power (α,β)-property case for arbitrary n≥2, and we extend the derivation case to the generalized derivation case.Corollary 3.
Letn>1 be an integer, let A=N∩AlgL be a CSL subalgebra of a von Neumann algebra N acting on a Hilbert space H, and let α,β be any automorphisms on A. If δ is a linear map on A such that (1)δAn=δAβAn-1+∑j=1n-1αAjδAβAn-j-1for any A∈A, where A0=I, then δ is an (α,β)-derivation.Corollary 4.
Letn>1 be an integer, let A=N∩AlgL be a CSL subalgebra of a von Neumann algebra N acting on a Hilbert space H, and let α,β be any automorphisms on A. If δ and d are two linear maps on A such that (2)δAn=δAβAn-1+∑j=1n-1αAjdAβAn-j-1for any A∈A, where A0=I, then d is an (α,β)-derivation and δ is a generalized (α,β)-derivation on A with respect to d.


## 2. Proof of Theorems1 and 2

Throughout the paper, we suppose thatN is a von Neumann algebra acting on a Hilbert space H, L is a CSL whose projections are contained in N, and A=N∩AlgL is the CSL subalgebra of the von Neumann algebra N.Lemma 5 (see [12]).
Suppose thatA=N∩AlgL is a CSL subalgebra of the von Neumann algebra N. Denote Q1(H), or Q1 simply, by the orthogonal projection onto the linear span of the set PAP⊥x:P∈L,A∈A,x∈H and Q2(H), or Q2 simply, by the orthogonal projection onto the linear span of the set P⊥A∗Px:P∈L,A∈A,x∈H, and Q=Q1(H)∨Q2(H). Then, one has the following:(i)
Q,Q1∈LatA. And Q1, Q2, Q∈L′∩N⊆A, where L′ is the commutant of L. Q1 commutes with Q2. Furthermore, Q⊥AQ=QAQ⊥=0 for any A∈A, so that A=QAQ⊕Q⊥AQ⊥.(ii)
IfQ≠I, then Q⊥AQ⊥ is a von Neumann algebra on Q⊥H.Proposition 6.
LetA=N∩AlgL be a CSL subalgebra of a von Neumann algebra N. Suppose that σ is an automorphism on A with the property that, for any projection P∈A, there are AP,BP∈A, depending on P, such that σ(P)=APPBP. In such case, a Jordan σ-derivation on A is a σ-derivation.Proof.
LetA be a CSL subalgebra of a von Neumann algebra N. Suppose that δ is a Jordan σ-derivation on A. For any A,B in A, (3)δA2=δAA+σAδA.Since AB+BA=(A+B)2-A2-B2, (4)δAB+BA=δAB+σAδB+δBA+σBδA.Replacing B by AB+BA in (4), we have that (5)δABA=δABA+σAδBA+σABδA.
In the sequel of proof, we choose an arbitrary nontrivial projectionP in the set (L∪Q,Q1). Let P⊥=I-P. Then P⊥AP=0 and A=PAP+PAP⊥+P⊥AP⊥ for any A∈A.
Fromδ(P)=δ(P)P+σ(P)δ(P), we have that σ(P)δ(P)P=σ(P⊥)δ(P)P⊥=0. By the condition of σ, (6)δP=σPδPP⊥+σP⊥δPP=σPδPP⊥.For any A∈A, by (4), we have that (7)δPAP⊥=δPPAP⊥+PAP⊥P=δPPAP⊥+σPδPAP⊥+δPAP⊥P+σPAP⊥δP.Since δ(P)PAP⊥=σ(P)δ(P)P⊥PAP⊥=0 and σ(PAP⊥)δ(P)=0, (8)δPAP⊥=σPδPAP⊥+δPAP⊥P.It follows that σ(P)δ(PAP⊥)P=σ(P⊥)δ(PAP⊥)P⊥=0. Since σ(P⊥)δ(PAP⊥)P=0 by the condition of σ, (9)δPAP⊥=σPδPAP⊥P⊥.By the fact that (10)2δPAP=δPPAP+PAPP=δPPAP+σPδPAP+δPAPP+σPAPδP,we get that (11)σP⊥δPAP=0.Similarly,(12)2δP⊥AP⊥=δP⊥P⊥AP⊥+P⊥AP⊥P⊥=δP⊥P⊥AP⊥+σP⊥δP⊥AP⊥+δP⊥AP⊥P⊥+σP⊥AP⊥δP⊥.It follows that (13)δP⊥AP⊥P=0.By (3), δ(I)=0 and δ(P⊥)=-δ(P)=-σ(P)δ(P)P⊥. Let A,B∈A. Then by (11) and (13), we have that (14)δAPBP⊥=δPAPPBP⊥+PBP⊥PAP=δPAPPBP⊥+σPAPδPBP⊥+δPBP⊥PAP+σPBP⊥δPAP=δPAPPBP⊥+σPAPδPBP⊥=δAPBP⊥+σAδPBP⊥.Let A,B,T∈A, P∈L. Then we have that (15)δABPTP⊥=δABPTP⊥+σABδPTP⊥,δABPTP⊥=δAPBPTP⊥=δAPBPTP⊥+σAδPBPTP⊥=δABPTP⊥+σAδBPTP⊥=δABPTP⊥+σAδBPTP⊥+σBδPTP⊥.It follows that (δ(AB)-(δ(A)B+σ(A)δ(B)))PTP⊥=0. The arbitrariness of T and P gives that (16)δAB-δAB+σAδBQ1=0.Similarly, for any A,B∈A, we have that (17)δPAP⊥B=δPAP⊥P⊥BP⊥+P⊥BP⊥PAP⊥=δPAP⊥P⊥BP⊥+σPAP⊥δP⊥BP⊥+δP⊥BP⊥PAP⊥+σP⊥BP⊥δPAP⊥=δPAP⊥P⊥BP⊥+σPAP⊥δP⊥BP⊥=δPAP⊥B+σPAP⊥δB.Let A,B,T∈A, P∈L. Then we have that (18)δPTP⊥AB=δPTP⊥AB+σPTP⊥δAB,δPTP⊥AB=δPTP⊥AP⊥B=δPTP⊥AP⊥B+σPTP⊥AP⊥δB=δPTP⊥AB+σPTP⊥AδB=δPTP⊥A+σPTP⊥δAB+σPTP⊥σAδB.It follows that (19)σPTP⊥δAB-δAB+σAδB=0,PTP⊥·σ-1δAB-δAB+σAδB=0.So (20)σ-1δAB-δAB+σAδB∗·P⊥T∗P=0by involution. The arbitrariness of T and P gives that (21)σ-1δAB-δAB+σAδB∗·Q2=0,(22)σQ2δAB-δAB+σAδB=0.We now define a new linear map Δ:A→A by (23)ΔA=δA-σAδQ1-δQ1AA∈A.Then Δ is also a Jordan σ-derivation. Since δ(Q1)=σ(Q1)δ(Q1)Q1⊥ by (6), it follows that Δ(Q1)=0 and Δ(Q1⊥)=Δ(I-Q1)=0. Note that TQ1=Q1TQ1 and Q1⊥T=Q1⊥TQ1⊥ for each T∈A. Then(24)ΔTQ1=ΔQ1TQ1=ΔQ1TQ1+σQ1ΔTQ1+σQ1TΔQ1=σQ1ΔTQ1=σIΔTQ1-σQ1⊥ΔTQ1=σIΔTQ1=ΔTQ1,(25)ΔQ1⊥T=ΔQ1⊥TQ1⊥=ΔQ1⊥TQ1⊥+σQ1⊥ΔTQ1⊥+σQ1⊥TΔQ1⊥=σQ1⊥ΔTQ1⊥=σQ1⊥ΔT-σQ1⊥ΔTQ1=σQ1⊥ΔT.
Claim.IfQ1∨Q2=I, thenδis aσ-derivation.
For this, it suffices obviously to prove thatΔ is a σ-derivation. In fact, let A,B∈A. Then equalities (16) and (22) hold if δ is replaced by Δ. By (24), we have that (26)ΔABQ1-ΔABQ1+σAΔBQ1Q1⊥=0.By (16) and (26), we have that (27)ΔABQ1=ΔABQ1+σAΔBQ1.By (22), (28)σQ2ΔQ1⊥AB-ΔQ1⊥AB+σQ1⊥AΔB=0.Since Q1∨Q2=I,  σ(Q1⊥)=σ(Q1⊥Q2) and (29)σQ1⊥ΔQ1⊥AB-ΔQ1⊥AB+σQ1⊥AΔB=0.By (25), we get that (30)σQ1ΔQ1⊥AB-ΔQ1⊥AB+σQ1⊥AΔB=0.It follows that (31)ΔQ1⊥AB=ΔQ1⊥AB+σQ1⊥AΔB.Equation (14) gives that (32)ΔAQ1BQ1⊥=ΔAQ1BQ1⊥+σAΔQ1BQ1⊥.By (17), we have that (33)ΔQ1AQ1⊥B=ΔQ1AQ1⊥B+σQ1AQ1⊥ΔB.By (24), (25), (26), (27), (31), (32), and (33), (34)ΔAB=ΔABQ1+AQ1BQ1⊥+Q1AQ1⊥B+Q1⊥AB=ΔABQ1+σAΔBQ1+ΔAQ1BQ1⊥+σAΔQ1BQ1⊥+ΔQ1AQ1⊥B+σQ1AQ1⊥ΔB+ΔQ1⊥AB+σQ1⊥AΔB=ΔABQ1+Q1BQ1⊥+σAΔBQ1+ΔQ1BQ1⊥+ΔQ1AQ1⊥+ΔQ1⊥AB+σQ1AQ1⊥+σQ1⊥AΔB=ΔAQ1BQ1+Q1BQ1⊥+σAΔQ1BQ1+ΔQ1BQ1⊥+ΔQ1AQ1⊥+ΔQ1⊥AQ1⊥B+σQ1AQ1⊥+σQ1⊥AQ1⊥ΔB=ΔAQ1B+σAΔQ1B+ΔAQ1⊥B+σAQ1⊥ΔB=ΔAQ1B+σAΔQ1B+ΔAQ1⊥B+σAΔQ1⊥B=ΔAB+σAΔB.It follows that Δ is a σ-derivation, and so is δ. The proof ofClaim is complete.
For the general case, letQ=Q1(L)∨Q2(L). By Lemma 5, the algebra A can be written as the direct sum A=QAQ⊕Q⊥AQ⊥. We now define a new linear map Δ1:A→A by (35)Δ1A=δA-σAδQ-δQAA∈A.Then Δ1 is a Jordan σ-derivation and Δ1(Q)=Δ1(Q⊥)=0. For any A∈A, by (5) and the assumption of σ, (36)Δ1QAQ=σQΔ1AQ=AQQBQΔ1AQ=QAQQBQΔ1AQ∈QAQ,Δ1Q⊥AQ⊥=Δ1Q⊥Q⊥AQ⊥Q⊥=σQ⊥Δ1Q⊥AQ⊥Q⊥.Then Δ1(Q⊥AQ⊥) is of form Δ1(Q⊥AQ⊥)=QA1Q⊥+Q⊥A2Q⊥. By Lemma 5, QA1Q⊥=0. It follows that (37)Δ1Q⊥AQ⊥=Q⊥A2Q⊥∈Q⊥AQ⊥for any A∈A. Then Δ1 maps QAQ, Q⊥AQ⊥ onto themselves, respectively. So Δ1 can be written as the direct sum Δ11⊕Δ12, where Δ11, Δ12 are the restrictions of Δ1 on QAQ, Q⊥AQ⊥, respectively. And Δ11, Δ12 are both Jordan σ-derivations. Since (38)QAQ=T∈QNQ:Q-QPTQP=0foranyP∈L=QNQ∩AlgQL,we have that QAQ is a CSL subalgebra of the von Neumann algebra QNQ. For any P∈L, A∈A, and x∈H, we have that QAQ⊥=0 and (39)PAP⊥x=QPAP⊥x=PQAP⊥x=PQAQ-QPx=QPAQ-QPx.Since (40)Q1L=span¯PAP⊥x:P∈L,A∈A,x∈H,Q1QL=span¯QPAQ-QPx:P∈L,A∈A,x∈H,we have that Q1(L)=Q1(QL) and similarly Q2(L)=Q2(QL). It follows that Q1(QL)∨Q2(QL)=Q is the identity element of QAQ. By theClaim, we have that Δ11 is a σ-derivation. Notice that Q⊥AQ⊥ is von Neumann algebra and von Neumann algebra is a semiprime ring. By Theorem2 of [7], we know that Δ12 is also a σ-derivation. Consequently, Δ1 is a σ-derivation and so is δ, as required.Lemma 7.
Suppose thatA=N∩AlgL is CSL subalgebra of the von Neumann algebra N. Denote Q1(H), or Q1 simply, by the orthogonal projection onto the linear span of the set PAP⊥x:P∈L,A∈A,x∈H; Q2(H), or Q2 simply, by the orthogonal projection onto the linear span of the set P⊥A∗Px:P∈L,A∈A,x∈H, and Q=Q1(H)∨Q2(H). Then Q,Q1∈L′′, the double commutant of L.Proof.
LetB∈L′.
SinceL′=AlgL∩(AlgL)∗, we have that BQ1=Q1BQ1 and B∗Q1=Q1B∗Q1. It follows that BQ1=Q1B. Hence Q1∈L′′.
Similarly,B∗P⊥A∗Px=P⊥B∗P⊥A∗Px+PB∗P⊥A∗Px=P⊥B∗P⊥A∗Px for any P∈L, A∈A, x∈H. It follows that B∗Q2=Q2B∗Q2 and Q2B=Q2BQ2 by involution. Noting that BQ2=Q2BQ2. So BQ2=Q2B. Hence Q2∈L′′.
It follows thatQ=Q1+Q2-Q1Q2∈L′′.Lemma 8 (Theorem2.1, [13]).
LetL1, L2 be commutative subspace lattices on Hilbert spaces H1 and H2, respectively, and ϕ:AlgL1→AlgL2 be an isomorphism. Let M be a masa (maximal abelian self-adjoint subalgebra) contained in AlgL1. Then there is a bounded invertible operator Y:H1→H2 and an automorphism ρ:AlgL1→AlgL1 such that(i)
ρ(M)=M for all M∈M.(ii)
ϕ(A)=Yρ(A)Y-1 for any A∈Alg(L1).Remark 9 (Remark1, [13]).
Lemma8 holds for an isomorphism ϕ:A1→A2, where A1 and A2 are assumed to contain masas.Proof of Theorem1.
Suppose thatδ is a Jordan σ-derivation on A=N∩AlgL. We first note that any CSL subalgebra of a von Neumann algebra contains a masa M which contains L′′, the double commutant of L.
In fact, for a CSLL, L′=AlgL∩(AlgL)∗, which gives that L′ is a von Neumann algebra. It follows that L′∩L′′ is an abelian von Neumann algebra. Since L′′⊆L′, L′′=L′∩L′′ is an abelian von Neumann algebra.
As in the proof of Proposition6, we can choose an arbitrary nontrivial projection P in (L∪Q,Q1). Lemma 7 gives that Q,Q1∈L′′⊆M, the masa (maximal abelian self-adjoint subalgebra). Lemma 8 with Remark 9 following it gives that the automorphism σ and the projection P meet the property of Proposition 6. Using the proof of Proposition 6, we can similarly prove that a Jordan σ-derivation on A is a σ-derivation.Proof of Theorem2.
Suppose thatδ is a Jordan (α,β)-derivation on A=N∩AlgL. For any A∈A, (41)δA2=αAδA+δAβA,β-1∘δA2=β-1∘αAβ-1∘δA+β-1∘δAβ-1∘βA.Let σ=β-1∘α and Δ=β-1∘δ; then the above equality gives that Δ is a Jordan σ-derivation. It follows that Δ is a σ-derivation by Theorem 1. So (42)β-1∘δAB=β-1∘αAβ-1∘δB+β-1∘δAβ-1∘βB,δAB=αAδB+δAβBfor any A,B∈A, which gives that δ is an (α,β)-derivation on A.Example 10.
LetA be CSL subalgebra of von Neumann algebra. If there is an invertible operator Y∈A such that Y is not the identity operator I on H. Consider the mapping σ on A by σ(A)=Y-1AY. Then σ is an automorphism on A and σ meets the property in Proposition 6. Let δ=σ-idA, where idA is the identity function on A. It is easy to see that δ is a σ-derivation but not a derivation on A.
For example, let{e1,e2,e3,e4} be an orthonormal basis for C4. We use the notation [124] for the span of the first, second, and fourth of these vectors and so on. A4 is the collection of 4×4 matrices of the form (43)a0bc0def00g0000h,where a,b,…,h are complex numbers. The algebra is reflexive and its lattice consists of the subspaces {0,[1],[2],[12],[123],[124],[1234]}. A4 is a CSL subalgebra of a von Neumann algebra. Let Y=1011011100100001. Then Y is invertible and Y-1=10-1-101-1-100100001. Consider the mapping σ on A4 by σ(A)=Y-1AY; that is, (44)σ:a0bc0def00g0000h⟼a0a+b-ga+c-h0dd+e-gd+f-h00g0000h.It is obvious that σ is an automorphism on A4 and σ meets the property in Proposition 6. Define a mapping δ on A4 by (45)δ:a0bc0def00g0000h⟼00a-g+ba-h+c00d-g+ed-h+f00000000.It is easy to see that δ is a σ-derivation and but not a derivation on A4.
It shows that Proposition6 and Theorems 1 and 2 in the present paper generalize Theorem1.2 of [12].


## 3. Generalizednth Power (α,β)-Maps on CSL Subalgebras of Von Neumann Algebras

The following proposition is essentially due to [7]. For the sake of completeness, we outline the proof.Proposition 11.
Letn>1 be an integer, let A be a unital operator algebra on a Hilbert space H with unit I, and let α,β be automorphisms on A. If δ is a linear map on A such that (46)δAn=δAβAn-1+∑j=1n-1αAjδAβAn-j-1for any A∈A, where A0=I, then δ is a Jordan (α,β)-derivation.Proof.
Whenn=2, our assumption is δ(A2)=δ(A)β(A)+α(A)δ(A), and we reduce the general situation to this case. Substituting I for A in our assumed relation (46) gives that δ(I)=0. Next, replace A with A+zI for any integer z. The result of this substitution, using the additivity of δ and δ(I)=0, is (47)δA+zIn=δA+zIβA+zIn-1+∑j=1n-1αA+zIjδA+zIβA+zIn-j-1.Expanding (47), now using the additivity of α and β, we collect terms containing zk to write ∑k=0nzkPk(A)=0. Note that P0(A) is the relation assumed in the proposition, so it is zero, which leads to (48)∑k=1nzkPkA=0.We replace z by 1,2,…,n in turn in equality (48). Expressing the resulting system of n homogeneous equations, we see that the coefficient matrix of the system is a Vandermonde matrix (49)11⋯1222⋯2n⋯⋯⋯⋯nn2⋯nn.Since the determinant of the matrix is different from zero, it follows that the system has only trivial solution, so each Pk(A)=0. In particular, Pn-2(A)=0. We get that (50)nn-2δA2=n-1n-2δAβA+αAδA+n-1n-2δAβA+21αAδA+n-2n-3δAβA+32αAδA+n-3n-4δAβA+⋯+n-1n-2αAδA+δAβA+n-1n-2αAδA.It follows that δ(A2)=δ(A)β(A)+α(A)δ(A). The proof is complete.The proof of Corollary3 is as follows.Proof.
It follows from Proposition11 and Theorem 2.The proof of Corollary4 is as follows.Proof.
It is similar to the proof of Theorem3 in [7].



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