# Zero Triple Product Determined Matrix Algebras

Authors: Hongmei Yao; Baodong Zheng
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

LetA be an algebra over a commutative unital ring 𝒞. We say that A is zero triple
product determined if for every 𝒞-module X and every trilinear map {·,·,·}, the following
holds: if {x,y,z}=0 whenever xyz=0, then there exists a 𝒞-linear operator T:A3→X
such that x,y,z=T(xyz) for all x,y,z∈A. If the ordinary
triple product in the aforementioned definition is replaced by Jordan triple
product, then A is called zero Jordan triple
product determined. This paper mainly shows that matrix algebra Mn(B), n≥3, where
B is any commutative unital algebra even different
from the above mentioned commutative unital algebra 𝒞, is always zero triple product determined, and Mn(F), n≥3, where F
is any field with chF≠2, is also zero Jordan triple product determined.


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## Body

## 1. Introduction

Over the last couple of years, several papers characterizing bilinear maps on algebras through their action on elements whose certain product is zero have been written; see [1–6]. The philosophy in these papers is that certain classical problems concerning linear maps that preserve zero product, Jordan product, commutativity, and so forth can be sometimes effectively solved by considering bilinear maps that preserve certain zero product properties. For example, in [1], in order to determine whether a linear map preserving zero product (resp., zero Jordan product, zero Lie product) is “closed” to a homomorphism, the authors introduced the definitions of zero product (resp., zero Jordan product, zero Lie product) determined algebras. The core idea of these definitions is to answer the aforementioned questions by determining the bilinear maps preserving zero product (resp., zero Jordan product, zero Lie product). Furthermore, as the main task, they gave the positive answer that the matrix algebra Mn(B) of n×n matrices over a unital algebra B is zero product determined, and under some further restrictions on B, Mn(B) is still zero Jordan (resp., Lie) product determined.Meanwhile, in preserver problems, there have appeared many kinds of preserving product forms, such as the forms in papers [7–10]. Particularly, in [10], there appears a definition of Jordan triple product. This definition has applications not only in Banach algebra (see [11]) but also in generalized inverse of matrices (see [12]).Inspired by the aforementioned, this gives rise to a question: whether can we generalize the products in paper [1] to more generalized forms, such as triple product and Jordan triple product? And immediately, it follows another interesting preserver problem: whether a linear map preserving zero triple product (resp., zero Jordan triple product) is still “closed” to a homomorphism? In order to answer the aforementioned questions, we introduce the following definitions. Let 𝒞 be a (fixed) commutative unital ring and A an algebra over 𝒞. By A3, we denote the 𝒞-linear span of all elements of the form xyz, where x,y,z∈A. X is a 𝒞-module and we denote by {·,·,·}:A×A×A→X a 𝒞-trilinear map. Consider the following conditions:(a)
for allx,y,z∈A such that xyz=0, we have {x,y,z}=0;(b)
there exists a𝒞-linear map T:A3→X such that {x,y,z}=T(xyz) for all x,y,z∈A.Trivially, (b) implies (a). We call thatA is a zero triple product determined algebra if for every 𝒞-module X and every 𝒞-trilinear map {·,·,·}, (a) implies (b). And we say that A is a zero Jordan triple product determined algebra if we replace the above triple product by Jordan triple product (x∘y∘z=xyz+zyx).From the previous definitions, it is interesting to examine whether the matrix algebraMn(B) of n×n matrices over a unital algebra B is still zero triple product (resp., zero Jordan triple product) determined. If the unital algebra B has no further restrictions, this problem will be difficult. Therefore, the purpose of this paper is to characterize the zero triple product (resp., zero Jordan triple product) determined algebra under some additional restrictions on the unital algebra B. In Section 2, we show that the answer is “yes” for the triple product if B is any commutative unital algebra even different from 𝒞. The Jordan triple product case, treated in Section 3, is more difficult; we only show that the matrix algebra Mn(F), where F is any field with chF≠2 (chF stands for the characteristic of a field) is also zero Jordan triple product determined.Finally, we end this section by giving an equivalent condition of (b) in the previous definition which is more convenient to use:(b′) if xt,yt,zt∈A,t=1,…,m, are such that ∑t=1mxtytzt=0, then ∑t=1m{xt,yt,zt}=0.


## 2. Zero Triple Product Determined Matrix Algebras

In this part, we will consider the matrix algebraMn(B), where B is any commutative unital algebra even different from the (fixed) commutative unital ring 𝒞 mentioned in Section 1. By beij, where b∈B, we denote the matrix whose (i,j) entry is b and all other entries are 0.Theorem 2.1.
LetB be a commutative unital algebra, then Mn(B) is a zero triple product determined algebra for every n≥3.Proof.
In order to prove that (a) implies (b), we only prove that (a) implies the equivalent condition (b'). Set A=Mn(B), let X be a 𝒞-module, and let {·,·,·} be a 𝒞-trilinear map such that for all x,y,z∈A, xyz=0 implies {x,y,z}=0. Throughout the proof, a,b,andc denote arbitrary elements in B and i,j,k,l,r,andh denote arbitrary indices.
First, sinceaeijbeklcerh=0 if j≠k or r≠l, we get
(2.1){aeij,bekl,cerh}=0ifj≠korr≠l.
Second, (aeij+abeik)(bejl-ekl)celh=0, where j≠k, and consequently
(2.2){aeij,bejl,celh}={abeik,ekl,celh},
replacing a by ab and b by 1 in (2.2), we have
(2.3){aeij,bejl,celh}={abeij,ejl,celh},
and using (2.3) and (2.2), this yields that
(2.4){aeij,bejl,celh}={abei1,e1l,celh}.
Similarly, from aeij(bejl+bcejk)(celh-ekh)=0, where l≠k, we obtain
(2.5){aeij,bejl,celh}={aeij,bcej1,e1h}.
Finally, combining (2.5) and (2.4), it follows that
(2.6){aeij,bejl,celh}={aeij,bcej1,e1h}={abcei1,e11,e1h}.
Let xt,yt,zt∈A be such that ∑t=1mxtytzt=0; then we only need to show
(2.7)∑t=1m{xt,yt,zt}=0.
Writing
(2.8)xt=∑i=1n∑j=1naijteij,yt=∑k=1n∑l=1nbkltekl,zt=∑r=1n∑h=1ncrhterh,
it follows, by examining the (i,h) entry of ∑t=1mxtytzt, that for all i and h, we have
(2.9)∑t=1m∑j=1n∑l=1naijtbjltclht=0.
Note that
(2.10)∑t=1m{xt,yt,zt}=∑t=1m{∑i=1n∑j=1naijteij,∑k=1n∑l=1nbkltekl,∑r=1n∑h=1ncrhterh},
by (2.1), this summation reduces to
(2.11)∑t=1m{xt,yt,zt}=∑t=1m∑i=1n∑j=1n∑l=1n∑h=1n{aijteij,bjltejl,clhtelh},
and using (2.6) and (2.9), we obtain
(2.12)∑t=1m{xt,yt,zt}=∑t=1m∑i=1n∑j=1n∑l=1n∑h=1n{aijtbjltclhtei1,e11,e1h}=∑i=1n∑h=1n{∑t=1m∑j=1n∑l=1naijtbjltclhtei1,e11,e1h}=0.
Therefore, the result of this theorem holds.


## 3. Zero Jordan Triple Product Determined Matrix Algebras

In this part, we only consider the matrix algebraMn(F), where F is a field with chF≠2, and 𝒞 is still the (fixed) commutative unital ring mentioned in Section 1. Let beij, where b∈F, be the matrix whose (i,j) entry is b and all other entries are 0.Theorem 3.1.
LetF be a field with chF≠2; then Mn(F) is a zero Jordan triple product determined algebra for every n≥3.Proof.
SetA=Mn(F), let X be a 𝒞-module, and let {·,·,·} be a 𝒞-trilinear map such that for all x,y,z∈A, x∘y∘z=0 implies {x,y,z}=0. Let a,b,andc be arbitrary elements from F and i,j,k,l,r,andh denote arbitrary indices.
First, forj≠k,h≠korj≠k,l≠iorl≠r,h≠korl≠r,andl≠i, we have aeij∘bekl∘cerh=0, and so
(3.1){aeij,bekl,cerh}=0ifj≠k,h≠korj≠k,l≠iorl≠r,h≠korl≠r,l≠i.
For any z∈A and i≠k,j≠k, we have (aeij+abeik)∘(bejk-ekk)∘z=0 and z∘(aeij+abeik)∘(bejk-ekk)=0, which implies
(3.2){aeij,bejk,z}={abeik,ekk,z}ifi≠k,j≠k,(3.3){z,aeij,bejk}={z,abeik,ekk}ifi≠k,j≠k.
Similarly,
(3.4){bejk,aeij,z}={ekk,abeik,z}ifi≠k,j≠k,(3.5){z,bejk,aeij}={z,ekk,abeik}ifi≠k,j≠k.
For any z∈A and i≠k, we have (aeik-eii)∘(abeik+bekk)∘z=0, (aeik-ekk)∘(abeik+beii)∘z=0, z∘(aeik-eii)∘(abeik+bekk)=0, and z∘(aeik-ekk)∘(abeik+beii)=0, and consequently,
(3.6){aeik,bekk,z}={eii,abeik,z}ifi≠k,(3.7){aeik,beii,z}={ekk,abeik,z}ifi≠k,(3.8){z,aeik,bekk}={z,eii,abeik}ifi≠k,(3.9){z,aeik,beii}={z,ekk,abeik}ifi≠k.
Similarly,
(3.10){bekk,aeik,z}={abeik,eii,z}ifi≠k,(3.11){beii,aeik,z}={abeik,ekk,z}ifi≠k,(3.12){z,bekk,aeik}={z,abeik,eii}ifi≠k,(3.13){z,beii,aeik}={z,abeik,ekk}ifi≠k.
Since aeih∘(eii-ehh)∘(eii+ehh)=0 and (eii-ehh)∘(eii+ehh)∘aeih=0, if h≠i, it follows that
(3.14){aeih,eii,eii}={aeih,ehh,ehh}ifi≠h,(3.15){eii,eii,aeih}={ehh,ehh,aeih}ifi≠h.
Using (3.8), (3.11), (3.15), and (3.14), we arrive at
(3.16){eii,eii,aeih}={aeih,ehh,ehh}={ehh,ehh,aeih}={aeih,eii,eii}ifi≠h.
Using (3.7) and (3.12), we have
(3.17){aeij,beji,ejj}={abeij,eji,ejj}ifi≠j.
Since (aeij+beji)∘(aeij-beji)∘eii=0 if i≠j, then
(3.18){aeij,beji,eii}={beji,aeij,eii}ifi≠j.
Using (3.18) and (3.17), it follows that
(3.19){aeij,beji,eii}={abeji,eij,eii}ifi≠j.
Using (3.6), (3.8), (3.18), and (3.17), we obtain
(3.20){eii,aeij,beji}={baeji,eij,eii}={eij,baeji,eii}ifi≠j.
Similarly, using (3.10), (3.9), and (3.17), this yields
(3.21){eii,aeji,beij}={abeji,eijeii}ifi≠j.
Further, we claim that(3.22){aeii,beii,ceii}={12a∘b∘ceii,eii,eii}.
If i≠j, then ((1/2)abeii+aeij-(1/2)abejj)∘(beji-eii+ejj)∘(ceii+cejj)=0; consequently,
(3.23){12abeii+aeij-12abejj,beji-eii+ejj,ceii+cejj}=0,ifi≠j.
Using (3.8)–(3.13) and (3.16), we get
(3.24){aeij,beji,ceii+cejj}={12abeii,eii,ceii}+{12abejj,ejj,cejj},ifi≠j.
From (aeii+beji-beij-aejj)∘(beii-aeij+aeji-bejj)∘(ceii+cejj)=0, where i≠j, it follows that
(3.25){aeii+beji-beij-aejj,beii-aeij+aeji-bejj,ceii+cejj}=0.
Using (3.3), (3.4), (3.6)–(3.13), and (3.16), we arrive at
(3.26){aeii,beii,ceii}+{aejj,bejj,cejj}={beij,aeji,ceii+cejj}+{beji,aeij,ceii+cejj}.
Then, by (3.24), we have
(3.27){abeii,eii,ceii}+{abejj,ejj,cejj}={aeii,beii,ceii}+{aejj,bejj,cejj},wherei≠j.
Since n≥3, we choose k such that k≠i,j; applying (3.27), we get
(3.28){aeii,beii,ceii}+{aejj,bejj,cejj}+{aekk,bekk,cekk}+{aeii,beii,ceii}={abeii,eii,ceii}+{abejj,ejj,cejj}+{abekk,ekk,cekk}+{abeii,eii,ceii}=2{abeii,eii,ceii}+{aekk,bekk,cekk}+{aejj,bejj,cejj},
then
(3.29){aeii,beii,ceii}={abeii,eiiceii}.
Similarly, by (ceii+cejj)∘(beji-eii+ejj)∘((1/2)abeii+aeij-(1/2)abejj)=0 and (ceii+cejj)∘(beii-aeij+aeji-bejj)∘(aeii+beji-beij-aejj)=0, where i≠j, we get
(3.30){ceii,beii,aeii}={ceii,baeii,eii}.
Then, combining (3.29) and (3.30), we have {ceii,aeii,beii}={abceii,eii,eii}. Hence, the claim (3.22) holds.
Finally, we claim that(3.31){aeij,beji,ceij}={eii,eii,a∘b∘ceij},wherei≠j.
For i≠j, we have (-(1/2)bceij+b-1ceji+cejj)∘(-(1/2)abeii+(1/2)abejj+aeji)∘(eii-ejj+beij)=0, and consequently {-(1/2)bceij+b-1ceji+cejj,-(1/2)abeii+(1/2)abejj+aeji,eii-ejj+beij}=0. Using (3.8)–(3.13), (3.16), (3.17), and (3.19), this can be reduced to
(3.32){ab2ceij,eii,eii}+{abceij,eji,ejj}={cejj,12abejj,ejj}+{12bceij,aeji,beij}.
Replacing b by 1 and 2 in (3.32), respectively, we have
(3.33){aceij,eii,eii}+{aceij,eji,ejj}={cejj,12aejj,ejj}+{12ceij,aeji,eij},(3.34){4aceij,eii,eii}+{2aceij,eji,ejj}={cejj,aejj,ejj}+{ceij,aeji,2eij}.
Computing (3.34)–2(3.33), this yields
(3.35){2aceij,eii,eii}={ceij,aeji,eij},
then taking (3.35) into (3.33), we derive
(3.36){aceij,eji,ejj}={cejj,12aejj,ejj},
replacing a by ab in (3.36), we get
(3.37){abceij,eji,ejj}={cejj,12abejj,ejj},
then taking (3.37) into (3.32), it follows that
(3.38){ab2ceij,eii,eii}={12bceij,aeji,beij},
and replacing b by b+1 in (3.38), and using (3.38), (3.35), and (3.16), we get
(3.39){ceij,aeji,beij}={eii,eii,2abceij}={eii,eii,a∘b∘ceij},wherei≠j.
Therefore, (3.31) holds.
Letxt,yt,andzt be such that ∑t=1mxt∘yt∘zt=0; we have to prove that ∑t=1m{xt,yt,zt}=0. Writing
(3.40)xt=∑i=1n∑j=1naijteij,yt=∑k=1n∑l=1nbkltekl,zt=∑r=1n∑h=1ncrhterh,
it follows, by examining the (i,h) entry of ∑t=1mxt∘yt∘zt=0, that for all i,h, we have
(3.41)∑t=1m∑j=1n∑l=1n(aijtbjltclht+cijtbjltalht)=0.
First, by (3.1), we get
(3.42)∑t=1m{xt,yt,zt}=∑t=1m∑i=1n∑j=1n∑k≠jn∑r=1n{aijteij,bkiteki,crkterk}(3.43)+∑t=1m∑i=1n∑j=1n∑l=1n∑h=1n{aijteij,bjltejl,clhtelh}(3.44)+∑t=1m∑i=1n∑j=1n∑r≠in{aijteij,bjiteji,crjterj}.
The summation (3.42) can be written as
(3.45)∑t=1m∑i=1n∑j≠in∑k≠jn∑r=1n{aijteij,bkiteki,crkterk}+∑t=1m∑i=1n∑k≠in∑r=1n{aiiteii,bkiteki,crkterk},
using (3.4), (3.5), and (3.21), the first summation can be rewritten as
(3.46)∑t=1m∑i=1n∑j≠in∑k≠jn∑r=1n{aijteij,bkiteki,crkterk}=∑t=1m∑i=1n∑j≠in∑k≠jn∑r=1n{ejj,aijtbkitekj,crkterk}=∑t=1m∑i=1n∑j≠in∑k≠jn∑r≠jn{ejj,aijtbkitekj,crkterk}+∑t=1m∑i=1n∑j≠in∑k≠jn{ejj,aijtbkitekj,cjktejk}=∑t=1m∑i=1n∑j≠in∑k≠jn∑r≠jn{ejj,ejj,aijtbkitcrkterj}+∑t=1m∑i=1n∑j≠in∑k≠jn{aijtbkitcjktekj,ejk,ejj},
and using (3.8), (3.12), (3.9), (3.10), (3.4), (3.17), and (3.18), we split the second summation in three parts:
(3.47)∑t=1m∑i=1n∑k≠in∑r=1n{aiiteii,bkiteki,crkterk}=∑t=1m∑i=1n∑k≠in∑r=1n{aiitbkiteki,ekk,crkterk}=∑t=1m∑i=1n∑k≠in∑r≠kn{aiitbkiteki,ekk,crkterk}+∑t=1m∑i=1n∑k≠in{aiitbkiteki,ekk,ckktekk}=∑t=1m∑i=1n∑k≠in∑r≠kn{aiitbkiteki,crkterk,err}+∑t=1m∑i=1n∑k≠in{eii,aiitbkiteki,ckktekk}=∑t=1m∑i=1n∑k≠in∑r≠k,in{aiitbkiteki,crkterk,err}+∑t=1m∑i=1n∑k≠in{aiitbkiteki,cikteik,eii}+∑t=1m∑i=1n∑k≠in{eii,eii,aiitbkitckkteki}=∑t=1m∑i=1n∑k≠in∑r≠k,in{eii,aiitbkitcrkteri,err}+∑t=1m∑i=1n∑k≠in{aiitbkitcikteki,eik,eii}+∑t=1m∑i=1n∑k≠in{eii,eii,aiitbkitckkteki}=∑t=1m∑i=1n∑k≠in∑r≠k,in{eii,eii,aiitbkitcrkteri}+∑t=1m∑i=1n∑k≠in{eik,aiitbkitcikteki,eii}+∑t=1m∑i=1n∑k≠in{eii,eii,aiitbkitckkteki}.
Therefore, the summation (3.42) can be rewritten as
(3.48)∑t=1m∑i=1n∑j≠in∑k≠jn∑r≠jn{ejj,ejj,aijtbkitcrkterj}+∑t=1m∑i=1n∑j≠in∑k≠jn{aijtbkitcjktekj,ejk,ejj}+∑t=1m∑i=1n∑k≠in∑r≠k,in{eii,eii,aiitbkitcrkteri}+∑t=1m∑i=1n∑k≠in{eik,aiitbkitcikteki,eii}+∑t=1m∑i=1n∑k≠in{eii,eii,aiitbkitckkteki}.
The summation (3.43) can be written as
(3.49)∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h=1n{aijteij,bjltejl,clhtelh}+∑t=1m∑i=1n∑j≠in∑h=1n{aijteij,bjjtejj,cjhtejh}+∑t=1m∑i=1n∑j=1n∑h=1n{aijteij,bjiteji,cihteih},
using (3.2), (3.10), (3.8), (3.16), (3.17), (3.18), and (3.6), the first summation can be rewritten as
(3.50)∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h=1n{aijteij,bjltejl,clhtelh}=∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h=1n{aijtbjlteil,ell,clhtelh}=∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h≠ln{aijtbjlteil,ell,clhtelh}+∑t=1m∑i=1n∑j=1n∑l≠i,jn{aijtbjlteil,ell,clltell}=∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h≠ln{aijtbjlteil,clhtelh,ehh}+∑t=1m∑i=1n∑j=1n∑l≠i,jn{eii,aijtbjlteil,clltell}=∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h≠l,in{aijtbjlteil,clhtelh,ehh}+∑t=1m∑i=1n∑j=1n∑l≠i,jn{aijtbjlteil,cliteli,eii}+∑t=1m∑i=1n∑j=1n∑l≠i,jn{eii,eii,aijtbjltcllteil}=∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h≠l,in{aijtbjltclhteih,ehh,ehh}+∑t=1m∑i=1n∑j=1n∑l≠i,jn{aijtbjltcliteli,eil,eii}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{eii,eii,aijtbjltcllteil}+∑t=1m∑i=1n∑l≠in{eii,eii,aiitbiltcllteil}=∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h≠l,i,jn{ehh,ehh,aijtbjltclhteih}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{ejj,ejj,aijtbjltcljteij}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{aijtbjltcliteli,eil,eii}+∑t=1m∑i=1n∑l≠in{aiitbiltcliteli,eil,eii}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{eii,eii,aijtbjltcllteil}+∑t=1m∑i=1n∑l≠in{eii,eii,aiitbiltcllteil},
and using (3.6), (3.3), (3.8), (3.16), and (3.20), we rewrite the second summation as
(3.51)∑t=1m∑i=1n∑j≠in∑h=1n{aijteij,bjjtejj,cjhtejh}=∑t=1m∑i=1n∑j≠in∑h=1n{eii,aijtbjjteij,cjhtejh}=∑t=1m∑i=1n∑j≠in∑h≠in{eii,aijtbjjteij,cjhtejh}+∑t=1m∑i=1n∑j≠in{eii,aijtbjjteij,cjiteji}=∑t=1m∑i=1n∑j≠in∑h≠i,jn{eii,aijtbjjteij,cjhtejh}+∑t=1m∑i=1n∑j≠in{eii,aijtbjjteij,cjjtejj}+∑t=1m∑i=1n∑j≠in{eij,aijtbjjtcjiteji,eii}=∑t=1m∑i=1n∑j≠in∑h≠i,jn{eii,aijtbjjtcjhteih,ehh}+∑t=1m∑i=1n∑j≠in{eii,eii,aijtbjjtcjjteij}+∑t=1m∑i=1n∑j≠in{eij,aijtbjjtcjiteji,eii}=∑t=1m∑i=1n∑j≠in∑h≠i,jn{ehh,ehh,aijtbjjtcjhteih}+∑t=1m∑i=1n∑j≠in{eii,eii,aijtbjjtcjjteij}+∑t=1m∑i=1n∑j≠in{eij,aijtbjjtcjiteji,eii},
and then using (3.3), (3.2), (3.8), (3.6), (3.20), (3.22), and (3.31), the third summation is split in four parts:
(3.52)∑t=1m∑i=1n∑j=1n∑h=1n{aijteij,bjiteji,cihteih}=∑t=1m∑i=1n∑j=1n∑h≠in{aijteij,bjiteji,cihteih}+∑t=1m∑i=1n∑j=1n{aijteij,bjiteji,ciiteii}=∑t=1m∑i=1n∑j=1n∑h≠i,jn{aijteij,bjiteji,cihteih}+∑t=1m∑i=1n∑j≠in{aijteij,bjiteji,cijteij}+∑t=1m∑i=1n∑j≠in{aijteij,bjiteji,ciiteii}+∑t=1m∑i=1n{aiiteii,biiteii,ciiteii}=∑t=1m∑i=1n∑j=1n∑h≠i,jn{aijteij,bjitcihtejh,ehh}+∑t=1m∑i=1n∑j≠in{eii,eii,aijt∘bjit∘cijteij}+∑t=1m∑i=1n∑j≠in{eii,aijteij,bjitciiteji}+∑t=1m∑i=1n{12aiit∘biit∘ciiteii,eii,eii}=∑t=1m∑i=1n∑j=1n∑h≠i,jn{aijtbjitcihteih,ehh,ehh}+∑t=1m∑i=1n∑j≠in{eii,eii,aijt∘bjit∘cijteij}+∑t=1m∑i=1n∑j≠in{eij,aijtbjitciiteji,eii}+∑t=1m∑i=1n{12aiit∘biit∘ciiteii,eii,eii}.
Hence, the summation (3.43) can be rewritten as
(3.53)∑t=1m∑i=1n∑j=1n∑l≠i,jn∑h≠l,i,jn{ehh,ehh,aijtbjltclhteih}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{ejj,ejj,aijtbjltcljteij}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{aijtbjltcliteli,eil,eii}+∑t=1m∑i=1n∑l≠in{aiitbiltcliteli,eil,eii}+∑t=1m∑i=1n∑j≠in∑l≠i,jn{eii,eii,aijtbjltcllteil}+∑t=1m∑i=1n∑l≠in{eii,eii,aiitbiltcllteil}+∑t=1m∑i=1n∑j≠in∑h≠i,jn{ehh,ehh,aijtbjjtcjhteih}+∑t=1m∑i=1n∑j≠in{eii,eii,aijtbjjtcjjteij}+∑t=1m∑i=1n∑j≠in{eij,aijtbjjtcjiteji,eii}+∑t=1m∑i=1n∑j=1n∑h≠i,jn{ehh,ehh,aijtbjitcihteih}+∑t=1m∑i=1n∑j≠in{eii,eii,aijt∘bjit∘cijteij}+∑t=1m∑i=1n∑j≠in{eij,aijtbjitciiteji,eii}+∑t=1m∑i=1n{12aiit∘biit∘ciiteii,eii,eii}.
Finally, using (3.5), (3.7), (3.8), (3.10), (3.12), (3.16), and (3.21), the summation (3.44) can be written as
(3.54)∑t=1m∑i=1n∑j≠in∑r≠in{aijteij,bjiteji,crjterj}+∑t=1m∑i=1n∑r≠in{aiiteii,biiteii,criteri}=∑t=1m∑i=1n∑j≠in∑r≠in{ejj,aijteij,bjitcrjteri}+∑t=1m∑i=1n∑r≠in{aiiteii,biitcriteri,err}=∑t=1m∑i=1n∑j≠in∑r≠i,jn{ejj,aijteij,bjitcrjteri}+∑t=1m∑i=1n∑j≠in{ejj,aijteij,bjitcjjteji}+∑t=1m∑i=1n∑r≠in{aiitbiitcriteri,err,err}=∑t=1m∑i=1n∑j≠in∑r≠i,jn{ejj,ejj,aijtbjitcrjterj}+∑t=1m∑i=1n∑j≠in{aijtbjitcjjteij,eji,ejj}+∑t=1m∑i=1n∑r≠in{err,err,aiitbiitcriteri}.
Consequently, rewriting the summation (3.44) as
(3.55)∑t=1m∑i=1n∑j≠in∑r≠i,jn{ejj,ejj,aijtbjitcrjterj}+∑t=1m∑i=1n∑j≠in{aijtbjitcjjteij,eji,ejj}+∑t=1m∑i=1n∑r≠in{err,err,aiitbiitcriteri},
therefore, we get
(3.56)∑t=1m{xt,yt,zt}=(3.47)+(3.52)+(3.54).
Next, we only need to prove that(3.47)+(3.52)+(3.54)=0.
Replacing the indicesr,k,i,andj by i,j,l,andh, correspondingly, in the first summation of (3.48), and rewriting the summation as
(3.57)∑t=1m∑i=1n∑j≠in∑k≠jn∑r≠jn{ejj,ejj,aijtbkitcrkterj}=∑t=1m∑i=1n∑j=1n∑l=1n∑h≠i,j,ln{ehh,ehh,cijtbjltalhteih},
then adding (3.57) and the first, the seventh, the tenth summations of (3.53), together, this yields
(3.58)∑t=1m∑i=1n∑j=1n∑l=1n∑h≠i,j,ln{ehh,ehh,(aijtbjltclht+cijtbjltalht)eih}.
Replacing iandk by landi, respectively, in the fifth summation of (3.48), then by (3.16), this summation is further equal to
(3.59)∑t=1m∑i=1n∑l≠in{eii,eii,alltbiltciiteil},
and replacing r,k,andi by i,j,andl, correspondingly, in the third summation of (3.48), then using (3.16), the summation is further equal to
(3.60)∑t=1m∑i=1n∑j≠in∑l≠i,jn{eii,eii,alltbjltcijteil}.
Then adding (3.59), (3.60), and the fifth and the sixth summations of (3.53), together, this yields
(3.61)∑t=1m∑i=1n∑j=1n∑l≠i,jn{eii,eii,(aijtbjltcllt+alltbjltcijt)eil}.
Replacing k by j in the fourth summation of (3.48), then adding the 12th summation of (3.53), we get
(3.62)∑t=1m∑i=1n∑j≠in{eij,(aijtbjitciit+aiitbjitcijt)eji,eii}.
Replacing iandr by jandi, correspondingly, in the third summation of (3.55), then adding the 8th summation of (3.53), we have
(3.63)∑t=1m∑i=1n∑j≠in{eii,eii,(aijtbjjtcjjt+ajjtbjjtcijt)eij}.
Replacing r,j,andi by i,j,andl, respectively, in the first summation of (3.55), and rewriting this summation as
(3.64)∑t=1m∑i=1n∑j≠in∑l≠i,jn{ejj,ejj,aljtbjltcijteij},
then adding (3.64) and the second summation of (3.53), together, and by (3.16), we derive
(3.65)∑t=1m∑i=1n∑j≠in∑l≠i,jn{eii,eii,(aijtbjltcljt+aljtbjltcijt)eij}.
Replacing iandl by jandi, respectively, in the fourth summation of (3.53), then adding the second summation of (3.55), we arrive at
(3.66)∑t=1m∑i=1n∑j≠in{(aijtbjitcjjt+ajjtbjitcijt)eij,eji,ejj}.
Replacing i,j,andk by l,i,andj, respectively, in the second summation of (3.48), and rewriting this summation as
(3.67)∑t=1m∑i=1n∑j≠in∑l≠i,jn{alitbjltcijteji,eij,eii}+∑t=1m∑i=1n∑j≠in{ajitbjjtcijteji,eij,eii},
then using (3.19) and (3.18), the aforementioned summation can be rewritten as
(3.68)∑t=1m∑i=1n∑j≠in∑l≠i,jn{alitbjltcijteji,eij,eii}+∑t=1m∑i=1n∑j≠in{eij,ajitbjjtcijteji,eii}.
Adding the second summation of (3.68) and the ninth summation of (3.53), together, we get
(3.69)∑t=1m∑i=1n∑j≠in{eij,(aijtbjjtcjit+ajitbjjtcijt)eji,eii}.
By (3.36), we know that the first summation of (3.68) can be written as
(3.70)∑t=1m∑i=1n∑j≠in∑l≠i,jn{12alitbjltcijteii,eii,eii},
similarly, using (3.36), the third summation of (3.53) can be written as
(3.71)∑t=1m∑i=1n∑j≠in∑l≠i,jn{12aijtbjltcliteii,eii,eii},
and then adding (3.70) and (3.71), together, we have
(3.72)∑t=1m∑i=1n∑j≠in∑l≠i,jn{12(aijtbjltclit+alitbjltcijt)eii,eii,eii}.
Therefore, we derive
(3.73)(3.47)+(3.52)+(3.54)=(3.52)+(3.60)+(3.62)+(3.64)+the11thsummationof(47)+(3.61)+(3.65)+(3.68)+(3.71)+the13thsummationof(3.52).
Our first goal is to show(3.74)(3.52)+(3.60)+(3.62)+(3.64)+the11thsummationof(3.52)=0.
First, notice that by (3.16), the summation (3.61) can be rewritten as
(3.75)∑t=1m∑i≠ln∑j≠ln∑l=1n{ell,ell,(aijtbjltcllt+alltbjltcijt)eil}.
Then replacing l by h, we have
(3.76)∑t=1m∑i≠hn∑j≠hn∑h=1n{ehh,ehh,(aijtbjhtchht+ahhtbjhtcijt)eih}.
Second, notice that
(3.77)(3.64)+the11thsummationof(3.52)=∑t=1m∑i=1n∑j≠in∑l≠jn{eii,eii,(aijtbjltcljt+cijtbjltaljt)eij}.
Then, by (3.16), and replacing j by h, this yields
(3.78)∑t=1m∑i≠hn∑l≠hn∑h=1n{ehh,ehh,(aihtbhltclht+cihtbhltalht)eih}.
Third, using (3.16), and replacing j by h in the summation (3.63), we obtain
(3.79)∑t=1m∑i≠hn∑h=1n{ehh,ehh,(aihtbhhtchht+cihtbhhtahht)eih}.
Finally, rewriting (3.58) as
(3.80)∑t=1m∑i≠hn∑j≠hn∑l≠hn∑h=1n{ehh,ehh,(aijtbjltclht+cijtbjltalht)eih},
then, adding (3.76), (3.78), (3.79), and (3.80), together, it follows that
(3.81)∑t=1m∑i≠hn∑j=1n∑l=1n∑h=1n{ehh,ehh,(aijtbjltclht+cijtbjltalht)eih}=∑i≠hn∑h=1n{ehh,ehh,∑t=1m∑j=1n∑l=1n(aijtbjltclht+cijtbjltalht)eih},
and using (3.41), we get (3.74) holds.Next, we claim that(3.82)(3.61)+(3.65)+(3.68)+(3.71)+the13thsummationof(3.52)=0.
First, using (3.19) and (3.36), we rewrite the summations (3.62) and (3.69), respectively, as(3.83)∑t=1m∑i=1n∑j≠in{12(aijtbjitciit+aiitbjitcijt)eii,eii,eii},∑t=1m∑i=1n∑j≠in{12(aijtbjjtcjit+ajitbjjtcijt)eii,eii,eii},
and then, adding (3.83) and (3.72), this yields(3.84)∑t=1m∑i=1n∑j≠in∑l=1n{12(aijtbjltclit+alitbjltcijt)eii,eii,eii}.
Similarly, by (3.36), the summation (3.66) can be rewritten as(3.85)∑t=1m∑i=1n∑j≠in{12(aijtbjitcjjt+ajjtbjitcijt)ejj,ejj,ejj}.
Adding (3.85) and the 13th summation of (3.53), it implies(3.86)∑t=1m∑i=1n∑j=1n{12(aijtbjitcjjt+ajjtbjitcijt)ejj,ejj,ejj}.
Replacing iandj by landi, respectively, in the summation (3.86), then adding (3.84), and by (3.41), we get(3.87)∑t=1m∑i=1n∑j=1n∑l=1n{12(aijtbjltclit+alitbjltcijt)eii,eii,eii}=∑i=1n{∑t=1m∑j=1n∑l=1n12(aijtbjltclit+alitbjltcijt)eii,eii,eii}=0.
Therefore, (3.82) holds.Consequently, we get(3.47)+(3.52)+(3.54)=0, and using (3.56), we finish the proof of this theorem.



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*Source: 925092-2012-02-25.xml*