VECTOR RECURRENCES. by J. MATTHYS

Transcription

1 R941 Philips Res. Repts 31, , 1976 LINEAR VECTOR RECURRENCES by J. MATTHYS Abstract General solution methods for linear vector recurrences are considered; an algorithm based on partial fraction expansion is shown to be the most efficient way of solving this problem. 1. Introduction The aim of this paper is to give a practical solution method for linear vector recurrences. The required definitions and notations are given in sec. 2. In section 3 a fundamental theorem is proved, yielding an equivalence relation for linear vector recurrences, together with a solution method, based on the reduction of a given vector recurrence to an equivalent set of scalar recurrences. A more practical algorithm is derived in sec. 4. It is based on the partial fraction expansion of the inverted characteristic matrix of the recurrence. In sec. 5 the boundary conditions are considered, whereas in sec. 6, the intimate connection between infinite block Hankel matrices and linear vector recurrences is recalled, which leads to the Z-transform approach of the problem. A particular solution of the inhomogeneous recurrence is derived in sec. 7, and finally, in sec. 8. an example is given. 2. Definitions and notations We are concerned with systems of linear recurrence equations, wich may be written as a single vector recurrence where Yn and '1}n are d vectors and the Af constant dx d matrices. Without loss of generality, Ao and Ak are assumed to be non zero. A sequence {Yn} satisfying recurrence (1) for all integer values of n will be called a solution of the recurrence. If the recurrence is homogeneous, i.e. an arbitrary linear combination of solutions is equally a solution and, conversely, every solution can be written as a linear combination of some minimal (1) (2)

2 LINEAR VECTOR RECURRENCES 121 set of m linearly independent solutions. Such a set is called a fundamental system or basis and is said to define the general solution of the homogeneous recurrence. Similarly the general solution of the complete recurrence (1) is defined by a particular solution of (1) and the general solution of (2). As a result, our main interest lies in the solution' of (2), which is completely characterized by the polynomial matrix Rom = Ao C k + Al C t - l +.. "+ At (3) Ro(C) is called the characteristic matrix, and its determinant, denoted ao =1= 0, am =1= 0, M +!-lo ~ kd (4) is called the characteristic polynomial of the recurrence. It is always assumed that em is not identically zero. If Rom is identically singular, it is easily seen that there is either an infinity of solutions, or no solution at all, and hence that the problem is initially ill-stated. In the following sections the homogeneous recurrence is treated. 3. Equivalence of recurrences 3.1. Theorem 1. If the characteristic matrices of two recurrences are equivalent, their respective solutions are related by a nonsingular linear transformation. Proof Let S(C) denote the characteristic matrix of the recurrence I L: B) ZII_) = O. )=0 (5) Suppose S(C) = P(C) Rom Q(C), (6) where P(C) and Q(C) are unimodular polynomial matrices, and let Q(C) be written as Q(C) = L: Qv Cv. v=o Q (7) Then, if {Zn} is a solution of (5), {Yn} defined as 11 Yn = L: Qv Zn+v v=o (8)

3 122 J. MATI'HYS is a solution of (2). This is easily verified by direct substitution, taking into account the relations resulting from the identification of the coefficients in (6). Obviously the symmetrical result also holds by the reciprocal transformation q' z, = L Qv' Yn+v, v=o (9) where the Qv' are the coefficients of Q-l(C). This establishes the required oneto-one correspondence between the solutions Comments If Rom and S(C) are left equivalent, the solutions of both recurrences are identical Writing the ith equation of (2) at a point n + r, instead of n, does not affect the problem as such and leaves the solutions unaltered. However, this reformulation is reflected in the characteristic matrix by a premultiplication by diag(c"). Therefore it is sufficient to consider the characteristic matrix within a premultiplication by diag (cr,) or an equivalent matrix The equivalence of recurrences is now defined as follows. Definition 1. Two recurrences are said to be equivalent if their characteristic matrices are related to each other by an expression (6) where Q(C) is a unimodular polynomial matrix and P(C) is equivalent to some diag (cr,). It is thus sufficient to solve one recurrence, to find by transformation solutions of all equivalent recurrences. (8) the In particular S(C) may denote the Smith canonical form of Rom. Since the former is a diagonal matrix, the corresponding recurrence (5) consists of d independent scalar recurrences, the general solution of which is known 4): for each component of Zn there are as many linearly independent solutions as there are nontrivial zeros in the corresponding element of S(C). This gives for the recurrences of this equivalence class a total number of M linearly independent solutions, where M denotes the reduced degree of em Ahomogeneous recurrence can thus be solved by the following algorithm. (1) Determine the Smith canonical form of the characteristic matrix Rom. (2) Determine the fundamental solutions of the resulting scalar recurrences. (3) Transform these solutions by (8) to yield the general solution of the given recurrence.

4 LINEAR VECTOR RECURRENCES 123 This method has the disadvantage to require the computation of the Smith canonical form. In the following section an alternative set of fundamental solutions is derived, which leads to a procedure avoiding the detour by the Smith form Remark In the following Sm will always denote the Smith canonical form of Ro(C). 4. The classical approach We try to find solutions of the form Substitution into (2) yields (10) (11) and a nontrivial solution is possible only if Ro(Cl) to Cl' S(C) can be written as sm = diag [CC - Cl)'} aic)], with 'lij ;;::::Vj-l and a";(cl) =1= 0. Now (11) is equivalent to S(Cl) Q-l(Cl) x = 0 is singular. With respect which yields, for VI = '11 2 =... = 'lido = 0, Vdo+ 1 > 0, d - do linearly independent solution vectors, e.g. the last d - do columns of Q(Cl). If Vd = 1, this is exactly the multiplicity of Cl; for Vd > 1, additional solutions are found by the classical technique of confluence of distinct zeros. This gives the fundamental solutions e(cl) = with m L (7) cln-i x" 1 ~I = (m -I)! q/m-i) (Cl). i= 1,2, d m = 0, 1, vj-l We now endeavour to derive the XI directly. Substitution of (14) into (2) yields, after some elementary simplifications m (12) (13) (14) (15) (16)

5 124 I.MATrHYS On the other hand, let denote the unipolar component corresponding to Cl of Ro-l(C). Then the (weak) degree of H(C) is known 1) to be the multiplicity #1 of Cl as a zero of e('). Further, taking the first 'Vd - 1 derivatives of the product CC - we obtain the following equalities 1=1 Cl)Yd Rom e;-l(c), (17) Ro(C 1) H Yd = 0, 1 - RO/(Cl) H Yd + RO(Cl) H Yd -1 = 0, 1! (I8) o (19) the XI satisfy equation (16). Moreover, since the rank ofthis matrix is deg H(C), the required fundamental solutions of the recurrence are derived by (14) from some #1 linearly independent columns of matrix (19). If this is done for all the zeros Cl of em, the general solution is found. The computation of the HI is straightforward and only requires the inversion of the characteristic matrix. The procedure can be summarized as follows. (1) Inversion of Ro(C). (2) Expansion of Ro-1 m into partial fractions. (3) For each pole Cl the resultant coefficient matrices constitute the array (I9) of rank #10 (4) #1 linearly independent columns of (19) yield the basic solutions by (14). 5. Boundary conditions For scalar k-step recurrences a solution is uniquely determined by its values at k given points. The same is true for vector recurrences if only Ao and Ak

6 LINEAR VECTOR RECURRENCES 125 are nonsingular. Ifnot, the system determining the coefficients ofthe fundament solutions is rectangular, and only has a solution if rank ~ (1) (2) (M) Ynl Ynl Ynl Ynl ~ (1) (2) (M) Ynz Ynz Ynz Ynz Y (1) Y (2) Y (M) nt n" n" Yn" ~ =M (20) where the Yn(l) (i = 1, 2,..., M) denote the fundamental solutions, and YnJ U = 1,..., k) the boundary values. Hence, only M elements of the boundary condition vectors are arbitrary, the others being fixed by condition (20). This seems to contradiet the fact that, for nonsingular Ao, any set of vectors Yo, Yl'..., Yk-l yields a unique sequence {Yn} for n ~ 0, even if M < kd However, this "solution" cannot be extended to n < 0, unless the given vectors satisfy (20). Moreover, only the M effectively arbitrary elements of condition (20) have an influence on the solution for n ~ k. In general, it is easily seen that, if e(c) is given by (4), and Ita> = kd - M - Ito, a set of k successive vectors Yo, Yb..., Yk-l is uniquely extended into the future (n ;:;k) iff their elements satisfy Ita> conditions, and uniquely into the past (n <0) iff they satisfy Ito supplementary conditions, leaving M degrees of freedom for a universally valid solution. If we are only interested in the interval 0 ~ n < 00, the possibility of having M + Ito linearly independent initial values may be taken into account by considering also the basic solutions (14) corresponding to C1 = 0 (with the convention on = (ln.o)' 6. Block Hankel matrices and the Z-transform 6.1. In the scalar case, there exists a well-known relationship between infinite Hankel matrices and linear recurrences 3). Similarly infinite block Hankel matrices may be considered. H= (21) where the Y, are dx d matrices. Since the following results are classical, the theorems are stated without proof (see e.g. ref. I). Theorem 2 The infinite block Hankel matrix (21) has finite rank s iff there exists between its elements a linear recurrence

7 126 J.MATTHYS k LA, Y n -I = 0 n = k, k + 1,... (22) with a characteristic polynomial of degree s, where s is the smallest number having this property. Remark. In the notation of (4), s = M + P.o, the term P.o standing for the additionallinearly independent initial conditions mentioned in sec. 5. Theorem 3 The infinite block Hankel matrix (21) has finite rank s iff the power series C- 1 Yo + C- 2 Y 1 + C- 3 Y 2 -l:"... yields a rational matrix Z(C) of degree s. (23) 6.2. The possibility to generate solutions of a linear recurrence for n ~ 0 as coefficients of a power series is used in the Z-transform approach 5). The generating function is a column of (23) and is easily seen to be where k-l x(c) = Ro -l(c) L R,+1(C) y" k-i R,(C) = LA, Ck-I-I. (24) (25) Yn may now be considered as the coefficient of cn in the Taylor expansion of C- 1 x(c- 1 ): 1 k-l Yn = ---.nn [C- 1 s;-1(c- 1 ) L R,+1(C- 1 ) Y,)c=o n. (26) which, in general, is not very practical, or, alternatively as a coefficient of a Laurent series, which yields k-l = L Res W Ra.-l.(C) L R ' +1(C) y,). (27) By elementary algebra, this can again be written in terms of the fundamental solutions (14). In practice, of course, the residues.are computed directly from (27).

8 LINEAR VECI'OR RECURRENCES The nonhomogeneous recurrence To obtain the general solution of a non homogeneous recurrence' (1) it remains to find a particular solution. Often this is easily done by identification techniques; with the Z-transform approach a non-zero second member is taken directly into account by adding a supplementary term ex> e;-l(c) L c-n+k "In n=k (28) to x(c) as given by (24). An alternative particular solution can be found as follows. We first observe that the leading coefficient matrix can be made nonsingular without affecting the solutions, by successively reducing Ao to its normal form and writing the equations corresponding to the resulting zero rows at the point n + 1 instead of n. This yields, after a finite number of steps, the required nonsingular coefficient matrix. By an argument, which is completely similar to the scalar case (ref. 4, p. 212), a particular solution of the latter recurrence for n ~ 0 is n = 0, 1, 2,..., k - 1 (29) if, for n >j, Yn,J is the solution of k LAL Yn-I,J = 0 (30) initialized by y. - n,j -. o ~,Ao-l n = 0, 1,..., j - 1 n=j (31) 8. Example As an example we solve the recurrence 1 0 0] [-2 0 1]' [1 1 1]..:,:..[ n ] [ o 1 1 Yn Yn-l 'Yn- 2 = n - 1.'. o ' n 2, ".'! (32) The general solution 'of the homogeneous equations is found from the s~~cessive expressions

9 128 I.MATrHYS (33) em = C(C_1)3 (C -2) (34) 1 [C(C -1) (C-2) R~-1(C)=- 0 e(c) 0 3C C(C _1)2 -(3C + 1)(C- 1)2 as 1 [ [0-7 7] 1 [1-6 = C C (C - 1) [ (C - 1) ] 1 [ ] , o 2(C-2) (35) (36) A particular solution is found by direct substitution of a fifth degree polynomial vector in n, and identification of the coefficients, yielding [ _L ns + 2. n n 3 _ II n 2 ] n n 2 + 4/ n. _n n n + 5 (37) Acknowledgement The author wishes to thank: Dr. Y. Genin for many valuable discussions. MBLE Research Laboratory Brussels, January 1976 REFERENCES 1) V. Belevitch, Classical network theory, Holden-Day, San Francisco, ~ B. Dejon, SIAM J. numero Anal. 4, , j F. R. Gantm ach e r;' Matrizenrechnung, VEB Deutscher Verlag der Wissenschaften, Berlin, 1958/59, Vol.,1/11. 4) P. Henrici, Discrete variable methods in ordinary differential equations, WHey and Sons, New York, ) W. Kaplan, Operational methods for linear systems, Addison-Wesley, Reading, Massachussets, ) H. M. Sloate and T. A. Bickart, J. ACM 20,7-26, 1973.