Additive maps seemingly preservinggroup inverses

精品论文Additive maps seemingly preserving group inverses Jian-Ying Ronga, Xu-Qing Liub a. Department of Foundation Courses, Huaian College of Information Technology, Huaian 223003, PR China;b. Department of Computing Science, Huaiyin Institute of Technology, Huaian 223001, PR ChinaAbstract: Additive preserver problems on algebra structures over square matrices spaces are of considerable interest to many authors in the past decade, in which the notion additive map preserving group inverses is mentioned. In the present note, we first propose a suitably acceptable notion additive map seemingly preserving group inverses of matrices over fields with characteristic not 2 or 3 and then investigate the general characterization of this notion as a beneficial attempt.Keywords: Additive map, Preserving group inverses, Seemingly preserving group inversesAMS(2000) Classification: 15A04, 15A331 IntroductionLet F be a field of characteristic not 2 or 3. Denote by F the multiplicative group, Mn (F) the algebra of all n n matrices over F, GLn (F) the general linear group of Mn (F), Ei j the n n matrix with 1 in the (i, j) position and 0 elsewhere.For given A Mn (F), if there exists (and therefore must be unique) B Mn (F) satisfies(1) ABT A = A, (2) BT ABT = BT , (3) ABT = BT A,we call BT the group inverse of A, and write BT = A# , where BT is the transpose of B. For the sake of convenience, we writeDi j = Ei j + E ji for i, j, i , j throughout this note.In the past decade, linear and additive preserver problems are quite active and of considerable interest to some authors; e.g., cf. 3,4,5 and 1,2,6,7,8, respectively. A notion of additive map preserving group inverses, restated in a definition version in the following Definition 1, was considered in 1, also see among others. In this note, we suggest to consider another notion of additive map seemingly preserving group inverses stated in the following Definition 2 and try investigating its description.Definition 1.1 We call an additive map f : Mn (F) Mn (F) preserving group inverses of matrices, if f (BT ) = f (A)# for any BT = A# with A, B Mn (F). Definition 1.2 We call an additive map f : Mn (F) Mn (F) seemingly preserving group inverses of matrices, if f (B)T =f (A)# for any BT = A# with A, B Mn (F). We denote by F the set of all additive maps seemingly preserving group inverses of matrices. In the following section, we will give the characterization of F .2 Additive map seemingly preserving group inversesNote that throughout this note any f F is assumed to be additive. We first give several lemmas.Lemma 2.1 Assume f F . Then for any a, b F and mutually distinct i, j, k 1, n, the statements below hold, where1, n = 1, 2, , n (the same below), namely,(1) f (aEii )T = f (a1 Eii )# ;(2) f (a1 Eii b1 E j j )T = f (aEii bE j j )# ; This research was supported by Grants HGQ0637 and HGQN0725 and the “Green & Blue Project” Program for 2008 to Cultivate Young CoreInstructors from Huaiyin Institute of Technology, Jiangsu Province, P.R. China. Corresponding author. Email address: rjy98 (J. Rong); liuxuqing688 (X. Liu).3精品论文(3) f (In a E ji )T = f (In aEi j )# ;(4) f (Ei j E ji )T = f (Ei j E ji )# ;(5) f ( E j j + aE ji + bE jk )T = f (E j j + aEi j + b Ek j )# ;2(6) 1 f (a Eii aE j j aDi j )T = f (a1 Eii a1 E j j a1 Di j )# ;2(7) f (Eii + 1 E j j E ji )T = f (Eii + 2E j j 2Ei j )# ; (8) f (Eii + 2E j j Di j )T = f (2 Eii + E j j Di j )# ; (9) f (a2 bEii aDi j )T = f (bE j j a1 Di j )# ;(10) f (Eii + E j j D ji )T = f (Eii + E j j Di j )# ; (11) f (Eii E j j aE ji )T = f (Eii E j j aEi j )# ; (12) f (Di j )T = f (Di j )# ;(13) f (Di j Ekk )T = f (Di j Ekk )# . Lemma 2.2 Assume f F . Then f (aEii ) f (bE j j )T = f (bE j j ) f (aEii ) = 0 for any i, j, i , j, and a, b F.Proof. From (2) of Lemma 1 and f F , we havef (aEii bE j j ) = f (aEii bE j j ) f (a1 Eii b1 E j j )T f (aEii bE j j ).By the above two equations adding and the characteristic of F not 2, we obtainf (aEii ) f (b1 E j j )T f (bE j j ) + f (bE j j ) f (b1 E j j )T f (aEii ) + f (bE j j ) f (a1 Eii )T f (bE j j ) = 0.By the characteristic of F not 2 or 3, a F arbitrary and f additive, we getf (bE j j ) f (a1 Eii )T (bE j j ) = 0,(2.1)f (aEii ) f (b1 E j j )T (bE j j ) + f (bE j j ) f (b1 E j j )T f (aEii ) = 0.(2.2) Post-multiplying (2.2) by f (b1 E j j )T and noting that (2.1) holds,0 = f (aEii ) f (b1 E j j )T (bE j j ) f (b1 E j j )T = f (aEii ) f (b1 E j j )Tcombining (1) of Lemma 1. Similarly, f (b1 E j j )T f (aEii ) = 0. Lemma 2.3 Assume f F . Then for any i, j, i , j, and a F,f (aEi j ) = f (aEi j ) f (Eii )T f (Eii ) + f (Eii ) f (Eii )T f (aEi j ) = f (aEi j ) f (E j j )T f (E j j ) + f (E j j ) f (E j j )T f (aEi j ),and f (aEi j ) f (a E ji )T f (aEi j ) = f (aEii ) f (aEi j )T f (aEii ) = f (aE j j ) f (aEi j )T f (aE j j ) = 0.Proof. From (3) of Lemma 1 and f F , we get f (In aEi j ) = f (In aEi j ) f (In aE ji )T f (In aEi j ). By the above two equations adding, a F arbitrary, f additive, and the characteristic of F not 2, f (aEi j ) f (aE ji )T f (aEi j ) = 0 andf (aEi j ) = f (aEi j ) f (In )T f (In ) + f (In ) f (In )T f (aEi j ) f (In ) f (aE ji )T f (In ).Pre-multiplying and post-multiplying the last equation by f (E j j )T and combining Lemma 2 and (1) of Lemma 1, we havef ( E j j )T f (aEi j ) f (E j j )T f ( E j j )T f (E j j ) f (aEi j )T f (E j j ) f ( E j j )T = 0(2.3) Taking b = 0 in (5) of Lemma 1, we obtainf (aEi j ) = f (aEi j ) f (Eii )T f (Eii ) + f (Eii ) f (Eii )T f (a Ei j ) + f (Eii ) f (aEi j )T f (Eii )= f (aEi j ) f (E j j )T f (E j j ) + f (E j j ) f (E j j )T f (aEi j ) + f (E j j ) f (a Ei j )T f (E j j ).Pre-multiplying and post-multiplying the last equation by f (E j j )T , we havef (E j j )T f (aEi j ) f (E j j )T + f (E j j )T f (E j j ) f (aEi j )T f (E j j ) f ( E j j )T = 0,which combining equation (2.3) yields f (E j j ) f (aEi j )T f (E j j ) = 0. Similarly, we obtainf (Eii ) f (aEi j )T f (Eii ) = 0.Consequently, f (aEi j ) = f (aEi j ) f (Eii )T f (Eii ) + f ( Eii ) f (Eii )T f (aEi j ) = f (aEi j ) f (E j j )T f (E j j ) + f (E j j ) f (E j j )T f (aEi j ). Lemma 2.4 Assume f F . Then the following are equivalent:(i) f = 0;(ii) f (Di j ) = 0 for some i , j; (iii) f (Eii ) = 0 for some i. Proof.(i)(ii): It is clear.(ii)(iii): If f (Di j ) = 0, for some i, j, i , j, from (6) and (2) of Lemma 1, we have31212f (Eii E j j )T =f (Eii E j j Di j )T = f (Eii E j j Di j )# = f (Eii E j j )# = f (Eii E j j )T ,hence f (Eii ) = f (E j j ). By (1) and (9) of Lemma 1,f (Eii )T = f (Eii + Di j )T = f ( E j j Di j )# = f (E j j )# = f (E j j )T = f (Eii )T ,which implies f (Eii ) = 0.(iii)(i): By Lemma 3,f (aEi j ) = f (aEi j ) f (Eii )T f (Eii ) + f (Eii ) f ( Eii )T f (aEi j ),f (aE ji ) = f (aE ji ) f (Eii )T f (Eii ) + f (Eii ) f (Eii )T f (aE ji ).If f (Eii ) = 0 for some i, then f (aEi j ) = 0 and f (aE ji ) = 0 for any j, j , i, further, f (Di j ) = 0. By proving of (ii)(iii), f (E j j ) = 0 for any j , i, further, f (E j j ) = 0 for any j. Again from Lemma 3, we get f (aEi j ) = 0 for any i, j and therefore f = 0. .By 3, Lemma 2.1 and 9, Lemma 1, the proving of the following lemma is trivial.Lemma 2.5 Suppose A1 , , An Mn (F) satisfy the following conditions:i) Ai , 0 are symmetric matrices;iii) A2 = Ai ;iii) Ai A j = 0 for any i , j.i=1Then PnAi = In and there is P GLn (F) such that Ai = P(ai Eii )PT , wherePT P = diag(a1 , , a1 ), ai F, i = 1, , n. 1nRemark 2.1 Assume f F , f , 0. Let Ai = f (Eii ) f (Eii )T = f (Eii )T f (Eii ), i = 1, , n. According to the abovei1discussion, Ai nis a set of non-zero orthogonal idempotent symmetric n n matrices. In the following, we will use Ai to characterize f F .Lemma 2.6 If 0 , f F , then there exists P GLn (F) such that eitherf (Ei j ) = Pi j Ei j PT forany i, j,(2.4)orf (Ei j ) = P ji E ji PT forany i, j,(2.5)where i j = ji , ii j j = ai a j , il k j = i j kl for any i, j, k, l.Proof. By Remark 1 and Lemma 5, there exists P GLn (F) such that Ai = P(ai Eii )PT , i = 1, , n, where PT P =diag(a1 , , a1 ). Set f (Ei j ) = PS(i j) PT forany i, j. Since f (Eii ) = Ai f (Eii ) = f (Eii )Ai , we obtain S(ii) = Eii S(ii) = S(ii) Eii ,1nand therefor S(ii) = ii Eii holds for some ii F. Moreover, by f (Eii )Ai f (Eii )T = A2 = Ai , we have 2 = a2 . By (7) and (1)6of Lemma 1, we havei1 Tiiif (Eii + 2E j j 2Ei j ) = f (Eii + 2E j j 2Ei j ) f (Eii + 2 E j j E ji )f (Eii + 2E j j 2Ei j ),which combing Lemma 2 and Lemma 3 yields f (Ei j ) = f (Eii ) f (E ji )T f (E j j ) + f (E j j ) f (E ji )T f (Eii ), and further, we getS(i j) = ii j j a1 1( ji) T( ji) T(i j)( ji)i a j Eii SE j j + E j j SEii , which implies S= aEi j + bE ji . Similarly, S= xEi j + yE ji . LetD(i j) = S(i j) + S( ji) = (a + x)Ei j + (b + y)E ji ,then we obtain D(i j) = ii j j a1 a1 Eii D(i j) E j j + E j j D(i j) Eii = ii j j a1 a1 D(i j) , which combining f , 0 and Lemma 4 yieldsijijii j j = ai a j . Further, S(i j) = Eii S( ji) E j j + E j j S( ji) Eii , which means a = y, b = x. From Lemma 3, f (Ei j ) f (E ji )T f (Ei j ) = 0,i.e., (aEi j + bE ji )PT P(xEi j + yE ji )T PT P(aEi j + bE ji ) = 0. Together with the above equation and a = y, b = x, we obtaina2 b = 0, which implies either a = 0 and y , 0, or a , 0 and y = 0, since f , 0 by Lemma 4 again. Moreover, we havef (Ei j ) = P(i j Ei j )PT or f (Ei j ) = P( ji E ji )PT , and the identity in the sense that either (2.4) or (2.5) holds for any i, j will be given in the following. In fact, if there exist some i, j, s, t such thatf (Ei j ) = i j PEi j PT , f (Est ) = t s PEt s PT ,then we consider f (Eis ) and f (Esi ). Noting that (5) of Lemma 1 holds, if f (Eis ) = is PEis PT , then f (Esi ) = si PEsi PT , which contradicts f (Ess + Est + Esi )T = f (Ess + Et s + Eis )# ; if f (Eis ) = si PEsi PT , then f (Esi ) = is PEis PT , which contradicts f (Eii + Ei j + Eis )T = f (Eii + E ji + Esi )# . Thus we complete the proving of the identity.Without loss of generality, we assume (2.4) holds as follows. Clearly,i j = ji , ii j j = ai a j .(2.6) In the following, to complete the proving of Lemma 6, it suffices to show that il k j = i j kl , for all i, j, k, l, in three cases. If i, j, k, l. are mutually distinct Let A = f (E ji ) + f (Ek j ) + f (Elk ) + f (Eil ), B = f (Ei j ) + f (E jk ) + f (Ekl ) + f (Eli ), then by 1 (A + B)T = 1 (A + B)# , we have224 f (Ei j ) = f (Eil ) f (Eil )T f (Ei j ) + f (Eil ) f (Ekl )T f (Ek j ) + f (Ei j ) f (Ek j )T f (Ek j ) + f (Ei j ) f (Ei j )T f (Ei j ),which combining (2.6) yields il k j = i j kl . If two of i, j, k, l, are equal, without loos of generality, assuming k = l. LetA = f (Eii + Ei j + E j j + E jk ), B = f (Eii E ji + E j j 2Eki + Ek j ).It can be readily justified BT = A# , further, ABT = BT A. Moreover, ki j j = i j k j , which is further equivalent toik k j = i j kk . The left cases hold clearly. Remark 2.2 In Lemma 6, for any i, j, k, l,(i) if equation (2.4) holds, then f (Ei j ) f (Ek j )T f (Ekl ) = f (Eil );(ii) if equation (2.5) holds, then f (E ji ) f (E jk )T f (Elk ) = f (Eil ).Proof. The conclusion (i) follows directly from the proving of il k j = i j kl for any i, j, k, l. Similarly,we have obtain (ii). Let A = (ai j ), B = (bi j ) Mn (F), we define the Schur Product of A and B as A B = (ai j bi j ).Theorem 2.1 Assume f : Mn (F) Mn (F) is an additive map. Then f F if and only if f is one of the following form:i) f = 0;ii) f (A) = P(E A )PT ;iii) f (A) = P(E A )T PT for all A Mn (F),where P, i j are defined as in Lemma 6, E = (i j ), is a field endomorphism on F satisfying (1) = 1, A = (ai j ).Proof. The sufficiency holds clearly by direct operations. In the following, we need to prove the necessity. Suppose f , 0. By Lemma 6, we assumef (Ei j ) = P(i j Ei j )PTfor any i, j, without loss of generality. By lemma 3, f (aEi j ) f (aE ji )T f (aEi j ) = 0 andf (aEi j ) = f (aEi j ) f (Eii )T f (Eii ) + f (Eii ) f (Eii )T f (aEi j ) = f (aEi j ) f (E j j )T f (E j j ) + f (E j j ) f (E j j )T f (aEi j ),which combining f (aEi j ) = f (a 1)Ei j + f (Ei j ), for all i, j, i , j, yields f (aEi j ) = P(i j (a)i j Ei j )PT , i j (a) F. Clearly,i j (1) = 1. From (5) of Lemma 1, we seef (bE j j a1 Di j ) = f (bE j j a1 Di j ) f (a2 bEii aDi j )T f (bE j j a1 Di j ),further,f (a1 Ei j ) = f (a1 Ei j ) f (aEi j )T f (a1 Ei j ),(2.7)f (bE j j ) = f (a1 E ji ) f (aE ji )T f (bE j j ) + f (a1 E ji ) f (a2 bEii )T f (a1 Ei j ) + f (bE j j ) f (aEi j ) f (a1 Ei j ).(2.8) Furthermore, (2.7) impliesi ji j (a1 ) = 1 (a),which combining (2.8) and the above equation yieldsii (a2 b) = ji (a) j j (b)i j (a).(2.9) Taking a = 1 and noticing i j (1) = 1 for all i, j, we obtain ii (b) = j j (b). Thus, ii = j j since b is arbitrary in F. From (2.9),we get ii (a2 ) = i j (a) ji (a) = ii (a)2 and therefore ii (a2 ) = ii (a)2 if taking b = 1 and b = a1 , respectively. Moreover,taking a = c + 1, b = 1 in (2.9), we obtain 2ii (c) = i j (c) + ji (c). Consequently, i j (c) = ji (c) by directly deducing and thereby ii = i j = ji for all i, j, which means i j is independent of i, j. We write i j = . 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