Two-sided linear split quaternionic equations with

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Erbakan University, Konya, Turkey. b. Department of Mathematics, Akdeniz University, Antalya,. Turkey. Published online: 28 Nov 2013. To cite this article: Melek ...

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Two-sided linear split quaternionic equations with n unknowns a

b

Melek Erdoğdu & Mustafa Özdemir a

Department of Mathematics-Computer Sciences, Necmettin Erbakan University, Konya, Turkey. b

Department of Mathematics, Akdeniz University, Antalya, Turkey. Published online: 28 Nov 2013.

Click for updates To cite this article: Melek Erdoğdu & Mustafa Özdemir (2015) Two-sided linear split quaternionic equations with n unknowns, Linear and Multilinear Algebra, 63:1, 97-106, DOI: 10.1080/03081087.2013.851196 To link to this article: http://dx.doi.org/10.1080/03081087.2013.851196

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Linear and Multilinear Algebra, 2015 Vol. 63, No. 1, 97–106, http://dx.doi.org/10.1080/03081087.2013.851196

Two-sided linear split quaternionic equations with n unknowns Melek Erdog˘ dua∗ and Mustafa Özdemirb a Department of Mathematics-Computer Sciences, Necmettin Erbakan University, Konya, Turkey; b Department of Mathematics, Akdeniz University, Antalya, Turkey

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Communicated by F. Zhang (Received 4 December 2012; accepted 15 September 2013) In this paper, we investigate linear split quaternionic equations with the terms of the form axb. We give a new method of solving general linear split quaternionic equations with one, two and n unknowns. Moreover, we present some examples to show how this procedure works. Keywords: quaternion; split quaternion; quaternion equation AMS Subject Classifications: 15A66; 51B20

1. Introduction Sir William Rowan Hamilton discovered the quaternions in 1843. This discovery, extending complex numbers to higher dimension, is major contribution made to mathematical science. The set of quaternions can be represented as H = {q = q0 + q1 e1 + q2 e2 + q3 e3 : q0 , q1 , q2 , q3 ∈ R}, where ei · 1 = 1 · ei = ei , ∀i ∈ {1, 2, 3}, e1 · e1 = e2 · e2 = e3 · e3 = −1, e1 · e2 = −e2 · e1 = e3 , e2 · e3 = −e3 · e2 = e1 , e3 · e1 = −e1 · e3 = e2 . The quaternions form a noncommutative division algebra.[1] In 1849, James Cockle introduced the set of split quaternions, also known as coquaternions. The real algebra of split quaternions, denoted by H S , is a four-dimensional vector space over the real field R with a basis {1, e1 , e2 , e3 } and the product rules ei · 1 = 1 · ei = ei , for all i ∈ {1, 2, 3} e1 · e1 = −e2 · e2 = −e3 · e3 = −1,

(1)

e1 · e2 = −e2 · e1 = e3 , e2 · e3 = −e3 · e2 = −e1 , e3 · e1 = −e1 · e3 = e2 . (2) The split quaternions abide by the rules (1) and (2). For any a, b ∈ H S , we denote the product a · b by ab throughout the paper. ∗ Corresponding author. Email: [email protected] © 2013 Taylor & Francis

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M. Erdog˘ du and M. Özdemir

We observe that for every split quaternion a, there exists a unique quadruple (a0 , a1 , a2 , a3 ) ∈ R4 such that 3  a = a0 + ai ei .

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i=1

This is called the canonical representation of split quaternion 3a. We call the real number ai ei is called the vector a0 the scalar part of a and denoted by Sa . The split quaternion i=1 part of a and denoted by Va . It is clear from (2) that H S is noncommutative. In contrast to quaternion algebra, the set of split quaternions contains zero divisors, nilpotent elements and nontrivial idempotents [2–5]. Split quaternions is a recently developing topic. There are some studies related to geometric applications of split quaternions such as [3,6] and [4]. In addition, the relation between split quaternions and complexified mechanics is discussed in [7]. Particularly, the geometric and physical applications of quaternions require solving quaternionic equations. Therefore, there are many studies on quaternionic and split quaternionic equations. For example; [8], the method of rearrangements is used to solve linear quaternionic equations involving axb, [9], zeros of quaternionic two-sided polynomials are discussed and in, [5] De Moivre’s formula is used to find the roots of split quaternion as well. Moreover, the properties of split quaternion matrices are discussed in [10]. In this paper, we investigate linear split quaternionic equations and their systems. We introduce a new method solving linear split quaternionic equations with one unknown. This method allows us to reduce any equations with axb terms to a real system with real unknowns. Later, we expand this method to linear split quaternionic equations with two and n unknowns. Moreover, we solve some linear split quaternionic equations as examples.

2. The general linear split quaternionic equations with one unknown In this section, we provide a characterization for the existence of a split quaternion solution x for the equation n  ai xbi = p (3) i=1

where a1 , ..., an , b1 , ..., bn , p are split quaternions, in terms of a real matrix equation. The set of all m × n real matrices is denoted by Mm×n (R). If A = (ai j ) ∈ Mm×n (R), we denote the transpose of A by A T , and write ai j also as (A)i j . If m = n, we write Mn×n (R) simply as Mn (R). The following lemma will be used in the proof of the main result of this section. Lem m a 1

Let p ∈ H S with the canonical representation p = p0 +

3 

pjej.

(4)

j=1

In addition, assume that l1 , ..., l4 ∈ H S . For each i ∈ {1, ..., 4}, write li in the canonical representation: 3  (0) (m) li = li + l i em . (5) m=1

Linear and Multilinear Algebra

99

Let x0 , ..., x3 be real numbers. Then the following two statements are equivalent: 3 (i) j=0 l j+1 x j = p (ii) The column matrix x = (x0 , x1 , x2 , x3 )T ∈ M4×1 (R) satisfies Lx = p, where L ∈ M4 (R) is the matrix given by (L)i j = l i−1 for all i, j = 1, ..., 4, and j p = ( p 0 , p1 , p2 , p3 ) T .

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Proof

It follows from (4) and (5) that

statement (i) holds ⇐⇒ there exist x 0 , x1 , x2 , x3 ∈ R such that   3 3 3    (0) (i) pi ei l j+1 + l j+1 ei x j = p0 + j=0

i=1

i=1

⇐⇒ there exist x0 , x1 , x2 , x3 ∈ R such that   3 3 3 3     (0) (i) + l j+1 x j + l j+1 x j ei = p0 + pi ei j=0

i=1

j=0

i=1

⇐⇒ there exist x0 , x1 , x2 , x3 ∈ R such that 3 3   (0) (i) l j+1 x j = p0 and l j+1 x j = pi , ∀i = 1, 2, 3 j=0

j=0

⇐⇒ there exist x0 , x1 , x2 , x3 ∈ R such that ⎛ (0) (0) (0) (0) ⎞ ⎛ ⎞ ⎛ l1 l2 l3 l4 x0 ⎜ (1) (1) (1) (1) ⎟ ⎜ ⎜ ⎜ l1 l2 l3 l4 ⎟ ⎜ x 1 ⎟ ⎟=⎜ ⎜ (2) (2) (2) (2) ⎟ ⎝ ⎠ ⎝ ⎝ l1 l2 l3 l4 ⎠ x 2 (3) (3) (3) (3) x3 l1 l2 l3 l4

⎞ p0 p1 ⎟ ⎟ p2 ⎠ p3

⇐⇒ statement (ii) holds. Remark 2 Let a, b ∈ H S with canonical representations a = 3 b0 + i=1 bi ei . It follows from (1) and (2) that −e1 e2 Va Vb = −a1 b1 + a2 b2 + a3 b3 + a1 a2 b1 b2

 a0 + e3 a3 b3

.

3

i=1 ai ei

and b =

(6)

We denote SVa Vb = Va , Vb  L . Then from (6) we get Va , Vb  L = −a1 b1 + a2 b2 + a3 b3 and

(7)

−e1 e2 e3 Va Vb = Va , Vb  L + a1 a2 a3 . b1 b2 b3

Since ai , bi ∈ R for all i = 1, 2, 3, we obtain Va Vb = 2 Va , Vb  L − Vb Va . We now provide the main result of this section.

(8)

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M. Erdog˘ du and M. Özdemir

Theorem 3 Let p ∈ H S with the canonical representation given by (4), and let a1 , ..., an , b1 , ..., bn be split quaternions with the canonical representations: (0)

ai = ai

3 

+

( j)

(0)

ai e j and bi = bi

+

j=1

3  j=1

( j)

bi e j

(9)

for all i = 1, ..., n. Let x0 , ..., x3 be real numbers, and x = (x0 , x1 , x2 , x3 )T and p = ( p0 , p1 , p2 , p3 )T . Then the following statements are equivalent: 3

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(1) Equation (3) has the solution x = x0 + 3 (2) j=0 l j+1 x j = p, where l1 =

n 

ai bi , l2 =

i=1

lm =

n 

n 

∈ HS .

j=1 x j e j

(1)

(ai bi e1 −2ai bi ) and

i=1

(m−1)

(ai bi em−1 +2ai bi

) for m = 3, 4.

(10)

i=1

(3) Lx = p, where L ∈ M4 (R) is the matrix defined as follows: For all i, j = 1, ..., 4, (i−1) are the real coefficients in the canonical representations (5) of split (L)i j = l j quaternions l1 , ..., l4 given by (10). Proof

(0)

(1) ⇐⇒ (2) : Let k ∈ {1, ..., n}. From x = x 0 + Vx and bk = bk + Vbk , we get (0)

(0)

xbk = x0 bk + Vx (bk + Vbk ) = x0 bk + bk Vx + Vx Vbk . (0)

Then from (8) and bk = bk − Vbk , we obtain  

(0) xbk = x0 bk + bk Vx + 2 Vbk , Vx L − Vbk Vx = bk x0 + bk Vx + 2 Vbk , Vx L . Thus from (7), (9) and Vx =

3

j=1 x j e j ,

we get

      (1) (2) (3) ak xbk = ak bk x0 + ak bk e1 − 2ak bk x1 + ak bk e2 + 2ak bk x2 + ak bk e3 + 2ak bk x3 . Hence Equation (3) has a solution x if and only if     n   (1) (2) ai bi x0 + ai bi e1 − 2ai bi x1 + ai bi e2 + 2ai bi x2 i=1    (3) x3 = p. + ai bi e3 + 2ai bi Then from (10), the result follows. (2) ⇐⇒ (3): This follows directly from Lemma 1. Example 4



Consider the split quaternionic system: e1 xe2 + xe3 = 1 + e1 .

(11)

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101

It follows from (10) that l1 = e1 e2 + e3 = 2e3 , l2 = −e1 e2 e1 − e3 e1 = −2e2 , l3 = −e1 e2 e2 + 2e1 − e3 e2 = 0 and

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l4 = −e1 e2 e3 − e3 e3 + 2 = 0. Thus from Theorem 3, we deduce that the system (11) is equivalent to solving the matrix equation ⎛ ⎞⎛ ⎞ ⎛ ⎞ 0 0 0 0 x0 1 ⎜ 0 0 0 0 ⎟ ⎜ x1 ⎟ ⎜ 1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 0 −2 0 0 ⎠ ⎝ x2 ⎠ = ⎝ 0 ⎠ , 2 0 0 0 0 x3 which has no solution. So the system (11) has no solution in H S . Example 5

Consider the split quaternionic system: e1 xe2 + (1 − e2 )x(e1 + e3 ) = 1 − e1 + e2 + e3 .

(12)

From (10), we obtain l1 = e1 e2 + (1 − e2 )(e1 + e3 ) = 2e1 + 3e3 , l2 = −e1 e2 e1 − (1 − e2 )(e1 + e3 )e1 − 2(1 − e2 ) = −e2 , l3 = −e1 e2 e2 + 2e1 − (1 − e2 )(e1 + e3 )e2 = −e1 − 2e3 and l4 = −e1 e2 e3 − (1 − e2 )(e1 + e3 )e3 = −1. Thus from Theorem 3, we see that the system (12) is equivalent to solving the matrix equation: ⎛ ⎞⎛ ⎞ ⎛ ⎞ 0 0 0 −1 x0 1 ⎜ 2 0 −1 0 ⎟ ⎜ x1 ⎟ ⎜ −1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ (13) ⎝ 0 −1 0 0 ⎠ ⎝ x2 ⎠ = ⎝ 1 ⎠ . 3 0 −2 0 1 x3 By Cramer’s rule, we get −1 0 −1 2 −1 −1 x0 = 1 −1 0 , x1 = 0 1 0 1 0 −2 3 1 −2

2 0 −1 2 0 −1 , x2 = 0 −1 1 , x3 = − 0 −1 0 3 0 1 3 0 −2

that is, the matrix Equation (13) has a unique solution x = (−3, −1, −5, −1)T . Hence the system (12) has the unique solution x = −3 − e1 − 5e2 − e3 .

,

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M. Erdog˘ du and M. Özdemir

Example 6

Consider the split quaternionic system: e1 xe2 + (1 − e2 )x(e1 + e3 ) = 1 − e1 + e2 + e3 .

(14)

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Similarly with previous examples, this equation is equivalent to the matrix equation ⎛ ⎞⎛ ⎞ ⎛ ⎞ 0 −1 0 −1 x0 0 ⎜ 1 0 −1 0 ⎟ ⎜ x1 ⎟ ⎜ 1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ 0 −1 0 −1 ⎠ ⎝ x2 ⎠ = ⎝ 0 ⎠ . 3 0 −3 0 3 x3 This system has infinitely many solutions of the form:   x = (t + 1, r, t, −r )T : t, r ∈ R . Consequently, Equation (14) has also infinitely many solutions of the form: x = t + 1 + r e1 + te2 − r e3 depending on real parameters r and t.

3. The general split quaternionic equations with two unknowns Let r, s ∈ {1, 2}, (ar s )i , (br s )i , p and q be split quaternions with canonical representations: (0)

(ar s )i = (ar s )i

+

3 

( j)

(0)

(ar s )i e j and (br s )i = (br s )i

+

j=1

3  j=1

( j)

(br s )i e j

(15)

for all i = 1, 2, ..., Mr s and p = p0 +

3 

pjej,

q = q0 +

j=1

3 

qjej.

(16)

j=1

Consider the linear split quaternionic equation of the form: M11  i=1 M21 

(a11 )i x(b11 )i +

(a21 )i x(b21 )i +

i=1

M12  i=1 M22 

(a12 )i y(b12 )i = p,

(17a)

(a22 )i y(b22 )i = q,

(17b)

i=1

where x and y are the split quaternionic unknowns with the canonical representations x = x0 +

3  j=1

x j e j and y = y0 +

3  j=1

yjej.

Linear and Multilinear Algebra

103

Similar to the discussion in the previous section, we may rewrite the Equations (17a) and (17b) as follows; 3  k=0 3 

(l11 )k+1 xk +

(l21 )k+1 xk +

k=0

3  k=0 3 

(l12 )k+1 yk = p, (l22 )k+1 yk = q,

k=0

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where (lr s )1 =

Mr s 

(ar s )i (br s )i , (lr s )2 =

i=1

(lr s ) j =

Mr s 

Mr s 

(1)

(ar s )i (br s )i e1 − 2(ar s )i (br s )i ,

(18a)

i=1 ( j−1)

(ar s )i (br s )i e j−1 + 2(ar s )i (br s )i

,

for r, s = 1, 2, j = 3, 4.

(18b)

i=1

Also, for every k ∈ {0, 1, 2, 3} and r, s ∈ {1, 2}, write (lr s )k+1 in the canonical representation 3  ( j) (0) (lr s )k+1 = (lr s )k+1 + (lr s )k+1 e j . j=1

Thus, the system of Equations (17a) and (17b) is equivalent to the linear system      L 11 L 12 x p £(x, y) = = L 21 L 22 y q

(19)

where L r s , r, s = 1, 2, are 4 × 4 real matrices given by (i−1)

(L r s )i j = (lr s ) j

,

for r, s = 1, 2 and i, j = 1, 2, 3, 4,

(20)

and x = (x0 , x1 , x2 , x3 )T , y = (y0 , y1 , y2 , y3 )T , p = ( p0 , p1 , p2 , p3 )T and q = (q0 , q1 , q2 , q3 ) T . Example 7

Consider the linear split quaternionic system with unknowns x and y

(2 − e1 )x(1 + e1 ) + e3 xe1 + (1 + e2 )xe3 + e1 y(e1 + 1) = −7 + 5e1 − 5e2 − 5e3 , (1 + e2 )xe3 + ye2 + (1 − 2e1 )ye3 = 2 − 4e1 − 5e2 + 9e3 . These equations are of the form (17a) and (17b), where (a11 )1 = 2 − e1 , (b11 )1 = 1 + e1 , (a11 )2 = e3 , (b11 )2 = e1 , (a11 )3 = 1 + e2 , (b11 )3 = e3 , (a12 )1 = e1 , (b12 )1 = e1 + 1, etc. Using Equations (18a) and (18b), we find (l11 )1 = (l11 )2 = (l11 )3 =

3 

(a11 )i (b11 )i = 3 + e2 + e3 ,

i=1 3 

(1)

= −2 + 3e1 − e2 − e3 ,

i=1 3 

(2)

= −1 − e1 + e2 − 2e3 ,

(a11 )i (b11 )i e1 − 2(a11 )i (b11 )i (a11 )i (b11 )i e2 + 2(a11 )i (b11 )i

i=1

104

M. Erdog˘ du and M. Özdemir (l11 )4 = (l12 )1 = (l12 )2 = (l12 )3 =

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(l12 )4 = (l21 )1 = (l21 )2 = (l21 )3 = (l21 )4 = (l22 )1 = (l22 )2 = (l22 )3 = (l22 )4 =

3 

(3)

= 1 + e1 + 4e2 + e3 ,

i=1 1 

(1)

= −1 − e1 ,

i=1 1 

(2)

= e 2 + e3 ,

i=1 1 

(3)

= −e2 + e3 ,

i=1 1 

(1)

= −1 − e2 ,

i=1 1 

(2)

= −e1 + e3 ,

i=1 1 

(3)

= 1 + e2 ,

i=1 2 

(1)

= −e2 + 3e3 ,

i=1 2 

(2)

= −1 − e1 ,

i=1 2 

(3)

= 1 − e1 .

(a11 )i (b11 )i e3 + 2(a11 )i (b11 )i

i=1 1 

(a12 )i (b12 )i = −1 + e1 , (a12 )i (b12 )i e1 − 2(a12 )i (b12 )i (a12 )i (b12 )i e2 + 2(a12 )i (b12 )i (a12 )i (b12 )i e3 + 2(a12 )i (b12 )i

i=1 1 

(a21 )i (b21 )i = −e1 + e3 , (a21 )i (b21 )i e1 − 2(a21 )i (b21 )i (a21 )i (b21 )i e2 + 2(a21 )i (b21 )i (a21 )i (b21 )i e3 + 2(a21 )i (b21 )i

i=1 2 

(a22 )i (b22 )i = 3e2 + e3 , (a22 )i (b22 )i e1 − 2(a22 )i (b22 )i (a22 )i (b22 )i e2 + 2(a22 )i (b22 )i (a22 )i (b22 )i e3 + 2(a22 )i (b22 )i

i=1

Using above equations and relation (20), we get



L 11

L 21

⎞ ⎛ 3 −2 −1 1 −1 ⎜ 0 3 −1 1 ⎟ ⎜ 1 ⎟ ⎜ =⎜ ⎝ 1 −1 1 4 ⎠ , L 12 = ⎝ 0 1 −1 −2 1 0 ⎛ ⎞ ⎛ 0 −1 0 1 0 ⎜ −1 0 −1 0 ⎟ ⎜0 ⎟ ⎜ =⎜ ⎝ 0 −1 0 1 ⎠ , L 22 = ⎝ 3 1 0 1 0 1

⎞ 0 0 0 0 ⎟ ⎟, 1 −1 ⎠ 1 1 ⎞ 0 −1 1 0 −1 −1 ⎟ ⎟. −1 0 0 ⎠ 3 0 0 −1 −1 0 0

Linear and Multilinear Algebra Using relation (19), we have the linear system: ⎞⎛ ⎞ ⎛ ⎛ x0 3 −2 −1 1 −1 −1 0 0 ⎜ 0 3 −1 1 1 −1 0 0 ⎟ ⎜ x1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎜ ⎜ 1 −1 1 4 0 0 1 −1 ⎟ ⎜ x2 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎜ ⎜ 1 −1 −2 1 0 0 1 1 ⎟ ⎜ x3 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎜ ⎜ 0 −1 0 1 0 0 −1 1 ⎟ ⎜ y0 ⎟ = ⎜ ⎟⎜ ⎟ ⎜ ⎜ ⎜ −1 0 −1 0 0 0 −1 −1 ⎟ ⎜ y1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎜ ⎝ 0 −1 0 1 3 −1 0 0 ⎠ ⎝ y2 ⎠ ⎝ 1 0 1 0 1 3 0 0 y3

105

−7 5 −5 −5 2 −4 −5 9

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎠

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This system has unique solution: x =(1, 3, 2, 0)T ,

y = (0, 1, −2, 3)T .

So, the given linear split quaternionic system has also a unique solution: x = 1 + 3e1 + 2e2 ,

y = e1 − 2e2 + 3e3 .

4. The general linear split quaternionic equations with n unknowns Consider the linear split quaternionic system with n split quaternionic unknowns z 1 , z 2 , ..., z n ; M11  i=1 M21 

(a11 )i z 1 (b11 )i +

(a21 )i z 1 (b21 )i +

i=1

M12  i=1 M22 

(a12 )i z 2 (b12 )i + ... + (a22 )i z 2 (b22 )i + ... +

i=1

M1n  i=1 M2n 

(a1n )i z n (b1n )i = p1 , (a2n )i z n (b2n )i = p2,

i=1

.. . Mn1  i=1

(an1 )i z 1 (bn1 )i +

Mn2 

(an2 )i z 2 (bn2 )i + ... +

i=1

Mnn  (ann )i z n (bnn )i = pn, i=1

where all right and left coefficients of z i are split quaternions for i = 1, 2, ..., n. Similar to the discussion in previous section, this system can be written as ⎞⎛ ⎞ ⎛ ⎞ ⎛ z1 p1 L 11 L 12 · · · L 1n ⎜ L 21 L 22 · · · L 2n ⎟ ⎜ z2 ⎟ ⎜ p2 ⎟ ⎟⎜ ⎟ ⎜ ⎟ ⎜ £(z1, z2 , ..., zn ) = ⎜ . .. . . . . ⎟ ⎜ .. ⎟ = ⎜ .. ⎟ , ⎝ .. . . ⎠⎝ . ⎠ ⎝ . ⎠ . L n1 L n2 · · · L nn

zn

pn

where zi = (z i(0) , z i(1) , z i(2) , z i(3) )T and pi = ( pi(0) , pi(1) , pi(2) , pi(3) )T for i = 1, 2, ..., n. Here, for r, s ∈ {1, 2, ..., m}, L r s are 4 × 4 real matrices given by (i−1)

(L r s )i j = (lr s ) j

,

for all r, s, i, j ∈ {1, 2, ..., n},

106

M. Erdog˘ du and M. Özdemir

where (lr s )1 = (lr s )k =

Mr s  i=1 Mr s 

(ar s )i (br s )i , (lr s )2 =

Mr s 

(1)

(ar s )i (br s )i e1 − 2(ar s )i (br s )i ,

i=1 (k−1)

(ar s )i (br s )i ek−1 + 2(ar s )i (br s )i

for k = 3, 4.

i=1

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References [1] Kantor IL, Solodovnikov AS. Hypercomplex numbers: an elementary introduction to algebras. New York: Springer-Verlag; 1989. [2] Cockle J. On systems of algebra involving more than one imaginary. Phil. Mag. 1849;35:434– 435. [3] Kula L, Yaylı Y. Split quaternions and rotations in semi euclidean space E 24 . J. Korean Math. Soc. 2007;44:1313–1327. [4] Özdemir M, Ergin AA. Rotations with unit timelike quaternions in Minkowski 3-space. J. Geom. Phys. 2006;56:322–336. [5] Özdemir M. The roots of a split quaternion. Appl. Math. Lett. 2009;22:258–263. [6] Özdemir M, Ergin AA. Some geometric applications of split quaternions. Proc. 16th Int. Conf. Jangjeon Math. Soc. 2005;16:108–115. [7] Brody DC, Graefe EM. On complexified mechanics and coquaternions. J. Phys. A: Math. Theory. 2011;44:1–9. [8] Shpakıvsky VS. Linear quaternionic equations and their systems. Adv. Appl. Clifford Algebras. 2011;21:637–645. [9] Janovska D, Opfer G. The classification and computations of zeros of quaternionic two-sided polynomials. Numerische Mathematik. 2010;115:81–100. [10] Alagöz Y, Oral KH, Yüce S. Split quaternion matrices. Miskolc Math. Notes. 2012;13:223–232.