Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 245452, 7 pages
http://dx.doi.org/10.1155/2013/245452
Research Article
Characterizations of Nonlinear Lie Derivations of 𝐵(𝑋)
Donghua Huo,1,2 Baodong Zheng,1 and Hongyu Liu2
1
2
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
Department of Mathematics, Mudanjiang Normal College, Mudanjiang 157012, China
Correspondence should be addressed to Baodong Zheng; zbd@hit.edu.cn
Received 25 December 2012; Accepted 3 February 2013
Academic Editor: Chunrui Zhang
Copyright © 2013 Donghua Huo et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Let 𝑋 be an infinite dimensional Banach space, and Φ : 𝐵(𝑋) → 𝐵(𝑋) is a nonlinear Lie derivation. Then Φ is the form 𝛿 + 𝜏
where 𝛿 is an additive derivation of 𝐵(𝑋) and 𝜏 is a map from 𝐵(𝑋) into its center 𝑍𝐵(𝑋) , which maps commutators into the zero.
1. Introduction
The local actions of derivations of operator algebras are
still not completely understood. One is the well-known
local derivation problem. The other is to study conditions
under derivations of operator algebras that can be completely
determined by the action on some sets of operators. In recent
years there has been a great interest in the study of Lie
derivations of operator algebras (e.g., see [1–6]). In particular,
the structure of additive or linear Lie derivation on ring or
algebra has been studied by many authors (e.g., see [7–10]).
Let 𝑋 be a Banach space and let 𝐵(𝑋) be the algebra of
all bounded linear operators on 𝑋. Some progress has been
made towards understanding the Lie derivation recently.
In [11], it was shown that letting 𝑋 be a Banach space of
dimension greater than 2, if 𝛿 : 𝐵(𝑋) → 𝐵(𝑋) is a linear
map satisfying 𝛿([𝐴, 𝐵]) = [𝛿(𝐴), 𝐵] + [𝐴, 𝛿(𝐵)] for any
𝐴, 𝐵 ∈ 𝐵(𝑋) with 𝐴𝐵 = 0, then 𝛿 = 𝑑 + 𝜏, where 𝑑 is
a derivation of 𝐵(𝑋) and 𝜏 : 𝐵(𝑋) → 𝐶𝐼 is a linear map
vanishing at commutators [𝐴, 𝐵] with 𝐴𝐵 = 0. In [12], it
was shown that every nonlinear Lie derivation of triangular
algebras is the sum of an additive derivation and a map into
its center sending commutators to zero. Ji and Qi [13] proved
that letting T be a triangular algebra over a commutative ring
𝑅, if 𝛿 : T → T is an 𝑅-linear map satisfying 𝛿([𝑥, 𝑦]) =
[𝛿(𝑥), 𝑦] + [𝑥, 𝛿(𝑦)] for any 𝑥, 𝑦 ∈ T with 𝑥𝑦 = 0, then
𝛿 = 𝑑 + 𝜏, where 𝑑 is a derivation of T and 𝜏 : T → 𝑍(T) is
an 𝑅-linear map vanishing at commutators [𝑥, 𝑦] with 𝑥𝑦 = 0.
Bai and Du [14] proved that letting M be a von Neumann
algebra with no central summands of type 𝐼1 , if Φ : M → M
is a nonlinear Lie derivation, then Φ is of the form 𝜎+𝜏, where
𝜎 is an additive derivation of M and 𝜏 is a mapping of M into
its center 𝑍M which maps commutators into zero.
Nevertheless, it would seem desirable to investigate nonlinear Lie derivations of 𝐵(𝑋). The present investigation
attempts to do this by using the properties of nonlinear Lie
derivation and previous research. A map 𝛿 : 𝐵(𝑋) → 𝐵(𝑋)
is called an additive derivation if it is additive and satisfies
𝛿(𝐴𝐵) = 𝛿(𝐴)𝐵 + 𝐴𝛿(𝐵) for any 𝐴, 𝐵 ∈ 𝐵(𝑋). A linear map
𝛿 : 𝐵(𝑋) → 𝐵(𝑋) is called a Lie derivation if 𝛿([𝐴𝐵]) =
[𝛿(𝐴), 𝐵] + [𝐴, 𝛿(𝐵)] for any 𝐴, 𝐵 ∈ 𝐵(𝑋), where [𝐴, 𝐵] =
𝐴𝐵 − 𝐵𝐴 is the usual Lie product. More generally, a map
Φ : 𝐵(𝑋) → 𝐵(𝑋) (without the additive assumption) is a
nonlinear Lie derivation if Φ([𝐴𝐵]) = [Φ(𝐴), 𝐵] + [𝐴, Φ(𝐵)]
for any 𝐴, 𝐵 ∈ 𝐵(𝑋).
Recall that 𝐵(𝑋) is called prime algebra if 𝑎𝐵(𝑋)𝑏 = 0
implies 𝑎 = 0 or = 0. An operator 𝑃 ∈ 𝐵(𝑋) is an idempotent
provided that 𝑃2 = 𝑃. In this paper, we suppose that 𝐵(𝑋)
is a prime associate algebra. We prove that every nonlinear
Lie derivation of 𝐵(𝑋) is the sum of an additive derivation
and a map from 𝐵(𝑋) into its center sending commutators to
the zero. We want to mention here that there is no additive
assumed.
2. Result and Proof
Fix a nontrivial idempotent 𝑃1 ∈ 𝐵(𝑋) and let 𝑃2 = 𝐼 − 𝑃1 .
In what follows, we write A𝑖𝑗 = 𝑃𝑖 𝐵(𝑋)𝑃𝑗 for, 𝑗 = 1, 2. Then
2
Abstract and Applied Analysis
every operator 𝐴 ∈ 𝐵(𝑋) can be written as 𝐴 = ∑2𝑖,𝑗=1 𝐴 𝑖𝑗 .
Note that 𝐴 𝑖𝑗 notation denotes an arbitrary element of A𝑖𝑗 .
Definition 1. Letting 𝑆 be a set of 𝐵(𝑋), we call
𝑍𝑆 = {𝐴 ∈ 𝐵 (𝑋) | [𝐴, 𝐵] = 0, ∀𝐵 ∈ 𝑆}
(1)
the centralizer of 𝑆 about 𝐵(𝑋).
Clearly, 𝑍𝐵(𝑋) is the center of 𝐵(𝑋).
Lemma 2. Consider that 𝑍A𝑖𝑗 = A𝑖𝑗 + 𝑍𝐵(𝑋) , 1 ≤ 𝑖 ≠ 𝑗 ≤ 2.
Proof. It is clearly that A𝑖𝑗 + 𝑍𝐵(𝑋) ⊆ 𝑍A𝑖𝑗 , 1 ≤ 𝑖 ≠ 𝑗 ≤ 2.
Conversely, without loss of generality, we proof that
𝑍A12 ⊆ A12 + 𝑍𝐵(𝑋) . For any 𝑇 = Σ2𝑖,𝑗=1 𝑇𝑖𝑗 ∈ 𝑍A12 , we have
𝑇11 𝐴 12 + 𝑇21 𝐴 12 = 𝐴 12 𝑇22 + 𝐴 12 𝑇21 ,
∀𝐴 12 ∈ A12 . (2)
Thus
𝑇21 𝐴 12 = 𝑇21 𝑃1 𝐴𝑃2 = 0,
𝑇11 𝐴 12 = 𝐴 12 𝑇22 ,
∀𝐴 ∈ 𝐵 (𝑋) ,
∀𝐴 12 ∈ A12 .
(3)
(4)
From (3), noting that 𝐵(𝑋) is prime, we obtain 𝑇21 𝑃1 = 0, that
is, 𝑇21 = 0. By 𝐴 11 𝐴 12 ∈ A12 , we have
𝑇11 𝐴 11 𝐴 12 = 𝐴 11 𝐴 12 𝑇22 = 𝐴 11 𝑇11 𝐴 12 ,
(𝑇11 𝐴 11 − 𝐴 11 𝑇11 ) 𝑃1 𝐴𝑃2 = 0,
∀𝐴 ∈ 𝐵 (𝑋) .
(5)
Since 𝐵(𝑋) is prime, we have 𝑇11 𝐴 11 = 𝐴 11 𝑇11 . We write
𝑇11 = 𝑍1 𝑃1 , where 𝑍1 ∈ 𝑍𝐵(𝑋) . Since 𝐴 12 𝐴 22 ∈ A12 , we
have
𝑇11 𝐴 12 𝐴 22 = 𝐴 12 𝐴 22 𝑇22 = 𝐴 12 𝑇22 𝐴 22 ,
𝑃1 𝐴𝑃2 (𝐴 22 𝑇22 − 𝑇22 𝐴 22 ) = 0,
∀𝐴 ∈ 𝐵 (𝑋) .
(6)
Lemma 3. Consider that 𝑍𝐵(𝑋) = 𝑍A12 ∩ 𝑍A21 .
Proof. From Lemma 2, we have
𝑍A12 ∩ 𝑍A21 = (A12 + 𝑍𝐵(𝑋) ) ∩ (A21 + 𝑍𝐵(𝑋) ) = 𝑍𝐵(𝑋) .
(12)
Theorem 4. Let 𝑋 be an infinite dimensional Banach space,
and Φ : 𝐵(𝑋) → 𝐵(𝑋) is a nonlinear Lie derivation. Then
Φ is the form 𝛿 + 𝜏, where 𝛿 is an additive derivation of 𝐵(𝑋)
and 𝜏 is a map from 𝐵(𝑋) into its center 𝑍𝐵(𝑋) , which maps
commutators into the zero.
Proof. Now we organize the proof in a series of claims.
Claim 1. Consider that Φ(0) = 0; moreover, 𝑇𝑖 ∈ 𝐵(𝑋) such
that Φ(𝑃𝑖 ) − [𝑃𝑖 , 𝑇𝑖 ] ∈ 𝑍𝐵(𝑋) and Φ(𝐴 𝑖𝑗 ) = 𝑃𝑖 Φ(𝐴 𝑖𝑗 )𝑃𝑗 +
[𝐴 𝑖𝑗 , 𝑇𝑖 ], Φ(𝐴 𝑗𝑖 ) = 𝑃𝑗 Φ(𝐴 𝑗𝑖 )𝑃𝑖 + [𝐴 𝑗𝑖 , 𝑇𝑖 ] for arbitrarily 𝐴 𝑖𝑗 ∈
A𝑖𝑗 (1 ≤ 𝑖 ≠ 𝑗 ≤ 2).
Obviously Φ(0) = Φ([0, 0]) = [Φ(0), 0] + [0, Φ(0)] = 0.
Moreover, for arbitrarily 𝐴 𝑖𝑗 ,
Φ (𝐴 𝑖𝑗 ) = Φ ([𝑃𝑖 , 𝐴 𝑖𝑗 ]) = [Φ (𝑃𝑖 ) , 𝐴 𝑖𝑗 ] + [𝑃𝑖 , Φ (𝐴 𝑖𝑗 )]
= Φ (𝑃𝑖 ) 𝐴 𝑖𝑗 − 𝐴 𝑖𝑗 Φ (𝑃𝑖 ) + 𝑃𝑖 Φ (𝐴 𝑖𝑗 )
− Φ (𝐴 𝑖𝑗 ) 𝑃𝑖 .
(13)
In (13), the left and right sides are, respectively, multiplied by
𝑃𝑖 and 𝑃𝑗 , and we have
𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 𝐴 𝑖𝑗 = 𝐴 𝑖𝑗 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗 .
In (13), the left and right sides are respectively multiplying by
𝑃𝑗 and 𝑃𝑖 , and we arrive at
Since 𝐵(𝑋) is prime we have 𝑇22 𝐴 22 = 𝐴 22 𝑇22 . We write 𝑇22 =
𝑍2 𝑃2 , where 𝑍2 ∈ 𝑍𝐵(𝑋) . From (4) we have
(𝑍1 − 𝑍2 ) 𝑃1 𝐴𝑃2 = 𝑍1 𝑃1 𝐴𝑃2 − 𝑃1 𝐴𝑃2 𝑍2 𝑃2
= 𝑇11 𝑃1 𝐴𝑃2 − 𝑃1 𝐴𝑃2 𝑇22 = 0.
(14)
𝑃𝑗 Φ (𝐴 𝑖𝑗 ) 𝑃𝑖 = 0.
(15)
For arbitrarily 𝐴 𝑗𝑖 ,
(7)
So
Φ (𝐴 𝑗𝑖 ) = Φ ([𝐴 𝑗𝑖 , 𝑃𝑖 ]) = [Φ (𝐴 𝑗𝑖 ) , 𝑃𝑖 ] + [𝐴 𝑗𝑖 , Φ (𝑃𝑖 )]
= Φ (𝐴 𝑗𝑖 ) 𝑃𝑖 − 𝑃𝑖 Φ (𝐴 𝑗𝑖 ) + 𝐴 𝑗𝑖 Φ (𝑃𝑖 )
(𝑍1 − 𝑍2 ) 𝑃1 = 0.
(8)
− Φ (𝑃𝑖 ) 𝐴 𝑗𝑖 .
(16)
Thus
(𝑍1 − 𝑍2 ) 𝐴𝑃1 = 𝐴 (𝑍1 − 𝑍2 ) 𝑃1 = 0,
∀𝐴 ∈ 𝐵 (𝑋) . (9)
So
𝑍1 = 𝑍2 .
(10)
In the previous equation, the left and right sides are, respectively multiplied by 𝑃𝑗 and 𝑃𝑖 , and we have
𝐴 𝑗𝑖 𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 = 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗 𝐴 𝑗𝑖 .
(17)
This together with (14) shows that
It follows that
𝑇 = 𝑇12 + 𝑍1 𝑃1 + 𝑍2 𝑃2
= 𝑇12 + 𝑍1 ∈ A12 + 𝑍𝐵(𝑋) .
[𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 + 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗 , 𝐴 𝑖𝑗 ]
(11)
= [𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 + 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗 , 𝐴 𝑗𝑖 ] = 0.
(18)
Abstract and Applied Analysis
3
Let = 1, and let 𝑗 = 2. For every 𝐴 12 ∈ A12 , [𝑇11 +
𝑇12 , 𝐴 12 ] = [𝑇11 , 𝐴 12 ]; therefore
From Lemma 3, it is that
𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 + 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗 ∈ 𝑍𝐵(𝑋) .
(19)
[Φ (𝑇11 + 𝑇12 ) , 𝐴 12 ] + [𝑇11 + 𝑇12 , Φ (𝐴 12 )]
= [Φ (𝑇11 ) , 𝐴 12 ] + [𝑇11 , Φ (𝐴 12 )] .
Denote 𝑇𝑖 = 𝑃𝑖 Φ(𝑃𝑖 )𝑃𝑗 − 𝑃𝑗 Φ(𝑃𝑖 )𝑃𝑖 . Therefore
Φ (𝑃𝑖 ) − [𝑃𝑖 , 𝑇𝑖 ] = 𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 + 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗 ∈ 𝑍𝐵(𝑋) . (20)
This yields [Φ(𝑇11 + 𝑇12 ) − Φ(𝑇11 ), 𝐴 12 ] = 0. From Lemma 2,
Φ (𝑇11 + 𝑇12 ) − Φ (𝑇11 )
At the same time, [𝐴 𝑖𝑗 , 𝑇𝑖 ] = 𝑃𝑗 Φ(𝑃𝑖 )𝑃𝑖 𝐴 𝑖𝑗 − 𝐴 𝑖𝑗 𝑃𝑗 Φ(𝑃𝑖 )𝑃𝑖 .
From (14),
[𝐴 𝑖𝑗 , 𝑇𝑖 ] = 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑖 𝐴 𝑖𝑗 − 𝐴 𝑖𝑗 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑖
+ 𝑃𝑖 Φ (𝑃𝑖 ) 𝑃𝑖 𝐴 𝑖𝑗 − 𝐴 𝑖𝑗 𝑃𝑗 Φ (𝑃𝑖 ) 𝑃𝑗
(21)
= Φ (𝑃𝑖 ) 𝐴 𝑖𝑗 − 𝐴 𝑖𝑗 Φ (𝑃𝑖 ) .
Remark. It is clear that 𝑇 → [𝑇, 𝑇𝑖 ] is an additive derivation.
Without loss of generality, we can assume that Φ(𝑃1 ) ∈ 𝑍𝐵(𝑋)
and Φ(A𝑖𝑗 ) ⊆ A𝑖𝑗 (1 ≤ 𝑖 ≠ 𝑗 ≤ 2).
Claim 2. Consider that Φ(A𝑖𝑖 ) ⊆ A𝑖𝑖 + 𝑍𝐵(𝑋) (𝑖 = 1, 2).
Let 𝑖 = 1, for every 𝑇11 ∈ A11 , and write Φ(𝑇11 ) =
∑2𝑖,𝑗=1 𝐵𝑖𝑗 . From Φ(𝑃1 ) ∈ 𝑍𝐵(𝑋) and [𝑃1 , 𝐴] = [𝐴, 𝑃2 ], we can
get Φ(𝑃2 ) ∈ 𝑍𝐵(𝑋) :
0 = Φ ([𝑇11 , 𝑃2 ]) = [Φ (𝑇11 ) , 𝑃2 ] + [𝑇11 , Φ (𝑃2 )]
= Φ (𝑇11 ) 𝑃2 − 𝑃2 Φ (𝑇11 ) .
= 𝑃1 (Φ (𝑇11 + 𝑇12 ) −Φ (𝑇11 )) 𝑃2 + 𝑍
(22)
Φ (𝑃1 𝑇𝑃2 ) = Φ ([𝑃1 , 𝑇])
= 𝐵22 𝐴 22 − 𝐴 22 𝐵22 + 𝑇11 Φ (𝐴 22 ) − Φ (𝐴 22 ) 𝑇11 .
𝑃1 (Φ (𝑇11 + 𝑇12 ) − Φ (𝑇11 )) 𝑃2
= Φ (𝑃1 (𝑇11 + 𝑇12 ) 𝑃2 ) − Φ (𝑃1 (𝑇11 ) 𝑃2 )
= Φ (𝑃1 𝑇11 𝑃2 + 𝑃1 𝑇12 𝑃2 ) − Φ (𝑃1 𝑇11 𝑃2 )
(28)
= Φ (𝑃1 𝑇12 𝑃2 ) = Φ (𝑇12 ) .
It shows that Φ(𝑇11 + 𝑇12 ) − Φ(𝑇11 ) = Φ(𝑇12 ) + 𝑍. The rest of
the other hand is also the same reason.
Step 2. Φ is additive on A12 and A21 .
For 𝑇12 , 𝑆12 ∈ A12 . Because 𝑇12 + 𝑆12 = [𝑃1 + 𝑇12 , 𝑃2 + 𝑆12 ],
by Step 1, we know that
Φ (𝑇12 + 𝑆12 ) = [Φ (𝑃1 + 𝑇12 ) , 𝑃2 + 𝑆12 ]
+ [𝑃1 + 𝑇12 , Φ (𝑃2 + 𝑆12 )]
= [Φ (𝑃1 ) + Φ (𝑇12 ) , 𝑃2 + 𝑆12 ]
(29)
+ [𝑃1 + 𝑇12 , Φ (𝑃2 ) + Φ (𝑆12 )]
= Φ (𝑇12 ) + Φ (𝑆12 ) .
(23)
The rest is similar.
Thus, 𝐵22 𝐴 22 − 𝐴 22 𝐵22 = 0, and so 𝐵22 = 𝑃2 𝑍 for some ∈
𝑍𝐵(𝑋) . Hence
Φ (𝑇11 ) = 𝐵11 − 𝑃1 𝑍 + 𝑍 ∈ A11 + 𝑍𝐵(𝑋) .
(27)
Then
From this 𝑃1 Φ(𝑇11 )𝑃2 = 𝑃2 Φ(𝑇11 )𝑃1 = 0. So we have 𝐵12 =
𝐵21 = 0. For every 𝐴 22 ∈ A22 ,
0 = Φ ([𝑇11 , 𝐴 22 ]) = [Φ (𝑇11 ) , 𝐴 22 ] + [𝑇11 , Φ (𝐴 22 )]
(26)
for ∈ 𝑍𝐵(𝑋) .
Since Φ(𝑃1 ) ∈ 𝑍𝐵(𝑋) and [𝑃1 , 𝑇] = 𝑃1 𝑇𝑃2 for all 𝑇 ∈ A11 ⊕
A12 ⊕ A22 . We have that
= [Φ (𝑃1 ) , 𝑇] + [𝑃1 , Φ (𝑇)] = 𝑃1 Φ (𝑇) 𝑃2 .
Because 𝑃𝑗 Φ(𝐴 𝑖𝑗 )𝑃𝑖 = 0 so by (13) we get Φ(𝐴 𝑖𝑗 ) =
𝑃𝑖 Φ(𝐴 𝑖𝑗 )𝑃𝑗 + [𝐴 𝑖𝑗 , 𝑇𝑖 ].
In the same way, we can get Φ(𝐴 𝑗𝑖 ) = 𝑃𝑗 Φ(𝐴 𝑗𝑖 )𝑃𝑖 +
[𝐴 𝑗𝑖 , 𝑇𝑖 ].
(25)
(24)
The other case can be treated similarly.
Claim 3. Φ is almost additive map; that is, for every 𝑇, 𝑆 ∈
𝐵(𝑋), Φ(𝑇 + 𝑆) − Φ(𝑇) − Φ(𝑆) ∈ 𝑍𝐵(𝑋) .
We divided the proof into several steps.
Step 1. For every 𝑇𝑖𝑖 , 𝑇𝑗𝑖 , and 𝑇𝑖𝑗 (1 ≤ 𝑖 ≠ 𝑗 ≤ 2), Φ(𝑇𝑖𝑖 +𝑇𝑖𝑗 )−
Φ(𝑇𝑖𝑖 ) − Φ(𝑇𝑖𝑗 ) ∈ 𝑍𝐵(𝑋) , Φ(𝑇𝑖𝑖 + 𝑇𝑗𝑖 ) − Φ(𝑇𝑖𝑖 ) − Φ(𝑇𝑗𝑖 ) ∈ 𝑍𝐵(𝑋) .
Step 3. For any 𝑇11 ∈ A11 , 𝑇22 ∈ A22 , Φ(𝑇11 + 𝑇22 ) − Φ(𝑇11 ) −
Φ(𝑇22 ) ∈ 𝑍𝐵(𝑋) .
Let 𝐴 12 ∈ A12 , and [𝑇11 + 𝑇22 , 𝐴 12 ] = 𝑇11 𝐴 12 − 𝐴 12 𝑇22 .
For Step 2, we know that
[Φ (𝑇11 + 𝑇22 ) , 𝐴 12 ] + [𝑇11 + 𝑇22 , Φ (𝐴 12 )]
= Φ (𝑇11 𝐴 12 ) + Φ (−𝐴 12 𝑇22 )
= Φ ([𝑇11 , 𝐴 12 ]) + Φ ([𝑇22 , 𝐴 12 ])
= [Φ (𝑇11 ) , 𝐴 12 ] + [𝑇11 , Φ (𝐴 12 )]
+ [Φ (𝑇22 ) , 𝐴 12 ] + [𝑇22 , Φ (𝐴 12 )] .
(30)
4
Abstract and Applied Analysis
So [Φ(𝑇11 + 𝑇22 ) − Φ(𝑇11 ) − Φ(𝑇22 ), 𝐴 12 ] = 0. From Lemma 2,
So
(31)
Φ (𝑇11 + 𝑇22 + 𝑇12 ) − Φ (𝑇11 ) − Φ (𝑇22 ) − Φ (𝑇12 ) ∈ 𝑍𝐵(𝑋) .
(39)
On the other hand, from (27) we get that 𝑃1 (Φ(𝑇11 + 𝑇22 ) −
Φ(𝑇11 ) − Φ(𝑇22 ))𝑃2 = 0. So
Similarly, Φ(𝑇22 +𝑇11 +𝑇21 )−Φ(𝑇22 )−Φ(𝑇11 )−Φ(𝑇21 ) ∈ 𝑍𝐵(𝑋) .
Φ (𝑇11 + 𝑇22 ) − Φ (𝑇11 ) − Φ (𝑇22 ) ∈ A12 + 𝑍𝐵(𝑋) .
Φ (𝑇11 + 𝑇22 ) − Φ (𝑇11 ) − Φ (𝑇22 ) ∈ 𝑍𝐵(𝑋) .
(32)
Step 4. For any 𝑇𝑖𝑖 , 𝑆𝑖𝑖 ∈ A𝑖𝑖 (𝑖 = 1, 2), Φ(𝑇𝑖𝑖 + 𝑆𝑖𝑖 ) − Φ(𝑇𝑖𝑖 ) −
Φ(𝑆𝑖𝑖 ) ∈ 𝑍𝐵(𝑋) .
Let 𝑖 = 1. For any 𝐴 12 ∈ A12 , [𝑇11 + 𝑆11 , 𝐴 12 ] = 𝑇11 𝐴 12 +
𝑆11 𝐴 12 . From Step 2, it is that
[Φ (𝑇11 + 𝑆11 ) , 𝐴 12 ] + [𝑇11 + 𝑆11 , Φ (𝐴 12 )]
Step 6. For all 𝑇, 𝑆 ∈ 𝐵(𝑋), Φ(𝑇 + 𝑆) − Φ(𝑇) − Φ(𝑆) ∈ 𝑍𝐵(𝑋) .
Let 𝑇 ∈ 𝐵(𝑋), and write 𝑇 = 𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 . If we
know that
Φ (𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 ) − Φ (𝑇11 )
− Φ (𝑇12 ) − Φ (𝑇21 ) − Φ (𝑇22 ) ∈ 𝑍𝐵(𝑋) ,
then it is right.
For every 𝐴 12 ∈ A12 , [𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 , 𝐴 12 ] = [𝑇11 +
𝑇21 + 𝑇22 , 𝐴 12 ]. By Step 5, it shows that
= Φ (𝑇11 𝐴 12 ) + Φ (𝑆11 𝐴 12 )
= Φ ([𝑇11 , 𝐴 12 ]) + Φ ([𝑆11 , 𝐴 12 ])
[Φ (𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 ) , 𝐴 12 ]
(33)
+ [𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 , Φ (𝐴 12 )]
= [Φ (𝑇11 ) , 𝐴 12 ] + [𝑇11 , Φ (𝐴 12 )]
= [Φ (𝑇11 ) + Φ (𝑇21 ) + Φ (𝑇22 ) , 𝐴 12 ]
+ [Φ (𝑆11 ) , 𝐴 12 ] + [𝑆11 , Φ (𝐴 12 )] .
(34)
On the other hand, from (27) we get that 𝑃1 (Φ(𝑇11 + 𝑆11 ) −
Φ(𝑇11 ) − Φ(𝑆11 ))𝑃2 = 0. So
Φ (𝑇11 + 𝑆11 ) − Φ (𝑇11 ) − Φ (𝑆11 ) ∈ 𝑍𝐵(𝑋) .
(35)
So
[Φ (𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 )
−Φ (𝑇11 ) − Φ (𝑇21 ) − Φ (𝑇22 ) , 𝐴 12 ] = 0.
Step 5. Consider that Φ(𝑇𝑖𝑖 + 𝑇𝑗𝑗 + 𝑇𝑖𝑗 ) − Φ(𝑇𝑖𝑖 ) − Φ(𝑇𝑗𝑗 ) −
Φ(𝑇𝑖𝑗 ) ∈ 𝑍𝐵(𝑋) .
Let 𝑖 = 1, and let 𝑗 = 2. For any 𝐴 12 ∈ A12 , [𝑇11 + 𝑇22 +
𝑇12 , 𝐴 12 ] = [𝑇11 + 𝑇22 , 𝐴 12 ]. By Step 3, it is that
[Φ (𝑇11 + 𝑇22 + 𝑇12 ) , 𝐴 12 ] + [𝑇11 + 𝑇22 + 𝑇12 , Φ (𝐴 12 )]
= [Φ (𝑇11 ) + Φ (𝑇22 ) , 𝐴 12 ] + [𝑇11 + 𝑇22 , Φ (𝐴 12 )] .
(36)
So [Φ(𝑇11 + 𝑇22 + 𝑇12 ) − Φ(𝑇11 ) − Φ(𝑇22 ), 𝐴 12 ] = 0. From
Lemma 2, it follows that
Φ (𝑇11 + 𝑇22 + 𝑇12 ) − Φ (𝑇11 ) − Φ (𝑇22 )
= 𝑃1 (Φ (𝑇11 + 𝑇22 + 𝑇12 ) − Φ (𝑇11 ) − Φ (𝑇22 )) 𝑃2 + 𝑍
(37)
for some central element 𝑍 ∈ 𝑍𝐵(𝑋) . From (27) we get that
(38)
(42)
Because Φ(𝑇12 ) ∈ A12 , we clearly have
[Φ (𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 ) − Φ (𝑇11 )
−Φ (𝑇12 ) − Φ (𝑇21 ) − Φ (𝑇22 ) , 𝐴 12 ] = 0.
The other case can be treated similarly.
Φ (𝑇12 ) = 𝑃1 Φ (𝑇11 + 𝑇22 + 𝑇12 ) 𝑃2 .
(41)
+ [𝑇11 + 𝑇21 + 𝑇22 , Φ (𝐴 12 )] .
So [Φ(𝑇11 + 𝑆11 ) − Φ(𝑇11 ) − Φ(𝑆11 ), 𝐴 12 ] = 0. From Lemma 2,
Φ (𝑇11 + 𝑆11 ) − Φ (𝑇11 ) − Φ (𝑆11 ) ∈ A12 + 𝑍𝐵(𝑋) .
(40)
(43)
Similarly, [𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 , 𝐴 21 ] = [𝑇11 + 𝑇12 + 𝑇22 , 𝐴 21 ],
and we can obtain
[Φ (𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 ) − Φ (𝑇11 )
−Φ (𝑇12 ) − Φ (𝑇21 ) − Φ (𝑇22 ) , 𝐴 21 ] = 0.
(44)
From Lemma 3,
Φ (𝑇11 + 𝑇12 + 𝑇21 + 𝑇22 ) − Φ (𝑇11 )
− Φ (𝑇12 ) − Φ (𝑇21 ) − Φ (𝑇22 ) ∈ 𝑍𝐵(𝑋) .
(45)
For every 𝐴 ∈ 𝐵(𝑋), from Step 6, we define a mapping 𝜏
of 𝐵(𝑋) into 𝑍𝐵(𝑋) by
𝜏 (𝐴) = Φ (𝐴) − Φ (𝑃1 𝐴𝑃1 )
− Φ (𝑃1 𝐴𝑃2 ) − Φ (𝑃2 𝐴𝑃1 ) − Φ (𝑃2 𝐴𝑃2 ) .
(46)
Then let 𝛿(𝐴) = Φ(𝐴) − 𝜏(𝐴).
In the following, we will prove that 𝛿 and 𝜏 are desired
maps.
Abstract and Applied Analysis
5
Claim 4. 𝛿 is an additive derivation.
We divide the proof into the following steps.
Step 1. 𝛿 is an additive map.
We only need to show that 𝛿 is an additive on A𝑖𝑖 . Let
𝐴 𝑖𝑖 , 𝐴󸀠𝑖𝑖 ∈ A𝑖𝑖 ,
𝐴󸀠𝑖𝑖 )
𝛿 (𝐴 𝑖𝑖 +
− 𝛿 (𝐴 𝑖𝑖 ) −
𝛿 (𝐴󸀠𝑖𝑖 )
= Φ (𝐴 𝑖𝑖 + 𝐴󸀠𝑖𝑖 ) − 𝜏 (𝐴 𝑖𝑖 + 𝐴󸀠𝑖𝑖 ) − Φ (𝐴 𝑖𝑖 )
(47)
𝛿 (𝐴 𝑖𝑖 + 𝐴󸀠𝑖𝑖 ) − 𝛿 (𝐴 𝑖𝑖 ) − 𝛿 (𝐴󸀠𝑖𝑖 ) ∈ 𝑍𝐵(𝑋) ∩ A𝑖𝑖 = {0} , (48)
= 𝛿(𝐴 𝑖𝑖 ) +
𝛿(𝐴󸀠𝑖𝑖 ).
= [Φ (𝑇𝑖𝑖 ) , 𝑆𝑖𝑗 ] + [𝑇𝑖𝑖 , Φ (𝑆𝑖𝑗 )]
= [𝛿 (𝑇𝑖𝑖 ) , 𝑆𝑖𝑗 ] + [𝑇𝑖𝑖 , 𝛿 (𝑆𝑖𝑗 )]
(49)
= 𝛿 (𝐴 11 ) 𝐵11 + 𝐴 11 𝛿 (𝐵11 ) + 𝛿 (𝐴 11 ) 𝐵12 + 𝐴 11 𝛿 (𝐵12 )
+ 𝛿 (𝐴 12 ) 𝐵22 + 𝐴 12 𝛿 (𝐵22 ) + 𝛿 (𝐴 22 ) 𝐵22 + 𝐴 22 𝛿 (𝐵22 ) .
(54)
(55)
for all 𝐴 ∈ A11 ⊕ A12 ⊕ A22 .
Similarly, we can know that 𝛿 is an additive derivation on
A11 ⊕ A21 ⊕ A22 , so there exists an operator 𝑆󸀠 ∈ 𝐵(𝑋) such
that
𝛿 (𝐴) = 𝑆󸀠 𝐴 − 𝐴𝑆󸀠
= [𝛿 (𝑇𝑖𝑗 ) 𝑆𝑗𝑗 ] + [𝑇𝑖𝑗 𝛿 (𝑆𝑗𝑗 )]
(56)
for all 𝐴 ∈ A11 ⊕ A21 ⊕ A22 .
= 𝛿 (𝑇𝑖𝑗 ) 𝑆𝑗𝑗 + 𝑇𝑖𝑗 𝛿 (𝑆𝑗𝑗 ) .
So
(50)
On the other hand,
Step 4. Consider that 𝛿(𝑇𝑖𝑗 𝑆𝑗𝑖 ) = 𝛿(𝑇𝑖𝑗 )𝑆𝑗𝑖 +𝑇𝑖𝑗 𝛿(𝑆𝑗𝑖 ), for every
𝑇𝑖𝑗 ∈ A𝑖𝑗 , 𝑆𝑗𝑖 ∈ A𝑗𝑖 (1 ≤ 𝑖 ≠ 𝑗 ≤ 2).
Now, we write 𝑇󸀠 = ∑2𝑖,𝑗=1 𝑇𝑖𝑗󸀠 and 𝑆󸀠 = ∑2𝑖,𝑗=1 𝑆𝑖𝑗󸀠 . Then by
Step 3, we have that
𝑇󸀠 (𝐴 11 + 𝐵22 ) − (𝐴 11 + 𝐵22 ) 𝑇󸀠
𝛿 (𝑇𝑖𝑖 𝑆𝑖𝑖 𝐴 𝑖𝑗 ) = 𝛿 (𝑇𝑖𝑖 ) 𝑆𝑖𝑖 𝐴 𝑖𝑗 + 𝑇𝑖𝑖 𝛿 (𝑆𝑖𝑖 𝐴 𝑖𝑗 )
= 𝑆󸀠 (𝐴 11 + 𝐵22 ) − (𝐴 11 + 𝐵22 ) 𝑆󸀠 .
= 𝛿 (𝑇𝑖𝑖 ) 𝑆𝑖𝑖 𝐴 𝑖𝑗 + 𝑇𝑖𝑖 𝛿 (𝑆𝑖𝑖 ) 𝐴 𝑖𝑗 + 𝑇𝑖𝑖 𝑆𝑖𝑖 𝛿 (𝐴 𝑖𝑗 ) .
(51)
So (𝛿(𝑇𝑖𝑖 𝑆𝑖𝑖 ) − 𝛿(𝑇𝑖𝑖 )𝑆𝑖𝑖 − 𝑇𝑖𝑖 𝛿(𝑆𝑖𝑖 ))𝐴 𝑖𝑗 = 0. Note that 𝐵(𝑋) is
prime, so
𝛿 (𝑇𝑖𝑖 𝑆𝑖𝑖 ) = 𝛿 (𝑇𝑖𝑖 ) 𝑆𝑖𝑖 + 𝑇𝑖𝑖 𝛿 (𝑆𝑖𝑖 ) .
= 𝛿 (𝐴 11 + 𝐴 12 + 𝐴 22 ) (𝐵11 + 𝐵12 + 𝐵22 )
𝛿 (𝐴) = 𝑇󸀠 𝐴 − 𝐴𝑇󸀠
= [Φ (𝑇𝑖𝑗 ) , 𝑆𝑗𝑗 ] + [𝑇𝑖𝑗 , Φ (𝑆𝑗𝑗 )]
𝛿 (𝑇𝑖𝑖 𝑆𝑖𝑖 𝐴 𝑖𝑗 ) = 𝛿 (𝑇𝑖𝑖 𝑆𝑖𝑖 ) 𝐴 𝑖𝑗 + 𝑇𝑖𝑖 𝑆𝑖𝑖 𝛿 (𝐴 𝑖𝑗 ) .
𝛿 (𝐴) 𝐵 + 𝐴𝛿 (𝐵)
So 𝛿(𝐴)𝐵 + 𝐴𝛿(𝐵) = 𝛿(𝐴𝐵), and so 𝛿 is an additive derivation
on A11 ⊕A12 ⊕A22 . So from [15, Theorem 4.1] we can see that
𝛿 is a derivation. Furthermore, due to the main result in [6],
there exists an operator 𝑇󸀠 ∈ 𝐵(𝑋) such that
𝛿 (𝑇𝑖𝑖 𝑆𝑖𝑗 ) = Φ (𝑇𝑖𝑖 𝑆𝑖𝑗 ) = Φ ([𝑇𝑖𝑖 , 𝑆𝑖𝑗 ])
𝛿 (𝑇𝑖𝑗 𝑆𝑗𝑗 ) = Φ (𝑇𝑖𝑗 𝑆𝑗𝑗 ) = Φ ([𝑇𝑖𝑗 , 𝑆𝑗𝑗 ])
+ 𝛿 (𝐴 12 ) 𝐵22 + 𝐴 12 𝛿 (𝐵22 ) + 𝛿 (𝐴 22 ) 𝐵22 + 𝐴 22 𝛿 (𝐵22 ) .
(53)
+ (𝐴 11 + 𝐴 12 + 𝐴 22 ) 𝛿 (𝐵11 + 𝐵12 + 𝐵22 )
Step 2. For every 𝑇𝑖𝑖 , 𝑆𝑖𝑖 ∈ A𝑖𝑖 , 𝑇𝑖𝑗 , 𝑆𝑖𝑗 ∈ A𝑖𝑗 , 𝑆𝑗𝑗 ∈ A𝑗𝑗 (1 ≤
𝑖 ≠ 𝑗 ≤ 2), and 𝛿(𝑇𝑖𝑖 𝑆𝑖𝑗 ) = 𝛿(𝑇𝑖𝑖 )𝑆𝑖𝑗 + 𝑇𝑖𝑖 𝛿(𝑆𝑖𝑗 ), 𝛿(𝑇𝑖𝑗 𝑆𝑗𝑗 ) =
𝛿(𝑇𝑖𝑗 )𝑆𝑗𝑗 + 𝑇𝑖𝑗 𝛿(𝑆𝑗𝑗 ), 𝛿(𝑇𝑖𝑖 𝑆𝑖𝑖 ) = 𝛿(𝑇𝑖𝑖 )𝑆𝑖𝑖 + 𝑇𝑖𝑖 𝛿(𝑆𝑖𝑖 ).
Let 𝑇𝑖𝑗 , 𝑆𝑖𝑗 , 𝐴 𝑖𝑗 ∈ A𝑖𝑗 (𝑖 ≠ 𝑗),
= 𝛿 (𝑇𝑖𝑖 ) 𝑆𝑖𝑗 + 𝑇𝑖𝑖 𝛿 (𝑆𝑖𝑗 ) ,
= 𝛿 (𝐴 11 𝐵11 ) + 𝛿 (𝐴 11 𝐵12 ) + 𝛿 (𝐴 12 𝐵22 ) + 𝛿 (𝐴 22 𝐵22 )
From 𝛿(A𝑖𝑗 ) ⊆ A𝑖𝑗 , we have
Thus
and so 𝛿(𝐴 𝑖𝑖 +
𝛿 (𝐴𝐵) = 𝛿 [(𝐴 11 + 𝐴 12 + 𝐴 22 ) (𝐵11 + 𝐵12 + 𝐵22 )]
= 𝛿 (𝐴 11 ) 𝐵11 + 𝐴 11 𝛿 (𝐵11 ) + 𝛿 (𝐴 11 ) 𝐵12 + 𝐴 11 𝛿 (𝐵12 )
+ 𝜏 (𝐴 𝑖𝑖 ) − Φ (𝐴󸀠𝑖𝑖 ) + 𝜏 (𝐴󸀠𝑖𝑖 ) .
𝐴󸀠𝑖𝑖 )
Let 𝐴 = 𝐴 11 + 𝐴 12 + 𝐴 22 and let 𝐵 = 𝐵11 + 𝐵12 + 𝐵22 be in
A11 ⊕ A12 ⊕ A22 . By Steps 1 and 2, we have
(52)
Step 3. 𝛿 is an additive derivation on A11 ⊕ A12 ⊕ A22
and A11 ⊕ A21 ⊕ A22 .
(57)
For all 𝐴 11 ∈ A11 , for all 𝐵22 ∈ A22 , this shows that
(𝑇󸀠 − 𝑆󸀠 ) 𝐴 11 = 𝐴 11 (𝑇󸀠 − 𝑆󸀠 ) ,
(𝑇󸀠 − 𝑆󸀠 ) 𝐵22 = 𝐵22 (𝑇󸀠 − 𝑆󸀠 ) .
(58)
Multiplying the previous two equations by 𝑃2 and 𝑃1 from
the left respectively, we can know that
󸀠
󸀠
(𝑇21
− 𝑆21
) 𝐴 11 = 0,
󸀠
󸀠
(𝑇12
− 𝑆12
) 𝐵22 = 0.
(59)
6
Abstract and Applied Analysis
󸀠
󸀠
]) ∈ 𝑍𝐵(𝑋) and 𝐵21
𝐴󸀠12 ≠ 0, then 𝜆 1 = 𝜆 2 . So
From 𝜏([𝐴󸀠12 , 𝐵21
Similarly,
󸀠
󸀠
󸀠
󸀠
(𝑇11
− 𝑆11
) 𝐴 11 = 𝐴 11 (𝑇11
− 𝑆11
),
𝛿 (𝐴 12 𝐵21 ) = 𝛿 (𝐴 12 ) 𝐵21 + 𝐴 12 𝛿 (𝐵21 ) ,
(60)
󸀠
󸀠
󸀠
󸀠
− 𝑆22
) 𝐵22 = 𝐵22 (𝑇22
− 𝑆22
).
(𝑇22
𝛿 (𝐵21 𝐴 12 ) = 𝛿 (𝐵21 ) 𝐴 12 + 𝐵21 𝛿 (𝐴 12 )
(66)
for any 𝐴 12 ∈ A12 and 𝐵21 ∈ A21 . So
So
󸀠
󸀠
𝑇12
= 𝑆12
,
󸀠
󸀠
𝑇21
= 𝑆21
,
󸀠
󸀠
− 𝑆11
= 𝜆 1 𝑃1 ,
𝑇11
(61)
󸀠
󸀠
𝑇22
− 𝑆22
= 𝜆 2 𝑃2
for some 𝜆 1 , 𝜆 2 ∈ C.
For every 𝐴 12 ∈ A12 and 𝐵21 ∈ A21 , using (55), (56), and
(61), we get that
𝛿 (𝑆𝑗𝑖 𝑇𝑖𝑗 ) = 𝛿 (𝑆𝑗𝑖 ) 𝑇𝑖𝑗 + 𝑆𝑗𝑖 𝛿 (𝑇𝑖𝑗 ) .
(68)
Let 𝑇, 𝑆 ∈ 𝐵(𝑋), and write 𝑇 = ∑2𝑖,𝑗=1 𝑇𝑖𝑗 and 𝑆 = ∑2𝑖,𝑗=1 𝑆𝑖𝑗 ,
+ 𝛿 (𝑇21 𝑆12 ) + 𝛿 (𝑇21 𝑆11 ) + 𝛿 (𝑇22 𝑆21 ) + 𝛿 (𝑇22 𝑆22 )
= 𝑇󸀠 𝐴 12 𝐵21 − 𝐴 12 𝐵21 𝑇󸀠 − (𝑇󸀠 𝐴 12 − 𝐴 12 𝑇󸀠 ) 𝐵21
= 𝛿 (𝑇) 𝑆 + 𝑇𝛿 (𝑆) .
− 𝐴 12 (𝑆󸀠 𝐵21 − 𝐵21 𝑆󸀠 )
(69)
󸀠
󸀠
󸀠
= −𝐴 12 𝐵21 𝑇 + 𝐴 12 𝑇 𝐵21 − 𝐴 12 𝑆 𝐵21 + 𝐴 12 𝐵21 𝑆
Claim 7. 𝜏 sends the commutators into zero:
𝜏 ([𝐴, 𝐵]) = Φ ([𝐴, 𝐵]) − 𝛿 ([𝐴, 𝐵])
= (𝜆 2 − 𝜆 1 ) 𝐴 12 𝐵21 .
(62)
𝛿 (𝐵21 𝐴 12 ) − 𝛿 (𝐵21 ) 𝐴 12 − 𝐵21 𝛿 (𝐴 12 )
󸀠
(67)
𝛿 (𝑇𝑆) = 𝛿 (𝑇11 𝑆11 ) + 𝛿 (𝑇11 𝑆12 ) + 𝛿 (𝑇12 𝑆22 ) + 𝛿 (𝑇12 𝑆21 )
𝛿 (𝐴 12 𝐵21 ) − 𝛿 (𝐴 12 ) 𝐵21 − 𝐴 12 𝛿 (𝐵21 )
󸀠
so
𝛿 (𝑇𝑖𝑗 𝑆𝑗𝑖 ) = 𝛿 (𝑇𝑖𝑗 ) 𝑆𝑗𝑖 + 𝑇𝑖𝑗 𝛿 (𝑆𝑗𝑖 ) ,
󸀠
󸀠
= [Φ (𝐴) , 𝐵] + [𝐴, Φ (𝐵)] − 𝛿 ([𝐴, 𝐵])
= [𝛿 (𝐴) , 𝐵] + [𝐴, 𝛿 (𝐵)] − 𝛿 ([𝐴, 𝐵])
󸀠
= 𝑆 𝐵21 𝐴 12 − 𝐵21 𝐴 12 𝑆 − (𝑆 𝐵21 − 𝐵21 𝑆 ) 𝐴 12
= 0.
− 𝐵21 (𝑆󸀠 𝐴 12 − 𝐴 12 𝑆󸀠 )
The proof is complete.
= −𝐵21 𝐴 12 𝑆󸀠 + 𝐵21 𝑆󸀠 𝐴 12 − 𝐵21 𝑇󸀠 𝐴 12 + 𝐵21 𝐴 12 𝑇󸀠
Acknowledgments
= (𝜆 2 − 𝜆 1 ) 𝐵21 𝐴 12 .
Since at least one of 𝑃2 or 𝑃1 is of rank greater than 2, 𝐴󸀠12 ∈
󸀠
A12 and 𝐵21
∈ A21 exist such that
󸀠
𝐴󸀠12 𝐵21
= 0,
󸀠
𝐵21
𝐴󸀠12 ≠ 0
(63)
󸀠
𝐴󸀠12 = 0,
𝐵21
󸀠
𝐴󸀠12 𝐵21
≠ 0.
(64)
or
We suppose that
get
󸀠
𝐴󸀠12 𝐵21
= 0 and
󸀠
𝐵21
𝐴󸀠12
≠ 0. From (62), we
󸀠
󸀠
󸀠
] + [𝐴󸀠12 , Φ (𝐵21
)] − 𝛿 ([𝐴󸀠12 , 𝐵21
])
= [Φ (𝐴󸀠12 ) , 𝐵21
󸀠
󸀠
󸀠
] + [𝐴󸀠12 , 𝛿 (𝐵21
)] − 𝛿 ([𝐴󸀠12 , 𝐵21
])
= [𝛿 (𝐴󸀠12 ) , 𝐵21
󸀠
󸀠
󸀠
− 𝐵21
𝛿 (𝐴󸀠12 ) + 𝐴󸀠12 𝛿 (𝐵21
)
= 𝛿 (𝐴󸀠12 ) 𝐵21
󸀠
󸀠
󸀠
) 𝐴󸀠12 − 𝛿 (𝐴󸀠12 𝐵21
) + 𝛿 (𝐵21
𝐴󸀠12 )
− 𝛿 (𝐵21
󸀠
󸀠
= (𝜆 1 − 𝜆 2 ) 𝐴󸀠12 𝐵21
+ (𝜆 2 − 𝜆 1 ) 𝐵21
𝐴󸀠12
The authors would like to thank his tutor Professor Baodong
Zheng for his valuable suggestions and encouragement. This
work was supported by Mudanjiang city science and technology plan Projects (G2013n1446) and Mudanjiang Normal
Colllege Youth academic backbone Projects (G201209).
References
󸀠
󸀠
󸀠
]) = Φ ([𝐴󸀠12 , 𝐵21
]) − 𝛿 ([𝐴󸀠12 , 𝐵21
])
𝜏 ([𝐴󸀠12 , 𝐵21
󸀠
= (𝜆 2 − 𝜆 1 ) 𝐵21
𝐴󸀠12 .
(70)
(65)
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