Quaternion Equations and Quaternion Polynomial Matrices

Transcription

1 Quaternion Equations and Quaternion Polynomial Matrices by Liji Huang A Thesis submitted to the Faculty of Graduate Studies of The University of Manitoba in partial fulfilment of the requirements of the degree of MASTER OF SCIENCE Department of Mathematics University of Manitoba Winnipeg Copyright 01 by Liji Huang

2 Contents Abstract i Acknowledgments iii 1. Introduction History of Quaternions Background for Quaternion Equations Background for Quaternion Polynomial Matrices Quaternion Equations 9.1. Quaternions and Equivalence Classes General Quaternion Quadratic Equations Special Quaternion Equations and problems A quadratic equation Another quadratic equation A homogeneous quadratic equation A pair of quaternion equations Another pair of quaternion equations A least norm problem Another least norm problem

3 3. Quaternion Polynomial Matrices Generalized Inverse Leverrier-Faddeev Method Generalized Inversion by Interpolation A. Maple Codes 89 A.1. Overview A.. The Codes Bibliography 119

4 Abstract In this thesis, we mainly consider finding formula solutions to quaternion equations and calculating the generalized inverses of quaternion polynomial matrices. In the first chapter, we give a condensed review of the history of quaternions. The topic of the second chapter is solving quadratic and other quaternion equations. Many mathematicians have worked on solving various quaternion equations. However, due to the complex nature of quaternions, the current "best" non-numerical result is a special case of quadratic quaternion equation presented in (Au-Yeung, 003), which pales in comparison to what is known about equations in complex numbers. In this chapter, we present a new way to solve not only almost all of the quaternion equations that have been solved so far, including Au-Yeung s result, but also several more complicated ones. This method can also be easily implemented into computer algebra systems such as Maple. Examples will be included. The topic of the third chapter is calculating the generalized inverses of quaternion polynomial matrices. It has been shown that a real (Karampetakis, 1997) or complex (Stanimirović et al., 007) polynomial matrix has the generalized inverse that can be calculated using the Leverrier-Faddeev algorithm (Faddeeva, 1959). It is natural to ask whether a quaternion polynomial matrix has a generalized inverse, and if yes, how it can be calculated. In this chapter, we first give conditions that a quaternion polynomial matrix must satisfy in order to have the generalized inverse. We then i

5 Contents present an interpolation method that can be used to calculate the generalized inverse of a given quaternion polynomial matrix. In the appendix, we include our quaternion polynomial matrix Maple module as well as some examples. ii

6 Acknowledgments First and foremost, I would like to express my most sincere gratitude to my thesis advisor, Dr. Zhang Yang. I greatly appreciate his encouragement and dedicated support. Countless hours of discussion with Dr. Zhang helped me overcome many obstacles during the writing up of this thesis. Without his wisdom and motivation this thesis would not have been possible. I greatly appreciate the faculty and staff in the Department of Mathematics for their teaching and support during my two years of being a graduate student in mathematics. I greatly thank my thesis committee members, Dr. Guenter Krause, Dr. Xikui Wang and Dr. Joe Williams, for reading my thesis and giving me feedback. On behalf of myself and my wife, I would like to thank Dr. Zhang, the Faculty of Graduate Studies, the Faculty of Science, and the Department of Mathematics, for financially supporting me through my graduate studies. I would like to thank the University of Manitoba library system for providing a rich supply of online and offline publications. I would also like to thank mapleprimes.com and latex-community.org for all of the excellent technical support I received from iii

7 Acknowledgments them, and their continued voluntary service to current and future graduate students and academics all over the world. Finally, I would like to express my appreciation to my family and friends, for all of their support and love. iv

8 Dedicated to my wife Zifang.

9 1. Introduction 1.1. History of Quaternions Quaternions were first discovered by Sir William Rowan Hamilton of Ireland on October 16 th, 1843, in a "flash of genius", on the Brougham Bridge. It is uncommon that both the date and location of a major mathematical discovery are known, but we precisely know these because at the end of Hamilton s life, in a letter to his second son, Archibald Henry, he wrote: Every morning in the early part of [October 1843], on my coming down to breakfast, your (then) little brother, William Edwin, and yourself, used to ask me, Well, papa, can you multiply triplets? Whereto I was always obliged to reply, with a sad shake of the head: No, I can only add and subtract them. But on the 16th day of the same month which happened to be Monday, and a Council day of the Royal Irish Academy I was walking in to attend and preside, and your mother was walking with me along the Royal Canal, to which she had perhaps driven; and although she talked with me now and then, yet an undercurrent of thought was going on in my mind which gave at last a result, whereof it is not too much to say that I felt at once the importance. An electric circuit seemed to close; and a spark flashed forth the herald (as I foresaw immediately) of many long years to come of definitely directed thought and work by myself, if spared, and, at all 1

10 1.1 History of Quaternions events, on the part of others if I should even be allowed to live long enough distinctly to communicate the discovery. Nor could I resist the impulse unphilosophical as it may have been to cut with a knife on a stone of Brougham Bridge, as we passed it, the fundamental formula with the symbols i, j, k: i = j = k = ijk = 1 which contains the Solution of the Problem, but, of course, the inscription has long since mouldered away (Derbyshire, 006). It was a breakthrough discovery in the sense that after the step from one-dimensional real numbers to two-dimensional complex numbers, the next step was not the natural one to three-dimensional super-complex numbers but to four-dimensional ones, i.e., quaternions. Hamilton embarked on an eight-year journey to develop an algebra for bracketed triplets but ended up discovering quaternions in a moment of euphoria. The Dutch mathematician Van der Waerden called Hamilton s inspiration "the leap into the fourth dimension" (van der Waerden, 1985). The Scottish mathematical physicist Peter Guthrie Tait, the leading quaternions exponent of his time, claimed "...no figure, nor even model, can be more expressive or intelligible than a quaternion equation" (Tait, 1890). However, Tait s lifetime friend, William Thomson, 1 st Baron Kelvin, disliked quaternions and would have nothing to do with them (Smith and Wise, 1989). Tait s other great friend, the Scottish mathematical physicist James Clerk Maxwell, remained unconvinced of quaternion s value in actual calculation, as he put it "...quaternions...is a mathematical method, but it is a method of thinking, and not, at least for the present generation, a method of saving thought" (Maxwell, 1873). Furthermore, the American scientist Josiah Willard Gibbs did not find quaternions suited to his needs and rejected the quaternionic analysis in favor of his own in-

11 1.1 History of Quaternions vention: vector analysis (Hastings, 010; Gibbs et al., 190). The English electrical engineer Oliver Heaviside, who independently formulated vector analysis similar to Gibbs s system, wrote "...it is impossible to think in quaternions, you can only pretend to do it" (Nahin, 00). The British mathematician Arthur Cayley gave a poetic illustration on the matter "...I compare a quaternion formula to a pocketmap, a capital thing to put in one s pocket, but which for use must be unfolded; the formula, to be understood, must be translated into coordinates" (Cayley, 1894). Even the English mathematician Alexander McAulay, a quaternion advocate, stated that "...not much advance in physics has been made by the aid of quaternions" (McAulay, 189). Indeed, in the century that has passed since McAulay wrote, it can be argued that quaternions faded away to a minor role in mathematics and physics while the Gibbs-Heaviside vector algebra became the everyday work tool of all modern engineers and scientists. In 189, Heaviside, hinting subtly that quaternions are what he called "a positive evil of no inconsiderable magnitude", summarized the matter in this way, and probably the best way "...the invention of quaternions must be regarded as a most remarkable feat of human ingenuity...but to find out quaternions required a genius" (Nahin). Today, quaternions still play a role in quantum physics (Adler, 1995; Finkelstein et al., 196) and other branches of physics (Churchill, 1990). Quaternions are now used throughout the aerospace industry for attitude control of aircraft and spacecraft (Kuipers, 00). Some people claim that quaternions are an important aspect in modern computer graphics for calculations involving three-dimensional rotations because quaternions use less memory, compose faster, and are naturally suited for efficient interpolation (Hanson, 005). Although, like in the academic circle of physics over a century ago, there are controversies regarding the advantages of quaternion orientation representations in today s computer graphics and modeling community 3

12 1.1 History of Quaternions (Hanson, 005). Quaternions are also attracting more and more attention in mathematics. There has been a new surge of interest in quaternions in the mathematics world a quick search in the American Mathematical Society s database shows that more than half of the 33 publications containing the word quaternion or quaternionic in the title have been published after the year of

13 1. Background for Quaternion Equations 1.. Background for Quaternion Equations This section attempts to give a general and brief review on what has been done on quaternion equations,.1 offers a more detailed introduction to quaternions themselves. Let x be a quaternion indeterminate and let a i be quaternions. In 1941, Niven looked into the equation: x m + a 1 x m 1 + a x m + + a m = 0, (1.1) where a m 0. He showed that 1.1 has at least one quaternion root. In the following year, Niven completely solved the equation x m a m = 0. In 1944, Eilenberg and Niven showed that even the most general equation with a unique highest term, a 1 xa x xa m + φ (x) = 0, where φ (x) is a sum of finite number of monomials of the form b 1 xb x xb k, k < m, has at least one quaternion root. This is also known as the "fundamental theorem of algebra" for quaternions. In 1944, Johnson gave necessary and sufficient conditions for the equation xa 1 + a x + a 3 = 0 to have a quaternion solution. His result was extended by Tian to include xa 1 + a x + a 3 = 0 5

14 1. Background for Quaternion Equations and xa 1 x + a = 0, where x denotes the quaternion conjugate of x. Very little was done to obtain formula solutions of quaternion equations after Johnson s result, until, in 00, Huang and So solved the special quadratic equation: x + a 1 x + a = 0. (1.) Then in the following year, Au-Yeung solved another special quadratic equation, which included both Johnson s and Huang and So s results as special cases: x + a 1 x + xa + a 3 = 0. (1.3) Au-Yeung s paper was the latest non-numerical attempt to find the exact roots of a non-linear quaternion equation. In 010, Tian studied and solved two pairs of linear quaternion equations xa 1 y = a, ya x = a 1, (1.4) and xa 1 y = a, ya x = a 1, (1.5) where y is also a quaternion indeterminate. In.3, we present a new method to finding the solutions to 1.3, 1.4 and 1.5. In addition, we give solutions to several previously unsolved equations and least norm problems, including, for example, the "two-sided" quadratic homogeneous equation: x + a 1 xa = 0, 6

15 1. Background for Quaternion Equations and the least norm problem: min a 1x xa a 3. x H In A., we present some maple codes to solve these quaternion equations and problems. 7

16 1.3 Background for Quaternion Polynomial Matrices 1.3. Background for Quaternion Polynomial Matrices In 1955, Penrose introduced the term generalized inverse that exists for any matrix with complex elements. In 1965, Decell gave a method for computing the generalized inverse of a constant complex matrix based on Leverrier-Faddeev s algorithm (Faddeeva, 1959), which recursively computes coefficients of the characteristic polynomial of the matrix. This method was later extended by Karampetakis to real polynomial matrices. In 006, Stanimirović and Petković applied an interpolation method to the result of Karampetakis. Then in the following year, Stanimirović et al. extended Stanimirović and Petković s result to include complex polynomial matrices with interpolations at real number data points. In Chapter 3, we first give conditions that a quaternion polynomial matrix must satisfy in order to have a generalized inverse. Then we present an interpolation method that can be used to calculate the generalized inverse of a given quaternion polynomial matrix. In A., we include our Maple codes for calculating the generalized inverse of a given quaternion matrix or quaternion polynomial matrix. We also include our Maple codes that can be used to do Newton and Lagrange quaternionic interpolation, which are closely related to the interpolation of the generalized inverse of a quaternion polynomial matrix. 8

17 . Quaternion Equations.1. Quaternions and Equivalence Classes In this section, we will outline some definitions and properties of quaternions. First, recall the definition of quaternions: Definition.1.1. Let R denote the field of the real numbers. Let H be a fourdimensional vector space over R with basis 1, i, j and k. A quaternion is a vector x = x x 1 i + x j + x 3 k H with real coefficients x 0, x 1, x and x 3. Throughout this thesis, for convenience, if h H, then we will assume that h = h 0 + h 1 i + h j + h 3 k where h i R, 0 i 3. We denote the real part, h 0, by Re h and the imaginary part, h 1 i + h j + h 3 k, by Im h. Addition and multiplication in H can be defined as follows: Definition.1.. Let x, y H. Then quaternion addition is defined as: x + y = (x 0 + y 0 ) + (x 1 + y 1 ) i + (x + y ) j + (x 3 + y 3 ) k. 9

18 .1 Quaternions and Equivalence Classes Quaternion multiplication is defined as: xy =x 0 y 0 x 1 y 1 x y x 3 y 3 + (x 0 y 1 + x 1 y 0 + x y 3 x 3 y ) i + (x 0 y x 1 y 3 + x y 0 + x 3 y 1 ) j + (x 0 y 3 + x 1 y x y 1 + x 3 y 0 ) k. Remark.1.3. By direct calculation, we obtain: ii = jj = kk = 1, ij = ji = k, jk = kj = i, ki = ik = j. It is also worth noting that with the above equations, the multiplication defined in Definition.1. follows. Lemma.1.4. Quaternion multiplication is not commutative in general. H is a division ring with center R. Proof. By direct calculation. Next, conjugate and norm for quaternions are defined as follows: Definition.1.5. Let h H. Then h denotes the conjugate of h, h = h 0 h 1 i h j h 3 k, and h denotes the norm of h, h = hh = h 0 + h 1 + h + h 3. Lemma.1.6. Let x, y H. Then x is non-negative and is a norm on H, i.e., x = 0 x = 0, x + y x + y, xy = yx = x y. 10

19 .1 Quaternions and Equivalence Classes Proof. We will only show that x + y x + y. The rest of the lemma is true by direct calculation. Since x + y = (x 0 + y 0 ) + i (x 1 + y 1 ) + j (x + y ) + k (x 3 + y 3 ) ) = ( (x 0 + y 0 ) + (x 1 + y 1 ) + (x + y ) + (x 3 + y 3 ) = (x 0 + y 0 ) + (x 1 + y 1 ) + (x + y ) + (x 3 + y 3 ) =x 0 + x 1 + x + x 3 + y 0 + y 1 + y + y 3 + x 0 y 0 + x 1 y 1 + x y + x 3 y 3 = x + y + x 0 y 0 + x 1 y 1 + x y + x 3 y 3, it suffices to show that x 0 y 0 + x 1 y 1 + x y + x 3 y 3 x y x 0 y 0 + x 1 y 1 + x y + x 3 y 3 x y (x 0 y 0 + x 1 y 1 + x y + x 3 y 3 ) x y (x 0 y 0 + x 1 y 1 + x y + x 3 y 3 ) ( ( x 0 + x 1 + x + x3) y 0 + y1 + y + y3), which is true by the Cauchy-Schwartz inequality. Next, inverse and conjugacy class for quaternions are defined as follows: Lemma.1.7. For 0 h H, h 1, the inverse of h, is uniquely determined by h 1 = h h. Furthermore, we have h 1 = 1 h. Proof. By direct calculation. Definition.1.8. Two quaternions x and y are said to be similar, or in the same 11

20 .1 Quaternions and Equivalence Classes conjugacy class, if these exists a quaternion v 0 such that v 1 xv = y, and we write x y. Lemma.1.9. x y implies x = y. Proof. By Lemma.1.7, x y v 1 xv = y y = v 1 xv = 1 x v = x. v Next, we state and prove an important result regarding similar quaternions. Theorem x y if and only if Re x = Re y and Im x = Im y. Proof. We only prove the forward direction of the theorem. The backward direction is straightforward. If x = 0 or y = 0, then the claim is true trivially. Assuming x 0 and y 0, we will first show that x 0 = y 0. Let xv = vy where v 0. That is, (x 0 + x 1 i + x j + x 3 k) (v 0 + v 1 i + v j + v 3 k) (.1) = (v 0 + v 1 i + v j + v 3 k) (y 0 + y 1 i + y j + y 3 k). After expanding.1 and comparing coefficients, we get: x 0 v 0 x 1 v 1 x v x 3 v 3 = y 0 v 0 y 1 v 1 y v y 3 v 3 = A, x 0 v 1 + x 1 v 0 + x v 3 x 3 v = y 1 v 0 + y 0 v 1 + y 3 v y v 3 = B, x v 0 + x 0 v + x 3 v 1 x 1 v 3 = y v 0 + y 0 v + y 1 v 3 y 3 v 1 = C, (.) x 0 v 3 + x 3 v 0 + x 1 v x v 1 = y 0 v 3 + y 3 v 0 + y v 1 y 1 v = D, 1

21 .1 Quaternions and Equivalence Classes which is equivalent to: v 0 (x 0 y 0 ) = v 1 (x 1 y 1 ) + v (x y ) + v 3 (x 3 y 3 ), v 0 (x 1 y 1 ) = v 1 (x 0 y 0 ) + v (x 3 + y 3 ) v 3 (x + y ), v 0 (x y ) = v 1 (x 3 + y 3 ) v (x 0 y 0 ) + v 3 (x 1 + y 1 ), (.3) v 0 (x 3 y 3 ) = v 1 (x + y ) v (x 1 + y 1 ) v 3 (x 0 y 0 ). Multiplying the four equations of.3 by (x 0 + y 0 ), (x 1 + y 1 ), (x + y ) and (x 3 + y 3 ) respectively, then adding them, we get x 0 (v 1 y 1 + v y + v 3 y 3 ) = y 0 (v 1 x 1 + v x + v 3 x 3 ) x 0 (y 0 v 0 A) = y 0 (x 0 v 0 A) Ax 0 = Ay 0. If x 0 = y 0, we are done. Assume x 0 y 0. Then A = 0. Next, multiplying the four equations of.3 by (x 1 + y 1 ), (x 0 + y 0 ), (x 3 y 3 ) and (x y ) respectively, then adding them, we get x 0 (y 1 v 0 y v 3 + y 3 v ) = y 0 (x v 3 x 3 v + x 1 v 0 ) x 0 (B y 0 v 1 ) = y 0 (B x 0 v 1 ) Bx 0 = By 0. Since x 0 y 0, B = 0. Next, multiplying the four equations of.3 by (x + y ), (x 3 y 3 ), (x 0 + y 0 ) and 13

22 .1 Quaternions and Equivalence Classes (x 1 y 1 ) respectively, then adding them, we get x 0 (y v 0 + y 1 v 3 y 3 v 1 ) = y 0 (x v 0 + x 3 v 1 x 1 v 3 ) x 0 (C y 0 v ) = y 0 (C x 0 v ) Cx 0 = Cy 0. Since x 0 y 0, C = 0. Lastly, multiplying the four equations of.3 by (x 3 + y 3 ), (x y ), (x 1 y 1 ) and (x 0 + y 0 ) respectively, then adding them, we get x 0 (y 3 v 0 + y v 1 y 1 v ) = y 0 (x 3 v 0 + x 1 v x v 1 ) x 0 (D y 0 v 3 ) = y 0 (D x 0 v 3 ) Dx 0 = Dy 0. Since x 0 y 0, D = 0. That is, A = B = C = D = 0 v = 0 or x = y = 0. Contradiction. Therefore, x 0 = y 0. Then by Lemma.1.9, x 1+x +x 3 = y 1+y +y 3, that is, Im x = Im y. 14

23 . General Quaternion Quadratic Equations.. General Quaternion Quadratic Equations Let a i, b i, c H where i N. We write a i = a i0 + a i1 i + a i j + a i3 k and b i = b i0 + b i1 i + b i j + b i3 k for easier notation. In this section, we focus on the general quadratic equation: where k N..4 is equivalent to the real non-linear system: k x + a i xb i + c = 0, (.4) i=1 (x 0 + m 11 ) x 1 + m 1 x + m 13 x 3 = c 1 + x 0 n 1, (1) m 1 x 1 + (x 0 + m ) x + m 3 x 3 = c + x 0 n, () m 31 x 1 + m 3 x + (x 0 + m 33 ) x 3 = c 3 + x 0 n 3, (3) (.5) x 0 x 1 x x 3 + d 1 x 0 + d x 1 + d 3 x + d 4 x 3 + c 0 = 0, (4) where m ij, n i and d i are given by: M = (m ij ) 3 3 a k i0 b i0 a i1 b i1 + a i b i + a i3 b i3 a i0 b i3 a i1 b i a i b i1 a i3 b i0 a i0 b i a i1 b i3 + a i b i0 a i3 b i1 = a i0 b i3 a i1 b i a i b i1 + a i3 b i0 a i0 b i0 + a i1 b i1 a i b i + a i3 b i3 a i0 b i1 a i1 b i0 a i b i3 a i3 b i, i=1 a i0 b i a i1 b i3 a i b i0 a i3 b i1 a i0 b i1 + a i1 b i0 a i b i3 a i3 b i a i0 b i0 + a i1 b i1 + a i b i a i3 b i3 k ( a i0 b i1 a i1 b i0 a i b i3 + a i3 b i ) i=1 k N = (n i ) 3 1 = ( a i0 b i + a i1 b i3 a i b i0 a i3 b i1 ), i=1 k ( a i0 b i3 a i1 b i + a i b i1 a i3 b i0 ) i=1 15

24 . General Quaternion Quadratic Equations and k (a i0 b i0 a i1 b i1 a i b i a i3 b i3 ) i=1 k ( a i0 b i1 a i1 b i0 + a i b i3 a i3 b i ) D = (d i ) 4 1 = i=1. k ( a i0 b i a i1 b i3 a i b i0 + a i3 b i1 ) i=1 k ( a i0 b i3 + a i1 b i a i b i1 a i3 b i0 ) i=1 Treating x 0 as a constant, let ˇM be the coefficient matrix of (1), () and (3) of.5: x 0 + m 11 m 1 m 13 m 1 x 0 + m m 3. m 31 m 3 x 0 + m 33 Direct calculation yields: σ (x 0 ) = det ˇM =8x (m 11 + m + m 33 )x 0 + (m 11 m + m 11 m 33 m 1 m 1 m 13 m 31 m 3 m 3 + m m 33 ) x 0 + m 11 m m 33 m 11 m 3 m 3 m 1 m 1 m 33 + m 13 m 1 m 3 + m 1 m 3 m 31 m 13 m m 31. We continue the discussion by cases: Case 1. We assume that det ˇM 0. Applying Cramer s rule to (1), () and (3) of.5, we obtain x 1, x and x 3 as rational polynomials of x 0 as follows: x 1 = p 1 (x 0 ) σ (x 0 ), (.6) x = p (x 0 ) σ (x 0 ) (.7) 16

25 . General Quaternion Quadratic Equations and where x 3 = p 3 (x 0 ) σ (x 0 ), (.8) p 1 (x 0 ) =4n 1 x (n 1 (m + m 33 ) n m 1 n 3 m 13 c 1 ) x 0 + (m 1 c + m 13 c 3 c 1 (m + m 33 )) x 0 + n 1 (m m 33 m 3 m 3 ) x 0 + n (m 13 m 3 m 1 m 33 ) x 0 + n 3 (m 1 m 3 m 13 m ) x 0 + c 1 (m 3 m 3 m 33 m ) + c (m 1 m 33 m 13 m 3 ) + c 3 (m 13 m m 1 m 3 ), p (x 0 ) =4n x (n (m 11 + m 33 ) n 1 m 1 n 3 m 3 c ) x 0 + (m 1 c 1 + m 3 c 3 c (m 11 + m 33 )) x 0 + n 1 (m 3 m 31 m 1 m 33 ) x 0 + n (m 11 m 33 m 13 m 31 ) x 0 + n 3 (m 13 m 1 m 11 m 3 ) x 0 + c 1 (m 1 m 33 m 3 m 31 ) + c (m 13 m 31 m 11 m 33 ) + c 3 (m 11 m 3 m 13 m 1 ) and p 3 (x 0 ) =4n 3 x (n 3 (m 11 + m ) n 1 m 31 n m 3 c 3 ) x 0 + (m 31 c 1 + m 3 c c 3 (m 11 + m )) x 0 + n 1 (m 1 m 3 m m 31 ) x 0 + n (m 1 m 31 m 11 m 3 ) x 0 + n 3 (m 11 m m 1 m 1 ) x 0 + c 1 (m m 31 m 1 m 3 ) + c (m 11 m 3 m 1 m 31 ) + c 3 (m 1 m 1 m 11 m ). 17

26 . General Quaternion Quadratic Equations Substituting.6,.7 and.8 back into (4) of.5, we obtain: 0 =x 0 + d 1 x 0 + c 0 p 1 (x 0 ) + p (x 0 ) + p 3 (x 0 ) σ (x 0 ) + d p 1 (x 0 ) + d 3 p (x 0 ) + d 4 p 3 (x 0 ). σ (x 0 ) Multiplication of the above equation by σ (x 0 ) yields: 0 =σ (x 0 ) ( x 0 + d 1 x 0 + c 0 ) p1 (x 0 ) p (x 0 ) p 3 (x 0 ) (.9) + σ (x 0 ) (d p 1 (x 0 ) + d 3 p (x 0 ) + d 4 p 3 (x 0 ))..9 is an 8 th order equation of x 0. After solving.9 for a real solution x 0, we substitute its value back into.6,.7 and.8 to get x 1, x and x 3 respectively to obtain a solution to.4. The prime difficulty here is that, by Galois theory, there is no formula solution for general eighth degree equations (Morandi, 1996). Furthermore, there does not seem to be a general formula to factor.9. However, as we will show in the next section, under certain conditions,.9 can be solved. Case. We assume that det ˇM = 0. Then x 0 is known: it is equal to a (fixed) real root (which always exists) of the cubic polynomial σ (x 0 ). Let Ň be the 3 1 vector: c 1 + x 0 n 1 c + x 0 n. c 3 + x 0 n 3 It is clear that.5 has a solution only if the last row of the reduced row echelon form of the augmented matrix ( ˇM Ň ) is a zero row. Under these conditions, (1), () and (3) of.5, now viewed as a linear system 18

27 . General Quaternion Quadratic Equations with unknowns x 1, x and x 3, has infinitely many solutions of the form: x 1 = s 1 y + t 1 z + w 1, x = s y + t z + w, x 3 = s 3 y + t 3 z + w 3, (.10) where s i, t i and w i R, 1 i 3, and y, z R are parameters. We plug.10 into (4) of.5 and obtain: 0 = (s 1 y + t 1 z + w 1 ) (s y + t z + w ) (s 3 y + t 3 z + w 3 ) + d (s 1 y + t 1 z + w 1 ) + d 3 (s y + t z + w ) + d 4 (s 3 y + t 3 z + w 3 ) (.11) + x 0 + d 1 x 0 + c 0. The parameters y and z must satisfy.11, therefore.11 can be seen as a quadratic equation of, WLOG, y: y + f (z) y + g (z) = 0, (.1) where f (z) is a linear polynomial of z and g (z) is a quadratic polynomial of z. We can take any z R such that the discriminant of.1, f (z) 4g (z), a quadratic polynomial of z, is non-negative. We then solve for y = f(z)± f(z) 4g(z). Finally, we substitute the value of y and z back into to.10 to get x 1, x and x 3 to obtain a solution to.4. 19

28 .3 Special Quaternion Equations and problems.3. Special Quaternion Equations and problems In this section, we give solutions to some specific quaternion equations and least norm problems A quadratic equation In this subsection, we consider the quadratic equation that was studied and solved in (Au-Yeung, 003): t + αt + tβ + γ = 0, (.13) where α, β and γ H and t is a quaternion indeterminate. Here I present a somewhat simpler solution. Let x = t + α 0+β 0. Then.13 becomes: 0 = ( x α 0 + β 0 ) + α ( x α 0 + β 0 ) + ( x α 0 + β 0 ) β + γ = x + ax + xb + c, (.14) where a = Im α, b = Im β and c = ( ) α 0 +β 0 (α + β) α 0 +β 0 + γ. The advantage of.14 over.13 is that the left and right coefficients of the indeterminate are both pure imaginary..14 is equivalent to the real non-linear system: x 0 x 1 + ( a 3 + b 3 ) x + (a b ) x 3 = ( a 1 b 1 ) x 0 c 1, (1) (a 3 b 3 ) x 1 + x 0 x + ( a 1 + b 1 ) x 3 = ( a b ) x 0 c, () ( a + b ) x 1 + (a 1 b 1 ) x + x 0 x 3 = ( a 3 b 3 ) x 0 c 3, (3) (.15) x 1 + x + x 3 + (a 1 + b 1 ) x 1 + (a + b ) x + (a 3 + b 3 ) x 3 = x 0 + c 0. (4) 0

29 .3 Special Quaternion Equations and problems We first show how to solve.15 if a = b. When a = b,.15 becomes: x 0 (x 1 + a 1 ) = c 1, (1) x 0 (x + a ) = c, () x 0 (x 3 + a 3 ) = c 3, (3) (.16) x 1 + x + x 3 + a 1 x 1 + a x + a 3 x 3 = x 0 + c 0. (4) Case 1. x 1 = a 1, If c = c 0 R, then.16 can have solutions only if x 0 = 0 or x = a, is part of the solution. x 3 = a 3, 1. When x 0 = 0 is part of the solution, (4) of.16 becomes: x 1 + x + x 3 + a 1 x 1 + a x + a 3 x 3 = c 3 3 (x i + a i ) = c + a i = c a. (.17) i=1 i=1.17 can only hold if c a 0. In which case, x 1, x and x 3 can be any real numbers that satisfies.17. x 1 = a 1,. When x = a, is part of the solution, (4) of.16 becomes: x 3 = a 3, a 1 a a 3 = x 0 + c a c = x 0. 1

30 .3 Special Quaternion Equations and problems The above equation can only hold if and a c 0. In which case, x 0 = ± a c. Case. If c R, then x 0 = 0 can never be part of the solution. We obtain, using the method demonstrated in., a quartic equation for x 0 : x ( ) c 0 + a 1 + a + a 3 x 0 1 ( ) c c + c 3 = 0. Replacing x 0 by y, we obtain a quadratic equation for y: f (y) = y + ( ) c 0 + a 1 + a + a 1 ( ) 3 y c c + c 3 = 0. The solutions of f are: y = ± (c 0 + a 1 + a + a 3) + (c 1 + c + c 3) (c 0 + a 1 + a + a 3). Since c R, c 1 + c + c 3 > 0, and therefore (c 0 + a 1 + a + a 3) + (c 1 + c + c 3) (c 0 + a 1 + a + a 3) > 0 and (c 0 + a 1 + a + a 3) + (c 1 + c + c 3) (c 0 + a 1 + a + a 3) < 0. So (c0 + a 1 x 0 = ± + a + ( 3) a + c 1 + c + ( 3) c c0 + a 1 + a + ) a 3 (c 0 a ) (Im c) (c 0 a ) =,

31 .3 Special Quaternion Equations and problems and x 1 = c 1 x 0 a 1, x = c x 0 a, x 3 = c 3 x 0 a 3. From now on, we can and we will assume that a b. The determinant of the coefficient matrix of (1), () and (3) of.15, ˇM = x 0 a 3 + b 3 a b a 3 b 3 x 0 a 1 + b 1 a + b a 1 b 1 x 0 σ (x 0 ) = det ˇM =8x ( (a 1 b 1 ) + (a b ) + (a 3 b 3 ) ) x 0 =x 0 ( 4x 0 + (a 1 b 1 ) + (a b ) + (a 3 b 3 ) )..15 can have infinitely many solutions only if σ (x 0 ) = 0, this would mean that x 0 = 0. First, when x 0 = 0, (4) of.16 becomes x 1 + x + x 3 + a 1 x 1 + a x + a 3 x 3 = c 3 3 (x i + a i ) = c + a i = c a. i=1 i=1 The above equation can only hold if c a 0. Next, when x 0 = 0, the augmented matrix of (1), () and (3) of.15 is 0 a 3 + b 3 a b c 1 A = a 3 b 3 0 a 1 + b 1 c. a + b a 1 b 1 0 c 3 3

32 .3 Special Quaternion Equations and problems We will show that A is the augmented matrix of a consistent linear system if and only if 3 c i (a i b i ) = 0. i=1 We will show the forward direction first. Case 1. If a i b i 0 for 1 i 3. We apply Gaussian elimination to A and obtain ( =. means row equivalent) : a 3 b 3 0 a 1 + b 1 c A =. 0 a 3 + b 3 a b c i=1 c i(a i b i ) a 3 b 3 Since A is the augmented matrix of a consistent linear system, 3i=1 c i (a i b i ) = 0.. Case. If a 1 b 1 = 0, a i b i 0 for i =, 3, then 0 a 3 + b 3 a b c 1 A = a 3 b c. a + b 0 0 c 3 Since A is the augmented matrix of a consistent linear system, the last two rows give: (a 3 b 3 ) x 1 = c ( a + b ) x 1 = c 3 (a b ) c + (a 3 b 3 ) c 3 = 0 3 c i (a i b i ) = 0. i=1 4

33 .3 Special Quaternion Equations and problems Similarly, we can show the same result when a b = 0, a i b i 0 for i = 1, 3 and a 3 b 3 = 0, a i b i 0 for i = 1,. Case 3. If a 1 b 1 0, a i b i = 0 for i =, 3, then c 1 A = 0 0 a 1 + b 1 c. 0 a 1 b 1 0 c 3 Since A is the augmented matrix of a consistent linear system, the first row gives c 1 = 0, and therefore 3 i=1 c i (a i b i ) = 0. Similarly, we can show the same result when a b 0, a i b i = 0 for i = 1, 3 and a 3 b 3 0, a i b i = 0 for i = 1,. For the converse, we assume that 3 i=1 c i (a i b i ) = 0. Case 1. If a 3 b 3 0, then we apply Gaussian elimination to A to obtain: a 3 b 3 0 a 1 + b 1 c A =. 0 a 3 + b 3 a b c 1 3 i= c i(a i b i ) a 3 b 3 a 3 b 3 0 a 1 + b 1 c = 0 a 3 + b 3 a b c A is always the augmented matrix of a consistent linear system as it has the solution set x 1 = c +(a 1 b 1 )s a 3 b 3, x = c 1+(a b )s a 3 b 3, x 3 = s, where s is a parameter. In order for.15 to have a solution, we must 5

34 .3 Special Quaternion Equations and problems have c a 0 and s must be a solution of ( ) c + (a 1 b 1 ) s ( ) c1 + (a b ) s 0 = + + s c 0 a 3 b 3 a 3 b 3 ( ) ( ) c + (a 1 b 1 ) s c1 + (a b ) s + (a 1 + b 1 ) + (a + b ) + (a 3 + b 3 ) s a 3 b 3 a 3 b 3 ( 3 ) 0 = (a i b i ) s + O (s), (.18) i=1 where O (s) is a complicated linear polynomial of s. The right hand side of.18 is not identically equal to zero as the coefficient of s is nonzero, and thus it can have at most solutions and therefore.15 can have at most solutions. Case. If a 3 b 3 = 0 and a b = 0, then a 1 b 1 0 because a b. 3i=1 c i (a i b i ) = 0 implies that c 1 = 0 and we have: A = 0 0 a 1 + b 1 c. 0 a 1 b 1 0 c 3 This system is always consistent as it has the solution set x 1 = s, x = c a 1 b 1, x 3 = c 3 a 1 b 1, where s is a parameter. In order for.15 to have a solution, we must 6

35 .3 Special Quaternion Equations and problems have c a 0 and s must be a solution of 0 =s + ( c a 1 b 1 ) + ( c3 a 1 b 1 + (a 1 + b 1 ) s + (a + b ) ) c0 ( c a 1 b 1 ) ( ) c3 + (a 3 + b 3 ) a 1 b 1 0 = (a 1 b 1 ) s + O (s), (.19) where O (s) is a complicated linear polynomial of s. The right hand side of.19 is not identically equal to zero as the coefficient of s is nonzero, and therefore it can have at most solutions and therefore.15 can have at most solutions. Case 3. If a 3 b 3 = 0 and a b 0, then 0 0 a b c 1 A = 0 0 a 1 + b 1 c. a + b a 1 b 1 0 c 3 3i=1 c i (a i b i ) = 0 implies that c (a b ) = c 1 (a 1 b 1 ) and therefore the first two rows are consistent. Furthermore, this system is always consistent as it has the solution set x 1 = c 3+(a 1 b 1 )s a b, x = s, x 3 = c 1 a b, where s is a parameter. In order for.15 to have a solution, we must 7

36 .3 Special Quaternion Equations and problems have c a 0 and s must be a solution of ( ) c3 + (a 1 b 1 ) s ( ) 0 = + s c1 + c 0 a b a b ( ) ( ) c3 + (a 1 b 1 ) s c1 + (a 1 + b 1 ) + (a + b ) s + (a 3 + b 3 ) a b a b 0 = ( (a 1 b 1 ) + (a b ) ) s + O (s), (.0) where O (s) is a complicated linear polynomial of s. The right hand side of.0 is not identically equal to zero as the coefficient of s is nonzero, and therefore it can have at most solutions and therefore.15 can have at most solutions. This concludes our discussion for possible scenarios where.15 can have infinitely many solutions. When x 0 0, we have σ (x 0 ) 0. Applying Cramer s rule to (1), () and (3) of.15, we obtain x 1, x and x 3 as rational polynomials of x 0 as follows: x 1 = p 1 (x 0 ) σ (x 0 ), (.1) x = p (x 0 ) σ (x 0 ) (.) and x 3 = p 3 (x 0 ) σ (x 0 ), (.3) 8

37 .3 Special Quaternion Equations and problems where p 1 (x 0 ) = 4 (a 1 + b 1 ) x (a b 3 a 3 b c 1 ) x 0 + (b 1 a 1 ) ( a 1 + a + a 3 b 1 b b 3) x0 + (c (b 3 a 3 ) + c 3 (a b )) x 0 + (b 1 a 1 ) (c 1 (a 1 b 1 ) + c (a b ) + c 3 (a 3 b 3 )), p (x 0 ) = 4 (a + b ) x (a 3 b 1 a 1 b 3 c ) x 0 + (b a ) ( a 1 + a + a 3 b 1 b b 3) x0 + (c 1 (a 3 b 3 ) + c 3 (b 1 a 1 )) x 0 + (b a ) (c 1 (a 1 b 1 ) + c (a b ) + c 3 (a 3 b 3 )) and p 3 (x 0 ) = 4 (a 3 + b 3 ) x (a 1 b a b 1 c 3 ) x 0 + (b 3 a 3 ) ( a 1 + a + a 3 b 1 b b 3) x0 + (c 1 (b a ) + c (a 1 b 1 )) x 0 + (b 3 a 3 ) (c 1 (a 1 b 1 ) + c (a b ) + c 3 (a 3 b 3 )). Substituting.1,.3 and.3 back into (4) of.15, we obtain: 0 = x 0 c 0 + p 1 (x 0 ) + p (x 0 ) + p 3 (x 0 ) σ (x 0 ) + (a 1 + b 1 ) p 1 (x 0 ) + (a + b ) p (x 0 ) + (a 3 + b 3 ) p 3 (x 0 ). σ (x 0 ) 9

38 .3 Special Quaternion Equations and problems Multiplication of the above equation by σ (x 0 ) yields a 6 th degree equation of x 0 : 0 =16x ( a 1 + a + a 3 + b 1 + b + b 3 + c 0 ) x c 0 ( (a1 b 1 ) + (a b ) + (a 3 b 3 ) ) x 0 + ( a 1 + a + a 3 b 1 b b 3) x 0 + (8a b 3 8a 3 b 4c 1 ) c 1 x 0 + (8a 3 b 1 8a 1 b 3 4c ) c x 0 + (8a 1 b 8a b 1 4c 3 ) c 3 x 0 ( 3 c i (a i b i )). i=1 Replacing x 0 by y, we obtain a cubic equation of y: 0 =f (y) =16y ( a 1 + a + a 3 + b 1 + b + b 3 + c 0 ) y + 4c 0 ( (a1 b 1 ) + (a b ) + (a 3 b 3 ) ) y + ( a 1 + a + a 3 b 1 b b 3) y + ( (8a b 3 8a 3 b ) c 1 4c 1) y + ( (8a 3 b 1 8a 1 b 3 ) c 4c ) y + ( (8a 1 b 8a b 1 ) c 3 4c3) y ( 3 ) c i (a i b i ) i=1 =16y 3 + D y + D 1 y + D 0. (.4) We next show that f has exactly one strictly positive solution. Since x 0 0, D 0 = ( 3i=1 c i (a i b i ) ) < 0. By Descartes rule of signs, f has at least one strictly positive solution. Furthermore, f can have multiple positive solutions only if D < 0 and D 1 > 0. We show that this is impossible. 30

39 .3 Special Quaternion Equations and problems If D < 0, then c 0 < a 1 +a +a 3 +b 1 +b +b 3. Therefore, 4c 0 ( (a1 b 1 ) + (a b ) + (a 3 b 3 ) ) + ( a 1 + a + a 3 b 1 b b 3 + (a b 3 a 3 b ) + (a 3 b 1 a 1 b 3 ) + (a 1 b a b 1 ) < ( a 1 + a + a 3 + b 1 + b + b 3 + ( a 1 + a + a 3 b 1 b b 3 ) ( (a1 b 1 ) + (a b ) + (a 3 b 3 ) ) + (a b 3 a 3 b ) + (a 3 b 1 a 1 b 3 ) + (a 1 b a b 1 ) = ( (a 1 b 1 ) + (a b ) + (a 3 b 3 ) ) < 0. This gives: ) ) 4c 0 ((a 1 b 1 ) + (a b ) + (a 3 b 3 ) ) ( ) + a 1 + a + a 3 b 1 b b 3 < (a b 3 a 3 b ) (a 3 b 1 a 1 b 3 ) (a 1 b a b 1 ). Then D 1 can be simplified as follows: D 1 =4c 0 ((a 1 b 1 ) + (a b ) + (a 3 b 3 ) ) ( ) + a 1 + a + a 3 b 1 b b 3 + (8a b 3 8a 3 b ) c 1 4c 1 + (8a 3 b 1 8a 1 b 3 ) c 4c + (8a 1 b 8a b 1 ) c 3 4c 3 < (a b 3 a 3 b ) (a 3 b 1 a 1 b 3 ) (a 1 b a b 1 ) + (8a b 3 8a 3 b ) c 1 4c 1 + (8a 3 b 1 8a 1 b 3 ) c 4c + (8a 1 b 8a b 1 ) c 3 4c 3 = (a b 3 a 3 b c 1 ) (a 3 b 1 a 1 b 3 c ) (a 1 b a b 1 c 3 ) 0. Therefore, f has exactly one strictly positive solution z. We can plug x 0 = ± z into.1,.3 and.3 to get x 1, x and x 3 respectively. In all of the above discussion, solutions to.13 can be obtained by calculating 31

40 .3 Special Quaternion Equations and problems t 0 = x 0 α 0+β 0, t = x α 0+β 0 = ( ) x 0 α 0+β 0 + x1 i + x j + x 3 k, that is, t 1 = x 1, t = x, for each x. t 3 = x 3, We are now able to give a theorem that sums up our result so far in this subsection. Theorem.3.1. The solutions of the quadratic equation t + αt + tβ + γ = 0, (.5) where α, β and γ H and t is a quaternion indeterminate, can be obtained by formulas according to the following cases, let a = Im α, b = Im β and c = ( ) α 0 +β 0 (α + β) α 0+β 0 + γ. In the following, we call t 0 = s α 0+β 0, t 1 = p 1( s) σ( s), t = p ( s) σ( s), t 3 = p 3( s) σ( s), and t 0 = s α 0+β 0, t 1 = p 1( s) σ( s), t = p ( s) σ( s), t 3 = p 3( s) σ( s), where s is the unique positive root of.4, the standard solutions. 1. a = b. a) c R. i. c a > 0. t 0 = α 0+β 0 and (t 1, t, t 3 ) can be any real number triple that satisfies the equation 3 i=1 (t i + a i ) = c a. This is the only case where.5 has infinitely many solutions. 3

41 .3 Special Quaternion Equations and problems ii. a c 0. t 0 = ± a c α 0+β 0, t 1 = a 1, t = a, t 3 = a 3. b) c / R. t 0 = s α 0+β 0, t 1 = a 1 c 1, s t = a c s, or t 3 = a c, s t 0 = s α 0+β 0, where = (c 0 a ) (Im c) (c 0 a ). t 1 = a 1 + c 1, s t = a + c s, t 3 = a + c, s. a b. Standard solutions always exist. Furthermore, a) 3 i=1 c i (a i b i ) = 0 and c a 0. 33

42 .3 Special Quaternion Equations and problems i. a 3 b 3. Possible solutions must be of the form t 0 = α 0+β 0, t 1 = c +(a 1 b 1 )s a 3 b 3, t = c 1+(a b )s a 3 b 3, t 3 = s, where s is a real number that satisfies the quadratic equation.18 ii. a 3 = b 3 and a = b. Possible solutions must be of the form t 0 = α 0+β 0, t 1 = s, t = c a 1 b 1, t 3 = c 3 a 1 b 1, where s is a real number that satisfies the quadratic equation.19 iii. a 3 = b 3 and a b. Possible solutions must be of the form t 0 = α 0+β 0, t 1 = c 3+(a 1 b 1 )s a b, t = s, t 3 = c 1 a b, where s is a real number that satisfies the quadratic equation.0 b) Otherwise. No additional solutions exist. Example.3.. Find all the solutions of t + (1 i j k) t + t (1 + i 3j + k) + ( 1 3i + j k) = 0. (.6) 34

43 .3 Special Quaternion Equations and problems The following steps have been implemented into a single procedure in Maple. For details, see the AY procedure of A.. Let x = t + 1. Then.6 becomes x + ( i j k) x + x (i 3j + k) + ( i + 6j k) = 0. (.7) We have 3 c i (a i b i ) = ( ) ( 1) + 6 ( + 3) + ( 1) ( 1) = i=1 By Theorem.3.1, there are only standard solutions and we obtain a cubic polynomial of y = x 0: f (y) = 16y y 51y 5. f (y) has one positive root y = 13 ( ( sin 6 3 arctan ) ) π 19 6 = s. The two solutions to.6 are ( ) p 1 ( s) s 1 + σ ( s) i + p ( s) σ ( s) j + p 3 ( s) σ ( k i k s) and ( ) p 1 ( s) s 1 + σ ( s) i + p ( s) σ ( s) j + p 3 ( s) σ ( k i j 1.93k s) These answers are verified to be correct by direct claculation. 35

44 .3 Special Quaternion Equations and problems.3.. Another quadratic equation In this subsection, we consider another quadratic equation: x + axb + c = 0, (.8) where a and b H with a 0 = b 0 = 0, c R and x is a quaternion indeterminate. We assume that ab 0. Otherwise,.8 is reduced to x + c = 0, which was studied and solved in (Niven, 194). We also assume that c 0. Otherwise,.8 is reduced to x + axb = 0, which we will discuss in is equivalent to the real non-linear system: (x 0 + m 11 ) x 1 + m 1 x + m 13 x 3 = ( a b 3 + a 3 b ) x 0, (1) m 1 x 1 + (x 0 + m ) x + m 3 x 3 = (a 1 b 3 a 3 b 1 ) x 0, () m 31 x 1 + m 3 x + (x 0 + m 33 ) x 3 = ( a 1 b + a b 1 ) x 0, (3) (.9) x 0 x 1 x x 3 + d 1 x 0 + d x 1 + d 3 x + d 4 x 3 + c = 0, (4) where m ij and d i are given by: M = (m ij ) 3 3 a 1 b 1 + a b + a 3 b 3 a 1 b a b 1 a 1 b 3 a 3 b 1 = a 1 b a b 1 a 1 b 1 a b + a 3 b 3 a b 3 a 3 b, a 1 b 3 a 3 b 1 a b 3 a 3 b a 1 b 1 + a b a 3 b 3 and a 1 b 1 a b a 3 b 3 a b 3 a 3 b D = (d i ) 4 1 =. a 1 b 3 + a 3 b 1 a 1 b a b 1 We already discussed in. how to solve.8 if the determinant of the coefficient 36

45 .3 Special Quaternion Equations and problems x 0 + m 11 m 1 m 13 matrix, ˇM = m 1 x 0 + m m 3, is zero. We can and will assume m 31 m 3 x 0 + m 33 that det ˇM 0. σ (x 0 ) = det ˇM = (b 1 a 1 + a b + a 3 b 3 + x 0 ) ( 4x 0 ( a 1 + a + a 3) ( b 1 + b + b 3)) 0, therefore, 4x 0 (a 1 + a + a 3) (b 1 + b + b 3) 0. Applying Cramer s rule to (1), () and (3) of.9, we obtain x 1 as a rational polynomial of x 0, Similarly, and [x 0 (a 3 b a b 3 )] (4x 0 (a 1 + a + a x 1 = 3) (b 1 + b + b 3)) (b 1 a 1 + a b + a 3 b 3 + x 0 ) (4x 0 (a 1 + a + a 3) (b 1 + b + b 3)) x 0 (a 3 b a b 3 ) = = p 1 (x 0 ). (.30) b 1 a 1 + a b + a 3 b 3 + x 0 x = x 3 = x 0 (a 1 b 3 a 3 b 1 ) b 1 a 1 + a b + a 3 b 3 + x 0 = p (x 0 ) (.31) x 0 (a b 1 a 1 b ) b 1 a 1 + a b + a 3 b 3 + x 0 = p 3 (x 0 ). (.3) Substituting.30,.31 and.3 back into (4) of.9, then multiplying through by (b 1 a 1 + a b + a 3 b 3 + x 0 ), we obtain a quartic equation of x 0, f (x 0 ) = x (c 3 4 α ) x β (4c α) x cβ = 0, (.33) where α = ( a 1 + a + a 3) ( b 1 + b + b 3 ) and β = a 1 b 1 + a b + a 3 b 3. 37

46 .3 Special Quaternion Equations and problems The discriminant of f is f = 1 56 β (α β ) ( (α + 4c) (3α 4c) 3 104c 3 β ). Since α β 0 by the Cauchy Schwartz inequality, f < 0 if an only if (α + 4c) (3α 4c) 3 104c 3 β < 0. Once we solve f for x 0, we substitute its value back into.30,.31 and.3 to get x 1, x and x 3 respectively to obtain a solution to.8. Example.3.3. Find all the solutions of x + ( i 3j + k) x ( i + j k) + 1 = 0. (.34) The following steps have been implemented into a single procedure in Maple. For details, see the AXBc procedure of A...34 is equivalent to (x 0 7) x 1 5x + 3x 3 = x 0, (1) 5x 1 + (x 0 + 3) x 5x 3 = 5x 0, () 3x 1 5x + (x 0 + 1) x 3 = 7x 0, (3) (.35) x 1 + x + x 3 x 1 + 5x + 7x 3 = x (4) x When the determinant of the coefficient matrix, 5 x , is 0, 3 5 x that is, when σ (x 0 ) = 8x 3 0 1x 0 168x = 0,.35 has no solutions because x x 0 the last row of the reduced row echelon form of 5 x x x x 0 is not a zero row. Otherwise, by.33, we obtain a quartic polynomial of x 0, f (x 0 ) = x 4 0 4x f (x 0 ) has four positive roots x 0 = ± 1 84 ± Therefore, by.30,.31 and 38

47 .3 Special Quaternion Equations and problems.3, the four solutions to.34 are p 1 i + p j + p 3 k i + 3.7j k, p 1 i + p j + p 3 k i 6.34j 8.87k, i +.0j +.83k, + p i + p j + p k and i j k + p i + p j + p k These answers are verified to be correct by direct claculation. 39

48 .3 Special Quaternion Equations and problems.3.3. A homogeneous quadratic equation In this subsection, we consider the currently unsolved homogeneous quadratic equation: x + axb = 0, (.36) where a and b H and x is the quaternion unknown. We assume that a / R and b / R. Otherwise,.36 is reduced to x (x + ab) = 0 or (x + ab) x = is equivalent to the real non-linear system: (x 0 + m 11 ) x 1 + m 1 x + m 13 x 3 = x 0 n 1, (1) m 1 x 1 + (x 0 + m ) x + m 3 x 3 = x 0 n, () m 31 x 1 + m 3 x + (x 0 + m 33 ) x 3 = x 0 n 3, (3) (.37) x 0 x 1 x x 3 + d 1 x 0 + d x 1 + d 3 x + d 4 x 3 + c 0 = 0, (4) where m ij, n i and d i are given by: M = (m ij ) 3 3 a 0 b 0 a 1 b 1 + a b + a 3 b 3 a 0 b 3 a 1 b a b 1 a 3 b 0 a 0 b a 1 b 3 + a b 0 a 3 b 1 = a 0 b 3 a 1 b a b 1 + a 3 b 0 a 0 b 0 + a 1 b 1 a b + a 3 b 3 a 0 b 1 a 1 b 0 a b 3 a 3 b, a 0 b a 1 b 3 a b 0 a 3 b 1 a 0 b 1 + a 1 b 0 a b 3 a 3 b a 0 b 0 + a 1 b 1 + a b a 3 b 3 a 0 b 1 a 1 b 0 a b 3 + a 3 b N = (n i ) 3 1 = a 0 b + a 1 b 3 a b 0 a 3 b 1, a 0 b 3 a 1 b + a b 1 a 3 b 0 40

49 .3 Special Quaternion Equations and problems and a 0 b 0 a 1 b 1 a b a 3 b 3 a 0 b 1 a 1 b 0 + a b 3 a 3 b D = (d i ) 4 1 =. a 0 b a 1 b 3 a b 0 + a 3 b 1 a 0 b 3 + a 1 b a b 1 a 3 b 0 We already showed in. how to solve such an equation when the determinant of x 0 + m 11 m 1 m 13 the coefficient matrix, ˇM = m 1 x 0 + m m 3, is zero. We can m 31 m 3 x 0 + m 33 and will assume that det ˇM 0. Direct calculation yields: σ (x 0 ) = det ˇM =8x (m 11 + m + m 33 )x 0 + (m 11 m + m 11 m 33 m 1 m 1 m 13 m 31 m 3 m 3 + m m 33 ) x 0 + m 11 m m 33 m 11 m 3 m 3 m 1 m 1 m 33 + m 13 m 1 m 3 + m 1 m 3 m 31 m 13 m m 31. Applying Cramer s rule to (1), () and (3) of.37, we obtain x 1, x and x 3 as rational polynomials of x 0 as follows: x 1 = p 1 (x 0 ) σ (x 0 ), (.38) x = p (x 0 ) σ (x 0 ) (.39) and x 3 = p 3 (x 0 ) σ (x 0 ), (.40) 41

50 .3 Special Quaternion Equations and problems where p 1 (x 0 ) =4n 1 x (n 1 (m + m 33 ) n m 1 n 3 m 13 ) x 0 + n 1 (m m 33 m 3 m 3 ) x 0 + n (m 13 m 3 m 1 m 33 ) x 0 + n 3 (m 1 m 3 m 13 m ) x 0, p (x 0 ) =4n x (n (m 11 + m 33 ) n 1 m 1 n 3 m 3 ) x 0 + n 1 (m 3 m 31 m 1 m 33 ) x 0 + n (m 11 m 33 m 13 m 31 ) x 0 + n 3 (m 13 m 1 m 11 m 3 ) x 0 and p 3 (x 0 ) =4n 3 x (n 3 (m 11 + m ) n 1 m 31 n m 3 ) x 0 + n 1 (m 1 m 3 m m 31 ) x 0 + n (m 1 m 31 m 11 m 3 ) x 0 + n 3 (m 11 m m 1 m 1 ) x 0. Substituting.38,.39 and.40 back into (4) of.37, we obtain: 0 =x 0 + d 1 x 0 p 1 (x 0 ) + p (x 0 ) + p 3 (x 0 ) σ (x 0 ) + d p 1 (x 0 ) + d 3 p (x 0 ) + d 4 p 3 (x 0 ), σ (x 0 ) Multiplication of the above equation by σ (x 0 ) yields an 8 th degree polynomial of 4

51 .3 Special Quaternion Equations and problems x 0, 0 =f (x 0 ) (.41) =σ (x 0 ) ( x 0 + d 1 x 0 ) d 1 x 0 p (x 0 ) p 3 (x 0 ) + σ (x 0 ) (d p 1 (x 0 ) + d 3 p (x 0 ) + d 4 p 3 (x 0 ))..41 can be factored into the product of x 0, a cubic and a quartic polynomial of x 0, f (x 0 ) = x 0 h (x 0 ) g (x 0 ), where the cubic polynomial is h (x 0 ) =x a 0 b 0 x 0 (.4) ( a 0b 0 1 ( ) ( ) ) a 1 0 3a 1 3a 3a 3 b 0 3b 1 3b 3b 3 x B (a 0b 0 a 1 b 1 a b a 3 b 3 ), and the quartic polynomial is g (x 0 ) = x a 0 b 0 x Ax a 0b 0 Bx B, (.43) with A = 4a 0b 0 ( a 1 + a + a 3 a 0) ( b 1 + b + b 3 b 0 ) and B = ( a 0 + a 1 + a + a 3) ( b 0 + b 1 + b + b 3). Furthermore, g (x 0 ) can be factored in R as g (x 0 ) = g 1 (x 0 ) g (x 0 ), 43

52 .3 Special Quaternion Equations and problems where g 1 (x 0 ) = x 0 + sx B and g (x 0 ) = x 0 + tx B, with s = a 0 b 0 + (b 1 + b + b 3) (a 1 + a + a 3) and t = a 0 b 0 (b 1 + b + b 3) (a 1 + a + a 3). The discriminants of g 1 and g are g1 = b 3a 0 a 1b 0 a b 0 a 3b 0 b a 0 a 0b 1 + b 0 a 0 (b 1 + b + b 3) (a 1 + a + a 3) = (a 0 (b 1 + b + b 3) b 0 (a 1 + a + a3)) 0 and g = b 3a 0 a 1b 0 a b 0 a 3b 0 b a 0 a 0b 1 b 0 a 0 (b 1 + b + b 3) (a 1 + a + a 3) = (a 0 (b 1 + b + b 3) + b 0 (a 1 + a + a3)) 0. We continue the discussion by cases: Case 1. When a 0 = b 0 = 0. 44

53 .3 Special Quaternion Equations and problems Then h (x 0 ), g (x 0 ) and σ (x 0 ) can be simplified to: h (x 0 ) =x ( ) ( ) 3a a + 3a 3 3b 1 + 3b + 3b 3 x0 (.44) 1 ( ) ( ) a a + a 3 b 1 + b + b 3 (a1 b 1 + a b + a 3 b 3 ), g (x 0 ) = ( x 0 1 ( ) ( a a + a 3 b 1 + b + b3) ) (.45) and ( σ (x 0 ) = 4 (x 0 + a 1 b 1 + a b + a 3 b 3 ) x 0 1 ( ) ( ) ) a a + a 3 b 1 + b + b 3. (.46) Note that we already assumed that σ (x 0 ) 0, and therefore.45 does not contribute any roots. We next show that.44 always has three real solutions. The discriminant of h is h = 7 ( ) a a + a ( 3 b 1 + b + b3) 0. ( (a 1 b a b 1 ) + (a 1 b 3 a 3 b 1 ) + (a b 3 a 3 b ) ) So the solution to f (x 0 ) are 0 and the 3 real roots of.44. Case. Otherwise. g1 < 0 unless a 0 b 0 = (a 1 +a +a 3) (b 1 +b +b3), that is, a 0 = cb 0 and a 1 + a + a 3 = c (b 1 + b + b 3) for some 0 < c. Under these conditions, g < 0 and therefore g does not contribute any solutions. However, g 1 can be simplified to x 0 + c ( b 0 + b 1 + b + b 3) x c ( b 0 + b 1 + b + b 3). (.47) Similarly, g < 0 unless a (a 1 0 b 0 = +a +a 3) = c for some c < 0. Under (b1 +b +b3) these conditions, g1 < 0 and therefore g 1 does not contribute any solu- 45

54 .3 Special Quaternion Equations and problems tions. However g can also be simplified to.47. So the roots of f (x 0 ) are 0, the real root(s) of.4 and a 0(b 0 +b 1 +b +b 3) b 0 and is contributed by.47. which has multiplicity Once we obtain x 0, we substitute its value back into.38,.39 and.40 to get x 1, x and x 3 respectively to obtain a solution to.36. Example.3.4. Find all the solutions of x + ( i j + 11k) x (9 + 10i 4j + 10k) = 0. (.48) The following steps have been implemented into a single procedure in Maple. For details, see the AXB procedure of A...48 is equivalent to (x 0 07) x 1 19x 345x 3 = 1x 0, (1) 359x 1 + (x ) x 71x 3 = 7x 0, () 175x x + (x 0 17) x 3 = 141x 0, (3) (.49) x 1 + x + x 3 89x x 41x 3 = x 0 435x 0. (4) When the determinant of the coefficient matrix, x x x 0 17 is 0, that is, when σ (x 0 ) = 8x x x = 0,.49 has no solutions because the last row of the reduced row echelon form of x x x x x x 0 46

55 .3 Special Quaternion Equations and problems is not a zero row. Otherwise, by.4, we obtain a cubic polynomial of x 0, f (x 0 ) = x 3 34x x f (x 0 ) has one real root: x 0 = 3 =δ. ( ) ( ) Therefore, by.38,.39 and.40, the solutions to.48 are 0 and δ + p 1 (δ) σ (δ) i + p (δ) σ (δ) j + p 3 (δ) σ (δ) k i 7.35j k. This answer is verified to be correct by direct claculation. 47

56 .3 Special Quaternion Equations and problems.3.4. A pair of quaternion equations Two nonzero quaternions a and b are said to be semisimilar if the following system has a solution: yax = b, xby = a, (.50) where x and y are nonzero quaternion indeterminates..50 was studied and solved in (Tian, 010). In this subsection, we present a somewhat simpler solution..50 can be written as y = bx 1 a 1, y = b 1 x 1 a, this gives: bx 1 a 1 = b 1 x 1 a (.51) b a 1 = b 1 a a = b. Therefore, a and b must satisfy the following equation for.55 to have solutions, a 0 + a 1 + a + a 3 = b 0 + b 1 + b + b 3. (.5).51 is equivalent to xb = a x, which is equivalent to the real linear homogeneous system: m 11 x 1 + m 1 x + m 13 x 3 + m 14 x 0 = 0, m 1 x 1 + m x + m 3 x 3 + m 4 x 0 = 0, m 31 x 1 + m 3 x + m 33 x 3 + m 34 x 0 = 0, (.53) m 41 x 1 + m 4 x + m 43 x 3 + m 44 x 0 = 0. 48

57 .3 Special Quaternion Equations and problems where m ij is given by the 4 4 skew-symmetric matrix: M = (m ij ) 4 4 d (a 0 a 3 + b 0 b 3 ) (a 0 a + b 0 b ) (a 0 a 1 b 0 b 1 ) (a 0 a 3 + b 0 b 3 ) d (a 0 a 1 + b 0 b 1 ) (a 0 a b 0 b ), (a 0 a + b 0 b ) (a 0 a 1 + b 0 b 1 ) d (a 0 a 3 b 0 b 3 ) (a 0 a 1 b 0 b 1 ) (a 0 a b 0 b ) (a 0 a 3 b 0 b 3 ) d where the diagonal entries are d = a 0 a 1 a a 3 b 0 + b 1 + b + b 3. By.5, d = a 0 b 0 + a 0 b 0 = a 0 b 0. Therefore, M becomes: a 0 b 0 (a 0 a 3 + b 0 b 3 ) a 0 a + b 0 b a 0 a 1 b 0 b 1 a 0 a 3 + b 0 b 3 a 0 b 0 (a 0 a 1 + b 0 b 1 ) a 0 a b 0 b. (a 0 a + b 0 b ) a 0 a 1 + b 0 b 1 a 0 b 0 a 0 a 3 b 0 b 3 (a 0 a 1 b 0 b 1 ) (a 0 a b 0 b ) (a 0 a 3 b 0 b 3 ) a 0 b 0 We continue the discussion by cases. Case 1. If a 0 = b 0 or a 0 = b 0, then det M = 0. Therefore, the homogeneous system.53 has a solution set with an infinite number of solutions (x 0, x 1, x, x 3 ). Each such x gives a solution to.50, x = x, y = bx 1 a 1. Case. Otherwise, we show that.50 has no solutions. Let A = a 0 + a 1 + a + a 3 and B = b 0 + b 1 + b + b 3. Direct calculation yields: det M =a 0b 0 (A B) ( ) A B 4a 0 + 4b 0 + (a 0 + b 0 ) (a 0 b 0 ) (a 0 A b 0 B) (a 0 A + b 0 B). 49

58 .3 Special Quaternion Equations and problems A = B by.5 and therefore, det M =0 + (a 0 + b 0 ) (a 0 b 0 ) A (a 0 b 0 ) A (a 0 + b 0 ) = (a 0 + b 0 ) (a 0 b 0 ) A. Since a 0, A 0 and therefore det M 0. This implies that the homogeneous system.53 has only trivial solutions (x 0, x 1, x, x 3 ) = (0, 0, 0, 0). A contradiction. Therefore,.50 has no solutions. Example.3.5. Find all (x, y) such that the following system holds y ( + 5i + 1j 5k) x = + 4i 3j + 13k, x ( + 4i 3j + 13k) y = + 5i + 1j 5k. (.54) The following steps have been implemented into a single procedure in Maple. For details, see the SemiSim procedure of A.. Since Re ( + 4i 3j + 13k) = = Re ( + 5i + 1j 5k) and + 4i 3j + 13k = ( 3) + 13 = 198 = ( 5) = + 5i + 1j 5k,.54 has infinitely many solutions. First we solve the linear system 16x + 18x 3 x 0 = 0, 16x 1 18x 3 30x 0 = 0, 18x 1 18x 36x 0 = 0, x x 36x 3 = 0. 50

59 .3 Special Quaternion Equations and problems Direct calculation yields: x 0 = 8t + 9t 3, x 1 = 15t + 18t 3, x = t, x 3 = t 3, where t and t 3 are arbitrary real numbers such that x = x 0 + x 1 i + x j + x 3 k 0. Then we have a solution set with an infinite number of solutions of the form: x = x, y = ( + 5i + 1j 5k) x 1 ( + 4i 3j + 13k) 1. For example, when t = and t 3 = 3, we have: x = i + j + 3k, y = 11 1 i 1 j 3 k Direct calculation shows that (x, y) satisfies

60 .3 Special Quaternion Equations and problems.3.5. Another pair of quaternion equations Two nonzero quaternions a and b are said to be consemisimilar if the following system has a solution: yax = b, xby = a, (.55) where x and y are nonzero quaternion indeterminates..55 was studied and solved in (Tian, 010). In this subsection, we present a somewhat simpler solution. Since st = ts and s 1 = s 1 for all s, t H,.55 can be written as y = a 1 x 1 b, y = b 1 x 1 a, this gives: a 1 x 1 b = b 1 x 1 a (.56) a 1 b = b 1 a a = b. Therefore a and b must satisfy the following equation in order for.55 to have solutions, a 0 + a 1 + a + a 3 = b 0 + b 1 + b + b 3. (.57).56 is equivalent to x 1 ba 1 = ab 1 x 1. On the other hand, we have: a 1 = a 0 a 1 i a j a 3 k a 0 + a 1 + a + a 3 = a a 0 + a 1 + a + a 3 and b 1 = b 0 b 1 i b j b 3 k b 0 + b 1 + b + b 3 = b. b 0 + b 1 + b + b 3 5

61 .3 Special Quaternion Equations and problems By.57,.56 is equivalent to bax = xab xab = bax, which is equivalent to the following system: m 11 x 1 + m 1 x + m 13 x 3 + m 14 x 0 = 0, m 1 x 1 + m x + m 3 x 3 + m 4 x 0 = 0, m 31 x 1 + m 3 x + m 33 x 3 + m 34 x 0 = 0, (.58) m 41 x 1 + m 4 x + m 43 x 3 + m 44 x 0 = 0, where m ij is given by the 4 4 skew-symmetric matrix: M = (m ij ) a 0 b 3 + a 3 b 0 (a 0 b + a b 0 ) a b 3 a 3 b (a 0 b 3 + a 3 b 0 ) 0 a 0 b 1 + a 1 b 0 (a 1 b 3 a 3 b 1 ) =. a 0 b + a b 0 (a 0 b 1 + a 1 b 0 ) 0 a 1 b a b 1 (a b 3 a 3 b ) a 1 b 3 a 3 b 1 (a 1 b a b 1 ) 0 Direct calculation yields det M = 0. Therefore, the homogeneous system.58 has a solution set with an infinite number of solutions (x 0, x 1, x, x 3 ). Each such x gives a solution to.55: x = x, y = b 1 x 1 a. Example.3.6. Find all (x, y) such that the following system holds: y (1 5i + 8j + 15k) x = 17 + i 4j 7k, x (17 + i 4j 7k) y = 1 5i + 8j + 15k. (.59) The following steps have been implemented into a single procedure in Maple. For details, see the CSSim procedure of A.. 53

62 .3 Special Quaternion Equations and problems Since 17 + i 4j 7k = ( 4) + ( 7) = 915 =1 + ( 5) = 1 5i + 8j + 15k,.59 has infinitely many solutions. First we solve the linear system: 48x 11x 3 304x 0 = 0, 48x 1 44x x 0 = 0, 11x x 59x 0 = 0, 304x 1 160x + 59x 3 = 0, to obtain: x 0 = 31 0 t t 3, x 1 = t 1, x = t t 3, x 3 = t 3, where t 1 and t 3 are arbitrary real numbers such that x = x 0 + x 1 i + x j + x 3 k 0. Then we have a solution set with an infinite number of solutions of the form: x = x, y = (17 + i 4j 7k) 1 x 1 (1 5i + 8j + 15k). For example, when t 1 = 1 and t 3 = 3, we have: x = 19 + i + 13j + 3k, y = + 56 i 4 j 3 k Direct calculation shows that (x, y) satisfies

63 .3 Special Quaternion Equations and problems.3.6. A least norm problem The following least norm problem was proposed in (Tian, 010), min ax xb c, x H where a, b and c H and x is a quaternion indeterminate that minimizes the value of ax xb c. In this subsection, we present a solution. We consider: ax xb = c. (.60).60 is equivalent to the real linear homogeneous system: m 11 x 1 + m 1 x + m 13 x 3 + m 14 x 0 = c 1, m 1 x 1 + m x + m 3 x 3 + m 4 x 0 = c, m 31 x 1 + m 3 x + m 33 x 3 + m 34 x 0 = c 3, (.61) m 41 x 1 + m 4 x + m 43 x 3 + m 44 x 0 = c 0, where m ij is given by: a 0 b 0 a 3 b 3 a + b a 1 b 1 a 3 + b 3 a 0 b 0 a 1 b 1 a b M = (m ij ) 4 4 =. a b a 1 + b 1 a 0 b 0 a 3 b 3 a 1 + b 1 a + b a 3 + b 3 a 0 b 0 Direct calculation yields: det M = ( a 1 + a + a 3 b 1 b b 3 ) + (a 0 b 0 ) ( (a 0 b 0 ) + a 1 + a + a 3 + b 1 + b + b 3). We continue the discussion by cases: Case 1. a b. Then det M = 0. Therefore, we can find the least norm solution 55

64 .3 Special Quaternion Equations and problems y 1 y of.61 by calculating = M c where M is the Moore-Penrose y 3 y 0 c 1 pseudoinverse of M and c c = (Bjõrck, 1996). This means that c 3 c 0 x = y 0 + y 1 i + y j + y 3 k satisfies ax xb c az zb c for all z H. Case. Otherwise. Then det M 0. Therefore, we can solve the linear system y 1 y.61 to obtain a unique solution such that.61 holds. This means y 3 y 0 that x = y 0 + y 1 i + y j + y 3 k satisfies ax xb c = 0. Example.3.7. Solve the following least norm problem: min (5 10i 5j + k) x x (3 4i 4j 8k) ( 9 i + 10j k). (.6) x H The following steps have been implemented into a single procedure in Maple. For details, see the LN1 procedure of A.. Since 5 10i 5j + k 3 4i 4j 8k, there exists an exact solution. We solve 56

65 .3 Special Quaternion Equations and problems the linear system: x 1 + 6x 9x 3 6x 0 =, 6x 1 + x + 14x 3 x 0 = 10, 9x 1 14x + x x 0 =, 6x 1 + x 10x 3 + x 0 = 9, to obtain: x 0 = , x 1 = 18, 415 x = , x 3 = Direct calculation shows that x = i 1073j + 37 k satisfies

66 .3 Special Quaternion Equations and problems.3.7. Another least norm problem The following least norm problem was proposed in (Tian, 010): min ax xb c, (.63) x H where a, b and c H and x is a quaternion indeterminate that minimizes the value of ax xb c. In this subsection, we present a solution. We consider: ax xb c = 0. (.64).64 is equivalent to the real linear homogeneous system: m 11 x 1 + m 1 x + m 13 x 3 + m 14 x 0 = c 1, m 1 x 1 + m x + m 3 x 3 + m 4 x 0 = c, m 31 x 1 + m 3 x + m 33 x 3 + m 34 x 0 = c 3, (.65) m 41 x 1 + m 4 x + m 43 x 3 + m 44 x 0 = c 0, where m ij is given by: a 0 + b 0 a 3 + b 3 a b a 1 b 1 a 3 b 3 a 0 + b 0 a 1 + b 1 a b M = (m ij ) 4 4 =. a + b a 1 b 1 a 0 + b 0 a 3 b 3 a 1 b 1 a b a 3 b 3 a 0 b 0 Direct calculation yields: det M = ( (a 0 + b 0 ) + (a 1 b 1 ) + (a b ) + (a 3 b 3 ) ) ( a 0 + a 1 + a + a 3 b 0 b 1 b b3). We continue the discussion by cases: Case 1. a b. Then det M = 0. Therefore, we can find the least norm solution 58

67 .3 Special Quaternion Equations and problems y 1 y of.65 by calculating = M c where M is the Moore-Penrose y 3 y 0 c 1 pseudoinverse of M and c c = (Bjõrck, 1996). This means that c 3 c 0 x = y 0 + y 1 i + y j + y 3 k satisfies ax xb c az zb c for all z H. Case. Otherwise. Then det M 0. Therefore, we can solve the linear system y 1 y.65 to obtain a unique solution such that.65 holds. This means y 3 y 0 that x = y 0 + y 1 i + y j + y 3 k satisfies ax xb c = 0. Example.3.8. Solve the following least norm problem: min (6 8i + j + 5k) x x (6 + i + 5j 8k) ( 3 + i + j 5k). (.66) x H The following steps have been implemented into a single procedure in Maple. For details, see the LN procedure of A.. Since 6 8i + j + 5k 6 + i + 5j 8k, we can only find a least square solution. We calculate M c where M is the Moore-Penrose pseudoinverse of 59

68 .3 Special Quaternion Equations and problems M = and c 1 = to obtain: x 0 = 39 05, x 1 = 119, 7380 x = , x 3 = Direct calculation shows that x = i j 519 k satisfies

69 3. Quaternion Polynomial Matrices 3.1. Generalized Inverse Definition H [x] denotes the set of quaternion polynomials { n } a i x i a i H, a n 0, n N, i=0 where x commutes element-wise with H. H [x] m n denotes the space of m n matrices with entries from H [x]. Remark If no ambiguity arises, the indeterminate x will be omitted and we write p H [x] or A H [x] m n and so on. I m denotes the m m identity matrix. 0 m n denotes the m n zero matrix. When the dimension is unspecified, a matrix is of appropriate dimension. Definition Let A H [x] m n and n i=0 p i x i = p 0 +p 1 x+ +p n x n = p H [x] where p i H. We give a list of some notations as follows: p = n i=0 p i x i, A denotes the conjugate of A, ( A ) ij = (A ij). If A = P + Qj where P, Q C [x] m n, then χ A = denotes the complex adjoint of A. A T H [x] n m denotes the transpose of A, ( A T ) ij = A ji. ( P Q ) Q P C [x] m n 61

70 3.1 Generalized Inverse A H [x] n m denotes the conjugate transpose of A, (A ) ij = (A ji ). In particular, for all p H [x], p = p. A H [x] n m, a generalized inverse of A (Penrose, 1955), if it exists, denotes the solution of the following system of equations, AXA = A (3.1) XAX = X (3.) (AX) = AX (3.3) (XA) = XA (3.4) We first show some elementary properties that will be used often throughout this chapter. Lemma Let A H [x] m 1 n and B H [x] n m. Then (AB) = B A. Proof. By direct calculation. Corollary For A H [x] m n, let B = AA. Then B = B. Proof. B = (AA ) = (A ) A = AA = B. Lemma Let A H [x] m n have a generalized inverse A. Then (A ) = ( A ). Proof. We show that ( A ) satisfies the defining relations of a generalized inverse stated in Definition 3.1.3, A ( A ) A = ( AA A ) = A, ( A ) A ( A ) = ( A AA ) = ( A ), [ A ( A ) ] = [( A A ) ] = A A = ( A A ) = A ( A ), [( A ) A ] = [( AA ) ] = AA = ( AA ) = ( A ) A. 6

71 3.1 Generalized Inverse Therefore, (A ) = ( A ). Lemma (Adaption of (Penrose, 1955)) Let A H [x] m n have a generalized inverse A. Then relations satisfied by A include: A ( A ) A = A = A ( A ) A and A AA = A = A AA. Proof. First, A ( A ) A =A ( AA ) = A AA = A =A AA = ( A A ) A = A ( A ) A. Next, A AA = ( A A ) A = ( AA A ) = A = ( AA A ) = A ( A ) A = A ( AA ) = A AA. We next show that A is uniquely defined for quaternion polynomial matrices. Lemma (Adaption of (Penrose, 1955)) Let A H [x] m n have a generalized inverse A. Then A is unique. Proof. We first show that 3. and 3.3 are equivalent to XX A = X. (3.5) Substituting 3.3 into 3., we obtain X = XAX = X (AX) = XX A. Conversely, 3.5 implies that: AXX A = AX (AXX A ) = (AX) AXX A = (AX). 63

72 3.1 Generalized Inverse So AX = (AX) and therefore X = XX A = X (AX) = XAX. Similarly 3.1 and 3.4 can be replaced by the equation: XAA = A. (3.6) Therefore, 3.5 and 3.6 are equivalent to 3.1, 3., 3.3 and 3.4. Likewise the following two equations are also equivalent to 3.1, 3., 3.3 and 3.4, X A A = A (3.7) A X X = X. (3.8) Suppose A satisfies 3.5 and 3.6, B satisfies 3.7 and 3.8. Then A =A ( A ) A = A ( A ) (B A A) =A ( A ) A AB = A AB = A AA B B = A B B = B. Therefore, if it exists, A is uniquely defined. Due to the non-commutativity of quaternions, there are two types of eigenvalues: right eigenvalues and left eigenvalues. Right eigenvalues have been studied extensively (Brenner, 1951; Lee, 1949; Baker, 1999). We will work with right eigenvalues towards our main result. For convenience, we will just use the term eigenvalue from now on. Definition A H m m is unitary if AA = A A = I. Lemma (Zhang, 1997) A H m m is hermitian, that is, A = A, if and d 1 only if there exists a unitary matrix U H m m such that U. AU =.., where d i are the eigenvalues of A. d m Proof. Corollary 6. of (Zhang, 1997). 64

73 3.1 Generalized Inverse Adapting ideas from (Puystjens and de Smet, 1980), we will give conditions that quaternion polynomial matrices must satisfy to have generalized inverses. We need to prove some lemmas first. Lemma (Adaption of (Puystjens and de Smet, 1980)) Let A H [x] m n have the generalized inverse A. If U H m m is a unitary matrix, then (UA) = A U. Proof. We show that A U satisfies the defining relations of the generalized inverse stated in Definition 3.1.3, UA ( A U ) UA = UAA A = UA, A U (UA) A U = A AA U = A U, ( UAA U ) = U ( AA ) U = UAA U, ( A U UA ) = ( A A ) = A A = A U UA. Therefore, (UA) = A U. Lemma A H m n induces a homomorphism H [x] n 1 to H [x] m 1, that is, for all P, Q H [x] n 1, A (P + Q) = AP + AQ H [x] m 1. Proof. By direct calculation. Lemma (Adaption of (Puystjens and de Smet, 1980)) Let A H [x] m n have the generalized inverse A. Consider A as a homomorphism from H [x] n 1 to H [x] m 1. Then ImageA = ImageAA = ImageAA and ImageA = ImageA A = ImageA A. Proof. Let P H [x] n 1. Then AP = AA AP = A ( A A ) P = AA [( A ) P ], 65

74 3.1 Generalized Inverse where AP H [x] m 1 and ( A ) P H [x] m 1. Therefore, ImageA ImageAA and ImageA ImageAA. For any Q H [x] m 1, it is clear that AA Q = A (A Q) where A Q H [x] n 1 and that AA Q = A ( A Q ) where A Q H [x] n 1. Thus, ImageAA ImageA and ImageAA ImageA. Therefore, ImageA = ImageAA and ImageA = ImageAA, i.e., ImageA = ImageAA = ImageAA. Similarly we can show that ImageA = ImageA A = ImageA A. Lemma (Adaption of (Puystjens and de Smet, 1980)) If E H [x] m m is a symmetric projection, that is, E = E = E, then E H m m. Proof. If f 1,, f m are the entries in the first row of E, with f 1 = f 1, then m m f 1 f 1 + f i f i = f 1 f1 + f i f i = f 1. i= i= Since f 1 = f 1, the leading coefficient of f1 is a positive real number. Note that the leading coefficient of m i= f i f i is also a positive real number. Thus, deg ( ( ) { ) m f1 deg f1 = deg f1 + f i f i = max deg ( ( ) m )} f1, deg f i f i i= i= deg ( ) f1. This shows that f 1 H. Furthermore, 0 = deg f 1 = deg ( mi= f i f i ) and therefore f i H for all 1 i m. The same can be done for the other rows of E. Therefore, E H m m. Now we are ready to give conditions that quaternion polynomial matrices must satisfy in order to have generalized inverses. 66

75 3.1 Generalized Inverse Theorem (Adaption of (Puystjens and de Smet, 1980)) Let A H [x] m n. Then A has the generalized inverse A A 1 A if and only if A = U with U 0 0 H m m unitary and A 1 A 1 + A A a unit in H [x] r r with r min {m, n}. Moreover, A = A 1 (A 1 A 1 + A A ) 1 0 A (A 1 A 1 + A A ) 1 U. 0 Proof. If A has the generalized inverse A, then AA = AA AA = ( AA ) = ( AA ). By Lemma , AA H m m. AA is hermitian and hence, by Lemma , there exists a unitary matrix U H m m such that U AA U = D where D is diagonal. Since D = U AA UU AA U = U AA AA U = U AA U = D, the diagonal entries of D are either 1 or 0. Therefore, we can rearrange the rows of I r 0 U so that D = with r min {m, n}. 0 0 Set A = U A. By Lemma , A has its own generalized inverse A and A A = I r 0. Set A = A 1 A, for arbitrary quaternion polynomial matrices 0 0 A 3 A 4 A 1 H [x] r r, A H [x] r (n r), A 3 H [x] (m r) r and A 4 H [x] (m r) (n r). Since A = A A A = I r 0 A 1 A A 1 A =, we must have A A 1 A = 0 0 A 3 A and therefore A A A 1 A 1 = + A A 0. Similarly, A = B 1 0 for some B B 0 and B. 67

76 3.1 Generalized Inverse By Lemma , ImageA A = ImageA = ImageA A I r 0 = Image. 0 0 This implies the surjectivity of A 1 A 1 + A A on H [x] r 1. Therefore, A 1 A 1 + A A is a unit in H [x] r r and A = UA A 1 A = U. 0 0 Next, we have that: A =A ( A ) A = A (A ) A = A (A A ) A 1 0 = (A 1 A 1 + A A ) 1 0 A A 1 (A 1 A 1 + A A = ) 1 0 A (A 1 A 1 + A A ) 1, 0 which gives: A = A 1 (A 1 A 1 + A A ) 1 0 A (A 1 A 1 + A A ) 1 U. 0 We will show the converse by direct computation. A 1 A Let A = U be in H [x] m n with U H m m unitary and A 1 A 1 + A A a 0 0 unit in H [x] r r with r min {m, n}. Let B = A 1 (A 1 A 1 + A A ) 1 0 A (A 1 A 1 + A A ) 1 U. 0 We show that A U satisfies the defining relations of the generalized inverse stated in Definition 3.1.3, 68

77 3.1 Generalized Inverse the first condition: ABA =U 0 0 A (A 1 A 1 + A A ) 1 U U A 1 A A 1 (A 1 A 1 + A A =U ) 1 0 A 1 A 0 0 A (A 1 A 1 + A A ) A 1 A =U = A, 0 0 the second condition: BAB = A (A 1 A 1 + A A ) 1 U U A (A 1 A 1 + A A ) 1 U 0 = A (A 1 A 1 + A A ) A (A 1 A 1 + A A ) 1 U 0 = A (A 1 A 1 + A A ) 1 U = B, 0 the third condition: (AB) = U 0 0 A (A 1 A 1 + A A ) 1 U 0 I r 0 = = = AB,

78 3.1 Generalized Inverse the last condition: (BA) = A 1 (A 1 A 1 + A A ) 1 0 A (A 1 A 1 + A A ) 1 U A 1 A U I r 0 = = I r 0 = BA Therefore, B = A 1 (A1A 1 + AA ) 1 0 U = A. The uniqueness of A is the A (A 1A 1 + A A ) 1 0 result of Lemma Lemma Let A H m m be hermitian. Then the eigenvalues of A are real. Proof. Let λ H be an eigenvalue of A with corresponding eigenvector 0 X = x 1. such that AX = Xλ. Then X AX = X Xλ. So X AX = λ X X and x m therefore X Xλ = λ X X = (X Xλ) (x 1, x m ) x 1. x m λ = ( xi x i ) λ = (( xi x i ) λ ) = λ ( xi x i ) = λ ( xi x i ). Note that 0 x i x i R [x]. Division of the above equation by x i x i gives λ = λ λ R. Theorem (Zhang, 1997)(Cayley-Hamilton theorem for Quaternion Matrices) Let A H m m. f A (λ) = det (λi m χ A ) is called the characteristic polyno- 70

79 3.1 Generalized Inverse mial of A, where λ is a complex indeterminate. Then f A (A) = 0. Furthermore, f A (λ 0 ) = 0 if and only if λ 0 is an eigenvalue of A. Proof. Theorem 8.1 of (Zhang, 1997). Definition Let A H [x] m n have the generalized inverse A. For any x R, set B = AA. Then f B (λ) = det (λi m χ B ) is called the characteristic polynomial of B. By Lemma , λ can be assumed to be an indeterminate that enjoys the following: λ = λ and λ commutes element-wise with H [x]. Lemma Let A H [x] m n have the generalized inverse A. For any x R, set B = AA. Then f B (λ) = g (λ) where g (λ) (R [x]) [λ]. Proof. We first show that f B (λ) (R [x]) [λ]. Since det ( (λi m χ B ) T ) = det (λi m χ B ) = det ((λi m χ B ) ) det (λi m χ B ) = det (λi m χ B ) = det (λi m χ B ), we have that: det (λi m χ B ) = f B (λ) (R [x]) [λ]. (3.9) Next, we show that f B (λ) = g (λ) where g (λ) (C [x]) [λ]. Let B = P + Qi. For any fixed 1 i, j m, we have B ij = a + bi + cj + dk where a, b, c and d R [x]. Since B is hermitian, B ji = a bi cj dk and therefore P ij = a + bi, P ji = a bi and Q ij = c + di, Q ji = c di. So P T = P and Q = Q T. Therefore, P Q χ B = = P Q Q P Q P T 71

80 3.1 Generalized Inverse λi m P λi m χ B = Q Q. λi m P T Next, we have: I m I m 0 I m I m 0 I m I m I m 0 I m Q Q P T λi m = λi m P Q Q λi m P T and det I m I m = det I m 0 = 1. 0 I m I m I m Therefore, λi m P f B (λ) = det Q Q = det Q λi m P T λi m P P T λi m. Q Note that: Q λi m P T P T λi m Q = Q λi m P P T λi m, Q which implies that Q P T λi m is skew-symmetric. By (Muir, 003), the λi m P Q determinant of Q P T λi m, also called its Pfaffian, can be written as λi m P Q the square of a polynomial in its entries. Therefore, f B (λ) = g (λ) where g (λ) (C [x]) [λ]. Finally we show that g (λ) (R [x]) [λ]. Suppose otherwise. Then g (λ) = a (λ) + b (λ) i where a and b (R [x]) [λ] with b (λ) 0. By 3.9, g (λ) (R [x]) [λ]. So a (λ) = 0 and therefore f B (λ) = g (λ) = (b (λ) i) = b (λ) 7

81 3.1 Generalized Inverse where b (λ) (R [x]) [λ]. For a fixed x R, let λ R be large enough such that λ I m χ B C m m is diagonally dominant with non-negative diagonal entries and that (b (x)) (λ ) 0. Since λ I m χ B is also hermitian, λ I m χ B is positive definite (Horn and Johnson, 1990). But det (λ I m χ B ) = ((b (x)) (λ )) < 0. Contradiction. So, b = 0 and therefore f B (λ) = g (λ) where g (λ) (R [x]) [λ]. Corollary Let A H [x] m n have the generalized inverse A. For any x R, set B = AA and f B (λ) = g (λ). Then g (B) = 0. We will call g (λ) the generalized characteristic polynomial of A. Proof. g (λ) (R [α]) [λ] by Lemma , so χ g(b) = g (χ B ). Next, f B (χ B ) = 0 by the Cayley-Hamilton theorem for complex polynomial matrices (Horn and Johnson, 1990), so g (χ B ) = 0. Therefore 0 = g (χ B ) = χ g(b), that is, g (B) = 0. 73

82 3. Leverrier-Faddeev Method 3.. Leverrier-Faddeev Method Lemma Let A H [x] m n have the generalized inverse A. Set B = AA. Then B = (A ) A and B B = BB. Proof. First we show that (A ) A satisfies the defining relations of the generalized inverse stated in Definition 3.1.3, the first condition: B [ (A ) A ] B =AA (A ) A AA =AA (A ) A = AA = B, the second condition: [ (A ) A ] B [ (A ) A ] = (A ) A AA (A ) A = (A ) A (A ) A = (A ) A, the third condition: ( B [ (A ) A ]) = ( AA (A ) A ) = ( AA ) =AA = AA (A ) A = B [ (A ) A ], the last condition: ([ (A ) A ] B ) = ( (A ) A AA ) = ( (A ) A ) = (A ) A = (A ) A AA = [ (A ) A ] B. Therefore, (A ) A = (AA ) = B. Next, (AA ) AA = (A ) A AA = (A ) A = ( (A ) A ) = AA = AA ( A ) A = AA (AA ), 74

83 3. Leverrier-Faddeev Method which is, B B = BB. Lemma 3... Let A H [x] m n have the generalized inverse A. Set B = AA. Then ( B ) k = ( B k ) where k N. Proof. We show that ( B ) k satisfies the defining relations of the generalized inverse stated in Definition 3.1.3, the first condition: B k ( B ) k B k =B k ( B ) k 1 ( BB ) B k 1 =B k ( B ) k 1 B k 1 = = B k, the second condition: ( B ) k B k ( B ) k = ( B ) k 1 ( BB ) B k 1 ( B ) k = ( B ) k 1 B k 1 ( B ) k = = ( B ) k, the third condition: ( B k ( B ) k ) = [ (B ) ] k (B ) k = ( B ) k B k = B k ( B ) k, the last condition: ( (B ) ) k ( B k = (B ) k) ( (B ) ) k ( = B k B ) k ( ) = B k B k. Therefore, ( B ) k ( ) = B k where k N. Lemma (Adaption of (Penrose, 1955)) Let A H [x] m n, B H [x] p q and C H [x] m q. If A and B both exist, then the quaternion polynomial matrix equation AXB = C has a solution in H [x] n p if and only if AA CB B = C, in which case the general solution is X = A CB + Y A AY BB, 75

84 3. Leverrier-Faddeev Method where Y H [x] n p is arbitrary. Proof. Suppose X satisfies AXB = C, then C = AXB = AA AXBB B = AA CB B. Conversely, if C = AA CB B then A CB is a particular solution of AXB = C. For the general solution, we must solve AXB = 0. Any expression of the form X = Y A AY BB satisfies AXB = 0, and conversely if AXB = 0 then X = A CB + Y A AY BB. Theorem (Adaption of (Penrose, 1955)) Let A H [x] m n have the generalized inverse A. For any x R, set B = AA. Suppose the generalized characteristic polynomial of A is g (λ) = λ m + a 1 λ m a k λ m k + + a m 1 λ + a m, where a i R [x]. If k is the largest integer such that a k 0, then the generalized inverse of A is given by A = 1 a k A [ B k 1 + a 1 B k + + a k 1 I ]. If a i = 0 for all 1 i m, then A = 0. Proof. By Corollary 3.1.0, we have: 0 = B m + a 1 B m a k B m k + + a m 1 B + a m I m. If k is the largest integer such that a k 0 and define B 0 = I, we may write: B m k ( B k + a 1 B k a k 1 B + a k I ) = 0. This equation guarantees a solution of the matrix equation B m k X = 0 and hence, 76

85 3. Leverrier-Faddeev Method by Lemma 3..1 and 3..3, all solutions are given by: X = Y ( B m k) B m k Y = Y B BY, where Y H m m is arbitrary. In particular, there exists Y 1 such that: B k + a 1 B k a k 1 B + a k I = Y 1 B BY 1 = Y 1 AA Y 1 Left multiplication of the latter equation by A gives: A B k + a 1 A B k a k 1 A B + a k A = 0. By Lemma 3.1.7, this can be simplified to: A B k 1 + a 1 A B k + + a k 1 A = a k A, which gives: A = 1 a k A [ B k 1 + a 1 B k + + a k 1 I ]. If a i = 0 for all 1 i m, then A = 0 and therefore A = 0. Lemma (Adaption of (Kalman, 1978)) Let A H [x] m n have the generalized inverse A. For any real x R, set B = AA. Let λ 1,, λ m, where m m, be all the non-zero eigenvalues of B. Then for 1 k m, tr [( B k + a 1 B k a k 1 B )] = ka k, where the a i arise from the generalized characteristic polynomial of A: g (λ) = λ m + a 1 λ m a k λ m k + + a m 1 λ + a m. 77

86 3. Leverrier-Faddeev Method Proof. Let Y = yi where y R. We can write g (Y ) as: g (Y ) = (Y B) ( Y m 1 + (B + a 1 I) Y m + ( B + a 1 B + a I ) Y m ( B m 1 + a 1 B m + + a m I )). As long as y is not an eigenvalue of B, (yi B) = Y B is non-singular, so we can write: (Y B) 1 g (Y ) =Y m 1 + (B + a 1 I) Y m + ( B + a 1 B + a I ) Y m ( B m 1 + a 1 B m + + a m I ). Taking the traces gives: tr [ (Y B) 1 g (Y ) ] =my m 1 + tr [(B + a 1 I)] y m + tr [( B + a 1 B + a I )] y m tr ( B m 1 + a 1 B m + + a m I ). Let C = (Y B) 1 g (Y ). Since g (Y ) = g (yi) = g (y) I, C = g (y) (Y B) 1. Therefore, trc = g (y) tr [ (Y B) 1]. tr [ (Y B) 1] is the sum of the eigenvalue of (Y B) 1. We will show these eigenvalues are the fractions 1 y λ 1,, 1 y λ m. Let ζ be an eigenvalue of (Y B) 1 with corresponding eigenvector Z such that: (Y B) 1 Z = Zζ. 78

87 3. Leverrier-Faddeev Method ζ is real by Lemma , and hence (Y B) Z = Z 1 ζ ( BZ = Z y 1 ). ζ Therefore, y 1 ζ = λ i ζ = 1 y λ i for some 1 i m. Since g (y) = (y λ 1 ) (y λ ) (y λ m ), we have that g (y) = g (y) ( ) 1 y λ y λ m and trc = g (y) The derivative of g is also equal to: g (y) = my m 1 + a 1 (m 1) y m + + a m 1. Therefore, my m 1 + a 1 (m 1) y m + + a m 1 =my m 1 + tr [(B + a 1 I)] y m + + tr ( B m 1 + a 1 B m + + a m I ) Comparing the coefficient of y m k 1 on both sides, we obtain a k (m k) = tr [( B k + a 1 B k a k 1 B + a k I )] a k (m k) = tr [( B k + a 1 B k a k 1 B )] + tra k I ka k = tr [( B k + a 1 B k a k 1 B )] Lemma (Adaption of (Decell, 1965; Faddeeva, 1959)) Let A H [x] m n have 79

88 3. Leverrier-Faddeev Method the generalized inverse A. For any x R, set B = AA. Suppose the generalized characteristic polynomial of A is g (λ) = λ m + a 1 λ m a k λ m k + + a m 1 λ + a m, where a i R [x]. Define a 0 = 1. If p is the largest integer such that a p 0 and we construct the sequence A 0,, A p as follows: A 0 = 0 1 = q 0 B 0 = I... A p 1 = AA B p A p = AA B p 1 tra p 1 p 1 = q p 1 B p 1 = A p 1 q p 1 I tra p = q p p B p = A p q p I then q i (x) = a i (x), i = 0,, p. Proof. We will show the desired result by mathematical induction. For p i 0, let P i be the statement: q i (x) = a i (x). P 0 is the statement: q 0 = a 0, which is true by definition. 80

89 3. Leverrier-Faddeev Method Suppose P i holds for all 0 i k 1, it remains to show that P k holds. We have A k =BB k 1 =B (A k 1 q k 1 I) =B (B (A k q k I) q k 1 I) = =B k q 1 B k 1 q B k q k 1 B =B k + a 1 B k a k 1 B. Therefore, tr A k = tr [( B k + a 1 B k a k 1 B )], which by Lemma 3..5 is equal to ka k. So q k = tr A k k = a k. Thus, P k holds. Therefore, by the principle of mathematical induction, P i holds for all p i 0. That is, q i = a i, i = 0,, p. 81

90 3.3 Generalized Inversion by Interpolation 3.3. Generalized Inversion by Interpolation In this section we present a method to obtain the generalized inverse of a quaternion polynomial matrix by interpolation at real number data points. Lemma (See (Lam, 001) for more details) An element r H is a root of a nonzero polynomial f H [x] iff x r is a right divisor of f. The set of polynomials in H [x] having r as a root is the left ideal H [x] (x r). Proof. By direct calculation. Lemma (See (Lam, 001) for more details) Let f, g and h H [x], f = gh and r H be such that β = h (r) 0. Then f (r) = g ( βrβ 1) h (r). In particular, if r is a root of f but not of h, then βrβ 1 is a root of g. Proof. Let g = n i=0 a i x i. Then f = ( n i=0 a i x i ) h = n i=0 a i hx i. Therefore, n n f (r) = a i h (r) r i = a i βr i i=0 i=0 n n ( = a i βr i β 1 β = a ) i βrβ 1 i β i=0 i=0 ( n ( = a ) ) i βrβ 1 i β = g ( βrβ 1) h (r). i=0 The last conclusion follows since H has no zero-divisors. Theorem (Adaption of (Gordon and Motzkin, 1965)) Let f H [x] be of degree n. Then the roots of f lie in at most n conjugacy classes of H. Proof. We will show the desired result by mathematical induction. For n 1, let P n be the statement: 8

91 3.3 Generalized Inversion by Interpolation Let f H [x] be of degree n. Then the roots of f lie in at most n conjugacy classes. P 1 is the statement: Let f H [x] be of degree 1. Then the root of f lies in at most 1 conjugacy class. P 1 is trivially true. Suppose P n holds for all 1 n k 1, it remains to show that P k holds. Let f H [x] be of degree k with roots c d. By Lemma 3.3.1, we can write f = g (x c), where g has a root that is, by Lemma 3.3., conjugate to d. Invoking the inductive hypothesis, d lies in at most k 1 conjugacy classes. So, the roots of f lie in at most k conjugacy classes. Thus, P k holds. Therefore, by the principle of mathematical induction, P n holds for all n 1. Theorem (Adaption of (Bray and Whaples, 1983)) Let c 1,, c n be n pairwise non-conjugate elements of H. Then there is a unique polynomial g n H [x], monic of degree n, such that g n (c 1 ) = = g n (c n ) = 0. Moreover, c 1,, c n are the only roots of g n in H. Proof. We first show the existence of g n for all n 1 by mathematical induction. For n 1, let P n be the statement: Let c 1,, c n H be pairwise non-conjugate. Then there is a polynomial g n H [x], monic of degree n, such that g n (c 1 ) = = g n (c n ) = 0. Moreover, c 1,, c n are the only roots of g in H. P 1 is the statement: Let c 1 H. Then there is a polynomial g 1 H [x], monic of degree 1, such that g 1 (c 1 ) = 0. Moreover, c 1 is the only root of g in H. 83

92 3.3 Generalized Inversion by Interpolation P 1 is trivially true as g = x c 1. Suppose P n holds for all 1 n k 1, it remains to show that P k holds. Let c 1,, c k H be pairwise non-conjugate. Invoking the inductive hypothesis, there exists a monic polynomial g k 1 of degree k 1 with c,, c k as its only roots, that is, g k 1 (c 1 ) 0. Construct g k as follows, g k = ( x g k 1 (c 1 ) c 1 g k 1 (c 1 ) 1) g k 1. By Lemma 3.3., g k (c 1 ) = 0. Thus, P k holds. Therefore, by the principle of mathematical induction, for all n 1, P n holds. We next show that g n is unique for all n 1. For some fixed n, let g g n be monic of degree n, such that g (c 1 ) = = g (c n ) = 0 too. Then deg (g n g ) n 1 but g n g has roots c 1,, c n which lie in n conjugacy classes of H. This contradicts Theorem Therefore, g n is unique for all n 1. Lemma Let c 1,, c n+1 H be pairwise non-conjugate and let d 1,, d n+1 H. Then there is a unique lowest degree polynomial f H [x], of degree p n, such that f (c i ) = d i for all 1 i n + 1. Proof. For any 1 s n+1, let S = {1,, n + 1}\{s }. By Theorem 3.3.4, for any s S, we can find unique h S H [x], monic of degree n, such that h S (c s ) = 0 and that {c s s S} are the only roots of h S in H. So, h S (c s ) 0 and therefore 0 α S, we can construct a polynomial g S of degree n such that g S (c α ) = as 1 α = s, below: g S = h S (c s ) 1 h S. We can then construct the polynomial f, of degree at most n, such that f (c i ) = d i 84

93 3.3 Generalized Inversion by Interpolation for all 1 i n + 1 as below: f = n+1 s =1 d s g S. We next show that f is unique. Suppose we have f H [x] of degree p p n, such that f f and that f (c i ) = d i for all 1 i n + 1 too. Then f f 0 is of degree at most p n. But f f has roots c 1,, c n+1 which lie in n+1 conjugacy classes of H. This contradicts Theorem Therefore, f is unique. Definition The degree of a given quaternion polynomial matrix A H [x] m n is deg A = max {deg (A ij ) 1 i m, 1 j n}. Lemma Let A H [x] m n have the generalized inverse A. Then deg A (m 1) deg A. Proof. By Theorem 3..4, deg A deg ( A (AA ) m ) deg (A m 1 ) (m 1) deg A. Remark The upper bound mentioned in Lemma is rarely achieved in practice, because some elements of AA (and other matrices) do not have maximal degree. Lemma Let c 1,, c k+1 H be pairwise non-conjugate and let A 1,, A k+1 H n m. Then there is a unique lowest degree matrix A H [x] n m of degree p k, such that A (c i ) = A i for all 1 i k + 1. Proof. For any 1 n 1 n and 1 m 1 m, by Lemma 3.3.5, there is a lowest degree polynomial A n1 m 1 determined by the values c 1,, c k+1 and (A 1 ) n1 m 1,, (A k+1 ) n1 m 1. In fact, for any 1 s k + 1, let S = {1,, k + 1} \ {s }. Then A n1 m 1 = k+1 s =1 (A s ) n1 m 1 g S, 85

94 3.3 Generalized Inversion by Interpolation 0 α S, where g S (c α ) =. Since n 1 and m 1 are chosen randomly, a lowest degree 1 α = s matrix A that satisfies A (c i ) = A i for all 1 i k + 1 is determined by A = k+1 s =1 A s g S. Next we show that A is unique. Suppose A A of degree p p also satisfies A (c i ) = A i for all 1 i k + 1. Then for some 1 n n and 1 m m, (A A ) n m 0. But (A A ) n m, of degree at most p k, has roots c 1,, c k+1 which lie in k+1 conjugacy classes of H. This contradicts Theorem Therefore, A is unique. Let A H [x] m n satisfy conditions of Theorem For any x R, set B = AA. Let p be the largest integer such that a p 0. We construct the sequence A 0,, A p as follows: A 0 = 0 1 = q 0 B 0 = I... A p 1 = AA B p A p = AA B p 1 tra p 1 p 1 = q p 1 B p 1 = A p 1 q p 1 I tra p = q p p B p = A p q p I Theorem In the above setting, let k = (m 1) deg A and c 1,, c k+1 R be k + 1 distinct real numbers such that q p (c s ) 0 for any 1 s k + 1. Let S = {1,, k + 1} \ {s }. Then A = k+1 s =1 A (c s ) g S where A (c s ) = 1 q p (c s ) A (c [ s ) B (c s ) p 1 q 1 (c s ) B (c s ) p q p 1 (c s ) I ] 86

95 3.3 Generalized Inversion by Interpolation and 0 α S, g S (c α ) = 1 α = s. Proof. By Theorem 3..4, Lemma 3..6 and Lemma Example Find, by interpolation, the generalized inverse of the quaternion polynomial matrix 14x i + 70j + 56k 56 8i 70j + 70k 8j 56k 14x 56 8i 14j 56k x 43i 10j 8k 8 + 4i + 10j 10k 4j + 8k x i + j + 8k A = 3x 3 + 3i 15j 1k 1 + 6i + 15j 15k 6j + 1k 3x i + 3j + 1k 4x 4 + 4i 0j 16k i + 0j 0k 8j + 16k 4x i + 4j + 16k The following steps have been implemented into a single procedure in Maple. For details, see the LFI procedure of A.. As mentioned in Remark 3.3.8, here we only need two data points because direct calculation shows that deg A =. Let c 1 = 0 and c = 1. We have: i + 70j + 56k 56 8i 70j + 70k 8j 56k 56 8i 14j 56k 43i 10j 8k 8 + 4i + 10j 10k 4j + 8k 8 31i + j + 8k A (c 1 ) = 3 + 3i 15j 1k 1 + 6i + 15j 15k 6j + 1k 1 + 1i + 3j + 1k 4 + 4i 0j 16k i + 0j 0k 8j + 16k i + 4j + 16k and i + 70j + 56k 56 8i 70j + 70k 8j 56k 4 8i 14j 56k 4 43i 10j 8k 8 + 4i + 10j 10k 4j + 8k 6 31i + j + 8k A (c ) = i 15j 1k 1 + 6i + 15j 15k 6j + 1k 9 + 1i + 3j + 1k 8 + 4i 0j 16k i + 0j 0k 8j + 16k 1 + 8i + 4j + 16k By the algorithm stated in Lemma 3..6, we calculate and obtain: A (c 1 ) = A (0) 1 = i 8j 34k i 96j + 81k i + 16j + 54k i + 168j + 7k i + 46j 38k i 93j 149k 5 76i 7j + 04k i 96j + 7k i 176j + 9k i + 68j + 194k i + 1j 04k i + 16j 7k 140 1i + 8j + 34k i + 96j 81k i 16j 54k i 168j 7k 87

96 3.3 Generalized Inversion by Interpolation and A (c ) = A (1) 1 = i 44j 330k i 8j + 15k i + 78j + 90k 9 110i + 104j + 10k i + 406j 40k i + 17j 39k 76 8i 13j + 144k i 176j + 19k i 160j + 300k i 0j + 150k i + 60j 180k i + 80j 40k 15 13i + 44j + 330k i + 8j 15k i 78j 90k i 104j 10k By Theorem , we have: A = A (c s ) g S = A (0) (1 x) + A (1) x s =1 1 = (1 + 10i 16j + 1k) x i 8j 34k ( 66 55i + 88j 66k) x i 96j + 81k ( i 0j 0k) x i + 46j 38k (44 88i + 110j + 110k) x i 93j 149k (16j + 8k) x i 176j + 9k ( 88j 44k) x i + 68j + 194k ( 1 10i + 16j 1k) x 140 1i + 8j 34k ( i 88j + 66k) x i + 96j 81k ( i 48j + 36k) x i + 16j + 54k ( i 64j + 48k) x i + 168j + 7k ( i 60j 60k) x 5 76i 7j + 04k ( i 80j + 80k) x i 96j + 7k. (48j + 4k) x i + 1j 04k (64j + 3k) x i + 16j 7k ( 36 30i + 48j 36k) x i 16j 54k ( 48 40i + 64j 48k) x i 168j 7k Direct calculation shows that A satisfies the defining relations of the generalized inverse stated in Definition

97 A. Maple Codes A.1. Overview At the early stage of my graduate studies, I realized the need of utilizing Maple when trying to solve quaternion quadratic equations. There are existing Maple packages dedicated to quaternions (Carter, 007; Harder, 010), but not quaternion polynomials or quaternion polynomial matrices. Therefore, I started writing my own Maple module. During my two years of graduate studies, I gradually extended this module to have over thirty procedures. These procedures enable the user to do basic quaternion and quaternion polynomial matrix operations such as additions and multiplications; to do Lagrange and Newton interpolations of a list of quaternion pairs; to obtain the generalized inverse of a quaternion polynomial matrix; to solve multiple quaternion equations and quaternion least norm problems; and so on. In A the codes of the procedures are provided with descriptions and examples. 89

98 A. The Codes A.. The Codes Q := module() option package; export M, *, ^, eval, Qdef, Qrand, QPrand, Mrand, PMrand, Qreal, Qimag, Qconj, Mconj, Qnorm, Qinv, AXXB, sortcollect, ZeroP, QMFNorm, PRotate, LagrangeP, AXBc, NewtonP, QAdjointMatrix, LFI, NoSim1, NoSim, Meval, AXB, SemiSim, CSSim, LN1, LN; local QMultiplyScalar, QMatrixMultiply, QScalarMultiply; M := proc(a,b) #M is the multiplication operator for quaternions and quaternion polynomials. local a1 := Qreal(a), b1 := coeff(a, I), c1 := coeff(a, J), d1 := coeff(a, K), a := Qreal(b), b := coeff(b, I), c := coeff(b, J), d := coeff(b, K); return sortcollect (a1*a - b1*b - c1*c - d1*d + (a1*b + b1*a + c1*d - d1*c)*i + (a1*c - b1*d + c1*a + d1*b)*j + (a1*d + b1*c - c1*b + d1*a)*k, x); end proc; 90

99 A. The Codes QMatrixMultiply := proc(a::matrix, B::Matrix) #Local procedure for quaternion matrix multiplication. local nrowsa := LinearAlgebra:-RowDimension(A), ncolsa := LinearAlgebra:-ColumnDimension(A), ncolsb := LinearAlgebra:-ColumnDimension(B), AB := Matrix(1..nrowsA, 1..ncolsB), i, j; for i from 1 to nrowsa do for j from 1 to ncolsb do AB(i, j) := simplify(add(m(a(i, k), B(k, j)), k = 1.. ncolsa)); end do; end do; end proc; QMultiplyScalar := proc(a::matrix, B) #Local procedure to multiply a quaternion matrix by a quaternion scalar. local nrowsa := LinearAlgebra:-RowDimension(A), ncolsa := LinearAlgebra:-ColumnDimension(A), AB := Matrix(1..nrowsA, 1..ncolsA), i, j; for i from 1 to nrowsa do for j from 1 to ncolsa do AB(i, j) := simplify(m(a(i, j), B)); end do; end do; 91

100 A. The Codes end proc; QScalarMultiply := proc(a, B::Matrix) #Local procedure to multiply a quaternion scalar by a quaternion matrix. local nrowsb := LinearAlgebra:-RowDimension(B), ncolsb := LinearAlgebra:-ColumnDimension(B), AB := Matrix(1..nrowsB, 1..ncolsB), i, j; for i from 1 to nrowsb do for j from 1 to ncolsb do AB(i, j) := simplify(m(a, B(i, j))); end do; end do; end proc; * := proc(m, n) #Newly defined multiplication operator for quaternions related elements. options remember; if type(m, Matrix) and type(n, Matrix) then QMatrixMultiply(m, n); elif type(m, Matrix) then QMultiplyScalar(m, n); elif type(n, Matrix) then QScalarMultiply(m, n); else M(m, n); 9

101 A. The Codes end if; end proc; ^ := proc(m, n::integer) #Newly defined exponential operator to include quaternion raised to positive integer powers. options remember; local i, p := 1; if type(m, freeof({i, J, K})) = false and n > 1 then for i from 1 to n do p := M(p, m); end do; return p; else m^n; end if; end proc; Qdef := proc(a, b, c, d) #Defines a quaternion a + bi + cj + dk. a + b*i + c*j + d*k; end proc: Qrand := proc(m::integer, n::integer) #Generates a random quaternion with all integer coordinates between m and n. Qdef(rand(m..n)(), rand(m..n)(), rand(m..n)(), rand(m..n)()); 93

102 A. The Codes end proc: QPrand := proc (m,n,k) #Generates a random quaternion polynomial of degee k, with coefficients Qrand (m, n). local j := k; return convert([seq(qrand(m, n)*x^(j - i), i = 0.. k)], + ) end proc; Mrand := proc(m, n, p, q) #Generates a random p q Matrix with quaternion entries Qrand (m, n). local M := Matrix(1.. p, 1.. q), i, j; for i from 1 to p do for j from 1 to q do M(i, j) := Qrand(m, n); end do; end do; end proc: PMrand := proc(m, n, k1, k, p, q) #Generates a random p q Matrix with random quaternion polynomial entries. local M := Matrix(1.. p, 1.. q), i, j; for i from 1 to p do for j from 1 to q do M(i, j) := QPrand(m, n, rand(k1..k)()); 94

103 A. The Codes end do; end do; end proc: Qreal := proc(a) #Calculates the real part of a quaternion. subs(i = 0, J = 0, K = 0, a); end proc: Qimag := proc(a) #Calculates the imaginary part of a quaternion. a - Qreal(a); end proc: Qconj := proc(f) #Calculates the conjugate of a quaternion or a quaternion polynomial. sortcollect(qreal(f) - Qimag(f), x); end proc: Mconj := proc(a) #Calculates the conjugate of a quaternion or quaternion polynomial matrix. local nrowsa:=linearalgebra:-rowdimension(a), ncolsa:=linearalgebra:-columndimension(a), AC := Matrix(1..nrowsA, 1..ncolsA), i, j; for i from 1 to nrowsa do for j from 1 to ncolsa do 95

104 A. The Codes AC(i, j) := Qconj(A(i, j)) end do; end do; end proc: sortcollect := proc (A, a) #Returns A as a polynomial in standard form of variable a. return sort(collect(a, a), a) end proc; eval := proc(n, n) #Calculates N (n) where N (x) is a polynomial of x. options remember; local i, f := 1, m := N; if has(n, x) = false then expand(n); elif is(expand(n), + ) = true then map(eval,expand(n),n); else for i from 1 to degree(n, x) do f := M(f, n); end do; M(coeff(N, x, degree(n, x)), f); end if; end proc; 96

105 A. The Codes Meval := proc(n, n) #Calculates N (n) where N (x) is a quaternion polynomial matrix of x. options remember; local i, j, m := Matrix(LinearAlgebra:-RowDimension(N), LinearAlgebra:-ColumnDimension(N)); for i from 1 to LinearAlgebra:-RowDimension(m) do for j from 1 to LinearAlgebra:-ColumnDimension(m) do m(i, j) := eval(n(i, j), n); end do; end do; end proc; PRotate := proc(m::integer) #Generates a block matrix. local M := Matrix(1.. *m, 1.. *m), i; for i from 1 to m do M(i, i + m) := -1; end do; for i from m + 1 to *m do M(i, i - m) := 1; end do; end proc; 97

106 A. The Codes Qnorm := proc(a) #Calculates the norm of a quaternion. (Qreal(a)^ + coeff(a, I)^ + coeff(a, J)^ + coeff(a, K)^)^(1/); end proc: Qinv := proc(a) #Calculates the inverse of a quaternion or a quaternion polynomial. sortcollect(qreal(a) - Qimag(a), x)/sortcollect ((Qreal(a)^ + coeff(a,i)^ + coeff(a, J)^ + coeff(a, K)^), x) end proc: NoSim1 := proc(a::list) #Ensures that a list of quaternion is ready for zero polynomial interpolation. options remember; local i, j, k, B := A, n := nops(b); for i from to n do for j from 1 to i - 1 do if Qreal(B[j]) = Qreal(B[i]) and Qnorm(B[j]) = Qnorm(B[i]) then for k from j + 1 to i - 1 do if Qreal(B[k]) = Qreal(B[i]) and Qnorm(B[k]) = Qnorm(B[i]) then B := subsop(i = z, B); next; end if; end do; end if; 98

107 A. The Codes end do; end do; return remove(t -> t = z, B); end proc; ZeroP := proc(b::list) #Calculates an zero polynomial for a list of quaternions. options remember; local A := [op({op(nosim1(b))})], i, n := nops(a), f; f[1] := x - op(1, A); for i from to n do f[i] := M((x - M(M(eval((f[i - 1]), op(i, A)), op(i, A)), Qinv(eval((f[i - 1]), op(i, A))))), f[i - 1]); end do; return sortcollect(f[n], x); end proc; NoSim := proc(a::listlist) #Ensures that a list of quaternion pairs is ready for Lagrange Interpolation. options remember; local i, j, k, B := A, C, n1 := nops(b), n; for i from to n1 do for j from 1 to i - 1 do if A[j][1] = A[i][1] then 99

108 A. The Codes B := subsop([i,1] = z, B); end if; end do; end do; C := remove(t -> t[1] = z, B); n := nops(c); for i from 3 to n do for j from 1 to i- do if Qreal(C[j][1]) = Qreal(C[i][1]) and Qnorm(C[j][1]) = Qnorm(C[i][1]) then for k from j + 1 to i - 1 do if Qreal(C[k][1]) = Qreal(C[i][1]) and Qnorm(C[k][1]) = Qnorm(C[i][1]) then C := subsop([i, 1] = z, C); next; end if; end do; end if; end do; end do; return remove(t -> t[1] = z, C); end proc; LagrangeP := proc(b::listlist) #Calculates the Lagrange polynomial for a list of quaternion pairs. local A := NoSim(B), n := nops(a), b, c, i, j, f, g; options remember; 100

109 A. The Codes for i from 1 to n do b[i] := op(1, op(i, A)); c[i] := op(, op(i, A)); end do; for i from 1 to n do f[i] := ZeroP([seq(b[j], j in {seq(1..n)} minus {i})]); end do; for i from 1 to n do g[i] := M(Qinv(eval(f[i], b[i])), f[i]); end do; return sortcollect(add(m(c[i], g[i]), i = 1..n), x); end proc: QAdjointMatrix := proc(a::matrix) #Calculates the adjoint matrix of a quaternion matrix. local nrowsa := LinearAlgebra:-RowDimension(A), ncolsa := LinearAlgebra:-ColumnDimension(A), A1 := Matrix(1..nrowsA, 1..ncolsA), A1C := Matrix(1..nrowsA, 1..ncolsA), A := Matrix(1..nrowsA, 1..ncolsA), AC := Matrix(1..nrowsA, 1..ncolsA), AM := Matrix(1..*nrowsA, 1..*ncolsA), i, j; for i from 1 to nrowsa do for j from 1 to ncolsa do A1(i, j) := A(i, j) - coeff(a(i, j), J)*J - coeff(a(i, j), K)*K; 101

110 A. The Codes A1C(i, j) := Qconj(A1(i, j)); A(i, j) := coeff(a(i, j), J) + coeff(a(i, j), K)*I; AC(i,j) := Qconj(A(i, j)); AM(i, j) := A1(i, j); end do; end do; for i from 1 to nrowsa do for j from ncolsa + 1 to *ncolsa do AM(i, j) := A(i, j - ncolsa); end do; end do; for i from nrowsa + 1 to *nrowsa do for j from 1 to ncolsa do AM(i, j) := -AC(i - nrowsa, j); end do; end do; for i from nrowsa + 1 to *nrowsa do for j from ncolsa + 1 to *ncolsa do AM(i, j) := A1C(i - nrowsa, j - ncolsa); end do; end do; for i from 1 to *nrowsa do 10

111 A. The Codes for j from 1 to *ncolsa do AM(i, j) := subs(i = ii, AM(i, j)); end do; end do; return AM; end proc; QMFNorm := proc(a::matrix) #Calculates the frobinius norm of a quaternion matrix. local nrowsa := LinearAlgebra:-RowDimension(A), ncolsa := LinearAlgebra:-ColumnDimension(A), i, j, N := 0; for i from 1 to nrowsa do for j from 1 to ncolsa do N := N + coeff(a[i, j], I)^ + coeff(a[i, j], J)^ + coeff(a[i, j], K)^ + (Qreal(A[i, j]))^; end do; end do; return (expand(n))^(1/); end proc; NewtonP := proc(b::listlist) #Calculates the Newton polynomials and the divided difference for a list of quaternion pairs. local g, i, j, d, b, c, A := NoSim(B), n := nops(a); options remember; 103

112 A. The Codes for i from 1 to n do b[i] := op(1, op(i, A)); c[i] := op(, op(i, A)); end do; d[0] := c[1]; g[0] := 1; for i from 1 to n - 1 do g[i] := ZeroP([seq(b[j], j = 1..i)]); d[i] := M(c[i + 1] - c[1] - add(m(d[j], eval(g[j], b[i + 1])), j = 1..i - 1), Qinv(eval(g[i], b[i + 1]))); end do; return [seq(g[i], i = 0..n - 1)], [seq(d[i], i = 0..n - 1)]; end proc; LFI := proc (A::Matrix) #Calculates the generalized inverse of a quaternion polynomial matrix. local nrowsa := LinearAlgebra:-RowDimension(A), H := LinearAlgebra:-Transpose(Mconj(A)), P := QMatrixMultiply(A, H), a, B, AA, i, j, k := 0, LFI; a[0] := 1; B[0] := LinearAlgebra:-IdentityMatrix(nrowsA); for i from 1 to nrowsa do AA[i] := QMatrixMultiply(P, B[i - 1]); a[i] := -LinearAlgebra:-Trace(AA[i])/i; B[i] := AA[i] + QScalarMultiply(a[i], B[0]); 104

113 A. The Codes end do; for j from 1 to nrowsa do if a[j] <> 0 then k := j; end if; end do; if k = 0 then LFI := 0; else LFI := [(-1/a[k]), QMatrixMultiply(H, B[k - 1])]; end if; end proc: AXXB:= proc (A, B, C) #Calculates the solutions of x + Ax + xb + C = 0. local a0 := Qreal(A), a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b0 := Qreal(B), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), c := ((a0 + b0)/)^ - M((A + B), ((a0 + b0)/)) + C, c0 := Qreal(c), c1 := coeff(c, I), c := coeff(c, J),c3 := coeff(c, K), one := -4*(a1 + b1)*x^3 + 4*(a*b3 - a3*b - c1)*x^ + (b1 - a1)*(a1^ + a^ + a3^ - b1^ - b^ - b3^)*x 105

114 A. The Codes + *(c*(b3 - a3) + c3*(a - b))*x + (b1 - a1)*(c1*(a1 - b1) + c*(a - b) + c3*(a3 - b3)), two := -4*(a + b)*x^3 + 4*(a3*b1 - a1*b3 - c)*x^ + (b - a)*(a1^ + a^ + a3^ - b1^ - b^ - b3^)*x + *(c1*(a3 - b3) + c3*(b1 - a1))*x + (b - a)*(c1*(a1 - b1) + c*(a - b) + c3*(a3 - b3)), three := -4*(a3 + b3)*x^3 + 4*(a1*b - a*b1 - c3)*x^ + (b3 - a3)*(a1^ + a^ + a3^ - b1^ - b^ - b3^)*x + *(c1*(b - a) + c*(a1 - b1))*x + (b3 - a3)*(c1*(a1 - b1) + c*(a - b) + c3*(a3 - b3)), d := 8*x^3 + *((a1 - b1)^ + (a - b)^ + (a3 - b3)^)*x, f := 16*x^3 + 8*(a1^ + a^ + a3^ + b1^ + b^ + b3^ + *c0)*x^ + 4*c0*((a1 - b1)^ + (a - b)^ + (a3 - b3)^)*x + (a1^ + a^ + a3^ - b1^ - b^ - b3^)^*x + (8*(a*b3 - a3*b)*c1-4*c1^)*x + (8*(a3*b1 - a1*b3)*c - 4*c^)*x + (8*(a1*b - a*b1)*c3-4*c3^)*x - (a1*c1 + a*c + a3*c3 - b1*c1 - b*c - b3*c3)^, Z := evalf(op(select(t -> evalf(t) > 0, map(re + Im*ii, simplify(evalc([solve(f,x)])))))); return map(t -> 106

115 A. The Codes (t + subs(x = t,one/d)*i + subs(x = t,two/d)*j + subs(x = t,three/d)*k - (a0 + b0)/), {sqrt(z), -sqrt(z)}); end proc: AXBc:= proc (A, B, c) #Calculates the solutions of x + AxB + c = 0. local a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), one := x*(a3*b - a*b3), two := x*(a1*b3 - a3*b1), three := x*(a*b1 - a1*b), a := (a1^ + a^ + a3^)*(b1^ + b^ + b3^), b := a1*b1 + a*b + a3*b3, d := *x + b, f := x^4 + (c - (3/4)*a)*x^ + (1/4)*b*(4*c - a)*x + (1/4)*c*b^, sol := fsolve(f, x, real); return map(t -> (t + subs(x = t,one/d)*i + subs(x = t,two/d)*j + subs(x = t,three/d)*k), {sol}); end proc: AXB:= proc (A, B) 107

116 A. The Codes #Calculates the solutions of x + AxB = 0. local a0 := Qreal(A), a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b0 := Qreal(B), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), m11 := a0*b0 - a1*b1 + a*b + a3*b3, m1 := a0*b3 - a1*b - a*b1 - a3*b0, m13 := -a0*b - a1*b3 + a*b0 - a3*b1, m1 := -a0*b3 - a1*b - a*b1 + a3*b0, m := a0*b0 + a1*b1 - a*b + a3*b3, m3 := a0*b1 - a1*b0 - a*b3 - a3*b, m31 := a0*b - a1*b3 - a*b0 - a3*b1, m3 := -a0*b1 + a1*b0 - a*b3 - a3*b, m33 := a0*b0 + a1*b1 + a*b - a3*b3, d := 8*x^3 + 4*(m11 + m + m33)*x^ + *(m11*m + m11*m33 - m1*m1 - m13*m31 - m3*m3 + m*m33)*x + m11*m*m33 - m11*m3*m3 - m1*m1*m33 + m13*m1*m3 + m1*m3*m31 - m13*m*m31, n1 := -a0*b1 - a1*b0 - a*b3 + a3*b, n := -a0*b + a1*b3 - a*b0 - a3*b1, n3 := -a0*b3 - a1*b + a*b1 - a3*b0, one := 4*n1*x^3 + *(n1*(m + m33) - n*m1 - n3*m13)*x^ 108

117 A. The Codes + n1*(m*m33 - m3*m3)*x + n*(m13*m3 - m1*m33)*x + n3*(m1*m3 - m13*m)*x, two := 4*n*x^3 + *(n*(m11 + m33) - n1*m1 - n3*m3)*x^ + n1*(m3*m31 - m1*m33)*x + n*(m11*m33 - m13*m31)*x + n3*(m13*m1 - m11*m3)*x, three := 4*n3*x^3 + *(n3*(m11 + m) - n1*m31 - n*m3)*x^ + n1*(m1*m3 - m*m31)*x + n*(m1*m31 - m11*m3)*x + n3*(m11*m - m1*m1)*x, ab := (a0^ + a1^ + a^ + a3^)*(b0^ + b1^ + b^ + b3^), h := x^3 + *a0*b0*x^ + ((4/3)*a0^*b0^ - (1/1)*(a0^ - 3*a1^ - 3*a^ - 3*a3^) *(b0^ - 3*b1^ - 3*b^ - 3*b3^))*x + (1/4)*ab*(a0*b0 - a1*b1 - a*b - a3*b3), g := x^4 + *a0*b0*x^3 + (1/)*(4*a0^*b0^ - (a1^ + a^ + a3^ - a0^)*(b1^ + b^ + b3^ - b0^))*x^ + (1/)*a0*b0*ab*x + (1/16)*ab^, solg := fsolve(g, x, real), solh := fsolve(h, x, real); 109

118 A. The Codes if a0 = 0 and b0 = 0 then return map(t -> (t + subs(x = t,one/d)*i + subs(x = t,two/d)*j + subs(x = t,three/d)*k), {solh}); else return map(t -> (t + subs(x = t, one/d)*i + subs(x = t, two/d)*j + subs(x = t, three/d)*k), {solg, solh}); end if; end proc: SemiSim:= proc (B, A) yax = #Calculates (x, y) such that xby = B. A local a0 := Qreal(A), a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b0 := Qreal(B), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), Z; if Qnorm(A) <> Qnorm(B) then return "No Solution" elif abs(a0) <> abs(b0) then return "No Solution" else Z := LinearAlgebra:-LinearSolve( Matrix([[a0^ - b0^, -a0*a3 - b0*b3, a0*a + b0*b, a0*a1 - b0*b1, 0], 110

119 A. The Codes [a0*a3 + b0*b3, a0^ - b0^, -a0*a1 - b0*b1, a0*a - b0*b, 0], [-a0*a - b0*b, a0*a1 + b0*b1, a0^ - b0^, a0*a3 - b0*b3, 0], [-a0*a1 + b0*b1, -a0*a + b0*b, -a0*a3 + b0*b3, a0^ - b0^, 0]]), free = t ); end if; return sort(z[4] + Z[1]*I + Z[]*J + Z[3]*K, [I, J, K]) and simplify(m(b, M(Qinv(Z[4] + Z[1]*I + Z[]*J + Z[3]*K),Qinv(A)))); end proc: CSSim:= proc (B, A) yax = #Calculates (x, y) such that xby = B. A #Calculates x,y such that local a0 := Qreal(A), a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b0 := Qreal(B), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), Z; if Qnorm(A) <> Qnorm(B) then return "No Solution" else Z := LinearAlgebra:-LinearSolve( Matrix([[0, a0*b3 + a3*b0, -a0*b - b0*a, a*b3 - a3*b, 0], [-a0*b3 - a3*b0, 0, a0*b1 + a1*b0, -a1*b3 + a3*b1, 0], [a0*b + a*b0, -a0*b1 - a1*b0, 0, a1*b - a*b1, 0], [-a*b3 + a3*b, a1*b3 - a3*b1, -a1*b + a*b1, 0, 0]]), 111

120 A. The Codes free = t ); end if; return simplify(z[4] + Z[1]*I + Z[]*J + Z[3]*K) and simplify(m(m(qinv(b),qinv(qconj(z[4] + Z[1]*I + Z[]*J + Z[3]*K))),A)) end proc: LN1:= proc (A, B, C) #Calculates the solution of min ax xb c. x H local a0 := Qreal(A), a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b0 := Qreal(B), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), c0 := Qreal(C), c1 := coeff(c, I), c := coeff(c, J), c3 := coeff(c, K), Z; if Qnorm(A) = Qnorm(B) and Qreal(A) = Qreal(B) then Z := LinearAlgebra:-MatrixMatrixMultiply(LinearAlgebra:-MatrixInverse( Matrix([[a0 - b0, -a3 - b3, a + b, a1 - b1], [a3 + b3, a0 - b0, -a1 - b1, a - b], [-a - b, a1 + b1, a0 - b0, a3 - b3], [-a1 + b1, -a + b, -a3 + b3, a0 - b0]]), method = pseudo), Matrix(4, 1, {(1, 1) = c1, (, 1) = c, (3, 1) = c3, (4, 1) = c0})); else Z := LinearAlgebra:-LinearSolve( Matrix([[a0 - b0, -a3 - b3, a + b, a1 - b1, c1], [a3 + b3, a0 - b0, -a1 - b1, a - b, c], 11

121 A. The Codes [-a - b, a1 + b1, a0 - b0, a3 - b3, c3], [-a1 + b1, -a + b, -a3 + b3, a0 - b0, c0]]), free = t ); end if; return simplify(z[4] + Z[1]*I + Z[]*J + Z[3]*K); end proc: LN:= proc (A,B,C) #Calculates the solution of min ax xb c. x H local a0 := Qreal(A), a1 := coeff(a, I), a := coeff(a, J), a3 := coeff(a, K), b0 := Qreal(B), b1 := coeff(b, I), b := coeff(b, J), b3 := coeff(b, K), c0 := Qreal(C), c1 := coeff(c, I), c := coeff(c, J), c3 := coeff(c, K), Z; if Qnorm(A) = Qnorm(B) and Qreal(A) = Qreal(B) then Z := LinearAlgebra:-MatrixMatrixMultiply(LinearAlgebra:-MatrixInverse( Matrix([[a0 + b0, -a3 + b3, a - b, a1 - b1], [a3 - b3, a0 + b0, -a1 + b1, a - b], [-a + b, a1 - b1, a0 + b0, a3 - b3], [-a1 - b1, -a - b, -a3 - b3, a0 - b0]]), method = pseudo), Matrix(4, 1, {(1, 1) = c1, (, 1) = c, (3, 1) = c3, (4, 1) = c0})); else Z := LinearAlgebra:-LinearSolve( Matrix([[a0 + b0, -a3 + b3, a - b, a1 - b1, c1], [a3 - b3, a0 + b0, -a1 + b1, a - b, c], 113

122 A. The Codes [-a + b, a1 - b1, a0 + b0, a3 - b3, c3], [-a1 - b1, -a - b, -a3 - b3, a0 - b0, c0]]), free = t ); end if; return simplify(z[4] + Z[1]*I + Z[]*J + Z[3]*K); end proc: end module: 114

123 A. The Codes 115

124 A. The Codes 116