On Linear Split Quaternionic Equations Melek Erdo¼ gdu
y
Mustafa Özdemirz
June 21, 2013
Abstract The main purpose of this paper is to examine the linear split quaternionic systems with n unknowns. We investigate these kind of systems with left and right coe¢ cients, seperately. Then, we give the classi…cations of these systems of equations by using complex adjoint matrix of split quaternion matrix. Besides, a method of …nding solution of these systems is given for each case. Finally, the examples with 2; 3 and 4 unknowns are given. Keywords :Quaternion, Split Quaternion, Quaternion Matrix.
1
Introduction
Sir William Rowan Hamilton …rst described the quaternions in 1843. This description is a kind of extension complex numbers to higher spatial dimensions. So the set of quaternions can be represented as H = fq = q1 + q2 i + q3 j + q4 k; q1 ; q2 ; q3 ; q4 2 Rg where i2 = j 2 = k 2 = 1 and ijk = 1. The set of quaternions is a member of noncommutative division algebra, [3]. After his creation of quaternions, James Cockle introduced coquaternions in 1849, which can be represented as b = fq = q1 + q2 i + q3 j + q4 k; q0 ; q1 ; q2 ; q3 2 Rg: H
Here, the imaginary units satisfy the relations i2 = 1; j 2 = k 2 = 1 and ijk = 1: Due to the division of imaginary units into positive and negative terms, the coquaternions came to be called split quaternions. The set of split quaternions is noncommutative, too. Contrary to quaternion algebra, the set of split quaternions contains zero divisors, nilpotent elements and nontrivial idempotents, [4], [8], [9], [6]. There are a lot of works associated with quaternion matrices. Zhang has given a brief survey on quaternions and matrix of quaternions which is presented in the study [2]. Because of noncommutativity of quaternions, the theory eigenvalues of a quaternion matrix becomes one of the special topics about quaternion matrices. So, right eigenvalues for quaternionic matrices with a topological approach has been discussed by Baker in [1]. On the other hand, a study on left eigenvalues of a quaternionic matrix has been done by Huang and So [5]. Split quaternions is a recently developing topic. There are works on geometric applications of split quaternions such as [8] and [6]. On the other hand, a study on split quaternion matrices,[12], gives some properties of split quaternion matrices and the de…nition of complex adjoint matrix of a split quaternion matrix. Department of Mathematics, Necmettin Erbakan University, Konya, TURKEY (
[email protected]). Author z Department of Mathematics, Akdeniz University, Antalya, TURKEY (
[email protected]). y Corresponding
1
The present article is concerned with the linear split quaternionic equations of n unknowns with left and right coe¢ cients, seperately. We …nd when these kind of systems have unique solution, in…nitely solution or no solution. For this purpose, a brief summary of split quaternion and some essential properties of split quaternion matrices are given to necessary background. Then, we give the classi…cations of these kind of system of equations by using complex adjoint matrix of the split quaternion matrix: Finally, we give the way of …nding solution of the system. Also, the examples with 2; 3 and 4 unknowns are given.
2 2.1
Split Quaternions and Matrix of Split Quaternions Split Quaternions
The set of split quaternions can be represented as b = fq = q1 + q2 i + q3 j + q4 k; q1 ; q2 ; q3 ; q4 2 Rg H
where i2 = 1; j 2 = k 2 = 1 and ijk = 1. We write any split quaternion in the form q = (q1 ; q2 ; q3 ; q4 ) = Sq + Vq where Sq = q1 denotes the scalar part of q and Vq = q2 i + q3 j + q4 k denotes vector part of q: If Sq = 0 then q is called pure split quaternion and the set of pure quaternions can be identi…ed with Minkowski 3 space. Here, the Minkowski 3 space is Euclidean space with Lorentzian inner product hu; viL =
u1 v1 + u2 v2 + u3 v3
where u = (u1 ; u2 ; u3 ); v = (v1 ; v2 ; v3 ) 2 E3 and denoted by E31 . And the rotations in Minkowski 3 space can be stated with split quaternions such as expressing the Euclidean rotations using quaternions, [8]. b is denoted by q and it is The conjugate of a split quaternions q = q1 + q2 i + q3 j + q4 k 2 H q = Sq
Vq = q1
q2 i
q3 j
q4 k:
b the sum and product of split quaternions p and q are For p; q 2 H, p + q = Sp + Sq + V p + V q ;
~ q + Sq V ~p+V ~ p ^L V ~ q; pq = Sp Sq + hVp ; Vq iL + Sp V
respectively. Here h , iL and ^L denote Lorentzian inner and vector product and are de…ned as hu; viL = u ^L v =
u1 v1 + u2 v2 + u3 v3 ; e1 u1 v1
e2 u2 v2
e3 u3 v3
;
for vectors u = (u1 ; u2 ; u3 ) and v = (v1 ; v2 ; v3 ) of Minkowski 3 space, respectively. And norm of the split quaternion q is de…ned by q p Nq = jqqj = jq02 + q12 q22 q32 j:
If N q = 1 then q is called unit split quaternion and q0 = q=Nq is a unit split quaternion for N q 6= 0: And the product Iq = qq = qq = q12 + q22 q32 q42 determines the character of a split quaternion. A split quaternion is spacelike, timelike or lightlike (null) if Iq < 0; Iq > 0 or Iq = 0; respectively. For further information, see [8] and [9]. 2
b the following properties are satis…ed Theorem 1 [12] For any p; q; r 2 H, i) qq = qq; _
ii) jc = cj, 8c 2 C; iii) q 2 = Sq2
2
kVq k + 2Sq Vq ;
iv) (pq) = q p; v) (pq)r = p(qr); vi) pq 6= qp in general ; vii) q = q if and only if q 2 R; viii) If q02 + q12 6= q22 + q32 then q
1
=
q
2;
kqk
b there exists a unique representation of the form q = c1 + c2 j such that c1 ; c2 2 C: ix) 8q 2 H Theorem 2 [12] Every split quaternion can be represented by a 2
2.2
2 complex matrix.
Split Quaternion Matrices
b with The set of m n matrices with split quaternion entries, which is denoted by Mm n (H); b and ordinary matrix addition and multiplication is a ring with unity. For A = (ast ) 2 Mm n (H) b q 2 H, right and left scalar multiplication are de…ned as Aq = (ast q)
and
qA = (qast );
respectively. For A 2 Mm
Mm
b
n (H);
B 2 Mn
b and p; q 2 H; b the followings are satis…ed:
r (H)
q(AB) = (qA)B; (Aq)B = A(qB); (pq)A = p(qA):
b is left (right) vector space over H: b
n (H)
b b is the conjugate of A; For A = (ast ) 2 Mm n (H); A = (ast ) 2 Mm n (H) T T b is the transpose of A; A = (A) 2 Mn m (H) b is conjugate transpose A = (ats ) 2 Mn m (H) b of A: For a square split quaternion matrix A 2 Mn n (H); if AA = A A then A is called normal matrix; if A = A then A is called Hermitian matrix; if AA = I then A is called unitary matrix. b if AB = BA = I then A is called invertible matrix and B is called the inverse For B 2 Mn n (H); of A. 3
b and B 2 Mn
Theorem 3 [12] For any A 2 Mm
b
n (H)
i) (A)T = (AT );
s (H);
the followings are satis…ed:
ii) (AB) = B A ; 1
iii) If A and B are invertible then (AB) iv) If A is invertible then (A ) 1
v) A
1
vi) AT
1)
6= (A
6= A
1
1
= (A
=B
1
A
1
;
) ;
in general; 1 T
in general;
vii) AB 6= A B in general; T
viii) (AB) 6= B T AT in general. b
Proposition 4 [12] Let A; B 2 Mn
n (H);
if AB = I then BA = I:
b where A1 ; A2 2 Mn 1 A1 A2 @ A A2 A1
De…nition 5 [12] Let A = A1 + A2 j 2 Mn 0
n (H)
is called the complex adjoint matrix of A and denoted by b
Theorem 6 [12] Let A; B 2 Mn i)
In
= I2n ;
ii)
A+B
iii)
AB
= =
A
+
A
6= (
The 2n
2n matrix
A.
then the followings are satis…ed:
B;
A B;
iv) If A is invertible then ( v)
n (H);
n (C):
A)
A)
1
=
A
1
;
in general.
b and 2 H. b If holds the equation Ax = x [Ax = x ] for some De…nition 7 Let A 2 Mn (H) nonzero split quaternion column vector x; then is called the left [right] eigenvalue of A: The set b : Ax = x; for some x 6= 0g is called left spectrum of A: of the left eigenvalues l (A) = f 2 H b : Ax = x ; for some x 6= 0g is called Similarly, the set of the right eigenvalues r (A) = f 2 H right spectrum of A: Theorem 8 [12] Let A 2 Mn i) A is invertible;
b
n (H);
then the followings are equivalent:
ii) Ax = 0 has unique solution 0; iii) j
Aj
= 6 0 i.e.
A
is an invertible matrix;
iv) A has no zero eigenvalue. More precisely if Ax = x or Ax = x for some split quaternion and some split quaternion column vector x 6= 0; then 6= 0:
4
b and A be the complex adjoint matrix of A. We de…ne the q De…nition 9 Let A 2 Mn (H) determinant of A by jAjq = j A j where j A j is the usual determinant of the complex matrix A : Theorem 10 [12] Let A; B 2 Mn
b
n (H);
then
i) A is invertible if and only if jAjq 6= 0; ii) jABjq = jAjq jBjq consequently, if A is an invertible matrix then A
1
1 q
= jAjq ;
iii) jP AQjq = jAjq for some elementary matrices P and Q: b Theorem 11 [10] For every A 2 Mn (H),
r (A)
where ( A.
A)
=f 2C:
Ay
\C= (
A );
= y; for some y 6= 0g; is spectrum of the complex adjoint matrix of
b where A1 ; A2 Theorem 12 [10] Let A = A1 + A2 j 2 Mn (H); = 1 + 2 j where 1 ; 2 2 C; is a left eigenvalues of A if such that maxfkx1 k1 ; kx2 k1 g > 0 and 2 32 (A1 x1 1 In ) (A2 2 In ) 4 54 x2 (A2 2 In ) (A1 1 In )
2 Mn (C). Then the split quaternion and only if there exists x1 ; x2 2 Cn 3
2
5=4
0 0
3
5:
Here, kyk1 = maxfkyi k : i = 1; 2; :::; ng where y = (y1 ; y2 ; :::; y2 ) is a split quaternion column vector. Corollary 13 [10] Let A be a n right eigenvalues.
n split quaternion matrix. A has at most 2n distinct complex
Corollary 14 Let A be a n n split quaternion matrix. A has exactly n right eigenvalues which are complex with nonnegative imaginary parts. Those eigenvalues are called standard eigenvalues of A: Theorem 15 Let A be a n
n split quaternion matrix. Then jAjq =
n Q
t=1
2
j tj
where t are standard eigenvalues of A and j t j is the module of the complex number 1; 2; :::; n. Proof. Since the eigenvalues of
A
are
1; n Q
t t
=
1;
n Q
j tj
t=1
t=1
which is equal to jAjq :
n;
2 ; :::;
n b jAj = Q j t j2 Remark 16 For A 2 Mn (H); q
0:
t=1
5
2 ; :::; 2
n,
t
for t =
then the determinant of
A
is
3
Linear Split Quaternionic Equations
In this part, we examine linear split quaternionic equations with left and right coe¢ cients, seperately. Using fundamental properties of split quaternions and their matrices, we convert the linear split quaternionic systems with n unknowns to complex systems with 2n unknows, for each case. So, a method of solving split quatenionic systems is given. Furthermore, we give some examples to show how this method works with 2, 3 and 4 unknowns.
3.1
Linear Split Quaternionic Equations with Left Coe¢ cients
Consider the following linear split quaternionic equations a11 x1 + a12 x2 + ::: + a1n xn = b1 ; a21 x1 + a22 x2 + ::: + a2n xn = b2 ; .. . an1 x1 + an2 x2 + ::: + ann xn = bn ; b for all i; j = 1; 2; :::; n where xi are split quaternionic unknowns for i = 1; 2; :::; n. Here aij 2 H b for i = 1; 2; :::; n: This system of equation can be rewritten as and bi 2 H Ax = b
where
We may write
2
6 b x=6 A = (aij ) 2 Mn (H); 6 4
x1 x2 .. . xn
(1)
2
3
6 7 6 7 7 and b = 6 4 5
b1 b2 .. . bn
3
7 7 7: 5
A = A1 + A2 j; x = y + zj and b = c + dj: Here A1 ; A2 2 Mn (C) and y; z; c and d are complex column vectors. So, equation 1 is equivalent to the following complex system with 2n complex unknowns; A1 y + A2 z = c; A1 z + A2 y = d: This system can be written as follows; 2 A1 4 A2
Let us denote
e= x
A2 A1 "
y z
32 54 #
y z
3
2
5=4
e= and b
"
c d c
d
3
5:
#
:
So, the linear equation of the form Ax = b with n unknowns is equivalent to complex system e with 2n complex unknowns. Thus, we have the following theorem: e=b Ax
6
b x and b be split quaternionic column vectors. Theorem 17 Let A 2 Mn (H);
i. The linear system of equations e = 2n: Rank( A ) = Rank( A ; b)
Ax
=
b has unique solution if and only if
ii. The linear system of equations Ax = b has no solution if and only if Rank(
A)
6= Rank(
e
A ; b):
iii. The linear system of equations Ax = b has in…nitely solution if and only if e = t < 2n: In this case the solution depends on 2n t complex Rank( A ) = Rank( A ; b) parameters. Here (
e denotes the augmented matrix of
A ; b)
A
e with b:
Remark 18 If jAjq 6= 0 then the unknowns of the system Ax = b can be found as e det( rA (b)) ; jAjq
yr =
zr =
det(
r+n e A (b))
jAjq
e denotes the matrix obtained from A by replacing its rth column for r = 1; 2; :::; n. Here rA (b) e with the column b: Therefore the solution of equation 1 is obtained by using cramer rule for its complex adjoint matrix.
Corollary 19 The linear split quaternionic system of equations Ax = b has unique solution if and only if all complex right eigenvalues of A are nonzero. Proof. The linear split quaternionic system of equations Ax = b has unique solution if and only if jAjq = j A j = 6 0 by theorem 17. From theorem 11, we know that r (A)
This means j
3.2
Aj
\C= (
A ):
6= 0 if and only if all complex right eigenvalues of A are nonzero.
Linear Split Quaternionic Equations with Right Coe¢ cients
Consider the following linear split quaternionic equations x1 a11 + x2 a21 + ::: + xn an1 = b1 ; x1 a12 + x2 a22 + ::: + xn an2 = b2 ; .. . x1 a1n + x2 a2n + ::: + xn ann = bn ; b for all i; j = 1; 2; :::; n where xi are split quaternionic unknowns for i = 1; 2; :::; n. Here aij 2 H b for i = 1; 2; :::; n: This system of equation can be rewritten as and bi 2 H
(2)
xA= b
where
b x= A = (aij ) 2 Mn (H);
x1
x2
::: xn
7
and b =
b1
b2
::: bn
:
Similarly with previous part, we may write A = A1 + A2 j; x = y + zj and b = c + dj: Here A1 ; A2 2 Mn (C) and y; z; c and d are complex row vectors. So, equation 2 is equivalent to the following complex system with 2n complex unknowns; yA1 + zA2 = c; yA2 + zA1 = d: This system can be written as follows; y
2 4
z
A1
A2
A2
A1
3
5=
c
d
:
Taking transpose of both sides of this equation, we get 3T 2 T 3 2 T 3 2 A1 A2 y c 5 4 5=4 5: 4 zT dT A2 A1 Let us denote
x=
"
#
yT zT
and b =
"
cT dT
#
:
So, the linear equation of the form xA= b with n split quaternionic unknowns is equivalent to complex sytem
T Ax
= b with 2n complex unknowns. Thus, we have the following theorem:
b b and x be split quaternionic row vectors. Theorem 20 Let A = A1 + A2 j 2 Mn (H); i. The linear system of equations xA Rank(
T A)
T A ; b)
= Rank(
=
b has unique solution if and only if
= 2n:
ii. The linear system of equations xA = b has no solution if and only if Rank( iii. The linear system of equations xA T A)
Rank( = Rank( parameters. Here (
T A ; b)
T A ; b)
=
Aj
=
= t < 2n: In this case the solution depends on 2n
T A
zr = t r A ) (b)
T A ; b):
T A
t complex
with b:
6= 0 then the unknowns of the system xA= b can be found as yr =
for r = 1; 2; :::; n. Here (
6= Rank(
b has in…nitely solution if and only if
denotes the augmented matrix of
Remark 21 If jAjq = j
T A)
det(( TA )r (b)) ; jAjq T r+n (b)) A)
det((
jAjq
denotes the matrix obtained from
T A
by replacing its rth column
with the column b: Therefore the solution of equation 2 is obtained by using cramer rule for its complex adjoint matrix.
8
EXAMPLES In this part, we give examples having no solution, unique solution and in…nitely many solutions for 2; 3 and 4 unknowns. Example 22 Consider the following linear system with split quaternionic left coe¢ cients (i + k)x1 + (1 + j)x2 = 1 + k; ( 1 + j k)x1 + (i + j)x2 = 1 i; where x1 and x2 are split quaternionic unknowns. This system can be written of the form Ax = b as follows; i+k 1 j x1 1+k = : 1+j k i+j x2 1 j Here 2 2 split quaternion matrix A and the split quaternionic column vector b can be rewritten as following form A = A1 + A2 j =
b= b1 + b2 j = We …nd 2 A
i 6 1 =6 4 i 1+i
1 i i 1 i 1 i 1 1
We obtain that Rank( solution.
A)
i 1 1 i 1 1
+
3 3 2 1 1 7 6 1 7 e = 6 1 7 and ( 7, b 4 i 5 1 5 1 i
= 3 and Rank(
i
+
1
i 1
1 1
i
j;
j:
2
i 6 1 e 6 A ; b) = 4 i 1+i
1 i i 1 i 1 i 1 1
1 1 1 i
3 1 1 7 7: i 5 1
e = 4: By theorem 17, the given equation has no
A ; b)
Example 23 Consider the following linear system with split quaternionic left coe¢ cients x1 + jx2 + (1 k)x3 = i; (1 + i)x1 + (j + k)x2 = j + k; (1 + j)x1 ix2 = 1 + k;
where x1 , x2 and x3 are split quaternionic unknowns. This system can be written of the form Ax = b as follows; 2 32 3 2 3 1 j 1 k x1 i 4 1+i j+k 0 5 4 x2 5 = 4 j + k 5 : 1+j i 0 x3 1+k Here 3 3 split quaternion matrix A and the split quaternionic column vector b can be rewritten as following form 2 3 2 3 1 0 1 0 1 i A = 4 1 + i 0 0 5 + 4 0 1 + i 0 5 j; 1 i 0 1 0 0 2
3 2 3 i 0 b= 4 0 5 + 4 1 + i 5 j: 1 i 9
We …nd 2
A
6 6 6 =6 6 6 4 2
6 6 6 e ( A ; b) = 6 6 6 4
3
1 0 1+i 0 1 i 0 1 0 1 i 1 0
1 0 1 0 0 1+i 0 1 0 i 1 0 0 1 i 0 0 1 i
i 0 0 1 0 0
1 0 1+i 0 1 i 0 1 0 1 i 1 0
1 0 1 0 0 1+i 0 1 0 i 1 0 0 1 i 0 0 1 i
i i 0 0 0 1 1 0 0 1 i 0 i
7 7 7 7 and 7 7 5 3
7 7 7 7: 7 7 5
e = 5: By theorem 16, this system has in…nitely many solutions Here Rank( A ) = Rank( A ; b) e can be found by long and tedious e=b depending on one complex parameter. The solution of A x computations. Using this solution, we …nd 2 3 2 3 2 3 2 3 x1 2i 2 i 2i + 2j k 4 x2 5 = 4 1 i 5 + 4 2i 5 j = 4 1 i 2k 5 x3 (1 + t)i t (1 + t)i + tj where t is a complex parameter.
Example 24 Consider the following linear system with split quaternionic left coe¢ cients ix1 + jx2 + (1 k)x3 + (j + k)x4 (1 + k)x1 + (2j + k)x2 + z3 (1 + j)x1 ix2 + (j k)x4 (i + j)x1 + kx2 + kx3
= i 8j; = 4 + 4j; = 8k; = 8i;
where x1 , x2 ; x3 and x4 are split quaternionic unknowns. This system can be written of the form Ax = b as follows; 2 32 3 2 3 i j 1 k j+k x1 i 8j 6 1 + k 2j + k 6 7 6 7 1 0 7 6 7 6 x2 7 = 6 4 + 4j 7 : 4 1+j i 0 j k 5 4 x3 5 4 8k 5 i+j k k 0 x4 8i e = Here Rank( A ) = Rank( A ; b) and tedious computations, we …nd 2 3 2 x1 11 + 11i 6 x2 7 6 9 + 6i 6 7 6 4 x3 5 = 4 34 + 10i x4 12 15i
8: By theorem 17, this system has unique solution. With long 3
2
3 2 14 + 5i 11 + 11i 14j + 5k 7 6 11 2i 7 6 9 + 6i 11j 2k 7+6 7 6 5 4 19 + 27i 5 j = 4 34 + 10i + 19j + 27k 17 2i 12 15i + 17j 2k
10
3
7 7: 5
Example 25 Consider the following linear system with split quaternionic coe¢ cients x1 i + x2 (1 + j) = 1 + k; x1 ( 1 + j) x2 i = 1 i; where x1 and x2 are split quaternionic unknowns. This system can be written of the form xA= b as follows; i 1+j x1 x2 1 i : = 1+k 1+j i We …nd that 2 3 2 3 1 i 1 0 1 6 7 6 1 i 1 0 7 T 6 7 ,b = 6 1 i 7 : A =4 0 4 5 5 i 1 i 1 0 1 0 1 i
Since jAjq = det A = 5 6= 0, this system has unique solution by theorem 20. Using remark 21, we …nd the solution of the given linear equation is x1 x2
=
1=5 2i=5 4=5 2i=5
+
4=5 3i=5 1=5 + 3i=5
j=
1=5 4=5
2i=5 2i=5
4j=5 3k=5 j=5 + 3k=5
:
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11
