Linear equations in quaternionic variables Drahoslava Janovsk a, - PowerPoint PPT Presentation

Linear equations in quaternionic variables Drahoslava Janovsk´ a, Institute of Chemical Technology, Prague Gerhard Opfer, University of Hamburg Hamburg - Harburg, September 12, 2008

Outline: • Basic definitions • Quaternionic linear mappings • Linear equations in one quaternionic variable • Sylvester’s equation in quaternions • Linear equations of general type in one quaternionic variable • Linear systems in quaternionic variables • Conclusions

Basic definitions for quaternions H := R 4 . . . the skew field of quaternions Let y = ( y 1 , y 2 , y 3 , y 4 ) ∈ H . x = ( x 1 , x 2 , x 3 , x 4 ) , Then x + y = ( x 1 + y 1 , x 2 + y 2 , x 3 + y 3 , x 4 + y 4 ) and xy = ( x 1 y 1 − x 2 y 2 − x 3 y 3 − x 4 y 4 , x 1 y 2 + x 2 y 1 + x 3 y 4 − x 4 y 3 , x 1 y 3 − x 2 y 4 + x 3 y 1 + x 4 y 2 , x 1 y 4 + x 2 y 3 − x 3 y 2 + x 4 y 1 ) . • The first component x 1 . . . the real part of x , denoted by ℜ x. • The second component x 2 . . . the imaginary part of x , denoted by ℑ x. • x = ( x 1 , 0 , 0 , 0) will be identified with x 1 ∈ R • x = ( x 1 , x 2 , 0 , 0) will be identified with x 1 + i x 2 ∈ C • The conjugate of x will be defined by x = ( x 1 , − x 2 , − x 3 , − x 4 ) � x 2 1 + x 2 2 + x 2 3 + x 2 • The absolute value of x will be defined by | x | = 4 x • The inverse quaternion is defined as x − 1 = | x | 2 for x ∈ H \{ 0 }

Classes of equivalence Two quaternions x and y are called equivalent, x ∼ y , if there is h ∈ H \{ 0 } such that y = h − 1 xh . [ x ] = { y ∈ H : y = h − 1 xh for h ∈ H \{ 0 }} . . . equivalence class of x Two quaternions x and y are equivalent if and only if Lemma. ℜ x = ℜ y | x | = | y | . and a ∈ R = ⇒ [ a ] = { a } c ∈ C = ⇒ c ∈ [ c ] , c ∈ [ c ] Let x = ( x 1 , x 2 , x 3 , x 4 ). Then Corollary. � x 2 2 + x 2 3 + x 2 x = ( x 1 , 4 , 0 , 0) = x 1 + | x v | i ∈ [ x ] � is the only complex element in [ x ] with non negative imaginary part.

Quaternionic linear mappings A mapping L : H → H is called a quaternionic linear mapping over R if L ( γx + δy ) = γL ( x ) + δL ( y ) for all x, y ∈ H , γ, δ ∈ R . Let us remark that L (0) = 0 , L ( αx ) � = αL ( x ) , L ( xα ) = L ( x ) α for α ∈ H \ R The four unit vectors in H are denoted by e 1 := (1 , 0 , 0 , 0) =: 1 , e 2 := (0 , 1 , 0 , 0) =: i , e 3 := (0 , 0 , 1 , 0) =: j , e 4 := (0 , 0 , 0 , 1) =: k .

A quaternionic linear mapping L is called singular Definition if there is x � = 0 with L ( x ) = 0 . The mapping L is called non singular if it is not singular. Theorem Let L be a quaternionic linear mapping. Then, there exists a matrix M ∈ R 4 × 4 such that L ( x ) = Mx , where x, L ( x ) have to be identified with the corresponding column vectors x , Mx and (with the same identification) M := ( L ( e 1 ) , L ( e 2 ) , L ( e 3 ) , L ( e 4 )) . A quaternionic linear mapping L is singular if and only if Lemma det M = 0 .

Let a := ( a 1 , a 2 , a 3 , a 4 ) ∈ H . Definition Let us introduce two mappings τ 1 , τ 2 : H → R 4 × 4 by   a 1 − a 2 − a 3 − a 4   a 2 a 1 − a 4 a 3   τ 1 ( a ) := ( ae 1 , ae 2 , ae 3 , ae 4 ) := A :=  ,  a 3 a 4 a 1 − a 2 a 4 − a 3 a 2 a 1   a 1 − a 2 − a 3 − a 4   a 4 − a 3 a 2 a 1 τ 2 ( a ) := ( e 1 a, e 2 a, e 3 a, e 4 a ) := �   A :=  .  a 3 − a 4 a 1 a 2 a 4 a 3 − a 2 a 1 Denote by H R all matrices of the type A , by H P all matrices of the type � A . H P are called pseudo-quaternions. Remark H R is isomorphic to H .

Let a = ( a 1 , a 2 , a 3 , a 4 ) ∈ H , b = ( b 1 , b 2 , b 3 , b 4 ) ∈ H , col ( a ) = ( a 1 , a 2 , a 3 , a 4 ) T Lemma be the column vector consisting of the four components of a . Then for a, b, c ∈ H we have (1) τ 2 ( ab ) = τ 2 ( b ) τ 2 ( a ) , (2) τ 1 ( a ) τ 2 ( b ) = τ 2 ( b ) τ 1 ( a ) , (3) col ( ab ) = τ 1 ( a ) col ( b ) = τ 2 ( b ) col ( a ) , (4) col ( abc ) = τ 1 ( a ) τ 1 ( b ) col ( c ) = τ 2 ( c ) τ 2 ( b ) col ( a ) , (5) col ( abc ) = τ 1 ( a ) τ 2 ( c ) col ( b ) .

Let L (1) ( x ) := axb for arbitrary a, b, x ∈ H . Theorem Then, there is a matrix M ∈ R 4 × 4 such that L (1) ( x ) := axb = Mx , � and M = τ 1 ( a ) τ 2 ( b ) . Let L ( m ) ( x ) := � m j =1 a ( j ) xb ( j ) for arbitrary a ( j ) , b ( j ) , x ∈ H be given. Corollary Then, there is a matrix M ∈ R 4 × 4 such that m � a ( j ) xb ( j ) = Mx , where L ( m ) ( x ) := j =1 m � M j := τ 1 ( a ( j ) ) τ 2 ( b ( j ) ) . M := M j , and j =1 The singular cases are exactly those with vanishing determinant of M . Apart from special cases it seems, however, to be impossible to characterize the singular cases just by the given quaternions a ( j ) , b ( j ) , j = 1 , 2 , . . . , m .

Linear equations in one variable Let m ∈ N , a = ( a (1) , a (2) , . . . , a m ) , b = ( b (1) , b (2) , . . . , b ( m ) ) ∈ H m , a ( j ) , b ( j ) ∈ H \{ 0 } , j = 1 , . . . , m , e ∈ H be given ? x ∈ H : m � a ( j ) xb ( j ) = e L ( m ) ( x ) := j =1 L ( m ) : H → H is the quaternionic linear mapping . Assumption: a (1) = b ( m ) = 1, a, b, c, d, x ∈ H L (1) ( x ) := x , L (2) ( x ) := ax + xb , L (3) ( x ) := ax + cxd + xb , . . . c, d ∈ H \ { R } ⇒ = cxd . . . middle terms

Lemma Let b = ( b 1 , b 2 , b 3 , b 4 ) , c = ( c 1 , c 2 , c 3 , c 4 ). Equation bxc = e has a unique solution for all choices of e if and only if bc � = 0. In this case the solution is x = b − 1 ec − 1 Corollary Let a, b, c, d, e be given quaternions, abcd � = 0. Let � L := axb + cxd , and solve � L ( x ) = e , x ∈ H . The equation � L ( x ) = e has a unique solution for all choices of e if and only if 4 � ℜ ( a − 1 c ) + ℜ ( bd − 1 ) � = 0 (( a − 1 c ) 2 j − ( bd − 1 ) 2 j ) � = 0 , or j =2 where the subscript j defines the component number.

Sylvester’s equation in quaternions: L (2) ( x ) := ax + xb = e , a / ∈ R , b / ∈ R Let L ( 2 ) ( x ) := ax + xb , a, b / ∈ R . Then, L (2) is singular if and only if (1) | a | = | b | and ℜ ( a ) + ℜ ( b ) = 0 . or in other words if and only if a ∼ − b . The nullspace of L (2) is  { 0 } if (1) is not valid ,  N = if (1) is valid and a, b ∈ R , H  2 − dim subspace of H ∈ R or b / ∈ R . if (1) is valid and a /

The equivalent matrix representation: L (2) ( x ) := ax + xb = Mx , where   s 1 − s 2 − s 3 − s 4   s 2 s 1 − d 4 d 3   M = τ 1 ( a ) τ 2 (1) + τ 1 (1) τ 2 ( b ) =   s 3 d 4 s 1 − d 2 s 4 − d 3 d 2 s 1 and s = a + b = ( s 1 , s 2 , s 3 , s 4 ) , d = a − b = ( d 1 , d 2 , d 3 , d 4 ) . The conditions (1) are equivalent to det M = 0 , where 1 ( | s | 2 + d 2 4 ) + ( s 2 d 2 + s 3 d 3 + s 4 d 4 ) 2 . det( M ) = s 2 2 + d 2 3 + d 2

Theorem Let a = ( a 1 , a 2 , a 3 , a 4 ) , b = ( b 1 , b 2 , b 3 , b 4 ) . The equation L ( 2 ) ( x ) = e has a unique solution for all choices of e if and only if 4 � ( a 2 j − b 2 a 1 + b 1 � = 0 or j ) � = 0 . j =2 In this case the solution is x = f − 1 ( e + a − 1 eb ) , f l := 2 ℜ b + a + | b | 2 a − 1 if a � = 0 l or r , f r := 2 ℜ a + b + | a | 2 b − 1 if b � = 0 . x = ( e + aeb − 1 ) f − 1 Remark Numerical computations: If | a | > 0 but close to zero, avoid the first formula containing a − 1 . If | a | ≤ | b | the use of the second formula is recommended, otherwise use the first one.

Linear equations of general type in one quaternionic variable Theorem Let L ( m ) ( x ) := � m j = 1 a ( j ) xb ( j ) for arbitrary a ( j ) , b ( j ) , x ∈ H be given. Then L ( m ) ( x ) = e is equivalent to the (4 × 4) − matrix equation � n � � τ 1 ( a ( j ) ) τ 2 ( b ( j ) ) col ( x ) = col ( e ) . j =1 Corollary The linear function L ( m ) is singular if and only if � n � � τ 1 ( a ( j ) ) τ 2 ( b ( j ) ) det = 0 . j =1

Banach’s fixed point theorem = ⇒ sufficient condition for nonsingularity m � a ( j ) xb ( j ) with m ≥ 3, a ( j ) , b ( j ) ∈ H , Let L ( m ) ( x ) := Theorem j = 1 a ( j ) b ( j ) � = 0 for all j = 1 , 2 , . . . , m. If there is j 0 , 1 ≤ j 0 ≤ m, such that � m j � = j 0 | a ( j ) || b ( j ) | κ := < 1 , | a ( j 0 ) || b ( j 0 ) | then L ( m ) is nonsingular. m � a ( j ) xb ( j ) with m ≥ 3, a ( j ) , b ( j ) ∈ H , Let L ( m ) ( x ) := Corollary j = 1 a ( j ) b ( j ) � = 0 for all j = 1 , 2 , . . . , m. If there is a constant k > 0 and an index j 0 , 1 ≤ j 0 ≤ m, such that | a ( j ) || b ( j ) | ≤ k for all j � = j 0 and | a ( j 0 ) || b ( j 0 ) | ≥ k 2 , then L ( m ) is nonsingular if k > m − 1. For k = m − 1 this is not necessarily true.