`Journal of Applied MathematicsVolume 2012, Article ID 925092, 18 pageshttp://dx.doi.org/10.1155/2012/925092`
Research Article

## Zero Triple Product Determined Matrix Algebras

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

Received 9 August 2011; Accepted 20 December 2011

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 [16]. 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 [710]. 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