1. Inverse of a Tridiagonal Matrix Let Us Consider the N-By-N Nonsingular Tridiagonal Matrix T   A1 B1    C1 A2 B2  

Pré-Publica¸cõesdo Departamento de Matemática Universidade de Coimbra Preprint Number 05–16

ON THE EIGENVALUES OF SOME TRIDIAGONAL MATRICES

C.M. DA FONSECA

Abstract: A solution is given for a problem on eigenvalues of some symmetric tridiagonal matrices suggested by William Trench. The method used is generalizable to other problems. Keywords: Tridiagonal matrices, eigenvalues, recurrence relations, Chebyshev polynomials. AMS Subject Classification (2000): 15A18, 65F15, 15A09, 15A47, 65F10.

1. Inverse of a tridiagonal matrix Let us consider the n-by-n nonsingular tridiagonal matrix T   a1 b1    c1 a2 b2   .. ..  T =  c2 . .  .  .. ..   . . bn−1  cn−1 an In [4], Usmani gave an elegant and concise formula for the inverse of the tridiagonal matrix T : ½ i+j −1 (−1) bi . . . bj−1θi−1φj+1/θn, if i ≤ j (T )ij = i+j (1.1) (−1) cj . . . ci−1θj−1φi+1/θn, if i > j , where θi’s verify the recurrence relation

θi = aiθi−1 − bi−1ci−1θi−2 , for i = 2, . . . , n , with initial conditions θ0 = 1 and θ1 = a1, and φi’s verify the recurrence relation

φi = aiφi+1 − biciφi+2 , for i = n − 1,..., 1 , with initial conditions φn+1 = 1 and φn = an. Observe that θn = det T .

Received 25 August, 2005. This work was supported by CMUC - Centro de Matem´aticada Universidade de Coimbra.

1 2 C.M. DA FONSECA

In [6], W.F. Trench proposed and solved the problem of ﬁnding eigenvalues and eigenvectors of the classes of symmetric matrices:

A = [min{i, j}]i,j=1,...,n and

B = [min{2i − 1, 2j − 1}]i,j=1,...,n . A. Kovaˇcechas presented a diﬀerent proof of this problem [2]. These two matrices are in fact particular cases of a more general matrix

C = [min{ai − b, aj − b}]i,j=1,...,n , with a > 0 and a 6= b. It is very interesting that, under the above conditions, C is always invertible and its inverse is a tridiagonal matrix.

Proposition 1.1. The tridiagonal matrix of order n

 a  1 + a−b −1    −1 2 −1   ......  Tn =  . . .   −1 2 −1  −1 1

1 is the inverse of aC.

Proof : Notice that θi’s verify the recurrence relation θi = 2θi−1 − θi−2, for i = 2, . . . , n − 1, and θn = θn−1 − θn−2, with initial conditions θ0 = 1 and 2a−b (i+1)a−b a θ1 = a−b . Then θi = a−b , for i = 1, . . . , n − 1, and θn = a−b. The φi’s verify the recurrence relation φi = 2φi+1 − φi+2, for i = n − 1,..., 2, with initial conditions φn+1 = 1 and φn = 1. Therefore, since φi = 1, for a i = n + 1,..., 2, and φ1 = a−b. Consequently, the inverse of Tn is the symmetric matrix such that

ai−b −1 i+j j−i a−b ai − b (Tn )ij = (−1) (−1) a = a−b a for i ≤ j. EIGENVALUES OF SOME TRIDIAGONAL MATRICES 3 2. Eigenpairs of a particular tridiagonal matrix According to the initial section the problem of ﬁnding the eigenvalues of C is equivalent to describing the spectra of a tridiagonal matrix. Here we give a general procedure to locate the eigenvalues of the matrix Tn from Proposition 1.1. Let us consider the set of polynomials {Qk(x)} deﬁned by the recurrence relation given by Q0(x) = 1 and Q1(x) = (ax + 1)Q0(x) ,

Qk(x) = (ax + 2)Qk−1(x) − Qk−2(x) , for k = 2, . . . , n − 1 , and µ ¶ 2a − b Q (x) = ax + Q (x) − Q (x) . n a − b n−1 n−2

Note that each polynomial Qk(x), for k = 0, . . . , n, is of degree k. The last recurrence relation has the following matricial form:    2a−b      Qn−1(x) a−b −1 Qn−1(x) 1  Q (x)   −1 2 −1   Q (x)   0   n−2  1    n−2     .   . . .   .   .  x  .  = −  ......   . +Qn(x)  .  .   a       Q1(x) −1 2 −1 Q1(x) 0 Q0(x) −1 1 Q0(x) 0 ¡ ¢ ¡ ¢ ax ax Since Qk(x) = Uk 2 + 2 − Uk−1 2 + 2 , for k = 0, . . . , n − 1, and µ ¶ a ³ ³ax ´ ³ax ´´ Q (x) = ax + 1 + U + 1 − U + 1 − n a − b n−1 2 n−2 2 ³ ³ax ´ ³ax ´´ − U + 2 − U + 1 n−2 2 n−3 2 ³ax ´ ³ax ´ = Un + 1 − Un−1 + 1 − 2 µ 2 ¶ a ³ ³ax ´ ³ax ´´ − 1 − U + 1 − U + 1 , a − b n−1 2 n−2 2 where Uk(x), for k = 0, . . . , n, are the Chebyshev polynomials of second kind 1 of degree k, the zeros of Qn(x) are exactly the eigenvalues of −aC, i.e., the (real) values which satisfy the equality ¡ ¢ ¡ ¢ U ax + 1 − U ax + 1 a n ¡2 ¢ n−1 ¡2 ¢ pn(x) := ax ax = 1 − . (2.1) Un−1 2 + 1 − Un−2 2 + 1 a − b 4 C.M. DA FONSECA

1 In general, (2.1) means that the eigenvalues of −aC are the intersections a of the graph of pn(x) with the line y = 1 − a−b. As a ﬁrst consequence consider the case when a¡= 1 and¢ b = 0.¡ The¢ eigen- x x values of −A are the solutions of the equation Un 2 + 1 −Un−1 2 + 1 = 0, which are, for k = 0, . . . , n − 1, µ ¶ 2k + 1 λ = 2 cos π − 2 . k 2n + 1

The value of an eigenvector associated to λk follows immediately: £ ¤t Qn−1(λk) ··· Q1(λk) Q0(λk) . Hence we proved the following: Theorem 2.1 ([2, 6]). The matrix A of order n, n ≥ 3, has the eigenpairs (λk, vk) given by 1 λ = (1 − cos (r ))−1 and v = [sin(jr )]t , k 2 k k k j=1,...,n where 2k + 1 r = π , k 2n + 1 for k = 0, . . . , n − 1. 1 If a = 2 and b = 1, then the eigenvalues of −2B are solutions of the equation Un (x + 1) − Un−2 (x + 1) = 0, which are, for k = 0, . . . , n − 1, µ ¶ 2k + 1 cos π − 1 . 2n 3. Location of eigenvalues

Since pn(x) deﬁned in (2.1) is strictly increasing, even if it is impossible to evaluate exactly the eigenvalues of C, one can locate them. For example, if b < 0, then 1 − a > 0, and each eigenvalue λ is located between the zeros ¡ ¢ a−b ¡ ¢ k ¡ ¢ ¡ ¢ ax ax ax ax of Un 2 + 1 −Un−1 2 + 1 and the zeros of Un−1 2 + 1 −Un−2 2 + 1 , i.e., lies in the intervals ¸ µ µ ¶ ¶ µ µ ¶ ¶· 2 2k + 1 2 2k − 1 cos π − 1 , cos π − 1 , a 2n + 1 a 2n − 1 ¡ ¡ ¢ ¢ 2 1 for k = n − 1,..., 1, and λ0 is on the right side of a cos 2n+1 π − 1 . If a 1 − a−b < 0, i.e., b > 0, one can make an analogous consideration. EIGENVALUES OF SOME TRIDIAGONAL MATRICES 5

Let us consider the matrix T6 from Proposition 1.1 with a = −b = 2, i.e.,   3 −1 0 0 0 0  2   −1 2 −1 0 0 0     0 −1 2 −1 0 0    .  0 0 −1 2 −1 0   0 0 0 −1 2 −1  0 0 0 0 −1 1 The eigenvalues of this matrix are located in the intervals ]−0.2514892, −0.0405070[ ]−0.6453951, −0.3451392[ ]−1.1205366, −0.8576851[ ]−1.5680647, −1.4154150[ ]−1.8854560, −1.8412535[ . and one is greater than −0.0290581. In fact, they are approximately

λ0 = −0.0220986 λ1 = −0.2058355 λ2 = −0.5715577 λ3 = −1.0510977 λ4 = −1.5262645 λ5 = −1.8731458 .

-2 -1.5 -1 -0.5 0

1 The intersection of the graphs y = p6(x) and y = 2 . Figure 1 6 C.M. DA FONSECA 4. A matrix of maximums

In the second section we have considered the matrix [min{i, j}]i,j. What happens if instead of the minimum we have the maximum? We note that the inverse of C must be tridiagonal because the upper and the lower triangular parts of C have rank 1 form. Theorem 4.1. For a positive integer n, consider the tridiagonal matrix of order n   −1 1    1 −2 1   ......  M =  . . .  . (4.1)  1 −2 1  1 1 −1 + n Then M is invertible and the inverse is −1 M = [max{i, j}]i,j=1,...,n .

Proof¡: From¢ (1.1) we have θi = −2θi−1 − θi−2, for i = 2, . . . , n − 1 and 1 θn = −1 + n θn−1 −θn−2, with initial conditions θ0 = 1 and θ1 = −1. Then i n−1 1 θi = (−1) , for i = 0, . . . , n − 1, and θn = (−1) n = det M. The φi’s verify the recurrence relation φi = −2φi+1 − φi+2, for i = n − 1,..., 2, with 1 initial conditions φn+1 = 1 and φn = −1 + n, and φ1 = −φ2 − φ3. Then n−i+1 1 φi = (−1) (i − 1)n, for i = 2, . . . , n + 1. Finally, the inverse of M is the symmetric matrix such that (−1)i−1(−1)n−j j (M −1) = (−1)i+j n = j for i ≤ j , ij n−1 1 (−1) n i.e., −1 M = [max{i, j}]i,j=1,...,n .

Let us consider again the recurrence relation of Qk(x) already deﬁned, with a = 1 and b = n + 1. This recurrence relation is equivalent to         Q0(x) −1 1 Q0(x) 0          Q1(x)   1 −2 1   Q1(x)   0   .   ......   .   .  x  .  =  . . .   . +Qn(x)  .  .         Qn−2(x) 1 −2 1 Qn−2(x) 0 1 Qn−1(x) 1 −1 + n Qn−1(x) 1 EIGENVALUES OF SOME TRIDIAGONAL MATRICES 7

Therefore one can located the eigenvalues of the matrix M using the argu- ments of the last section. Note that £ ¤t Q0(λk) Q1(λk) ··· Qn−1(λk) is an eigenvector of M associated to the eigenvalue λk. References [1] C.F. Fischer, R. Usmani, Properties of some tridiagonal matrices and their application to boundary value problems, SIAM J. Numer. Anal. 6 (1969) 127-142. [2] A. Kovaˇcec,WMY 2000 and PARIS, August 8, 1900 (A Celebration and A Dedication), Pre-Print 00 − 21, Department of Mathematics, University of Coimbra. [3] J.W. Lewis, Inversion of tridiagonal matrices, Numer. Math. 38 (1982) 333-345. [4] R. Usmani, Inversion of a tridiagonal Jacobi matrix, Linear Algebra Appl. 212/213 (1994) 413-414. [5] R. Usmani, Inversion of Jacobi’s tridiagonal matrix, Comput. Math. Appl., 27 (1994) 59-66. [6] W.F. Trench, Eigenvalues and eigenvectors of two symmetric matrices, IMAGE Bulletin of the International Linear Algebra Society 22 (1999) 28-29.

C.M. da Fonseca Departamento de Matematica,´ Universidade de Coimbra, 3001 - 454 Coimbra, Portugal E-mail address: [email protected]