1. Inverse of a tridiagonal matrix

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1 Pré-Publicções do Deprtmento de Mtemátic Universidde de Coimbr Preprint Number ON THE EIGENVALUES OF SOME TRIDIAGONAL MATRICES CM DA FONSECA Abstrct: A solution is given for problem on eigenvlues of some symmetric tridigonl mtrices suggested by Willim Trench The method used is generlizble to other problems Keywords: Tridigonl mtrices, eigenvlues, recurrence reltions, Chebyshev polynomils AMS Subject Clssifiction 2000): 15A18, 65F15, 15A09, 15A47, 65F10 1 Inverse of tridigonl mtrix Let us consider the n-by-n nonsingulr tridigonl mtrix T 1 b 1 c 1 2 b 2 T = c 2 b n 1 c n 1 In [4], Usmni gve n elegnt nd concise formul for the inverse of the tridigonl mtrix T : T 1 ) ij = where θ i s verify the recurrence reltion n { 1) i+j b i b j 1 θ i 1 φ j+1 /θ n, if i j 1) i+j c j c i 1 θ j 1 φ i+1 /θ n, if i > j, θ i = i θ i 1 b i 1 c i 1 θ i 2, for i = 2,, n, 11) with initil conditions θ 0 = 1 nd θ 1 = 1, nd φ i s verify the recurrence reltion φ i = i φ i+1 b i c i φ i+2, for i = n 1,, 1, with initil conditions φ n+1 = 1 nd φ n = n Observe tht θ n = det T Received 25 August, 2005 This work ws supported by CMUC - Centro de Mtemátic d Universidde de Coimbr 1

2 2 CM DA FONSECA In [6], WF Trench proposed nd solved the problem of finding eigenvlues nd eigenvectors of the clsses of symmetric mtrices: A = [min{i, j}] i,j=1,,n nd B = [min{2i 1, 2j 1}] i,j=1,,n A Kovčec hs presented different proof of this problem [2] These two mtrices re in fct prticulr cses of more generl mtrix C = [min{i b, j b}] i,j=1,,n, with > 0 nd b It is very interesting tht, under the bove conditions, C is lwys invertible nd its inverse is tridigonl mtrix Proposition 11 The tridigonl mtrix of order n T n = is the inverse of 1 C Proof : Notice tht θ i s verify the recurrence reltion θ i = 2θ i 1 θ i 2, for i = 2,, n 1, nd θ n = θ n 1 θ n 2, with initil conditions θ 0 = 1 nd θ 1 = 2 Then θ i = i+1), for i = 1,, n 1, nd θ n = The φ i s verify the recurrence reltion φ i = 2φ i+1 φ i+2, for i = n 1,, 2, with initil conditions φ n+1 = 1 nd φ n = 1 Therefore, since φ i = 1, for i = n + 1,, 2, nd φ 1 = Consequently, the inverse of T n is the symmetric mtrix such tht i b Tn 1 ) ij = 1) i+j 1) j i = i b for i j

3 EIGENVALUES OF SOME TRIDIAGONAL MATRICES 3 2 Eigenpirs of prticulr tridigonl mtrix According to the initil section the problem of finding the eigenvlues of C is equivlent to describing the spectr of tridigonl mtrix Here we give generl procedure to locte the eigenvlues of the mtrix T n from Proposition 11 Let us consider the set of polynomils {Q k x)} defined by the recurrence reltion given by Q 0 x) = 1 nd Q 1 x) = x + 1)Q 0 x), nd Q k x) = x + 2)Q k 1 x) Q k 2 x), for k = 2,, n 1, Q n x) = x + 2 b ) Q n 1 x) Q n 2 x) b Note tht ech polynomil Q k x), for k = 0,, n, is of degree k The lst recurrence reltion hs the following mtricil form: 2 Q n 1 x) 1 Q n 1 x) 1 Q n 2 x) x Q 1 x) = Q n 2 x) Q 1 x) +Q nx) 0 0 Q 0 x) 1 1 Q 0 x) 0 Since Q k x) = U x k 2 + 2) U x k ), for k = 0,, n 1, nd Q n x) = x ) x2 ) x2 )) U n 1 b + 1 U n x ) x )) U n U n x ) x ) = U n U n ) x2 ) x2 )) U n 1 b + 1 U n 2 + 1, where U k x), for k = 0,, n, re the Chebyshev polynomils of second kind of degree k, the zeros of Q n x) re exctly the eigenvlues of 1 C, ie, the rel) vlues which stisfy the equlity p n x) := U x n 2 + 1) U x n ) U x n ) U x n ) = 1 b 21)

4 4 CM DA FONSECA In generl, 21) mens tht the eigenvlues of 1 C re the intersections of the grph of p n x) with the line y = 1 As first consequence consider the cse when = 1 nd b = 0 The eigenvlues of A re the solutions of the eqution U x n 2 + 1) U x n ) = 0, which re, for k = 0,, n 1, λ k = 2 cos ) 2k + 1 2n + 1 π 2 The vlue of n eigenvector ssocited to λ k follows immeditely: [ Qn 1 λ k ) Q 1 λ k ) Q 0 λ k ) ] t Hence we proved the following: Theorem 21 [2, 6]) The mtrix A of order n, n 3, hs the eigenpirs λ k, v k ) given by where for k = 0,, n 1 λ k = cos r k)) 1 nd v k = [sinjr k )] t j=1,,n, r k = 2k + 1 2n + 1 π, If = 2 nd b = 1, then the eigenvlues of 1 2B re solutions of the eqution U n x + 1) U n 2 x + 1) = 0, which re, for k = 0,, n 1, ) 2k + 1 cos 2n π 1 3 Loction of eigenvlues Since p n x) defined in 21) is strictly incresing, even if it is impossible to evlute exctly the eigenvlues of C, one cn locte them For exmple, if b < 0, then 1 > 0, nd ech eigenvlue λ k is locted between the zeros of U x n 2 + 1) U x n ) nd the zeros of U x n ) U x n ), ie, lies in the intervls ] ) ) 2 2k + 1 cos 2n + 1 π 1, 2 ) )[ 2k 1 cos 2n 1 π 1, for k = n 1,, 1, nd λ 0 is on the right side of 2 cos 1 2n+1 π) 1 ) If 1 < 0, ie, b > 0, one cn mke n nlogous considertion

5 EIGENVALUES OF SOME TRIDIAGONAL MATRICES 5 Let us consider the mtrix T 6 from Proposition 11 with = b = 2, ie, The eigenvlues of this mtrix re locted in the intervls ] , [ ] , [ ] , [ ] , [ ] , [ nd one is greter thn In fct, they re pproximtely λ 0 = λ 1 = λ 2 = λ 3 = λ 4 = λ 5 = The intersection of the grphs y = p 6 x) nd y = 1 2 Figure 1

6 6 CM DA FONSECA 4 A mtrix of mximums In the second section we hve considered the mtrix [min{i, j}] i,j Wht hppens if insted of the minimum we hve the mximum? We note tht the inverse of C must be tridigonl becuse the upper nd the lower tringulr prts of C hve rnk 1 form Theorem 41 For positive integer n, consider the tridigonl mtrix of order n M = ) n Then M is invertible nd the inverse is M 1 = [mx{i, j}] i,j=1,,n Proof : From 11) we hve θ i = 2θ i 1 θ i 2, for i = 2,, n 1 nd θ n = 1 + n) 1 θn 1 θ n 2, with initil conditions θ 0 = 1 nd θ 1 = 1 Then θ i = 1) i, for i = 0,, n 1, nd θ n = 1) n 1 1 n = det M The φ i s verify the recurrence reltion φ i = 2φ i+1 φ i+2, for i = n 1,, 2, with initil conditions φ n+1 = 1 nd φ n = n, nd φ 1 = φ 2 φ 3 Then φ i = 1) n i+1 i 1) 1 n, for i = 2,, n + 1 Finlly, the inverse of M is the symmetric mtrix such tht ie, M 1 ) ij = 1) i+j 1)i 1 1) n j j n 1) n 1 1 n M 1 = [mx{i, j}] i,j=1,,n = j for i j, Let us consider gin the recurrence reltion of Q k x) lredy defined, with = 1 nd b = n + 1 This recurrence reltion is equivlent to Q 0 x) 1 1 Q 0 x) 0 Q 1 x) x Q n 2 x) = Q 1 x) Q n 2 x) +Q nx) 0 0 Q n 1 x) Q n n 1 x) 1

7 EIGENVALUES OF SOME TRIDIAGONAL MATRICES 7 Therefore one cn locted the eigenvlues of the mtrix M using the rguments of the lst section Note tht [ Q0 λ k ) Q 1 λ k ) Q n 1 λ k ) ] t is n eigenvector of M ssocited to the eigenvlue λ k References [1] CF Fischer, R Usmni, Properties of some tridigonl mtrices nd their ppliction to boundry vlue problems, SIAM J Numer Anl ) [2] A Kovčec, WMY 2000 nd PARIS, August 8, 1900 A Celebrtion nd A Dediction), Pre-Print 00 21, Deprtment of Mthemtics, University of Coimbr [3] JW Lewis, Inversion of tridigonl mtrices, Numer Mth ) [4] R Usmni, Inversion of tridigonl Jcobi mtrix, Liner Algebr Appl 212/ ) [5] R Usmni, Inversion of Jcobi s tridigonl mtrix, Comput Mth Appl, ) [6] WF Trench, Eigenvlues nd eigenvectors of two symmetric mtrices, IMAGE Bulletin of the Interntionl Liner Algebr Society ) CM d Fonsec Deprtmento de Mtemátic, Universidde de Coimbr, Coimbr, Portugl E-mil ddress: cmf@mtucpt