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Journal of Applied Mathematics

Volume 2012 (2012), Article ID 925092, 18 pages

http://dx.doi.org/10.1155/2012/925092

## Zero Triple Product Determined Matrix Algebras

^{1}Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China^{2}College of Science, Harbin Engineering University, Harbin 150001, China

Received 9 August 2011; Accepted 20 December 2011

Academic Editor: Xianhua Tang

Copyright © 2012 Hongmei Yao and Baodong Zheng. 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.

#### Abstract

Let be an algebra over a commutative unital ring . We say that is zero triple
product determined if for every -module and every trilinear map , the following
holds: if whenever , then there exists a -linear operator
such that for all . If the ordinary
triple product in the aforementioned definition is replaced by Jordan triple
product, then is called zero Jordan triple
product determined. This paper mainly shows that matrix algebra , , where
*B* is any commutative unital algebra even different
from the above mentioned commutative unital algebra , is always zero triple product determined, and , , where *F*
is any field with ch, is also zero Jordan triple product determined.

#### 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 of matrices over a unital algebra is zero product determined, and under some further restrictions on , 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 be a (fixed) commutative unital ring and an algebra over . By , we denote the -linear span of all elements of the form , where . is a -module and we denote by a -trilinear map. Consider the following conditions:(a)for all such that , we have ;(b)there exists a -linear map such that for all .

Trivially, (b) implies (a). We call that is a zero triple product determined algebra if for every -module and every -trilinear map , (a) implies (b). And we say that is a zero Jordan triple product determined algebra if we replace the above triple product by Jordan triple product ().

From the previous definitions, it is interesting to examine whether the matrix algebra of matrices over a unital algebra is still zero triple product (resp., zero Jordan triple product) determined. If the unital algebra 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 . In Section 2, we show that the answer is “yes” for the triple product if is any commutative unital algebra even different from . The Jordan triple product case, treated in Section 3, is more difficult; we only show that the matrix algebra , where is any field with ( 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:

() if , are such that , then .

#### 2. Zero Triple Product Determined Matrix Algebras

In this part, we will consider the matrix algebra , where is any commutative unital algebra even different from the (fixed) commutative unital ring mentioned in Section 1. By , where , we denote the matrix whose entry is and all other entries are 0.

Theorem 2.1. *Let be a commutative unital algebra, then is a zero triple product determined algebra for every .*

*Proof. *In order to prove that (a) implies (b), we only prove that (a) implies the equivalent condition (). Set , let be a -module, and let be a -trilinear map such that for all , implies . Throughout the proof, denote arbitrary elements in and denote arbitrary indices.

First, since if or , we get
Second, , where , and consequently
replacing by and by 1 in (2.2), we have
and using (2.3) and (2.2), this yields that
Similarly, from , where , we obtain
Finally, combining (2.5) and (2.4), it follows that
Let be such that ; then we only need to show
Writing
it follows, by examining the entry of , that for all and , we have
Note that
by (2.1), this summation reduces to
and using (2.6) and (2.9), we obtain
Therefore, the result of this theorem holds.

#### 3. Zero Jordan Triple Product Determined Matrix Algebras

In this part, we only consider the matrix algebra , where is a field with , and is still the (fixed) commutative unital ring mentioned in Section 1. Let , where , be the matrix whose entry is and all other entries are 0.

Theorem 3.1. *Let be a field with ; then is a zero Jordan triple product determined algebra for every .*

*Proof. *Set , let be a -module, and let be a -trilinear map such that for all , implies . Let be arbitrary elements from and denote arbitrary indices.

First, for , we have , and so
For any and , we have and , which implies
Similarly,
For any and , we have , , , and , and consequently,
Similarly,
Since and , if , it follows that
Using (3.8), (3.11), (3.15), and (3.14), we arrive at
Using (3.7) and (3.12), we have
Since if , then
Using (3.18) and (3.17), it follows that
Using (3.6), (3.8), (3.18), and (3.17), we obtain
Similarly, using (3.10), (3.9), and (3.17), this yields

Further, we claim that
If , then ; consequently,
Using (3.8)–(3.13) and (3.16), we get
From , where , it follows that
Using (3.3), (3.4), (3.6)–(3.13), and (3.16), we arrive at
Then, by (3.24), we have
Since , we choose such that ; applying (3.27), we get
then
Similarly, by and , where , we get
Then, combining (3.29) and (3.30), we have . Hence, the claim (3.22) holds.

Finally, we claim that
For , we have , and consequently . Using (3.8)–(3.13), (3.16), (3.17), and (3.19), this can be reduced to
Replacing by 1 and 2 in (3.32), respectively, we have
Computing (3.34)–2(3.33), this yields
then taking (3.35) into (3.33), we derive
replacing by in (3.36), we get
then taking (3.37) into (3.32), it follows that
and replacing by in (3.38), and using (3.38), (3.35), and (3.16), we get
Therefore, (3.31) holds.

Let be such that ; we have to prove that . Writing
it follows, by examining the entry of , that for all , we have
First, by (3.1), we get
The summation (3.42) can be written as
using (3.4), (3.5), and (3.21), the first summation can be rewritten as
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:
Therefore, the summation (3.42) can be rewritten as
The summation (3.43) can be written as
using (3.2), (3.10), (3.8), (3.16), (3.17), (3.18), and (3.6), the first summation can be rewritten as
and using (3.6), (3.3), (3.8), (3.16), and (3.20), we rewrite the second summation as
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:
Hence, the summation (3.43) can be rewritten as
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
Consequently, rewriting the summation (3.44) as
therefore, we get

Next, we only need to prove that .

Replacing the indices by , correspondingly, in the first summation of (3.48), and rewriting the summation as
then adding (3.57) and the first, the seventh, the tenth summations of (3.53), together, this yields
Replacing by , respectively, in the fifth summation of (3.48), then by (3.16), this summation is further equal to
and replacing by , correspondingly, in the third summation of (3.48), then using (3.16), the summation is further equal to
Then adding (3.59), (3.60), and the fifth and the sixth summations of (3.53), together, this yields
Replacing by in the fourth summation of (3.48), then adding the 12th summation of (3.53), we get
Replacing by , correspondingly, in the third summation of (3.55), then adding the 8th summation of (3.53), we have
Replacing by , respectively, in the first summation of (3.55), and rewriting this summation as
then adding (3.64) and the second summation of (3.53), together, and by (3.16), we derive
Replacing by , respectively, in the fourth summation of (3.53), then adding the second summation of (3.55), we arrive at
Replacing by , respectively, in the second summation of (3.48), and rewriting this summation as
then using (3.19) and (3.18), the aforementioned summation can be rewritten as
Adding the second summation of (3.68) and the ninth summation of (3.53), together, we get
By (3.36), we know that the first summation of (3.68) can be written as
similarly, using (3.36), the third summation of (3.53) can be written as
and then adding (3.70) and (3.71), together, we have
Therefore, we derive

Our first goal is to show
First, notice that by (3.16), the summation (3.61) can be rewritten as
Then replacing by , we have