LEVERRIER-FADEEV ALGORITHM AND CLASSICAL ORTHOGONAL POLYNOMIALS

This is a reprint from Revista de la Academia Colombiana de Ciencias Vol 8 (106 (004, 39-47 LEVERRIER-FADEEV ALGORITHM AND CLASSICAL ORTHOGONAL POLYNOMIALS by J Hernández, F Marcellán & C Rodríguez

LEVERRIER-FADEEV ALGORITHM AND CLASSICAL ORTHOGONAL POLYNOMIALS by J Hernández, F Marcellán & C Rodríguez 1 To Professor Jairo Charris Castañeda, as a tribute of our mathematical friendship Resumen Hernández, J, F Marcellán & C Rodríguez: Leverrier Fadeev Algorithm and Classical Orthogonal Polynomials Rev Acad Colomb Cienc 8 (106: 39 47, 004 ISSN 0370-3908 Usando propiedades estructurales de los polinomios ortogonales clásicos (Hermite, Laguerre, Jacobi y Bessel, se implementa el algoritmo de Leverrier-Fadeev para obtener el polinomio característico de una matriz cuadrada de elementos complejos Palabras clave: Polinomio característico, funciones de transferencia, polinomios ortogonales, funcionales lineales clásicos Abstract Using structural properties of classical orthogonal polynomials (Hermite, Laguerre, Jacobi, and Bessel, an implementation of Leverrier-Fadeev algorithm to obtain the characteristic polynomial of a square matrix with complex entries is presented Key words: Characteristic Polynomial, Transfer Functions, Orthogonal Polynomials, Classical Linear Functionals 1 Introduction For a given matrix A C n n an algorithm attributed to Leverrier, Fadeev, and others, allows the simultaneous determination of the characteristic polynomial of A and the adjoint matrix of si A, wherei denotes theidentitymatrixinc n n Indeed, if p(s =det(si A =s n + a n k s k denotes the characteristic polynomial of A and 1 Departamento de Matemáticas, Universidad Carlos III de Madrid, Avenida de la Universidad 30, 8911, Leganés, Spain jhbenite@mathuc3mes, pacomarc@inguc3mes, teijeiro@mathuc3mes AMS Classification 000: Primary 4C0, 33C4 39

40 n Ã(s =Adj(sI A =s I + s k B n k 1 denotes the adjoint matrix of si A, and taking into account Ã(s =p(s(si A 1, then the coefficients (a k and the matrices (B k canbe generated from a 1 = tr A, B 1 = A + a 1 I, a k = 1 k tr (AB k 1, B k = a k I + AB k 1, (11 for k =,,n 1 Here tr A denotes the trace of the matrix A Notice that (11 can be read as follows (See [3] (si AÃ(s =p(si, dp(s ds =trã(s (1 Despite the little value from a numerical point of view, this algorithm is useful for theoretical purposes as well as for the applications in linear control theory More 1 precisely, is the transfer function of a continuous time linear system with n inputs and n outputs p(sã(s The algorithm takes into account the representation of the characteristic polynomial and the adjoint matrix in terms of the canonical basis { s k} n in the linear space of polynomials with complex coefficients and degree at most n From a computational point of view the accuracy of the algorithm using an orthogonal polynomial system is improved For some particular cases of orthogonal polynomials S Barnett [1] gave an implementation of the algorithm The key idea is the relation (1 as well as the expression of the derivative of the polynomial P k,k=1,,n, in terms of the family {P k } n The aim of our contribution is to present a general approach for families of classical orthogonal polynomials (Hermite, Laguerre, Jacobi, and Bessel taking into account a characterization of such families obtained in [4] Indeed it allows to give an expression of P k as a linear combination of P k+1,p k, and P k 1 Thuswecanshowavery simple implementation of the Leverrier algorithm, where parameters associated with the three-term recurrence relation play the main role The structure of the paper is the following In the section we summarize the basic properties of classical orthogonal polynomials In the section 3 we present the adapted version of Leverrier algorithm for bases of classical orthogonal polynomials, and we analyze it for each family of classical orthogonal polynomials In the section 4, some examples are tested Classical Orthogonal Polynomials Let u be a linear functional in the linear space IP of polynomials with complex coefficients If, denotes the duality bracket then c n = u, x n is said to be the moment of order n associated with the linear functional u The linear functional u is said to be quasi-definite [] if the principal submatrices of the Hankel matrix H =(c i+j i,j=0 are non-singular In such a case, there exists a unique sequence of monic polynomials {P n } n=0 such that (i u, x k P n =0, k =0, 1,,n 1 (ii u, x n P n 0 (iii degp n = n The sequence {P n } n=0 is said to be a sequence of monic orthogonal polynomials (SMOP with respect to u It is very well known that {P n } n=0 satisfies a three-term recurrence relation xp n (x =P n+1 (x+β n P n (x+γ n P (x, (1 n =1,, with γ n 0 The converse is also true and this result is due to several authors despite the fact is known as Favard s Theorem [] If q(x denotes a polynomial, then a new linear functional ũ = q(xu can be introduced as follows ũ, p(x = u, p(xq(x ( for every p IP On the other hand, as for a distribution, the derivative of the linear functional u, Du, is given by Du, p(x = u, p (x, p IP Definition 1 A linear functional u is said to be classical if there exist polynomials φ, ψ, with deg φ and deg ψ =1 suchthat D(φu =ψu (3 Up to a linear change of variables, four cases appear

41 (i φ(x = 1 This leads to Hermite linear functional with ψ(x = x (ii φ(x = x This leads to Laguerre linear functional with ψ(x = x + α +1 (iii φ(x =x 1 This yields the Jacobi linear functional with ψ(x = (α + β +x + β α (iv φ(x =x This yields the Bessel linear functional with ψ(x =(α +x + Theorem (see [4] If {P n } n=0 is the SMOP associated with u, then the following statements are equivalent (i u is a classical linear functional (ii {Q n } n=0, withq n = P n+1 n +1,isaSMOP (iii P n = Q n + r n Q + s n Q n (iv φ(xq n = a n P n+ + b n P n+1 + c n P n,withc n 0 Tabla 1 Coefficients in the three-term recurrence relation (1 β n Hermite 0 γ n n Laguerre n + α +1 n(n + α Jacobi Bessel β α (n + α + β(n + α + β + α (n + α(n + α + 4n(n + α(n + β(n + α + β (n + α + β 1(n + α + β (n + α + β +1 4n(n + α (n + α 1(n + α (n + α +1 Tabla Coefficients in the relation of the Theorem (iii r n s n Hermite 0 0 Laguerre n 0 Jacobi Bessel n(α β (n + α + β(n + α + β + 4n (n + α(n + α + 4n(n 1(n + α(n + β (n + α + β 1(n + α + β (n + α + β +1 4n(n 1 (n + α 1(n + α (n + α +1 3 Leverrier-Fadeev Algorithm Let {P n } n=0 be a sequence of monic orthogonal polynomials If we expand the characteristic polynomial p(s and the adjoint matrix Ã(s ofamatrixa Cn n in terms of the above basis in the linear space of polynomials with complex coefficients, then we get: p(s =P n (s+ â n k P k (s, (31

4 n Ã(s =P (si + P k (s ˆB n k 1 (3 From the first identity in (1 ( n (si A P (si + P k (s ˆB n k 1 = P n (si + â n k P k (si (33 Taking into account the three-term recurrence relation (1 for the family {P n } n=0, (33 becomes P n (si + â n k P k (si =[P n (s+β P + γ P n ] I P (sa + n (P k+1 (s+β k P k (s+γ k P k 1 (s ˆB n n k 1 P k (sa ˆB n k 1 Equating coefficients of P k in the previous expression we get A ˆB 0 = â 1 I + β ˆB0 + ˆB 1, A ˆB 1 = â I + γ ˆB0 + β n ˆB1 + ˆB, A ˆB n k 1 = â n k I + γ k+1 ˆBn k + β k ˆBn k 1 + ˆB n k, k =1,,,n 3, (34 A ˆB = â n I + γ 1 ˆBn + β 0 ˆB, with ˆB 0 = I Inamatrixform ˆB A ˆB 0 = M ˆB ˆB 0 where M = J n [0 â] J n is the Jacobi matrix of dimension n associated with the SMOP {P n } n=0 i e β 0 γ 1 0 0 1 β 1 γ J n = 0 0 γ 0 0 1 β and â = â n â â 1 In the literature, the matrix M is called the comrade matrix of A with respect to the orthogonal system {P n } n=0 His characteristic polynomial is p(s In particular, we get tr A = β j â 1 j=0 On the other hand, from the second relation in (1 for n =, 3, we have n P n(s+ â n k P k(s =np (s+ P k (str ˆB n k 1 (3 If {P n } n=0 is a classical family then, from theorem (iii, we get P k (s = P k+1 (s k +1 + r P k (s k k P k 1 + s (s k,k=, 3, k 1

43 Thus, substitution in (3 yields P n(s+ â n k P k(s = P n n(s+r n 1 P n (s+s n P n (s+ n ( P + k+1 (s k +1 + r P k (s P k 1 k + s (s k tr k k 1 ˆB n k 1 + k= ( +tr ˆB P 1(s+tr ˆB P (s n + r 1 P 1(s Finally, equating the coefficients of P k in both hand sides we get (n 1â 1 = nr +tr ˆB 1, (n â = ns + r n tr ˆB 1 +tr ˆB, kâ n k = s k+1 tr ˆB n k + r k tr ˆB n k 1 +tr ˆB n k, k =1,,,n 3 (36 Thus, in order to obtain (â k and(ˆb k we will proceed as follows First Step â 1 = n (β r tr A (37 Indeed, taking traces in the first equation of (34, and (36 { tr A = nâ 1 + nβ +tr ˆB 1, (n 1â 1 = nr +tr ˆB 1, and (37 follows Second Step ˆB 1 = A ˆB 0 +â 1 I β ˆB0 (38 Thus, for k =1,,,n 3, (n kâ n k =(β k r k tr ˆB n k 1 + (γ k+1 s k+1 tr ˆB n k tr n k 1, (311 as well as ˆB n k = A ˆB n k 1 +â n k I γ k+1 ˆBn k β k ˆBn k 1 These results follow from the expressions in (34 and (36 for k =1,,n 3 Finally, taking traces in the last equation of (34 we get nâ n = β 0 tr ˆB + γ 1 tr ˆB n tr Third Step â =(γ s tr ˆB 0 + (β n r n tr ˆB 1 tr 1 (39 As a conclusion we get Theorem 31 (i For k =0, 1,,n 1, Indeed, from the second equation in (34 and (36 { tr 1 = n (γ â +β n tr ˆB 1 +tr ˆB, tr ˆB = (n â ns r n tr ˆB 1, (n kâ n k =(β k r k tr ˆB n k 1 + (γ k+1 s k+1 tr ˆB n k tr n k 1, (31 and (39 follows Fourth Step ˆB = A ˆB 1 +â I γ ˆB0 β n ˆB1 (310 with the convention ˆB 1 =0, r 0 =0, s 1 =0 (ii For k =1,,,n 1 ˆB n k = A ˆB n k 1 +â n k I γ k+1 ˆBn k β k ˆBn k 1 (313

44 The implementation of the algorithm is as follows DATA: {β k }, {γ k} n k=1, {r k}, {s k} n k=1 Initial Condition: ˆB 1 =0, ˆB 0 = I 1 From ˆB n k and ˆB n k 1 taking into account (31 we get â n k From (313 we get ˆB n k END and, for each family of monic orthogonal polynomials, is given in below 3 Laguerre Case According to Theorem 31 we get (i (n kâ n k =[(k + α +1 k]tr ˆB n k 1 + (k+1(k+α+1tr ˆB n k tr n k 1 ie (n kâ n k =(k + α + 1tr ˆB n k 1 + (k +1(k + α + 1tr ˆB n k tr n k 1 31 Hermite Case According to Theorem 31 we get (i (n kâ n k = k+1 tr ˆB n k tr n k 1 (ii ˆB n k = A ˆB n k 1 +â n k I k +1 ˆB n k (314 In particular, taking traces in (ii and using (i we get tr ˆB n k = kâ n k This is formula (31 in [1] Furthermore, substituting in (i we get (n kâ n k = (k +1(k + â n k tr ie tr n k 1 = n k 1, (31 (k +1(k + â n k (n kâ n k 4k(k +1(k +1+α(k +1+β (k + α + β + 1(k + α + β + (k + α + β +3 (ii ˆB n k = A ˆB n k 1 +â n k I (k +1(k + α +1ˆB n k (k + α +1ˆB n k 1 (316 Taking traces in (ii and using (i we get tr ˆB n k = kâ n k ktr ˆB n k 1 Thus we deduce (n kâ n k =(k + α + 1tr ˆB n k 1 + (k + α +1 [(k +1â n k 1 tr ˆB ] n k 1 tr n k 1, ie (n kâ n k =(k + α +1(k +1â n k 1 tr n k 1 (317 Up to a normalization this is the formula (33b in [1], when α =0 33 Jacobi Case According to Theorem 31 we get ( β α (n kâ n k = (k + α + β(k + α + β + k(α β tr (k + α + β(k + α + β + ˆB n k 1 + ( 4(k +1(k +1+α(k +1+β(k +1+α + β (k + α + β + 1(k + α + β + (k + α + β +3 tr ˆB n k tr n k 1 = β α k + α + β + tr ˆB n k 1 + 4(k +1(k +1+α(k +1+β (k + α + β + (k + α + β +3 tr ˆB n k tr n k 1 On the other hand, if α = β then we are in the Gegenbauer case The linear functional is symmetric and thus get 4(k +1(k +1+α (n kâ n k = (k +α + (k +α +3 tr ˆB n k tr n k 1 = k +1 k +α +3 tr ˆB n k tr n k 1, (318

4 or, equivalently tr ˆB 4k(k +1(k +1+α n k = kâ n k + (k +α + 1(k +α + (k +α +3 tr ˆB n k, ie tr ˆB k(k +1 n k = kâ n k + (k +α + 1(k +α +3 ˆB n k Notice that the symmetry of the linear functional yields an important simplification in our algorithm Furthermore ˆB n k = A ˆB 4(k +1(k + α +1 (k +α +1 n k 1 +â n k I (k +α + 1(k +α + (k +α +3 ˆB n k = A ˆB n k 1 +â n k I (k +1(k +α +1 (k +α + 1(k +α +3 ˆB n k This is, up to the corresponding normalization, the formula (30 in [1] for α =0 (319 34 Bessel Case According to Theorem 31 we get (n kâ n k = ie = α 4k (k + α(k + α + tr ˆB 4(k + 1(k + α +1 n k 1 (k + α + 1(k + α + (k + α +3 tr ˆB n k tr k + α + tr ˆB 4(k +1 n k 1 (k + α + (k + α +3 tr ˆB n k tr n k 1, 1 (n kâ n k + k +1+ α tr ˆB n k 1 +tr together with ˆB n k = A ˆB n k 1 +â n k I + 4 Example n k 1 + ( k +1+ α n k 1 k +1 tr ˆB n k =0, (30 (k + α +3 4(k +1(k + α +1 (k + α + 1(k + α + (k + α +3 ˆB α n k + (k + α(k + α + ˆB n k 1 (31 Consider A = 1 4 1 1 0 4 1 1 3 1 4 1 6 which has characteristic polynomial a(s =s 4 s 3 +9s 7s + We apply the algorithm for each basis 41 Hermite Basis From (31, a 1 = tr A =, and from (314 4 4 1 4 B 1 = a 1 I + A = 4 1 1 7 3 1 4 1 1 Using (31, we get and from (314 B = a I 3 B 0 + AB 1 = Using again (31 and from (314 a =3 1 tr (AB 1=1, 7 1 10 9 17 33 3 9 3 7 7 3 a 3 = a 1 1 3 tr (AB = 9,

46 B 3 = a 3 I B 1 + AB = 1 Finally 8 0 1 1 4 11 49 14 1 11 39 11 3 8 33 7 a 4 = 1 4 a 1 4 tr (AB 3= 9 4 Hence, the characteristic polynomial of A is given by (31 as a(s =H 4 (s H 3 (s+1h (s 9 H 1(s+ 9 4 H 0(s 4,, Laguerre Basis We consider the family {L 0 n} n=0 From (317, a 1 =16 tr A =11, and from (316 4 1 4 B 1 = a 1 I 7B 0 + A = 4 4 1 1 3 1 4 1 10 Using (317, we get a = 9 a 1 1 tr (AB 1=36, and from (316 B = a I 9B 0 B 1 + AB 1 4 17 14 11 = 1 1 13 13 1 13 16 9 3 3 18 Using again (317 a 3 = 4 3 a 1 3 tr (AB =3, and from (316 B 3 = a 3 I 4B 1 3B + AB 7 4 7 = 4 6 1 10 3 7 4 9 4 11 Finally a 4 = 1 4 a 3 1 4 tr (AB 3=7 Hence, the characteristic polynomial of A is given by (31 as a(s =L 0 4(s+11L 0 3(s+36L 0 (s+3l 0 1(s+7L 0 0(s 43 Jacobi Basis We consider the family P n = P n (0,0 (Legendre Polynomials From (318, a 1 = tr A =, and from (319 B 1 = a 1 I + A = Using (318, we get and from (319 4 4 1 4 4 1 1 7 3 1 4 1 1 a = 6 7 1 tr (AB 1= 69 7, B = a I 9 3 B 0 + AB 1 = Using again (318 13 1 10 9 47 33 3 9 133 3 7 7 3 a 3 = 1 tr B 1 1 3 tr (AB = 10, and from (319 B 3 = a 3 I 4 1 B 1 + AB Finally = 1 3 10 3 16 19 71 19 1 17 1 4 14 49 10 a 4 = 1 1 tr B 1 4 tr (AB 3= 6 Hence, the characteristic polynomial of A is given by (31 as a(s =P 4 (s P 3 (s+ 69 7 P (s 10P 1 (s+ 6 P 0(s Now, We consider the family U n = P ( 1, 1 n (Chebyshev Polynomials of the second kind From (318, a 1 = tr A =, and from (319 B 1 = a 1 I + A = Using (318, we get 4 4 1 4 4 1 1 7 3 1 4 1 1 a = 3 4 1 tr (AB 1= 39 4,

47 and from (319 B = a I 1 4 B 0 + AB 1 = 1 10 9 19 33 3 9 3 3 7 7 1 Using again (318 a 3 = 1 9 tr B 1 1 3 tr (AB = 19, and from (319 Finally B 3 = a 3 I 1 4 B 1 + AB = 1 4 1 4 31 0 6 7 93 4 1 3 71 19 0 6 13 a 4 = 1 16 tr B 1 4 tr (AB 3= 3 8 Hence, the characteristic polynomial of A is given by (31 as a(s =U 4 (s U 3 (s+ 39 4 U (s 19 U 1(s+ 3 8 U 0(s 44 Bessel Basis We consider the family B n = Bn 0 From (30, a 1 = 1 tr A = 6, and from (31 4 1 4 B 1 = a 1 I + A = 6 4 1 1 8 3 1 4 1 0 Using (30, we get a = 1 1 6 tr B 1 1 tr (AB 1= 10 7, and from (31 B = a I + 1 3 B 0 + AB 1 = 33 3 9 9 11 38 7 6 8 168 6 8 3 3 Using again (30 a 3 = 1 30 tr B 1 1 6 tr B 1 3 tr (AB = 89 1, and from (31 Finally B 3 = a 3 I + 1 1 B 1 + AB = 1 3 1 1 3 34 43 17 3 17 43 141 7 6 31 116 10 a 4 = 1 1 tr B 1 4 tr B 3 1 4 tr (AB 3= 84 Hence, the characteristic polynomial of A is given by (31 as a(s =B 4 (s 6B 3 (s+ 10 7 B (s 89 1 B 1(s+ 84 B 0(s Acknowledgements The work of the first author has been supported by Fundación Universidad Carlos III de Madrid The work of the second author has received the financial support from Dirección General de Investigación (Ministerio de Ciencia y Tecnología of Spain, grant BFM003-0633- C03-0 References [1] S Barnett, Leverrier s algorithm for orthogonal polynomials bases, Linear Alg and Appl 36 (1996, 4 63 [] T S Chihara, An introduction to orthogonal polynomials, Gordon and Breach, New York, 1978 [3] J S Frame, Matrix functions and applications IV: Matrix functions and constituent matrices, IEEE Spectrum 1 (1964, 13 131 [4] F Marcellán, A Branquinho, & J C Petronilho, Classical orthogonal polynomials: A functional approach, Acta Appl Math 34 (1994, 83 303