The inverse of a tridiagonal matrix

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1 Linear Algebra and its Applications 325 (2001) The inverse of a tridiagonal matrix Ranjan K. Mallik Department of Electrical Engineering, Indian Institute of Technology, Hauz Khas, New Delhi , India Received 6 March 1998; accepted 1 September 2000 Submitted by R.A. Brualdi Abstract In this paper, explicit formulae for the elements of the inverse of a general tridiagonal matrix are presented by first extending results on the explicit solution of a second-order linear homogeneous difference equation with variable coefficients to the nonhomogeneous case, and then applying these extended results to a boundary value problem. A formula for the characteristic polynomial is obtained in the process. We also establish a connection between the matrix inverse and orthogonal polynomials. In addition, the case of a cyclic tridiagonal system is discussed Elsevier Science Inc. All rights reserved. AMS classification: 15A09; 39A10 Keywords: Tridiagonal matrix; Inverse; Second-order linear difference equation; Variable coefficients; Explicit solution; Orthogonal polynomials 1. Introduction Tridiagonal matrices [1 4] are connected with different areas of science and engineering, including telecommunication system analysis [5] and finite difference methods for solving partial differential equations [4,6,7]. In many of these areas, inversions of tridiagonal matrices are necessary. Efficient algorithms [8], indirect formulae [1,9 12], and direct expressions in some special cases [4,7] for such inversions are known. Bounds on the elements of the inverses of diagonally dominant tridiagonal matrices have also been obtained [13]. However, explicit formulae for the Tel.: ; fax: address: rkmallik@ee.iitd.ernet.in (R.K. Mallik) /01/$ - see front matter 2001 Elsevier Science Inc. All rights reserved. PII:S (00)

2 110 R.K. Mallik / Linear Algebra and its Applications 325 (2001) elements of a general tridiagonal matrix inverse, which can give a better analytical treatment to a problem, are not available in the open literature [1]. In this paper, we present explicit formulae for the elements of the inverse of a general tridiagonal matrix. The approach is based on linear difference equations [14,15], and is as follows. Results on the explicit solution of a second-order linear homogeneous difference equation with variable coefficients [16] are extended to the nonhomogeneous case. Then these extended results are applied to a boundary value problem to obtain the desired formulae. A formula for the characteristic polynomial is obtained in the process. We also establish a connection between the matrix inverse and orthogonal polynomials. In addition, the case of a cyclic tridiagonal system is discussed. 2. Results on the second-order linear homogeneous difference equation with variable coefficients In this section, we will review some results on the second-order linear homogeneous difference equation with variable coefficients [16]. Let N denote the set of natural numbers. A set S q (L, U), whereq,l,u N, has been defined in [16] as the set of all q-tuples with elements from {L, L + 1,...,U} arranged in ascending order so that no two consecutive elements are present,thatis, S q (L, U) {L, L + 1,...,U} if U L and q = 1, (1a) { (k 1,...,k q ): k 1,...,k q {L, L + 1,...,U}; k l k l 1 2 for l = 2,...,q } U L + 2 if U L + 2and2 q, 2 (1b) otherwise. (1c) For example, S 1 (2, 6) ={2, 3, 4, 5, 6}, S 2 (2, 6) ={(2, 4), (2, 5), (2, 6), (3, 5), (3, 6), (4, 6)}, S 3 (2, 6) ={(2, 4, 6)}, S 4 (2, 6) =. The following results have been proved in [16]: For U L,1 q (U L + 2)/2, ( ) U L q + 2 (U L q + 2)! S q (L, U) = = q q!(u L 2q + 2)!. (2) For U L, q 2,

3 R.K. Mallik / Linear Algebra and its Applications 325 (2001) S q (L, U + 1)=S q (L, U) { (k 1,...,k q 1,k q ): k q = U + 1; (k 1,...,k q 1 ) S q 1 (L, U 1) }. (3) We add a proposition which is similar to result (3). Proposition 1. For U L, q 2, S q (L 1,U)=S q (L, U) { (k 1,k 2,...,k q ): k 1 = L 1; (k 2,...,k q ) S q 1 (L + 1,U) }. (4) Proof. If either U = L, q 2orU L + 1, q> (U L + 3)/2, then Eq. (4) holds trivially. For U L + 1, 2 q (U L + 3)/2, we get, from (1a) (1c), S q (L 1,U)= { (k 1,...,k q ): k 1,...,k q {L 1,L,...,U}; k l k l 1 2forl = 2,...,q }. (5) The right-hand side of (5) can be expressed as the union of two disjoint sets, one containing q-tuples with k 1 /= L 1 and the other containing q-tuples with k 1 = L 1, which we denote as S q0 (L 1,U)and S q1 (L 1,U), respectively. Therefore, where S q (L 1,U)= S q0 (L 1,U) S q1 (L 1,U), (6) S q0 (L 1,U)= { (k 1,...,k q ): k 1,...,k q {L, L + 1,...,U}; k l k l 1 2forl = 2,...,q }, S q1 (L 1,U)= { (k 1,...,k q ): k 1 = L 1; k 2,...,k q {L, L + 1,...,U}; k l k l 1 2forl = 2,...,q }. (7) It is clear from (1a) (1c) that S q0 (L 1,U)= S q (L, U). (8) In the set S q1 (L 1,U),sincek 1 = L 1, we have k 2 (L 1) 2, which implies that k 2,...,k q {L + 1,L+ 2,...,U}. Therefore, S q1 (L 1,U)= { (k 1,k 2,...,k q ): k 1 = L 1; (k 2,...,k q ) S q 1 (L + 1,U) }. (9) Combining Eqs. (6), (8) and (9), we get (4).

4 112 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Consider the second-order linear homogeneous difference equation y n+2 = A n y n+1 + B n y n, n 1, (10) with integral index n, variable complex coefficients A n and B n, B n /= 0, and complex initial values y 1,y 2.Define,fork 2, σ k B k. (11) A k 1 A k It has been shown in [16] that the explicit solution of difference equation (10), which is an expression for y n+2 in terms of only coefficients A 1,...,A n, B 1,...,B n,and initial values y 1,y 2,isgivenby where y n+2 = C n y 2 + D n y 1, n 0, (12) C 0 = 1, C 1 = A 1, (13a) D 0 = 0, D 1 = B 1, D 2 = B 1 A 2 (13b) and C n = (A 1 A n ) 1 + n/2 (σ k1 σ kq ) (k 1,...,k q ) S q (2,n) for n 2, (14a) D n = B 1 (A 2 A n ) 1 + (n 1)/2 (k 1,...,k q ) S q (3,n) (σ k1 σ kq ) for n 3, (14b) where S q (L, U) is defined by (1a) (1c), and σ k is defined by (11). 3. Solution of the second-order linear nonhomogeneous difference equation with variable coefficients We now focus on the second-order linear nonhomogeneous difference equation y n+2 = A n y n+1 + B n y n + x n+2, n 1, (15) with integral index n, variable complex coefficients A n and B n, B n /= 0, complex forcing term x n+2, and complex initial values y 1,y 2.

5 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Let σ k = B k /(A k 1 A k ) for k 2 as in (11). For n 0, i 1, define a quantity E n (i), which is a function of the A k sandtheσ k s, as (n i+1)/2 E n (i) (A i A n ) 1 + (σ k1 σ kq ) (k 1,...,k q ) S q (i+1,n) if i = 1,...,n 1, n 2, A n if i = n, n 1, 1 if i = n + 1, n 0, 0 otherwise. (16) Therefore, E n (i) is nontrivial only for n i 1, i 1. Note that E n (1) = C n for n 0, (17) where C n is given by (13a) and (14a). Two recurrences for E n (i) are established by the following two propositions. Proposition 2. For n i 1,i 1, E n+2 (i) = A n+2 E n+1 (i) + B n+2 E n (i). (18) Proof. Using the definition of E n (i) in (16) and σ i in (11), the right-hand side of (18) can be written for n = i 1, i 1as A i+1 E i (i) + B i+1 E i 1 (i)=a i+1 A i + B i+1 =(A i A i+1 )(1 + σ i+1 ) =E i+1 (i) (19) and for n = i, i 1as A i+2 E i+1 (i) + B i+2 E i (i)=(a i+2 A i+1 A i )(1 + σ i+1 ) + B i+2 A i =(A i A i+1 A i+2 )(1 + σ i+1 + σ i+2 ) =E i+2 (i), (20) which imply that (18) holds for n = i 1,i, i 1. Now consider the case when n i + 1, i 1. Again, using (16) and noting the fact that B n+2 = A n+1 A n+2 σ n+2,weget A n+2 E n+1 (i) + B n+2 E n (i) ( = (A i A n+2 ) 1 + (n i+2)/2 (k 1,...,k q ) S q (i+1,n+1) (σ k1 σ kq )

6 114 R.K. Mallik / Linear Algebra and its Applications 325 (2001) (n i+3)/2 + σ n+2 + q=2 ( = (A i A n+2 ) (n i+2)/2 q=2 (n i+3)/2 q=2 (k 1,...,k q 1 ) S q 1 (i+1,n) n+2 k 1 =i+1 σ k1 (k 1,...,k q ) S q (i+1,n+1) (k 1,...,k q 1 ) S q 1 (i+1,n) (σ k1 σ kq ) (σ k1 σ kq 1 )σ n+2 ) (σ k1 σ kq 1 )σ n+2 ). (21) Now n i + 2 n i + 3 if (n i + 2) is even, 2 = 2 n i if(n i + 2) is odd. 2 If (n i + 2) is odd, by (1a) (1c), (22) S (n i+3)/2 (i + 1,n+ 1) = S (n i+2)/2 +1 (i + 1,n+ 1) =. Then, using (3), we get S q (i + 1,n+ 2)= S q (i + 1,n+ 1) { (k 1,...,k q ): k q = n + 2; (k 1,...,k q 1 ) S q 1 (i + 1,n) }, (23) for q = 2,..., (n i + 3)/2. Putting together the summation terms of (21) using (22) and (23), we obtain A n+2 E n+1 (i) + B n+2 E n (i) ( = (A i A n+2 ) 1 + = E n+2 (i) (n i+3)/2 (k 1,...,k q ) S q (i+1,n+2) (σ k1 σ kq ) ) (24) for n i + 1, i 1. A combination of (19), (20) and (24) yields (18). Proposition 3. For 1 i n 1,n 2, E n (i) = A i E n (i + 1) + B i+1 E n (i + 2). (25)

7 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Proof. From the definition of E n (i) in (16), the right-hand side of (25) can be written for i = n 1as A n 1 E n (n) + B n E n (n + 1)=A n 1 A n + B n =(A n 1 A n )(1 + σ n ) =E n (n 1) (26) and for i = n 2as A n 2 E n (n 1) + B n 1 E n (n)=(a n 2 A n 1 A n )(1 + σ n ) + B n 1 A n =(A n 2 A n 1 A n )(1 + σ n + σ n 1 ) =E n (n 2), (27) which imply that (25) holds for i = n 2,n 1. Now consider the case when 1 i n 3. Using (16) and (11) we obtain A i E n (i + 1) + B i+1 E n (i + 2) ( (n i)/2 = (A i A n ) 1 + ( = (A i A n ) 1 + (n i+1)/2 + q=2 n k 1 =i+1 (n i+1)/2 + q=2 (k 1,...,k q ) S q (i+2,n) (k 2,...,k q ) S q 1 (i+3,n) (n i)/2 σ k1 + q=2 (k 2,...,k q ) S q 1 (i+3,n) (σ k1 σ kq ) + σ i+1 σ i+1 (σ k2 σ kq ) (k 1,...,k q ) S q (i+2,n) ) (σ k1 σ kq ) ) σ i+1 (σ k2 σ kq ). (28) Now n i n i + 1 if (n i) is even, 2 = 2 n i + 1 if(n i) is odd. 2 If (n i) is odd, then by (1a) (1c), S (n i+1)/2 (i + 2,n)= S (n i)/2 +1 (i + 2,n)=. Then, using Proposition 1, we get (29) S q (i + 1,n)=S q (i + 2,n) { (k 1,...,k q ): k 1 = i + 1; (k 2,...,k q ) S q 1 (i + 3,n) }, (30)

8 116 R.K. Mallik / Linear Algebra and its Applications 325 (2001) for q = 2,..., (n i + 1)/2. Putting together the summation terms of (28) using (29) and (30), we obtain (25). The following proposition provides the explicit solution of the difference equation (15), which consists of an expression for y n+2 in terms of only coefficients A 1,...,A n, B 1,...,B n, initial values y 1,y 2, and forcing terms x 3,...,x n+2. Proposition 4. The solution of difference equation (15) with initial values y 1,y 2 is given by n+2 y n+2 = C n y 2 + D n y 1 + E n (k 1)x k, n 1, (31) k=3 where E n (k) is defined by (16), and C n and D n by (13a), (13b), (14a), and (14b). Proof. The solution of (15) with initial values y 1,y 2 can be expressed as y n+2 = H n + P n, n 1, (32) where H n is the solution of the homogeneous equation, andp n the particular solution. From the results of Section 2 (Eqs. (10) and (12)), we have H n = C n y 2 + D n y 1, n 1. (33) It is also known that the particular solution P n satisfies the recurrence P n+2 = A n+2 P n+1 + B n+2 P n + x n+4, n 1. (34) Using (15), (13a), (13b), (14a) and (14b), we get y 3 = C 1 y 2 + D 1 y 1 + x 3, y 4 = C 2 y 2 + D 2 y 1 + (x 4 + A 2 x 3 ), where C 1 = A 1, C 2 = A 1 A 2 + B 2, D 1 = B 1, D 2 = B 1 A 2, which in turn implies that (35) (36) P 1 = x 3, P 2 = x 4 + A 2 x 3. (37) We now proceed to prove the validity of the expression n+2 P n = E n (k 1)x k (38) k=3 for n 1. It is clear from (37) and the definition of E n (k) in (16) that expression (38) holds for n = 1, 2. Let (38) hold for P n and P n+1 for n 1. Then, from (34), Proposition 2 and (16), we get

9 R.K. Mallik / Linear Algebra and its Applications 325 (2001) n+3 n+2 P n+2 =A n+2 E n+1 (k 1)x k + B n+2 E n (k 1)x k + x n+4 n+2 = k=3 n+2 = k=3 k=3 k=3 [A n+2 E n+1 (k 1) + B n+2 E n (k 1)]x k + A n+2 x n+3 + x n+4 E n+2 (k 1)x k + E n+2 (n + 2)x n+3 + E n+2 (n + 3)x n+4. (39) Thus, n+4 P n+2 = E n+2 (k 1)x k, (40) k=3 which agrees with (38). By mathematical induction, expression (38) is valid for all n 1. From (32), (33) and (38), we obtain (31) An example Consider the case when we need to find y 6 in the difference equation (15) in terms of coefficients A 1,A 2,A 3,A 4,B 1,B 2,B 3,B 4, initial values y 1,y 2, and forcing terms x 3,x 4,x 5,x 6. From Proposition 4, we have y 6 = C 4 y 2 + D 4 y E 4 (k 1)x k. k=3 Eqs. (14a) and (14b) imply that ( C 4 = (A 1 A 2 A 3 A 4 ) 1 + k 1 S 1 (2,4) ( D 4 = (B 1 A 2 A 3 A 4 ) 1 + k 1 S 1 (3,4) σ k1 + σ k1 ). (k 1,k 2 ) S 2 (2,4) Now, from the definition of S q (L, U) in (1a) (1c), we get S 1 (2, 4) ={2, 3, 4}, S 2 (2, 4) ={(2, 4)}, S 1 (3, 4) ={3, 4}. ) (σ k1 σ k2 ),

10 118 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Therefore, C 4 = (A 1 A 2 A 3 A 4 )(1 + σ 2 + σ 3 + σ 4 + σ 2 σ 4 ), D 4 = (B 1 A 2 A 3 A 4 )(1 + σ 3 + σ 4 ), where, from (11), σ 2 = B 2, σ 3 = B 3, σ 4 = B 3. A 1 A 2 A 2 A 3 A 3 A 4 In addition, from (16), we obtain E 4 (2) = (A 2 A 3 A 4 )(1 + σ 3 + σ 4 ), E 4 (3) = (A 3 A 4 )(1 + σ 4 ), E 4 (4) = A 4, E 4 (5) = 1. Therefore, by substituting the expressions for σ 2, σ 3 and σ 4 in C 4, D 4, E 4 (2), E 4 (3), E 4 (4), E 4 (5), we finally have the desired solution, which is y 6 =(A 1 A 2 A 3 A 4 + B 2 A 3 A 4 + B 3 A 1 A 4 + B 4 A 1 A 2 + B 2 B 4 )y 2 + (B 1 A 2 A 3 A 4 + B 1 B 3 A 4 + B 1 B 4 A 2 )y 1 + (A 2 A 3 A 4 + B 3 A 4 + B 4 A 2 )x 3 + (A 3 A 4 + B 4 )x 4 + A 4 x 5 + x A boundary value problem and the tridiagonal matrix inverse Consider the difference equation (15) over the finite index interval [1,K + 2], K 2, with zero boundary values, that is, the recurrence y n+2 = A n y n+1 + B n y n + x n+2, 1 n K, (41) with boundary values y 1,y K+2 such that y 1 = y K+2 = 0. (42) The K equations in (41) can be written in matrix form as B 1 A 1 1 B 2 A B K 1 A K 1 1 B K A K 1 y 1 y 2. y K+1 y K+2

11 = R.K. Mallik / Linear Algebra and its Applications 325 (2001) x 3... (43) x K+2 Using boundary conditions (42), the solution of the difference equation (41) can be expressed as 1 A 1 1 y 2 B 2 A x 3. = (44) y K+1 0 B K 1 A K 1 1 x K+2 B K A K However, from Proposition 4, the quantities y 3,...,y K+2 of recurrence (41) can be expressed in terms of y 1,y 2 as i 1 y i+1 = C i 1 y 2 + D i 1 y 1 + E i 1 (j + 1)x j+2, 2 i K + 1. (45) j=1 As a result of boundary conditions (42), substitution of i = K + 1 in (45) gives K y K+2 = C K y 2 + E K (j + 1)x j+2 = 0, (46) j=1 and therefore, when C K /= 0, K [ y 2 = E ] K(j + 1) x j+2. (47) C K j=1 Substituting (47) in (45) and setting y 1 = 0 we obtain y i+1 = K j=1 i 1 C i 1 E K (j + 1) C K x j+2 + E i 1 (j + 1)x j+2, 2 i K, C K /= 0. (48) j=1 j=1 In other words, i 1 [ y i+1 = E i 1 (j + 1) C ] i 1E K (j + 1) x j+2 C K + K j=i [ C ] i 1E K (j + 1) x j+2, 2 i K, C K /= 0. (49) C K

12 120 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Using (16) and (17), Eqs. (47) and (49) can be written together as y i+1 = K [ j=1 E i 1 (j + 1) E i 1(1)E K (j + 1) E K (1) ] x j+2, 1 i K, E K (1) /= 0. (50) Thus the solution of boundary value problem (41), (42) is given by (47) and (49) or alternatively by (50). This solution also gives the inverse of a tridiagonal matrix. Denoting the K K tridiagonal matrix derived from the K (K + 2) matrix on theleft-handside of (43) (byremovingthefirst and (K + 2)th columns) as A 1 1 B 2 A T K = , K 2, (51) 0 B K 1 A K 1 1 B K A K and its inverse as R K = TK 1 = [ ] K R i,j i,j=1, (52) we obtain, from (44), (47) and (49), an expression for R i,j, the element in the ith row and jth column of T 1 K.Itisgivenby R i,j =E i 1 (j + 1) C i 1E K (j + 1) C K for j = 1,...,i 1, i= 2,...,K, = C i 1E K (j + 1) C K for j = i,...,k, i = 1,...,K, C K /= 0, (53) where E i (j) is defined in (16), and C i in (13a) and (14a). Alternatively, from (44) and (50), we get R i,j = E i 1 (j + 1) E i 1(1)E K (j + 1) E K (1) for i, j = 1,...,K, E K (1) /= 0. (54) The following proposition gives an expression for the determinant of the tridiagonal matrix T K. Proposition 5. For K 2, det(t K ) = ( 1) K C K, (55) where T K is given by (51) and C K by (14a).

13 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Proof. From (51), it is clear that if we find det(t K ) by using the Kth row of T K followed by its Kth column, we obtain det(t K ) = A K det(t K 1 ) + B K det(t K 2 ), K 4. (56) It is simple to show that ([ A1 det(t 2 ) = det B 2 ]) 1 = A A 1 A 2 (1 + σ 2 ) = C 2, 2 (57a) A det(t 3 )=det B 2 A B 3 A 3 = A 1 A 2 A 3 (1 + σ 2 + σ 3 ) = C 3. (57b) Since C K = E K (1), it is governed by the second-order recursion C K = A K C K 1 + B K C K 2, K 4 (58) by Proposition 2. Therefore, ( 1) K C K = A K ( 1) K 1 C K 1 + B K ( 1) K 2 C K 2, K 4. (59) Recursions (56) and (59) for det(t K ) and ( 1) K C K, respectively, are identical, and from (57a) and (57b) we find that they have the same initial values ( 1) 2 C 2 and ( 1) 3 C 3. Hence, det(t K ) = ( 1) K C K for K 2. If it so happens that det(t K ) = 0, then Proposition 5 implies that C K = 0. In that case, the matrix T K is not invertible, and Eq. (53) for the elements of T 1 K does not hold. Let Δ j,i denote the minor corresponding to the jth row and ith column of T K. Then, from (53) and Proposition 5, we have Δ j,i =( 1) (K i j) [C K E i 1 (j + 1) C i 1 E K (j + 1)] for j = 1,...,i 1, i= 2,...,K, = ( 1) (K i j) C i 1 E K (j + 1) for j = i,...,k, i = 1,...,K, (60) or, alternatively, from (54) and Proposition 5, Δ j,i = ( 1) (K i j) [E K (1)E i 1 (j + 1) E i 1 (1)E K (j + 1)] for i, j = 1,...,K. (61)

14 122 R.K. Mallik / Linear Algebra and its Applications 325 (2001) The tridiagonal matrix inverse We now consider a nonsingular K K tridiagonal matrix Φ K (K 2), with complex entries, given by α 1 β 1 γ 2 α 2 β 2 0 Φ K = , 0 γ K 1 α K 1 β K 1 γ K α K β 1,...,β K 1 /= 0, γ 2,...,γ K /= 0. (62) Without loss of generality, let β K be defined as β K 1, and A i α i, 1 i K, β i B i γ i β i, 2 i K, β K (63a) (63b) such that A 1,...,A K, B 2,...,B K are entries of T K in (51). We can express Φ K as Φ K = diag(β 1,...,β K )T K, and therefore, the inverse Ψ K of Φ K can be expressed as Ψ K = ΦK 1 = [ ( ) ] K 1 ψ i,j i,j=1 = T 1 K diag 1,...,, (64) β 1 where, from (52) and (54), ψ i,j = R i,j β j = 1 [ E i 1 (j + 1) E ] i 1(1)E K (j + 1), β j E K (1) E K (1) /= 0, (65a) such that (using (16) and (63b)) ( ) E n (m)=( 1) n m+1 αm αn β m 1 + (n m+1)/2 β n (k 1,...,k q ) S q (m+1,n) if m = 1,...,n 1, n= 2,...,K, (σ k1 σ kq )

15 R.K. Mallik / Linear Algebra and its Applications 325 (2001) = α n β n if m = n, n = 1,...,K, =1 ifm = n + 1, n= 0,...,K, =0 otherwise (65b) and (using (11) and (63b)) σ k = β k 1γ k. (65c) α k 1 α k The set S q (m + 1,n)in (65b) is defined by (1a) (1c). It is known that for the matrix Ψ K in (64), there exist four sequences {u i } K i=1, {v i } K i=1, {r i} K i=1, {s i} K i=1 [9,10,17] such that ψ i,j = { ui v j if i>j, We can express Ψ K as r i s j if i j. (66) ( (Φ ) T 1 ) T Ψ K = K, (67) where Φ T K can be obtained from Φ K by interchanging the positions of β j and γ j+1 in (62) for j = 1,...,K 1. Defining, without loss of generality, γ K+1 as γ K+1 1, (68) and interchanging β j and γ j+1 for j = 1,...,K in formula (65b) for E n (m), we obtain a quantity F n (m) expressed as ( ) F n (m)=( 1) n m+1 αm α n γ m (n m+1)/2 γ n+1 (k 1,...,k q ) S q (m+1,n) (σ k1 σ kq ) if m = 1,...,n 1, n= 2,...,K, = α n γ n+1 if m = n, n = 1,...,K, =1 ifm = n + 1, n= 0,...,K, =0 otherwise, (69) where σ k, which is invariant under the interchange of β k 1 and γ k, is given by (65c). The elements of Ψ K can alternatively be written in terms of F n (m) as ψ i,j = 1 [ F j 1 (i + 1) F ] j 1(1)F K (i + 1), F K (1) /= 0. (70) γ i+1 F K (1)

16 124 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Observe from (65a) and (69) that, for m n, (β m β n ) F n (m) = E n (m) (γ m+1 γ n+1 ). (71) Therefore, combining (65a), (70) and (71), we obtain F j 1(1)F K (i + 1) γ i+1 F K (1) = E j 1(1)E K (i + 1) (γ j+1 γ i ) if i>j, ψ i,j = β i E K (1) (β j β i 1 ) (72) F K (1) /= 0, E i 1(1)E K (j + 1) β j E K (1) if i j, E K (1) /= 0. Thus, (72), along with (65b) and (69), gives an explicit formula for the element in the ith row and jth column of the inverse of the tridiagonal matrix Φ K in (62). We see that (72) is consistent with the structure (66) of Ψ K. Without loss of generality, substituting r i = E i 1 (1), s j = E K(j + 1) β j E K (1), u i = F K(i + 1) γ i+1 F K (1) = s (γ 2 γ i ) i (β 1 β i 1 ), v j = F j 1 (1) = r j (β 1 β j 1 ) (γ 2 γ j ), we obtain from Proposition 2 the recurrences r i = 1 (α i 1 r i 1 + γ i 1 r i 2 ), β i 1 v j = 1 γ j (α j 1 v j 1 + β j 2 v j 2 ) (73) (74a) (74b) for sequences {r i } and {v j }, respectively, and from Proposition 3 the recurrences s j = 1 β j (α j+1 s j+1 + γ j+2 s j+2 ), (75a) u i = 1 (α i+1 u i+1 + β i+1 u i+2 ) (75b) γ i+1 for sequences {s j } and {u i }, respectively. Recurrences (74a), (74b), (75a) and (75b) agree with those in Theorem 2 of [9]. Note that (66) combined with (73) can also be used to obtain the tridiagonal matrix inverse.

17 R.K. Mallik / Linear Algebra and its Applications 325 (2001) The quantities β K and γ K+1 which have been defined to be 1 cancel out due to multiplication in the expression for ψ i,j in (72) as well as in the expressions for u i,s j in (73). Hence, we can also consider β K and γ K+1 to be arbitrary nonzero complex quantities in the definition of E n (m) in (65b) and F n (m) in (69) An example Consider the 3 3 tridiagonal matrix Φ 3 = α 1 β 1 0 γ 2 α 2 β 2, β 1,β 2 /= 0, γ 2,γ 3 /= 0, 0 γ 3 α 3 which is the matrix in (62) with K = 3. To obtain the inverse of this matrix, we first obtain the sequences {u i } 3 i=1, {v i} 3 i=1, {r i} 3 i=1, {s i} 3 i=1 using (73), and then substitute their expressions in (66). We need the expressions for σ 2 and σ 3 which, from (65c), are given by σ 2 = β 1γ 2, σ 3 = β 2γ 3. α 1 α 2 α 2 α 3 Now, from (73) and (65b), we have r 1 = E 0 (1) = 1, r 2 = E 1 (1) = α 1 β 1, and r 3 = E 2 (1) = α 1α 2 β 1 β 2 (1 + σ 2 ) = 1 β 1 β 2 (α 1 α 2 β 1 γ 2 ), s 1 = E 3(2) β 1 E 3 (1) α 2 α 3 (1 + σ 3 ) β = 2 β ( 3 β 1 α ) 1α 2 α 3 (1 + σ 2 + σ 3 ) β 1 β 2 β 3 (α 2 α 3 β 2 γ 3 ) = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), s 2 = E 3(3) β 2 E 3 (1)

18 126 R.K. Mallik / Linear Algebra and its Applications 325 (2001) ( α ) 3 β 3 = ( β 2 α ) 1α 2 α 3 (1 + σ 2 + σ 3 ) β 1 β 2 β 3 β 1 α 3 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), s 3 = E 3(4) β 3 E 3 (1) 1 = ( β 3 α ) 1α 2 α 3 (1 + σ 2 + σ 3 ) β 1 β 2 β 3 β 1 β 2 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ). By interchanging the positions of β k 1 and γ k,k= 2, 3, 4inr i, we obtain v i,and by doing the same in s i, we obtain u i. Thus, from (73), v 1 = F 0 (1) = 1, v 2 = F 1 (1) = α 1 γ 2, and v 3 = F 2 (1) = α 1α 2 γ 2 γ 3 (1 + σ 2 ) = 1 γ 2 γ 3 (α 1 α 2 β 1 γ 2 ), u 1 = F 3(2) γ 2 F 3 (1) α 2 α 3 (1 + σ 3 ) γ = 3 γ ( 4 β 1 α ) 1α 2 α 3 (1 + σ 2 + σ 3 ) γ 2 γ 3 γ 4 (α 2 α 3 β 2 γ 3 ) = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), u 2 = F 3(3) γ 3 F 3 (1)

19 R.K. Mallik / Linear Algebra and its Applications 325 (2001) ( α ) 3 γ 4 = ( γ 3 α ) 1α 2 α 3 (1 + σ 2 + σ 3 ) γ 2 γ 3 γ 4 γ 2 α 3 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), u 3 = F 3(4) γ 4 F 3 (1) 1 = ( γ 4 α ) 1α 2 α 3 (1 + σ 2 + σ 3 ) γ 2 γ 3 γ 4 γ 2 γ 3 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ). Note that the expressions for r 1,r 2,r 3, s 1,s 2,s 3, u 1,u 2,u 3, v 1,v 2,v 3 do not contain β 3 or γ 4. From (66), the diagonal elements of the inverse of Φ 3 are then given by ψ 1,1 = u 1 v 1 = ψ 2,2 = u 2 v 2 = (α 2 α 3 β 2 γ 3 ) (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), α 1 α 3 (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), (α 1 α 2 β 1 γ 2 ) ψ 3,3 = u 3 v 3 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ). The upper diagonal elements are given by β 1 α 3 ψ 1,2 = r 1 s 2 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), ψ 1,3 = r 1 s 3 = β 1 β 2 (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), α 1 β 2 ψ 2,3 = r 2 s 3 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ). Each of the lower diagonal elements ψ i,j is obtained by interchanging β k 1 and γ k, k = 2, 3 in the corresponding upper diagonal element ψ j,i. Therefore, γ 2 α 3 ψ 2,1 = u 2 v 1 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ), ψ 3,1 = u 3 v 1 = γ 2 γ 3 (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ),

20 128 R.K. Mallik / Linear Algebra and its Applications 325 (2001) α 1 γ 3 ψ 3,2 = u 3 v 2 = (α 1 α 2 α 3 β 1 γ 2 α 3 β 2 γ 3 α 1 ). 5. Characteristic polynomial The tridiagonal matrix of (62) satisfies Φ K = diag(β 1,...,β K )T K, (76) where T K is given by (51) with A i = α i, 1 i K, B i = γ i, 2 i K, β K = 1, β i β i as in (63a), (63b). From (76) and Proposition 5, we get det(φ K ) = ( 1) K (β 1 β K )C K = ( 1) K (β 1 β K )E K (1). (77) Using (65b), this can be rewritten as K/2 det(φ K ) = (α 1 α K ) 1+ (σ k1 σ kq ), (78) (k 1,...,k q ) S q (2,K) where σ k = ((β k 1 γ k )/(α k 1 α k )) as in (65c). Now the characteristic polynomial of Φ K is given by det(λi K Φ K ) = ( 1) K det(φ K λi K ), (79) where I K denotes the K K identity matrix. By replacing α k by (λ α k ) for k = 1,...,K in (79), we get det(λi K Φ K )=(λ α 1 ) (λ α K ) K/2 1+ (k 1,...,k q ) S q (2,K) η k1 (λ) η kq (λ), (80a) where β k 1 γ k η k (λ) = (λ α k 1 )(λ α k ), (80b) and S q (2,K) is given by (1a) (1c). Thus, (80a) and (80b) give an explicit formula for the characteristic polynomial of a tridiagonal matrix. For example, when K = 4weget det (λi 4 Φ 4 )=(λ α 1 )(λ α 2 )(λ α 3 )(λ α 4 ) [1 + η 2 (λ) + η 3 (λ) + η 4 (λ) + η 2 (λ)η 4 (λ)] =(λ α 1 )(λ α 2 )(λ α 3 )(λ α 4 ) β 1 γ 2 (λ α 3 )(λ α 4 ) β 2 γ 3 (λ α 1 )(λ α 4 ) β 3 γ 4 (λ α 1 )(λ α 2 ) + β 1 γ 2 β 3 γ 4.

21 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Orthogonal polynomials Consider a set {p n (t)} n 0 of polynomials that are orthogonal on the interval [a,b] with respect to some nonnegative weight function w(t) [18]. The polynomial p n (t) has degree n, and satisfies the second-order homogeneous difference equation [18,19] β n p n (t) = (t α n )p n 1 (t) γ n p n 2 (t), n 1, (81) where β n,p n = 0, p 0 (t) is some nonzero constant, andp 1 (t) is a function of t, which may be identically zero. The difference equation (81) can be rewritten as p n (t) = (t α n) p n 1 (t) γ n p n 2 (t), n 1, (82) β n β n with initial values p 1 (t) and p 0 (t). Comparing (82) with the homogeneous difference equation (10), we can write y n+2 = p n (t), n = 1, 2,..., (83a) A n = (t α n), β n B n = γ n. β n (83b) From (12) (14b), (83a) and (83b), the solution of (82) is given by p n (t) = C n (t)p 0 (t) + D n (t)p 1 (t), n 0, (84) where C 0 (t) = 1, C 1 (t) = (t α 1), β 1 (85a) and D 0 (t) = 0, D 1 (t) = γ 1 β 1, D 2 (t) = γ 1(t α 2 ) β 1 β 2 C n (t)= (t α 1) (t α n ) β 1 β n n/2 1+ (k 1,...,k q ) S q (2,n) (85b) η k1 (t) η kq (t) for n 2, (86a) D n (t)= γ 1(t α 2 ) (t α n ) β 1 β n (n 1)/2 1+ (k 1,...,k q ) S q (3,n) η k1 (t) η kq (t) for n 3, (86b)

22 130 R.K. Mallik / Linear Algebra and its Applications 325 (2001) such that β k 1 γ k η k (t) = (t α k 1 )(t α k ) as in (80b). Note that C n (t) is a polynomial of degree n for n 0, while D n (t) is a polynomial of degree n 1forn 1. Thus, (84) is an expression for p n (t), n 0, interms of the recurrence coefficients α 1,...,α n,β 1,...,β n,γ 1,...,γ n and initial values p 1 (t), p 0 (t). Let p 0 (t) p(t) =.. (87) p K 1 (t) denote the K 1 vector of orthogonal polynomials, and Φ K the tridiagonal matrix in (62) of coefficients α 1,...,α K, β 1,...,β K 1, γ 2,...,γ K of the difference equation (81) for n = 1,...,K. We can rewrite (81) for n = 1,...,K as [19,20] (ti K Φ K )p(t) = β K p K (t)e K + γ 1 p 1 (t)e 1, (88) where e 1 and e K are the first and last columns, respectively, of I K.Ift is not an eigenvalue of Φ K, we get from (88) p(t) = β K p K (t)(φ K ti K ) 1 e K γ 1 p 1 (t)(φ K ti K ) 1 e 1. (89) Replacing each diagonal element α k by α k t in Φ K, we obtain the matrix Φ K ti K. Thus, each of the polynomials p 0 (t),...,p K 1 (t) can be expressed in terms of the inverse of the tridiagonal matrix Φ K ti K and the functions p K (t) and p 1 (t). The vector (Φ K ti K ) 1 e K is the Kth column of (Φ K ti K ) 1. Replacing α k by (t α k ), k = 1,...,K in (65b), we obtain from (72) the following expression for ψ n+1,k (t), the element in the (n + 1)th row and Kth column of (Φ K ti K ) 1 : ψ n+1,k (t) = (t α 1 ) (t α n ) β 1 β n β K (t α 1 ) (t α K ) β 1 β K 1+ n/2 1+ K/2 (k 1,...,k q ) S q (2,n) (k 1,...,k q ) S q (2,K) η k1 (t) η kq (t) η k1 (t) η kq (t). (90) Similarly, by replacing α k by (t α k ), k = 1,...,K in (69), we obtain from (72) the following expression for ψ n+1,1 (t), the element in the (n + 1)th row and first column of (Φ K ti K ) 1 :

23 ψ n+1,1 (t) = R.K. Mallik / Linear Algebra and its Applications 325 (2001) (t α n+2 ) (t α K ) γ n+3 γ K+1 γ n+2 (t α 1 ) (t α K ) γ 2 γ K+1 (K n 1)/2 1+ η k1 (t) η kq (t) (k 1,...,k q ) S q (n+3,k) K/2 1+ η k1 (t) η kq (t) (k 1,...,k q ) S q (2,K). (91) Now p n (t) is the (n + 1)th row of p in (89). Therefore, combining (89), (90) and (91), we obtain, for n = 0,...,K 1, n/2 (β n+1 β K ) 1+ (k 1,...,k q ) S q (2,n) p n (t) = K/2 (t α n+1 ) (t α K ) 1+ (k 1,...,k q ) S q (2,K) η k1 (t) η kq (t) η k1 (t) η kq (t) p K (t) + (K n 1)/2 (γ 1 γ n+1 ) 1+ η k1 (t) η kq (t) (k 1,...,k q ) S q (n+3,k) p K/2 1 (t), (t α 1 ) (t α n+1 ) 1+ η k1 (t) η kq (t) (k 1,...,k q ) S q (2,K) (92) where η k (t) is given by (80b). Thus, (92) is an expression for the K intermediate polynomials p n (t), 0 n K 1, in terms of the recurrence coefficients α 1,..., α K, β 1,...,β K, γ 1,...,γ K and boundary values p 1 (t), p K (t). We now show how the general formula for p n (t) in (84) can be used to obtain expressions for Legendre, Hermite and Chebyshev polynomials Legendre polynomials Legendre polynomials can be generated by the difference equation p n (t) = (2n 1) tp n 1 (t) n (n 1) p n 2 (t), n 1, (93) n with p 0 (t) = 1. The interval of orthogonality is [ 1, 1] and the weight function w(t) = 1. Comparing (93) with (82), we obtain α n = 0, β n = n (2n 1), γ (n 1) n =, n 1. (94) (2n 1)

24 132 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Since γ 1 = 0, we have D n (t) = 0. Therefore, p 1 (t) does not affect the solution of (93). Now, from (80b), (k 1) 2 η k (t) = (2k 3)(2k 1)t 2. (95) Using (86a), we can write the solution of (93) as p n (t)=c n (t) = (2n)! 2 n (n!) 2 n/2 t n + ( 1) q t n 2q Denote G n,q as G n,q = q (k 1,...,k q ) S q (2,n) i=1 Applying property (3) on S q (2,n)we get (k 1,...,k q ) S q (2,n) i=1 q (k i 1) 2. (96) (2k i 3)(2k i 1) (k i 1) 2 (2k i 3)(2k i 1). (97) S q (2,n)=S q (2,n 1) { (k 1,...,k q 1,k q ): k q = n; (k 1,...,k q 1 ) S q 1 (2,n 2) }, (98) which implies that G n,q in (97) follows the recurrence (n 1) 2 G n,q = G n 1,q + (2n 3)(2n 1) G n 2,q 1. (99) From (99), it can be shown by mathematical induction that (n!) 2 (2n 2q)! G n,q = (2n)!q!(n q)!(n 2q)!. (100) Combining (96), (97) and (100), we finally obtain p n (t) = n/2 ( 1) q (2n 2q)! 2 n q!(n q)!(n 2q)! tn 2q, (101) which is the expression for the Legendre polynomial of degree n. This can alternatively be written as p n (t) = 1 d n [( t 2 2 n n! dt n 1 ) n].

25 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Hermite polynomials Hermite polynomials can be generated by the recurrence p n (t) = tp n 1 (t) (n 1)p n 2 (t), n 1, (102) with p 0 (t) = 1. The interval of orthogonality is (, ) and the weight function w(t) = e (t2 /2). Comparing (102) with (82), we get α n = 0, β n = 1, γ n = (n 1), n 1. (103) Since γ 1 = 0, we have D n (t) = 0; hence, p 1 (t) does not affect the solution of (102). From (80b), (k 1) η k (t) = t 2. (104) We can then write the solution of (102) as n/2 q p n (t) = C n (t) = t n + ( 1) q t n 2q (k i 1). (105) Denoting G n,q as G n,q = (k 1,...,k q ) S q (2,n) i=1 (k 1,...,k q ) S q (2,n) i=1 q (k i 1), (106) and applying property (3) on S q (2,n)we obtain the recurrence G n,q = G n 1,q + (n 1)G n 2,q 1. (107) It can be shown by mathematical induction using this recurrence that n! G n,q = 2 q q!(n 2q)!. (108) Combining (105), (106) and (108), we get p n (t) = n/2 q=0 ( 1) q n! 2 q q!(n 2q)! tn 2q, (109) which is the expression for the Hermite polynomial of degree n. This can alternatively be written as p n (t) = ( 1) n e (t2 /2) dn [ e (t 2 /2) ] dt n Chebyshev polynomials Chebyshev polynomials can be generated by the difference equation

26 134 R.K. Mallik / Linear Algebra and its Applications 325 (2001) p n (t) = 2tp n 1 (t) p n 2 (t), n 1, (110) whose coefficients do not vary with n. The interval of orthogonality is [ 1, 1] and the weight function w(t) = 1/ 1 t 2. Here and α n = 0, β n = 1 2 = γ n, n 1, (111) η k (t) = 1 4t 2. (112) Since η k (t) does not depend on k, we can apply property (2) on S q (2,n)and S q (3,n) in (86a) and (86b). This results in followed by S q (2,n) = ( n q q n/2 C n (t) = (2t) n + ), S q (3,n) = ( 1) q ( n q q ( n q 1 q ) (2t) n 2q, ), (113) (114a) D n (t) = C n 1 (t). For the Chebyshev polynomials of the first kind,wehave p 0 (t) = 1, p 1 (t) = t, and therefore (114b) p n (t)=c n (t) + td n (t) =C n (t) tc n 1 (t) = 1 n/2 2 (2t)n + ( 1) q [( n q q ) + ( )] n q 1 q (2t) n 2q. (115) This can also be expressed as p n (t) = cos ( n cos 1 (t) ). For the Chebyshev polynomials of the second kind,wehave p 0 (t) = 1, p 1 (t) = 0, and therefore p n (t) = C n (t), wherec n (t) is given by (114a). This can also be expressed as p n (t) = sin( (n + 1) cos 1 (t) ) sin ( cos 1 (t) ) = sin ( (n + 1) cos 1 (t) ). 1 t 2

27 R.K. Mallik / Linear Algebra and its Applications 325 (2001) A cyclic tridiagonal system Consider the second-order linear difference equation over the finite index interval [1,K+ 2], K 2, given by β n y n+2 = α n y n+1 γ n y n + β n x n+2, 1 n K, (116) with variable complex coefficients β n,α n,γ n,whereβ n,γ n /= 0, complex forcing term x n+2, and complex boundary values y 1,y K+2. The boundary values are periodic with period K, that is, we have the boundary conditions y 1 = y K+1, y K+2 = y 2. (117) The K equations in (116) can be written in matrix form after applying the boundary conditions as Λ K y 2. = diag(β 1,...,β K ) y K+1 x 3. x K+2 where α 1 β 1 γ 1 γ 2 α 2 β 2 0 Λ K = , 0 γ K 1 α K 1 β K 1 β K γ K α K, (118a) β 1,...,β K /= 0, γ 1,...,γ K /= 0. (118b) The system ((118a), (118b)) of equations is a cyclic linear tridiagonal system [21], and the matrix Λ K in (118b) a cyclic tridiagonal matrix. The solution of system ((118a), (118b)), or alternatively, of the difference equation (116) with boundary conditions (117) for n = 1,...,K can be expressed as y 2. = Λ 1 K y K+1 β 1 x 3. β K x K+2. (119) Let Λ 1 K = [ ] K ω i,j i,j=1. (120) The difference equation (116) can be rewritten as y n+2 = A n y n+1 + B n y n + x n+2, 1 n K, (121a) as in (41) with A n = α n β n, B n = γ n β n. (121b)

28 136 R.K. Mallik / Linear Algebra and its Applications 325 (2001) From Proposition 4, the quantities y 3,...,y K+2 of recurrence (121a) can be expressed in terms of y 1,y 2 as i 1 y i+1 = C i 1 y 2 + D i 1 y 1 + E i 1 (j + 1)x j+2, 2 i K + 1, (122) j=1 where E n (m) is given by (65b), and from (13a), (13b), (14a), (14b) and (121b), C 1 = α 1, D 1 = γ 1, D 2 = γ 1α 2, (123a) β 1 β 1 β 1 β 2 ( ) n/2 C n = ( 1) n α1 αn 1+ (σ k1 σ kq ) β 1 β n (k 1,...,k q ) S q (2,n) for n 2, (123b) D n = ( 1) n γ ( ) (n 1)/2 1 α2 αn 1+ β 1 β 2 β n (k 1,...,k q ) S q (3,n) (σ k1 σ kq ) for n 3, (123c) such that S q (L, U) is defined by (1a) (1c), and σ k = ((β k 1 γ k )/(α k 1 α k )) as in (65c). Putting i = K in (122) and applying the boundary condition y K+1 = y 1 in the resulting equation, we get (1 D K 1 )y 1 C K 1 y 2 = K 1 j=1 E K 1 (j + 1)x j+2. Similarly, putting i = K + 1andy K+2 = y 2 in (122) results in K D K y 1 + (1 C K )y 2 = E K (j + 1)x j+2. j=1 We can then express y 1 and y 2 in terms of x 3,...,x K+2 as K 1 [ ] [ ] E 1 y1 1 DK 1 C K 1 (j + 1)x j+2 K 1 j=1 =. (124) y 2 D K 1 C K K E K (j + 1)x j+2 Substituting (124) in (122), we can express y i+1 as K y i+1 = ω i,j β j x j+2, (125) i=1 j=1

29 R.K. Mallik / Linear Algebra and its Applications 325 (2001) which implies from (119) and (120) that ω i,j is the element in the ith row and jth column of Λ 1. Thus, we can obtain an explicit formula for the inverse of a cyclic tridiagonal matrix by solving a boundary value problem with periodic boundary conditions. 8. Case of constant diagonals When α n = α, β n = β, γ n = γ, n = 1,...,K, (126) the tridiagonal matrix Φ K in (62) is Toeplitz, while the cyclic tridiagonal matrix Λ K in (118b) is circulant, which also implies that it is Toeplitz. The quantities C n, D n and E n (m) in (65b) and (123a) (123c) can now be expressed in terms of the Chebyshev polynomials of the second kind discussed in Section 6.3 as ( ) γ n m+1 ( ) α E n (m) = ( 1) n m+1 p n m+1 β 2, γβ ( ) γ n ( ) α C n = E n (1) = ( 1) n p n β 2, γβ ( ) γ n+1 ( ) α D n = ( 1) n p n 1 β 2, γβ where, using (114a), ( ) ( ) α α n n/2 p n 2 ( )( ) = 1 + ( 1) q n q γβ q γβ γβ q α 2. From (72), the elements of the inverse of the tridiagonal matrix Φ K in (62) under condition (126) can be expressed as ( ) ( ) α α ( 1) i j ( ) γ i j p j 1 2 p K i γβ 2 γβ ( ) if i>j, γβ β α p K 2 γβ ψ i,j = ( ) ( ) α α ( 1) j i ( ) j i p i 1 β 2 p K j γβ 2 γβ ( ) if i j, γβ γ α p K 2 γβ E K (1) /= 0. The inverse of the cyclic tridiagonal matrix Λ K in (118b) under condition (126) can also be obtained from E n (m), C n,d n using the approach discussed in Section 7.

30 138 R.K. Mallik / Linear Algebra and its Applications 325 (2001) Conclusions We have obtained explicit formulae for the elements of the inverse of a general tridiagonal matrix by deriving the explicit solution of a second-order linear nonhomogeneous difference equation with variable coefficients, and then applying the solution to a boundary value problem with zero boundary values. Using the formula for the determinant, we have got an expression for the characteristic polynomial. A connection between the matrix inverse and orthogonal polynomials has also been established. It has also been shown that how an application of the solution of a secondorder linear difference equation to a boundary value problem with periodic boundary conditions can yield the inverse of a cyclic tridiagonal matrix. In the simple case of a tridiagonal or a cyclic tridiagonal matrix with constant diagonals, the elements of the inverse can be expressed in terms of the Chebyshev polynomials of the second kind. Acknowledgments The author thanks Dr. Y.V.S.S. Sanyasi Raju, Department of Mathematics, Indian Institute of Technology, Chennai, for his valuable suggestions. He also expresses his gratitude to the anonymous referee for his/her constructive comments. References [1] G. Meurant, A review on the inverse of symmetric tridiagonal and block tridiagonal matrices, SIAM J. Matrix Anal. Appl. 13 (3) (1992) [2] S.-F. Xu, On the Jacobi matrix inverse eigenvalue problem with mixed given data, SIAM J. Matrix Anal. Appl. 17 (3) (1996) [3] A. Bunse-Gerstner, R. Byers, V. Mehrmann, A chart of numerical methods for structured eigenvalue problems, SIAM J. Matrix Anal. Appl. 13 (2) (1992) [4] C.F. Fischer, R.A. Usmani, Properties of some tridiagonal matrices and their application to boundary value problems, SIAM J. Numer. Anal. 6 (1) (1969) [5] L. Lu, W. Sun, The minimal eigenvalues of a class of block-tridiagonal matrices, IEEE Trans. Inform. Theory 43 (2) (1997) [6] G. D. Smith, Numerical Solution of Partial Differential Equations: Finite Difference Methods, second ed., Oxford University Press, New York, [7] R.M.M. Mattheij, M.D. Smooke, Estimates for the inverse of tridiagonal matrices arising in boundary-value problems, Linear Algebra Appl. 73 (1986) [8] G.H. Golub, C.F. Van Loan, Matrix Computations, third ed., Johns Hopkins University Press, Baltimore, MD, [9] T. Yamamoto, Y. Ikebe, Inversion of band matrices, Linear Algebra Appl. 24 (1979) [10] W.W. Barrett, A theorem on inverses of tridiagonal matrices, Linear Algebra Appl. 27 (1979) [11] W.W. Barrett, P.J. Feinsilver, Inverses of banded matrices, Linear Algebra Appl. 41 (1981)

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