Ths artcle was downloaded by: [Necmettn Erbaan Unversty] On: 24 March 2015, At: 05:44 Publsher: Taylor & Francs Informa Ltd Regstered n England and Wales Regstered Number: 1072954 Regstered offce: Mortmer House, 37-41 Mortmer Street, London W1T 3JH, UK Lnear and Multlnear Algebra Publcaton detals, ncludng nstructons for authors and subscrpton nformaton: http://www.tandfonlne.com/lo/glma20 Two-sded lnear splt quaternonc equatons wth n unnowns Mele Erdoğdu a & Mustafa Özdemr b a Department of Mathematcs-Computer Scences, Necmettn Erbaan Unversty, Konya, Turey. b Department of Mathematcs, Adenz Unversty, Antalya, Turey. Publshed onlne: 28 Nov 2013. Clc for updates To cte ths artcle: Mele Erdoğdu & Mustafa Özdemr (2015) Two-sded lnear splt quaternonc equatons wth n unnowns, Lnear and Multlnear Algebra, 63:1, 97-106, DOI: 10.1080/03081087.2013.851196 To ln to ths artcle: http://dx.do.org/10.1080/03081087.2013.851196 PLEASE SCROLL DOWN FOR ARTICLE Taylor & Francs maes every effort to ensure the accuracy of all the nformaton (the Content ) contaned n the publcatons on our platform. However, Taylor & Francs, our agents, and our lcensors mae no representatons or warrantes whatsoever as to the accuracy, completeness, or sutablty for any purpose of the Content. Any opnons and vews expressed n ths publcaton are the opnons and vews of the authors, and are not the vews of or endorsed by Taylor & Francs. The accuracy of the Content should not be reled upon and should be ndependently verfed wth prmary sources of nformaton. Taylor and Francs shall not be lable for any losses, actons, clams, proceedngs, demands, costs, expenses, damages, and other labltes whatsoever or howsoever caused arsng drectly or ndrectly n connecton wth, n relaton to or arsng out of the use of the Content. Ths artcle may be used for research, teachng, and prvate study purposes. Any substantal or systematc reproducton, redstrbuton, resellng, loan, sub-lcensng, systematc supply, or dstrbuton n any form to anyone s expressly forbdden. Terms &
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Lnear and Multlnear Algebra, 2015 Vol. 63, No. 1, 97 106, http://dx.do.org/10.1080/03081087.2013.851196 Two-sded lnear splt quaternonc equatons wth n unnowns Mele Erdoğdu a and Mustafa Özdemr b a Department of Mathematcs-Computer Scences, Necmettn Erbaan Unversty, Konya, Turey; b Department of Mathematcs, Adenz Unversty, Antalya, Turey Communcated by F. Zhang Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 (Receved 4 December 2012; accepted 15 September 2013) In ths paper, we nvestgate lnear splt quaternonc equatons wth the terms of the form axb. We gve a new method of solvng general lnear splt quaternonc equatons wth one, two and n unnowns. Moreover, we present some examples to show how ths procedure wors. Keywords: quaternon; splt quaternon; quaternon equaton AMS Subject Classfcatons: 15A66; 51B20 1. Introducton Sr Wllam Rowan Hamlton dscovered the quaternons n 1843. Ths dscovery, extendng complex numbers to hgher dmenson, s major contrbuton made to mathematcal scence. The set of quaternons can be represented as where H ={q = q 0 + q 1 e 1 + q 2 e 2 + q 3 e 3 : q 0, q 1, q 2, q 3 R}, e 1 = 1 e = e, {1, 2, 3}, e 1 e 1 = e 2 e 2 = e 3 e 3 = 1, e 1 e 2 = e 2 e 1 = e 3, e 2 e 3 = e 3 e 2 = e 1, e 3 e 1 = e 1 e 3 = e 2. The quaternons form a noncommutatve dvson algebra.[1] In 1849, James Cocle ntroduced the set of splt quaternons, also nown as coquaternons. The real algebra of splt quaternons, denoted by H S, s a four-dmensonal vector space over the real feld R wth a bass {1, e 1, e 2, e 3 } and the product rules e 1 = 1 e = e, for all {1, 2, 3} e 1 e 1 = e 2 e 2 = e 3 e 3 = 1, (1) e 1 e 2 = e 2 e 1 = e 3, e 2 e 3 = e 3 e 2 = e 1, e 3 e 1 = e 1 e 3 = e 2. (2) The splt quaternons abde by the rules (1) and (2). For any a, b H S, we denote the product a b by ab throughout the paper. Correspondng author. Emal: merdogdu@onya.edu.tr 2013 Taylor & Francs
98 M. Erdoğdu and M. Özdemr Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 We observe that for every splt quaternon a, there exsts a unque quadruple (a 0, a 1, a 2, a 3 ) R 4 such that a = a 0 + 3 a e. Ths s called the canoncal representaton of splt quaternon a. We call the real number a 0 the scalar part of a and denoted by S a. The splt quaternon 3 a e s called the vector part of a and denoted by V a. It s clear from (2) that H S s noncommutatve. In contrast to quaternon algebra, the set of splt quaternons contans zero dvsors, nlpotent elements and nontrval dempotents [2 5]. Splt quaternons s a recently developng topc. There are some studes related to geometrc applcatons of splt quaternons such as [3,6] and [4]. In addton, the relaton between splt quaternons and complexfed mechancs s dscussed n [7]. Partcularly, the geometrc and physcal applcatons of quaternons requre solvng quaternonc equatons. Therefore, there are many studes on quaternonc and splt quaternonc equatons. For example; [8], the method of rearrangements s used to solve lnear quaternonc equatons nvolvng axb, [9], zeros of quaternonc two-sded polynomals are dscussed and n, [5] De Movre s formula s used to fnd the roots of splt quaternon as well. Moreover, the propertes of splt quaternon matrces are dscussed n [10]. In ths paper, we nvestgate lnear splt quaternonc equatons and ther systems. We ntroduce a new method solvng lnear splt quaternonc equatons wth one unnown. Ths method allows us to reduce any equatons wth axb terms to a real system wth real unnowns. Later, we expand ths method to lnear splt quaternonc equatons wth two and n unnowns. Moreover, we solve some lnear splt quaternonc equatons as examples. 2. The general lnear splt quaternonc equatons wth one unnown In ths secton, we provde a characterzaton for the exstence of a splt quaternon soluton x for the equaton n a xb = p (3) where a 1,..., a n, b 1,..., b n, p are splt quaternons, n terms of a real matrx equaton. The set of all m n real matrces s denoted by M m n (R). If A = (a j ) M m n (R), we denote the transpose of A by A T, and wrte a j also as (A) j.ifm = n, we wrte M n n (R) smply as M n (R). The followng lemma wll be used n the proof of the man result of ths secton. Lemma 1 Let p H S wth the canoncal representaton p = p 0 + 3 p j e j. (4) In addton, assume that l 1,..., l 4 H S. For each {1,..., 4}, wrte l n the canoncal representaton: l = l (0) + 3 l (m) e m. (5) m=1
Lnear and Multlnear Algebra 99 Let x 0,..., x 3 be real numbers. Then the followng two statements are equvalent: () 3 j=0 l j+1 x j = p () The column matrx x = (x 0, x 1, x 2, x 3 ) T M 4 1 (R) satsfes Lx = p, where L M 4 (R) s the matrx gven by (L) j = l 1 j for all, j = 1,..., 4, and p = (p 0, p 1, p 2, p 3 ) T. Proof It follows from (4) and (5) that Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 statement () holds there exst x 0, x 1, x 2, x 3 R such that [ ] 3 l (0) j+1 + 3 l () j+1 e x j = p 0 + 3 p e j=0 there exst x 0, x 1, x 2, x 3 R such that [ ] 3 + l (0) j+1 x j + 3 3 l () j+1 x j e = p 0 + 3 p e j=0 j=0 there exst x 0, x 1, x 2, x 3 R such that 3 l (0) j+1 x j = p 0 and 3 l () j+1 x j = p, = 1, 2, 3 j=0 l (3) 1 l (3) 2 l (3) 3 l (3) 4 statement () holds. j=0 there exst x 0, x 1, x 2, x 3 R such that l (0) 1 l (0) 2 l (0) 3 l (0) 4 x 0 l (1) 1 l (1) 2 l (1) 3 l (1) 4 l (2) 1 l (2) 2 l (2) 3 l (2) x 1 x 4 2 = x 3 Remar 2 Let a, b H S wth canoncal representatons a = a 0 + 3 a e and b = b 0 + 3 b e. It follows from (1) and (2) that e 1 e 2 e 3 V a V b = a 1 b 1 + a 2 b 2 + a 3 b 3 + a 1 a 2 a 3 b 1 b 2 b 3. (6) We denote S Va V b = V a, V b L. Then from (6) we get p 0 p 1 p 2 p 3 V a, V b L = a 1 b 1 + a 2 b 2 + a 3 b 3 (7) and V a V b = V a, V b L + Snce a, b R for all = 1, 2, 3, we obtan We now provde the man result of ths secton. e 1 e 2 e 3 a 1 a 2 a 3 b 1 b 2 b 3 V a V b = 2 V a, V b L V b V a. (8).
100 M. Erdoğdu and M. Özdemr Theorem 3 Let p H S wth the canoncal representaton gven by (4), and let a 1,..., a n, b 1,..., b n be splt quaternons wth the canoncal representatons: a = a (0) + 3 a ( j) e j and b = b (0) + 3 b ( j) e j (9) for all = 1,..., n. Let x 0,..., x 3 be real numbers, and x = (x 0, x 1, x 2, x 3 ) T and p = (p 0, p 1, p 2, p 3 ) T. Then the followng statements are equvalent: (1) Equaton (3) has the soluton x = x 0 + 3 x j e j H S. (2) 3 j=0 l j+1 x j = p, where Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 l 1 = n a b, l 2 = n (a b e 1 2a b (1) ) and l m = n (a b e m 1 +2a b (m 1) ) for m = 3, 4. (10) (3) Lx = p, where L M 4 (R) s the matrx defned as follows: For all, j = 1,..., 4, (L) j = l ( 1) j are the real coeffcents n the canoncal representatons (5) of splt quaternons l 1,..., l 4 gven by (10). Proof (1) (2) : Let {1,..., n}. From x = x 0 + V x and b = b (0) Then from (8) and b = b (0) xb = x 0 b + V x (b (0) + V b ) = x 0 b + b (0) V x + V x V b. V b, we obtan + V b, we get xb = x 0 b + b (0) V x + 2 V b, V x L V b V x = b x 0 + b V x + 2 V b, V x L. Thus from (7), (9) and V x = 3 x j e j, we get ( a xb = a b x 0 + a b e 1 2a b (1) ) ( x 1 + Hence Equaton (3) has a soluton x f and only f n [ ( a b x 0 + ( + a b e 1 2a b (1) ) ] x 3 a b e 3 + 2a b (3) a b e 2 + 2a b (2) ) ( x 1 + = p. ) ( x 2 + a b e 2 + 2a b (2) a b e 3 + 2a b (3) ) x 2 ) x 3. Then from (10), the result follows. (2) (3): Ths follows drectly from Lemma 1. Example 4 Consder the splt quaternonc system: e 1 xe 2 + xe 3 = 1 + e 1. (11)
Lnear and Multlnear Algebra 101 It follows from (10) that l 1 = e 1 e 2 + e 3 = 2e 3, l 2 = e 1 e 2 e 1 e 3 e 1 = 2e 2, l 3 = e 1 e 2 e 2 + 2e 1 e 3 e 2 = 0 Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 and l 4 = e 1 e 2 e 3 e 3 e 3 + 2 = 0. Thus from Theorem 3, we deduce that the system (11) s equvalent to solvng the matrx equaton 0 0 0 0 x 0 1 0 0 0 0 x 1 0 2 0 0 x 2 = 1 0, 2 0 0 0 x 3 0 whch has no soluton. So the system (11) has no soluton n H S. Example 5 From (10), we obtan and Consder the splt quaternonc system: e 1 xe 2 + (1 e 2 )x(e 1 + e 3 ) = 1 e 1 + e 2 + e 3. (12) l 1 = e 1 e 2 + (1 e 2 )(e 1 + e 3 ) = 2e 1 + 3e 3, l 2 = e 1 e 2 e 1 (1 e 2 )(e 1 + e 3 )e 1 2(1 e 2 ) = e 2, l 3 = e 1 e 2 e 2 + 2e 1 (1 e 2 )(e 1 + e 3 )e 2 = e 1 2e 3 l 4 = e 1 e 2 e 3 (1 e 2 )(e 1 + e 3 )e 3 = 1. Thus from Theorem 3, we see that the system (12) s equvalent to solvng the matrx equaton: 0 0 0 1 x 0 1 2 0 1 0 x 1 0 1 0 0 x 2 = 1 1. (13) 3 0 2 0 x 3 1 By Cramer s rule, we get 1 0 1 x 0 = 1 1 0 1 0 2, x 2 1 1 1 = 0 1 0 3 1 2, x 2 0 1 2 = 0 1 1 3 0 1 that s, the matrx Equaton (13) has a unque soluton, x 3 = 2 0 1 0 1 0 3 0 2, x = ( 3, 1, 5, 1) T. Hence the system (12) has the unque soluton x = 3 e 1 5e 2 e 3.
102 M. Erdoğdu and M. Özdemr Example 6 Consder the splt quaternonc system: e 1 xe 2 + (1 e 2 )x(e 1 + e 3 ) = 1 e 1 + e 2 + e 3. (14) Smlarly wth prevous examples, ths equaton s equvalent to the matrx equaton 0 1 0 1 x 0 0 1 0 1 0 x 1 0 1 0 1 x 2 = 1 0. 3 0 3 0 x 3 3 Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 Ths system has nfntely many solutons of the form: { } x = (t + 1, r, t, r) T : t, r R. Consequently, Equaton (14) has also nfntely many solutons of the form: dependng on real parameters r and t. x = t + 1 + re 1 + te 2 re 3 3. The general splt quaternonc equatons wth two unnowns Let r, s {1, 2}, (a rs ), (b rs ), p and q be splt quaternons wth canoncal representatons: (a rs ) = (a rs ) (0) for all = 1, 2,..., M rs and + 3 (a rs ) ( j) e j and (b rs ) = (b rs ) (0) + 3 (b rs ) ( j) e j (15) p = p 0 + 3 p j e j, Consder the lnear splt quaternonc equaton of the form: M 11 q = q 0 + 3 q j e j. (16) (a 11 ) x(b 11 ) + (a 12 ) y(b 12 ) = p, (17a) M 21 M 12 M 22 (a 21 ) x(b 21 ) + (a 22 ) y(b 22 ) = q, (17b) where x and y are the splt quaternonc unnowns wth the canoncal representatons 3 3 x = x 0 + x j e j and y = y 0 + y j e j.
Lnear and Multlnear Algebra 103 Smlar to the dscusson n the prevous secton, we may rewrte the Equatons (17a) and (17b) as follows; Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 3 (l 11 ) +1 x + =0 3 (l 21 ) +1 x + =0 3 (l 12 ) +1 y = p, =0 3 (l 22 ) +1 y = q, =0 where M rs M rs (l rs ) 1 = (a rs ) (b rs ), (l rs ) 2 = (a rs ) (b rs ) e 1 2(a rs ) (b rs ) (1), (18a) M rs (l rs ) j = (a rs ) (b rs ) e j 1 + 2(a rs ) (b rs ) ( j 1), for r, s = 1, 2, j = 3, 4. (18b) Also, for every {0, 1, 2, 3} and r, s {1, 2}, wrte (l rs ) +1 n the canoncal representaton (l rs ) +1 = (l rs ) (0) +1 + 3 (l rs ) ( j) +1 e j. Thus, the system of Equatons (17a) and (17b) s equvalent to the lnear system ( )( ) ( ) L11 L (x, y) = 12 x p = L 21 L 22 y q where L rs, r, s = 1, 2, are 4 4 real matrces gven by (19) (L rs ) j = (l rs ) ( 1) j, for r, s = 1, 2 and, j = 1, 2, 3, 4, (20) and x = (x 0, x 1, x 2, x 3 ) T, y = (y 0, y 1, y 2, y 3 ) T, p = (p 0, p 1, p 2, p 3 ) T and q = (q 0, q 1, q 2, q 3 ) T. Example 7 Consder the lnear splt quaternonc system wth unnowns x and y (2 e 1 )x(1 + e 1 ) + e 3 xe 1 + (1 + e 2 )xe 3 + e 1 y(e 1 + 1) = 7 + 5e 1 5e 2 5e 3, (1 + e 2 )xe 3 + ye 2 + (1 2e 1 )ye 3 = 2 4e 1 5e 2 + 9e 3. These equatons are of the form (17a) and (17b), where (a 11 ) 1 = 2 e 1,(b 11 ) 1 = 1 + e 1, (a 11 ) 2 = e 3,(b 11 ) 2 = e 1,(a 11 ) 3 = 1 + e 2,(b 11 ) 3 = e 3,(a 12 ) 1 = e 1,(b 12 ) 1 = e 1 + 1, etc. Usng Equatons (18a) and (18b), we fnd (l 11 ) 1 = 3 (a 11 ) (b 11 ) = 3 + e 2 + e 3, (l 11 ) 2 = 3 (a 11 ) (b 11 ) e 1 2(a 11 ) (b 11 ) (1) = 2 + 3e 1 e 2 e 3, (l 11 ) 3 = 3 (a 11 ) (b 11 ) e 2 + 2(a 11 ) (b 11 ) (2) = 1 e 1 + e 2 2e 3,
104 M. Erdoğdu and M. Özdemr Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 (l 11 ) 4 = 3 (a 11 ) (b 11 ) e 3 + 2(a 11 ) (b 11 ) (3) = 1 + e 1 + 4e 2 + e 3, (l 12 ) 1 = 1 (a 12 ) (b 12 ) = 1 + e 1, (l 12 ) 2 = 1 (a 12 ) (b 12 ) e 1 2(a 12 ) (b 12 ) (1) = 1 e 1, (l 12 ) 3 = 1 (a 12 ) (b 12 ) e 2 + 2(a 12 ) (b 12 ) (2) = e 2 + e 3, (l 12 ) 4 = 1 (a 12 ) (b 12 ) e 3 + 2(a 12 ) (b 12 ) (3) = e 2 + e 3, (l 21 ) 1 = 1 (a 21 ) (b 21 ) = e 1 + e 3, (l 21 ) 2 = 1 (a 21 ) (b 21 ) e 1 2(a 21 ) (b 21 ) (1) = 1 e 2, (l 21 ) 3 = 1 (a 21 ) (b 21 ) e 2 + 2(a 21 ) (b 21 ) (2) = e 1 + e 3, (l 21 ) 4 = 1 (a 21 ) (b 21 ) e 3 + 2(a 21 ) (b 21 ) (3) = 1 + e 2, (l 22 ) 1 = 2 (a 22 ) (b 22 ) = 3e 2 + e 3, (l 22 ) 2 = 2 (a 22 ) (b 22 ) e 1 2(a 22 ) (b 22 ) (1) = e 2 + 3e 3, (l 22 ) 3 = 2 (a 22 ) (b 22 ) e 2 + 2(a 22 ) (b 22 ) (2) = 1 e 1, (l 22 ) 4 = 2 (a 22 ) (b 22 ) e 3 + 2(a 22 ) (b 22 ) (3) = 1 e 1. Usng above equatons and relaton (20), we get L 11 = L 21 = 3 2 1 1 0 3 1 1 1 1 1 4 1 1 2 1 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0, L 12 =, L 22 = 1 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 3 1 0 0 1 3 0 0,.
Lnear and Multlnear Algebra 105 Usng relaton (19), we have the lnear system: 3 2 1 1 1 1 0 0 x 0 7 0 3 1 1 1 1 0 0 x 1 5 1 1 1 4 0 0 1 1 x 2 5 1 1 2 1 0 0 1 1 x 3 0 1 0 1 0 0 1 1 y 0 = 5 2. 1 0 1 0 0 0 1 1 y 1 4 0 1 0 1 3 1 0 0 y 2 5 1 0 1 0 1 3 0 0 y 3 9 Ths system has unque soluton: Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 x =(1, 3, 2, 0) T, y = (0, 1, 2, 3) T. So, the gven lnear splt quaternonc system has also a unque soluton: x = 1 + 3e 1 + 2e 2, y = e 1 2e 2 + 3e 3. 4. The general lnear splt quaternonc equatons wth n unnowns Consder the lnear splt quaternonc system wth n splt quaternonc unnowns z 1, z 2,..., z n ; M 11 (a 11 ) z 1 (b 11 ) + (a 12 ) z 2 (b 12 ) +... + (a 1n ) z n (b 1n ) = p 1, M 21 M 12 M 22 M 1n M 2n (a 21 ) z 1 (b 21 ) + (a 22 ) z 2 (b 22 ) +... + (a 2n ) z n (b 2n ) = p 2, M n1 (a n1 ) z 1 (b n1 ) + (a n2 ) z 2 (b n2 ) +... + (a nn ) z n (b nn ) = p n, M n2 where all rght and left coeffcents of z are splt quaternons for = 1, 2,..., n. Smlar to the dscusson n prevous secton, ths system can be wrtten as L 11 L 12 L 1n z 1 p 1 L 21 L 22 L 2n z 2 (z 1, z 2,..., z n ) =.......... = p 2.., L n1 L n2 L nn z n p n M nn. where z = (z (0), z (1), z (2), z (3) ) T and p = (p (0), p (1), p (2), p (3) ) T for = 1, 2,..., n. Here, for r, s {1, 2,..., m}, L rs are 4 4 real matrces gven by (L rs ) j = (l rs ) ( 1) j, for all r, s,, j {1, 2,..., n},
106 M. Erdoğdu and M. Özdemr where M rs (l rs ) 1 = (a rs ) (b rs ),(l rs ) 2 = (a rs ) (b rs ) e 1 2(a rs ) (b rs ) (1) M rs M rs (l rs ) = (a rs ) (b rs ) e 1 + 2(a rs ) (b rs ) ( 1) for = 3, 4., References Downloaded by [Necmettn Erbaan Unversty] at 05:44 24 March 2015 [1] Kantor IL, Solodovnov AS. Hypercomplex numbers: an elementary ntroducton to algebras. New Yor: Sprnger-Verlag; 1989. [2] Cocle J. On systems of algebra nvolvng more than one magnary. Phl. Mag. 1849;35:434 435. [3] Kula L, Yaylı Y. Splt quaternons and rotatons n sem eucldean space E2 4. J. Korean Math. Soc. 2007;44:1313 1327. [4] Özdemr M, Ergn AA. Rotatons wth unt tmele quaternons n Mnows 3-space. J. Geom. Phys. 2006;56:322 336. [5] Özdemr M. The roots of a splt quaternon. Appl. Math. Lett. 2009;22:258 263. [6] Özdemr M, Ergn AA. Some geometrc applcatons of splt quaternons. Proc. 16th Int. Conf. Jangjeon Math. Soc. 2005;16:108 115. [7] Brody DC, Graefe EM. On complexfed mechancs and coquaternons. J. Phys. A: Math. Theory. 2011;44:1 9. [8] Shpaıvsy VS. Lnear quaternonc equatons and ther systems. Adv. Appl. Clfford Algebras. 2011;21:637 645. [9] Janovsa D, Opfer G. The classfcaton and computatons of zeros of quaternonc two-sded polynomals. Numersche Mathemat. 2010;115:81 100. [10] Alagöz Y, Oral KH, Yüce S. Splt quaternon matrces. Msolc Math. Notes. 2012;13:223 232.