Standard Polynomial Equations over Division Algebras

Transcription

1 Standard Polynomial Equations over Division Algebras Adam Chapman, Casey Machen Department of Mathematics, Michigan State University, East Lansing, MI Abstract Given a central division algebra D of degree d over a field F, we associate to any standard polynomial φ(z) = z n +c n z n + +c 0 over D a companion polynomial Φ(z) of degree nd with coefficients in F whose roots are exactly the conjugacy classes of the roots of φ(z) We explain how in case D is a quaternion algebra, all the roots of φ(z) can be recovered from the roots of Φ(z) On the way, we also generalize certain theorems that were known for H to any division algebra, such as the connection between the right eigenvalues of a matrix and the roots of its characteristic polynomial, and the connection between the roots of a standard polynomial and left eigenvalues of the companion matrix Keywords: Polynomial Equations, Division Algebras, Division Rings, Right Eigenvalues, Left Eigenvalues 200 MSC: primary 6K20; secondary 5A8, 5B33, R52 Introduction Ever since the first example of a noncommutative division algebra - the real quaternion algebra H - was discovered by Hamilton, the computational aspects of these algebras have interested mathematicians, especially solving polynomial equations and finding the left and right eigenvalues of a quaternionic matrix One of the earliest papers on this subject is [Niv4], where it was proved that every standard polynomial over the real quaternion algebra D over a addresses: adamchapman@yahoocom (Adam Chapman), machenc@gmailcom (Casey Machen) Preprint submitted to?? August 7, 205

2 real closed field, ie a polynomial of the form z n + c n z n + + c 0 with c 0,, c n D, has a root in D A good survey on polynomial equations over more general division rings can be found in [GM65] The current state of the arts when it comes to polynomial equations over quaternion algebras can be outlined as follows: In [HS02] a solution was given to any quadratic (ie n = 2) standard polynomial over H In [Abr09] it was generalized to any quaternion algebra over fields of characteristic not 2, and in [Cha5] to quaternion algebras over fields of characteristic 2 as well In [SPV0] it was shown that the roots of any standard polynomial of degree n over H are also roots of the companion polynomial, ie the characteristic polynomial of the companion matrix, which is a polynomial of degree 2n with coefficients in R, and in [JO0] it was shown how the roots of the original polynomial can be recovered form the roots of the companion polynomial In this paper we generalize the result from [SPV0] to any central division algebra, and in the special case of quaternion algebras it provides a method for finding all the roots of a given standard polynomial We make use of the theory of left and right eigenvalues, which was studied extensively for matrices over H and here we present some generalizations to any central division algebra 2 Right Eigenvalues of Matrices over Central Division Algebras By a central division algebra we mean a division algebra D which is of finite dimension over its center F, a field The dimension in this case is known to be d 2 for some positive integer d which we call the degree of D Let D be a central division algebra of degree d over F and let n be some positive integer We write M n (D) for the ring of n n matrices over D, and D n for the column vector space of dimension n over D The vector space D n is a left M n (D)-module and a D bi-module A right eigenvalue of a matrix A in M n (D) is an element λ D for which there exists a vector v D n satisfying Av = vλ A left eigenvalue is defined in a similar way, just with Av = λv The ring M n (D) can be identified with the ring of endomorphisms of D n as a right D-module Let K be a maximal subfield of D The ring D can be viewed as a vector space over K, and in particular [D : K] = d Therefore, D n (as a right D-module) can be identified with the vector space K nd Let g : D n K nd be a K-vector space isomorphism Note that g(vλ) = g(v)λ for 2

3 any v D n and λ K Via this identification, each element in M n (D) can be identified with an endomorphism of K dn This gives rise to an embedding f : M n (D) M nd (K) Note that g(av) = f(a)g(v) for any A M n (D) and v D n The fact that M n (D) embeds into M nd (K) is well-known The characteristic polynomial Φ A (z) of A is defined to be the determinant of f(a) zi where I is the identity matrix in M nd (K) The coefficients of Φ A (z) are known to be in F, not just in K, and they are independent of the choice of maximal subfield K of D Note that by saying that two elements d and d 2 in D are conjugates we mean that there exists some nonzero q D such that d = qd 2 q (Do not be confused with the complex conjugate or the real quaternion conjugate) The conjugacy class of an element in D is the set of all its conjugates It was shown in [Lee49] that the conjugacy classes of right eigenvalues of an n n matrix A over H = R+iR+jR+ijR are the conjugacy ( classes) of roots A A of the complex roots of the characteristic polynomial of 2 where A 2 A A and A 2 are the n n matrices over R(i)( = C) satisfying A = A + ja 2 We now generalize this fact to any central division Theorem 2 Let D be a central division F -algebra of degree d, and n be a positive integer Let A M n (D) An element λ D is a right eigenvalue of A if and only if Φ A (λ) = 0 Proof Let K be a maximal subfield of D containing λ Let g : D n K nd and f : M n (D) M nd (K) be as defined above Then f(a)g(v) = g(av) = g(vλ) = g(v)λ Assume λ is a right eigenvalue Then there exists a nonzero vector v D n such that Av = vλ Hence λ is an eigenvalue of f(a), and so a root of Φ A (z) Assume Φ A (λ) = 0 Then λ is an eigenvalue of f(a), so there exists a nonzero vector w K nd such that f(a)w = wλ Let v = g (w) Then Av = vλ, hence λ is a right eigenvalue of A Note that since Φ A (z) has coefficients in F, for each root λ of Φ A (z) all its conjugates are roots as well, so one can consider the roots of Φ A (z) as a collection of conjugacy classes Each such conjugacy class corresponds to a finite field extension of F, and so can be identified with an element in the algebraic closure F of F Therefore, in order to find all the roots of Φ A (z), one can solve it as a polynomial over F, and then for each root λ check 3

4 whether F (λ) is a subfield of D If it is, then λ is a root of Φ A (z) in D, and otherwise it is not 3 Standard Polynomials and Left Eigenvalues of the Companion Matrix Let φ(z) = z n + c n z n + + c 0 be a standard polynomial with coefficients c 0,, c n in D We want to find the roots of φ(z) in D, ie all the elements λ D satisfying φ(λ) = 0 Given such a polynomial, we define its companion matrix to be C φ = c 0 c c n 2 c n Theorem 3 The roots of φ(z) are exactly the left eigenvalues of C φ Furthermore, given such a left eigenvalue λ of the companion matrix C φ, the vector v = is a corresponding left eigenvector, ie a vector satis- λ fying C φ v = λv Proof The element λ D is a left eigenvalue of C φ if and only if there exists a nonzero vector v v = D n satisfying C φ v = λv This equality is equivalent to the system v n v 2 = λv v n = λv n c 0 v c n v n = v n 4

5 The first n equations mean that v is then becomes λ (c 0 + c λ + + c n + λ n )v = 0 v and the last equation By dividing by v from the right (and we can since v is a nonzero vector), we obtain that λ is a root of φ(z) In the opposite direction, it is straight-forward to see that for any λ D satisfying c 0 + c λ + + c n + λ n = 0, C φ λ = λ which means that λ is a left eigenvalue of C φ λ Corollary 32 Every left eigenvalue of C φ is also a right eigenvalue Proof Let λ be a left eigenvector Then is the corresponding λ eigenvector Now λ λ = λ λ Corollary 33 Every right eigenvalue of C φ is conjugate to some left eigenvalue of C φ Proof Assume λ is a right eigenvalue Then there exists a nonzero vector v v = D n v n 5

6 satisfying C φ v = vλ From this equality we obtain the system v 2 = v λ v n = v n λ c 0 v c n v n = v n Substituting the first n equations in the last one we obtain c 0 + c v λ + + c n v + v λ n = 0 Since v is nonzero, v is nonzero Take λ = v λv Then λ is a root of φ(z) and so a left eigenvalue of C φ Let the characteristic polynomial of C φ be denoted by Φ(z) We call it the companion polynomial of φ(z) Note that it is of degree nd and has coefficients in F Remark 34 It is not mentioned explicitly in [JO0], but the companion polynomial q(z) = z 2n + (c n + c n )z n + associated to any polynomial p(z) = z n + c n z n + is indeed the characteristic polynomial of the companion matrix This can be easily verified by a straight-forward computation: take the companion matrix C p, write it as A + ja 2 for the appropriate A, A 2 M n (R(i)) and compute the characteristic polynomial of ( ) A A 2 A 2 A Theorem 35 The roots of φ(z) are also roots of Φ(z), and each conjugacy class of roots of Φ(z) contains a root of φ(z) Proof By Theorem 3 the roots of φ(z) are left eigenvalues of the companion matrix By Corollary 32 those left eigenvalues are also right eigenvalues, and by Theorem 2 the right eigenvalues are roots of the companion polynomial The second assertion is immediate from Corollary 33 4 Standard Polynomials over Quaternion Algebras Let φ(z) be again a standard polynomial over a central division algebra D of degree d over F The algebra D is called a quaternion algebra if d = 2 6

7 Every element λ D has a characteristic polynomial, which means that λ d + b d λ d + + b 0 = 0 for some b d,, b 0 F Note that the characteristic polynomial of λ does not change if we conjugate λ by any q D Write C d (z),, C 0 (z) for the functions from D to F such that z d + C d (z)z d + + C 0 (z) is the characteristic polynomial of z Write φ(z) as ψ d (z)z d + +ψ 0 (z) using the functions C d (z),, C 0 (z) For example, if φ(z) = z 3 and d = 2 then φ(z) = z 2 z + c = ( C (z)z C 0 (z))z + c = C (z)z 2 C 0 (z)z = C (z)( C (z)z C 0 (z)) C 0 (z)z = (C (z) 2 C 0 (z))z + C (z)c 0 (z) and so ψ (z) = C (z) 2 C 0 (z) and ψ 0 (z) = C (z)c 0 (z) By Theorem 35, the roots of φ(z) are also roots of Φ(z) Assuming we have all the roots of Φ(z) at hand, we want to recover the roots of φ(z) The following statement is immediate: Proposition 4 Given a root λ of Φ(z), λ is a root of φ(z) if and only if λ is a root of the polynomial Ψ(z) = ψ d (λ)z d + + ψ 0 (λ) Assume d = 2 and fix a root λ of Φ(z) Then the polynomial Ψ(z) defined in the proposition above is either a linear polynomial or a constant It cannot be a nonzero constant, because that would mean that no element in the conjugacy class of λ is a root of φ, contradictory to Theorem 35 If Φ(z) is constantly zero then every element in the conjugacy class of λ is a root of φ(z) If Φ(z) is a linear polynomial, then the root of φ(z) in the conjugacy class of λ can be found simply by solving the linear equation Ψ(z) = 0 Acknowledgements We thank Uzi Vishne and Uriya First for the useful discussions Bibliography [Abr09] Marco Abrate, Quadratic formulas for generalized quaternions, J Algebra Appl 8 (2009), no 3, MR (200h:6038) [Cha5] Adam Chapman, Quaternion quadratic equations in characteristic 2, J Algebra Appl 4 (205), no 3, , 8 MR

8 [GM65] B Gordon and T S Motzkin, On the zeros of polynomials over division rings, Trans Amer Math Soc 6 (965), MR (33 #4050a) [HS02] Liping Huang and Wasin So, Quadratic formulas for quaternions, Appl Math Lett 5 (2002), no 5, MR (2003d:2003) [JO0] Drahoslava Janovská and Gerhard Opfer, A note on the computation of all zeros of simple quaternionic polynomials, SIAM J Numer Anal 48 (200), no, MR (20c:70) [Lee49] H C Lee, Eigenvalues and canonical forms of matrices with quaternion coefficients, Proc Roy Irish Acad Sect A 52 (949), MR (2,53i) [Niv4] Ivan Niven, Equations in quaternions, Amer Math Monthly 48 (94), MR (3,264b) [SPV0] R Serôdio, E Pereira, and J Vitória, Computing the zeros of quaternion polynomials, Comput Math Appl 42 (200), no 8-9, , Numerical methods and computational mechanics (Miskolc, 998) MR (2002f:3006) 8