Eigenvalues and eigenvectors of tridiagonal matrices

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1 Electronic Journal of Linear Algebra Volume 15 Volume ) Article Eigenvalues and eigenvectors of tridiagonal matrices Said Kouachi kouachisaid@caramailcom Follow this and additional works at: Recommended Citation Kouachi, Said 006), "Eigenvalues and eigenvectors of tridiagonal matrices", Electronic Journal of Linear Algebra, Volume 15 DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository For more information, please contact scholcom@uwyoedu

2 Volume 15, pp , April EIGENVALUES AND EIGENVECTORS OF TRIDIAGONAL MATRICES SAID KOUACHI Abstract This paper is continuation of previous work by the present author, where explicit formulas for the eigenvalues associated with several tridiagonal matrices were given In this paper the associated eigenvectors are calculated explicitly As a consequence, a result obtained by Wen- Chyuan Yueh and independently by S Kouachi, concerning the eigenvalues and in particular the corresponding eigenvectors of tridiagonal matrices, is generalized Expressions for the eigenvectors are obtained that differ completely from those obtained by Yueh The techniques used herein are based on theory of recurrent sequences The entries situated on each of the secondary diagonals are not necessary equal as was the case considered by Yueh Key words Eigenvectors, Tridiagonal matrices AMS subject classifications 15A18 1 Introduction The subject of this paper is diagonalization of tridiagonal matrices We generalize a result obtained in 5] concerning the eigenvalues and the corresponding eigenvectors of several tridiagonal matrices We consider tridiagonal matrices of the form α + b c a 1 b c a b A n = 0 0, 1) 0 cn a n 1 β + b where {a j } n 1 j=1 and {c j} n 1 j=1 are two finite subsequences of the sequences {a j} j=1 and {c j } j=1 of the field of complex numbers C, respectively, and α, β and b are complex numbers We suppose that a j c j = { d 1, if j is odd d, if j is even j =1,,, ) where d 1 and d are complex numbers We mention that matrices of the form 1) are of circulant type in the special case when α = β = a 1 = a = =0andall the entries on the subdiagonal are equal They are of Toeplitz type in the special case when α = β = 0 and all the entries on the subdiagonal are equal and those on the superdiagonal are also equal see U Grenander and G Szego 4]) When the Received by the editors 9 January 006 Accepted for publication 1 March 006 Handling Editor: Ludwig Elsner Laboratoire de mathematiques, Universite de badji, Mokhtar, Annaba, 300, Algeria kouachisaid@caramailcom) 115

3 Volume 15, pp , April S Kouachi entries on the principal diagonal are not equal, the calculi of the eigenvalues and the corresponding eigenvectors becomes very delicate see S Kouachi 6]) When a 1 = a = = c 1 = c = =1,b = andα = β = 0, the eigenvalues of A n has been constructed by J F Elliott ] and R T Gregory and D Carney 3] to be kπ λ k = +cos,k=1,,, n n +1 When a 1 = a = = c 1 = c = =1,b = andα =1andβ =0orβ =1,the eigenvalues has been reported to be and λ k = +cos kπ,k=1,,, n, n kπ λ k = +cos,k=1,,, n, n +1 respectively without proof W Yueh1] has generalized the results of J F Elliott ] and R T Gregory and D Carney 3] to the case when a 1 = a = = a, c 1 = c = = c and α =0,β= ac or α =0,β= ac or α = β = ac α = β = ac or α = β = ac He has calculated, in this case, the eigenvalues and their corresponding eigenvectors λ k = b + ac cos θ k,k=1,, n, where θ k = kπ n, kπ n n, k =1,, n respectively In S Kouachi5], we have generalized the results of W Yueh 1] to more general matrices of the form 1) for any complex constants satisfying condition n+1, k 1)π n+1, k 1)π and k 1)π a j c j = d,j=1,,, where d is a complex number We have proved that the eigenvalues remain the same as in the case when the a i s and the c i s are equal but the components of the eigenvector u k) σ) associated to the eigenvalue λ k, which we denote by u k) j σ),j=1,, n,are of the form u k) j σ) = d) 1 j a σ1 a σj 1 u k) d sinn j +1)θ k β sinn j)θ k 1,j=1,, n, d sin nθ k β sinn 1)θ k where θ k is given by formula d sin n +1)θ k d α + β)sinnθ k + αβ sin n 1) θ k =0,k=1,, n Recently in S Kouachi 6], we generalized the above results concerning the eigenvalues of tridiagonal matrices 1) satisfying condition ), but we were unable to calculate the corresponding eigenvectors, in view of the complexity of their expressions The

4 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 117 matrices studied by J F Elliott ] and R T Gregory and D Carney 3] are special cases of those considered by W Yueh1] which are, at their tour, special cases with regard to those that we have studied in S Kouachi 5] All the conditions imposed in the above papers are very restrictive and the techniques used are complicated and are not in general) applicable to tridiagonal matrices considered in this paper even tough for small n For example, our techniques are applicable for all the 7 7 matrices A 7 = i i 7 5 7i i and A 7 = i i i i 5 8i i 5 18i i 5 +i i 5 3 6, and guarantee that they possess the same eigenvalues and in addition they give their exact expressions formulas 15) lower) since condition ) is satisfied: λ 1,λ 4 =5± 3 ) ) + 3 ) 6 4 ) cos λ,λ 5 =5± 3 ) ) + 3 ) 6 4 ) cos λ 3,λ 6 =5± 3 ) ) + 3 ) 6 4 ) cos λ 7 = ), π 7 4π 7 6π 7 whereas the recent techniques are restricted to the limited case when the entries on the subdiagonal are equal and those on the superdiagonal are also equal and the direct calculus give the following characteristic polynomial P λ) =λ ) 6 35 λ ) λ 5 + ) ) λ ), ), ), ) λ λ 3 +

5 Volume 15, pp , April S Kouachi ) λ ) for which the roots are very difficult to calculate degree of P 5) If σ is a mapping not necessary a permutation) from the set of the integers from 1to n 1) into the set of the integers different of zero N, we denote by A n σ) the n n matrix α + b c σ a σ1 b c σ a σ b A n σ) = ) 0 cσn a σn 1 β + b and by n σ) = n σ) λi n its characteristic polynomial If σ = i, where i is the identity, then A n i) and its characteristic polynomial n i) will be denoted by A n and n respectively Our aim is to establish the eigenvalues and the corresponding eigenvectors of the matrices A n σ) The Eigenvalue Problem Throughout this section we suppose d 1 d 0In the case when α = β =0,thematrixA n σ) and its characteristic polynomial will be denoted respectively by A 0 n σ) and 0 n σ) and in the general case they will be denoted by A n and n We put where Y = d 1 + d +d 1d cos θ, 3) Y = b λ 31) In S Kouachi 6], we have proved the following result Theorem 1 When d 1 d 0, the eigenvalues of the class of matrices A n σ) on the form 11) are independent of the entries a i,c i,i=1,, n 1) and of the mapping σ provided that condition ) is satisfied and their characteristic determinants are given by n =d 1 d ) m 1 d 1d Y α β)sinm+1)θ+αβy αd 1 βd ) sin mθ sin θ, 4a) when n =m +1 is odd and d1d sinm+1)θ+αβ+d m 1 α+β)y ] d sin mθ+αβ 1 sinm 1)θ n =d 1 d ) d sin θ, 4b) when n =m is even

6 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 119 Proof Whenα = β = 0, formulas 4a) and 4b) become, respectively when n =m +1isoddand 0 n =d 1 d ) m Y sin m +1)θ, 4a0) sin θ sinm +1)θ + d sin mθ 0 n =d 1d ) m d 1, 4b0) sin θ when n =m is even Since the right hand sides of formulas 4a0) and 4b0) are independent of σ, then to prove that the characteristic polynomial of A 0 n σ), which we denote by 0 n, is also, it suffices to prove them for σ = i Expanding 0 n in terms of it s last column and using ) and 3), we get 0 n = Y 0 n 1 d 0 n,n=3,, 4a1) when n =m +1isoddand 0 n = Y 0 n 1 d 1 0 n,n=3,, 4b1) when n =m is even Then by writing the expressions of 0 n for n =m +1, m and m 1respectively, multiplying 0 m and 0 m 1 by Y and d 1 respectively and adding the three resulting equations term to term, we get 0 m+1 = Y d 1 d ) 0 m 1 d 1 d 0 m 3, 4a) We will prove by induction in m that formula 4a0) is true If n =m + 1is odd, for m =0andm = 1formula 4a0) is satisfied Suppose that it is satisfied for all integers <m, then from 4a) and using 3), we get 0 m+1 = Y d 1d ) m sinmθ cos θ sin m 1) θ sin θ Using the well known trigonometric formula sinη cos ζ =sinη + ζ)+sinη ζ), *) for η = mθ and ζ = θ, we deduce formula 4a0) When n =m is even, applying formula 4a1) for n =m +1, we get 0 m = 0 m+1 + d 0 m 1 Y By direct application of 4a0) two times, for n =m +1andn =m 1,tothe right hand side of the last expression, we deduce 4b0) If we suppose that α 0orβ 0, then expanding n in terms of the first and last columns and using the linear property of the determinants with regard to its columns, we get

7 Volume 15, pp , April S Kouachi Y c 0 0 n = 0 n α a Y E n 1 β E 1 n 1 + αβ 0 0 cn 0 0 a n Y where En 1 1 and E n 1 are the n 1) square matrices of the form 1), En 1 i = Y c i 0 0 a i Y 0 0 cn+i 3 0 a n+i 3 Y,i=1, Since all the entries a i s on the subdiagonal and c i s on the superdiagonal satisfy condition ), then using formulas 4a0) and 4b0) and taking in the account the order of the entries a i s and c i s, we deduce the general formulas 4a) and 4b) Before proceeding further, let us deduce from formula 4b) a proposition for the matrix B n σ) which is obtained from A n σ) by interchanging the numbers α and β Proposition When n is even, the eigenvalues of B n σ) are the same as A n σ) Let us see what formula 4) says and what it does not say It says that if a i,c i,i= 1,, n 1are other constants satisfying condition ) and α + b c a 1 b c a A b n = c n a n 1 β + b then the matrices A n, A n and A n σ) possess the same characteristic polynomial and hence the same eigenvalues Therefore we have this immediate consequence of formula 4) Corollary 3 The class of matrices A n σ), whereσ is a mapping from the set of the integers from 1 to n 1) into N are similar provided that all the entries on the subdiagonal and on the superdiagonal satisfy condition ) The components of the eigenvector u k) σ), k =1,, n associated to the eigenvalue λ k, k =1,, n, which we denote by u k) j, j =1,, n,aresolutionsofthe

8 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 11 linear system of equations α + k )u k) 1 + c σ1 u k) =0, a σ1 u k) 1 + k u k) + c σ u k) 3 =0, a σn 1 u k) n 1 + β + k)u k) n =0, 5) where k = Y is given by formula 3) and θ k, k =1,, n are solutions of d 1 d k α β)sinm +1)θ k + αβ k αd 1 βd ) sin mθk =0, 6a) when n =m +1isoddand d 1 d sinm +1)θ k + αβ + d α + β) ] k sin mθk + αβ d 1 sinm 1)θ k =0, 6b) d when n =m is even Since these n equations are linearly dependent, then by eliminating the first equation we obtain the following system of n 1) equations and n 1) unknowns, written in a matrix form as k c σ 0 0 a σ k 0 0 cσn a σn 1 β + k ) u k) u k) 3 u k) n = a σ1 u k) ) The determinant of this system is given by formulas 4) for α =0andn replaced by n 1and equal to k) n 1 =d 1d ) m 1 d 1 d sinm +1)θ k + d 1 β k ] sin mθk sin θ k, 8a) when n =m +1isoddand d 1 k β)sinmθ k β sinm 1)θ k n 1 =d 1d ) m 1 d, 8b) sin θ k k) when n =m is even, for all k =1,, n u k) j σ) = Γk) j σ),j,k=1,, n, 9) k) n 1

9 Volume 15, pp , April S Kouachi where Γ k) j σ) = k c σ 0 a σ1 u k) a σ k cσj 0 0 a σj k 0 0 aσj 1 0 c σj 0 k 0 aσj +1 cσn a σn 1 β + k ), j =,, n, k =1,, n By permuting the j first columns with the j 1)-th one and using the properties of the determinants, we get u k) j σ) = 1) j Λk) j where Λ k) j σ) is the determinant of the matrix C k) j σ) = σ),j=,, n, 10) n 1 T k) j 1 σ) 0 0 S k) n j σ) ), where T k) j 1 σ) = a σ1 u k) 1 k c σ a σ 0 0 cσj 0 0 k 0 0 a σj 1 is the supertriangular matrix of order j 1with diagonal a σ1 u k) 1,a σ,, a σj 1 ) and k c σj S k) n j σ) = a σj cσn a σn 1 β + k ),

10 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 13 is a tridiagonal matrix of order n j belonging to the class of the form 11) and satisfying condition ) Thus C k) j σ) = T k) S j 1 σ) k) n j σ) 11) = a σ1 a σj 1 u k) 1 k) n j,j=,, n and k =1,, n, where k) n j σ) is given by formulas 4) for α =0andn 1replaced by n j k) n j = when n is odd and k) n j = d 1 d ) n j 1 d 1d sin n j d 1 d ) n j 1 d 1 d ) n j 1 +1)θ k+d 1 β k) sin n j θ k sin θ k k β)sin n j+1 )θ k β d d 1 sin n j 1 sin θ k )θ k, when j is odd,, when j is even, 1a) d k β)sin n j 1 1 )θ k β sin n j 1 )θ k d sin θ k, when j is odd, d 1 d ) n j 1 d 1d sin n j +1)θ k+d β k) sin n j θ k sin θ k, when j is even, 1b) when n is even, for all j =,, n and k =1,, n Using formulas 9)-1), we get Finally u k) j u k) j σ) = 1) j 1 a σ1 a σj 1 u k) k) n j 1 σ) =µ j σ) u k) 1 when n is odd and u k) j σ) =µ j σ) u k) 1 k) n 1 d 1d sin n j +1)θ k+d 1 β k) sin n j θ k d 1d sin n+1 )θ k+d 1 β k) sin n 1 d1 d k β)sin n j+1 )θ k β d d 1 sin n j 1 )θ k d 1d sin n+1 )θ k+d 1 β k) sin n 1 k β)sin n j 1 )θ k β d 1 d sin n j 1 )θ k k β)sin n θ k β d 1 d sin n 1)θ k 1 d 1d sin n j +1)θ k+d β k) sin n j θ k d1d,j=,, n and k =1,, n 13) k β)sin n θ k β d 1 d sin n 1)θ k )θ k, when j is odd,, when j is even, )θ k 13a), when j is odd,, when j is even, 13b)

11 Volume 15, pp , April S Kouachi for all j =,, n and k =1,, n, whenn is even, where µ j σ) = d 1 d ) 1 j aσ1 a σj 1,j=,, n, ) k = Y and θ k are given respectively by 3) and formulas 6) 3 Special Cases From now on, we put ρ j σ) = where µ j σ) isgivenby ) 31 Case when n is odd If α = β =0, we have d 1 d ) n 1 µj σ), j =1,, n, ) Theorem 31 If α = β =0,the eigenvalues λ k σ), k =1,, n of the class of matrices A n σ) on the form 11)are independent of the entries a i,c i,i=1,, n 1) and of σ provided that condition ) is satisfied and they are given by b + d 1 + d +d 1d cos θ k, k =1,, m, λ k = b d 1 + d +d 1d cos θ k, k = m +1,, m, 14) The corresponding eigenvectors u k) σ) = given by u k) j σ) =ρ j σ) and u k) j σ) = b, k = n u k) 1 σ),, u k) n σ)) t,k=1,, n 1 are d 1 d sin n j +1)θ k + d n j 1 sin θ k, when is j odd, d1 d b λ k )sin n j+1 )θ k, when is j even, a σ1 a σj 1 d ) n j, when j is odd, 0, when j is even, 14a) 14b) j =1, n, whereρ j σ) is given by ) and θ k = kπ n+1, k =1,, m, k m)π n+1, k = m +1,, m Proof, We take a σ0 = a 0 = 1 The eigenvalues λ k, k =1,, m are trivial consequence of 4) by putting m +1)θ = kπ, k =1,, m and using 3) The

12 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 15 eigenvalue λ n is a consequence of 4) and 31) by putting Y = 0 Formula 14a) is a trivial consequence of 13a) by taking β = 0 and choosing u k) 1 = ) n 1 d 1 d d 1 d sin n +1 ) ] n 1 )θ k + d 1 sin θ k,k=1,, n 1 Concerning the n th eigenvector, we solve directly system 7) and choose u n) 1 = d ) n 1 If α = d and β = d 1 or β = d 1 and α = d, then using the trigonometric formula *) for η = m+1 )θ k and ζ = θ k, formula 6a) becomes m +1) k ± d 1 + d )) sin θ k cos θ k =0, 6a1) then we get Theorem 3 If α = d and β = d 1 or β = d 1 and α = d,the eigenvalues λ k σ),k=1,, n of the class of matrices A n σ) on the form 11)are independent of the entries a i,c i,i=1,, n 1) andofσ provided that condition ) is satisfied and they are given by b + d 1 + d +d 1d cos θ k, k =1,, m, λ k = b d 1 + d +d 1d cos θ k, k = m +1,, m, 15) b α + β), k = n t The corresponding eigenvectors u k) σ) = u k) 1 σ),, u k) n σ)), k =1,, n are given by d sin n j u k) +1)θ k +d 1 b + λ k ]sin n j θ k,j is odd, j σ) =ρ j σ) d d d1 1 b + λ k )sin n j+1 )θ k + d sin n j 1 ] )θ k,j is even, 15a) k =1, n 1 and u n) j σ) =ρ j σ) j =1,, n, whenα = d and β = d 1 and u k) j σ) =ρ j σ) 1, when j is odd, d d 1, when j is even, d sin n j +1)θ k +d 1 + b λ k ]sin n j θ k,j is odd, d d d1 1 + b λ k )sin n j+1 )θ k + d sin n j 1 ] )θ k,j is even, 15b)

13 Volume 15, pp , April S Kouachi k =1, n 1 and 1, when j is odd, u n) j σ) =ρ j σ) d d 1, when j is even, j =1,, n, whenβ = d 1 and α = d, j =1,, n, whereρ j σ) is given by ) and kπ n, k =1,, m, θ k = k m)π n, k = m +1,, m Proof Formula 15) is a simple consequence of 6a1) Using 13a), the expressions 15a) and 15b) are trivial by choosing u k) 1 = ) n 1 d 1 d d sin n +1 ) ] n 1 )θ k +d 1 b + λ k ]sin θ k, if α = d and β = d 1 and u k) 1 = ) n 1 d 1 d d sin n +1 ) ] n 1 )θ k +d 1 + b λ k ]sin θ k, when β = d 1 and α = d The last eigenvector is obtained by choosing u n) 1 σ) = ) n 1 d 1 d aσ1 a σj 1 formula When α = d and β = d 1 or α = d and β = d 1, then using the trigonometric cosη sin ζ =sinη + ζ) sinη ζ), **) for η = m+1 )θ k and ζ = θ k, formula 4a) becomes then we get k ± d d 1 )) cos m +1) θ k sin θ k =0, 6a) Theorem 33 If α = d and β = d 1 or β = d 1 and α = d,the eigenvalues λ k σ),k=1,, n of the class of matrices A n σ) on the form 11)are independent of the entries a i,c i,i=1,, n 1) andofσ provided that condition ) is satisfied and they are given by b + d 1 + d +d 1d cos θ k, k =1,, m, λ k = b d 1 + d +d 1d cos θ k, k = m +1,, m, 16) b α + β), k = n

14 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 17 t The corresponding eigenvectors u k) σ) = u k) 1 σ),, u k) n σ)), k =1,, n are given by 15a) and u n) 1) j 1 j σ) =ρ j σ), when j is odd, d d 1 1) j+ j =1,, n,, when j is even, when α = d and β = d 1 and 15b) and u n) 1) j 1 j σ) =ρ j σ), when j is odd, d d 1 1) j, when j is even, when β = d 1 and α = d,where ρ j σ) is given by ) and θ k = { k 1)π n, k =1,, m, k m) 1)π n, k = m +1,, m j =1,, n, Proof Formula 16) is trivial by solving 6a) Concerning the eigenvectors, following the same reasoning as in the case when α = d and β = d 1 or β = d 1 and α = d and since we use formula 13a) to find the components of the eigenvectors which depend only of β, we deduce the same results The last eigenvector is obtained by passage to the limit in formula 13a) when θ k tends to π and choosing the first component as in the previous case 3 Case when n is even If αβ = d, then using *) for η = mθ k and ζ = θ k, formula 6b) becomes d1 d cos θ k + αβ + d α + β) ] k sin mθk =0 Using 3), we get k α + β) k + d ] d 1 sin mθk =0, 6b1) which gives sin mθ k =0 and k α + β) k + d d 1 =0, 3) then we get Theorem 34 If αb = d,the eigenvalues λ k σ), k =1,, n of the class of matrices A n σ) on the form 11)are independent of the entries a i,c i,i=1,, n 1)

15 Volume 15, pp , April S Kouachi and of σ provided that condition ) is satisfied and they are given by b + d 1 + d +d 1d cos θ k, k =1,, m 1, b d 1 + d +d 1d cos θ k, k = m,, m, λ k = α + β)+ α β) +4d 1 17) b +, k = n 1, α + β) α β) +4d 1 b +, k = n t The corresponding eigenvectors u k) σ) = u k) 1 σ),, u k) n σ)),k=1,, n are given by b λ k β)sin n j 1 u k) )θ k β d 1 sin n j 1 d )θ k,j is odd, j σ) =ρ j σ) 1 d1 d1d d sin n j +1)θ k + d β b λ k ) ) sin n j θ k],jeven 17a) where ρ j σ), j =1,, n is given by ) and θ k = kπ n, k =1,, m 1, k m+1)π n, k = m,, m The eigenvectors u n 1) σ) and u n) σ) associated respectively with the eigenvalues λ n 1 and λ n are given by formula 13b), where θ k is given by 3), 31) and 3) Proof Formula 17) is a consequence of 6b1) The eigenvectors are a consequence of formula 13b) by choosing u k) 1 = b λ ) n 1 k β)sin n θ k β d 1 sin n d 1)θ k, when j is odd, d 1 d b λ k β)sin n θ k β d 1 sin n d 1)θ k, when j is even When α = β = ±d, then, using **), formula 6b) gives d 1 d cos mθ =0, 6b) then we have Theorem 35 If α = β = ±d,the eigenvalues λ k σ), k =1,, n of the class of matrices A n σ) on the form 11)are independent of the entries a i,c i,i=

16 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 19 1,, n 1) andofσ provided that condition ) is satisfied and they are given by b + d 1 + d +d 1d cos θ k, k =1,, m, λ k = b 18) d 1 + d +d 1d cos θ k, k = m +1,, n, t The corresponding eigenvectors u k) σ) = u k) 1 σ),, u k) n σ)), k =1,, n are given by b λ k d )sin n j 1 u k) )θ k d 1 sin n j 1 )θ k,j is odd, j σ) =ρ j σ) d 1 d sin n j +1)θ k + d d b λ k ) ] sin n j θ k,j is even, 18a) when α = β = d and b λ k + d )sin n j 1 u k) )θ k + d 1 sin n j 1 )θ k,j is odd, j σ) =ρ j σ) d 1 d sin n j +1)θ k + d + d b λ k ) ] sin n j θ k,j is even, 18b) when α = β = d where ρ j σ), j =1,, n is given by ) and θ k = k 1)π n, k =1,, m, k m 1)π n, k = m +1,, n Proof Formula 18) is a consequence of 6b)The eigenvectors are a consequence of formula 13b) by choosing u k) 1 = ) n 1 b λ k d )sin n θ k d 1 sin n 1)θ k, when j =l +1 d 1 d d1 d b λk d )sin n θ k d 1 sin n 1)θ ] k, when j =l when α = β = d u k) 1 = ) n 1 b λ k + d )sin n θ k + d 1 sin n 1)θ k, when j =l +1 d 1 d d1 d b λk + d )sin n θ k + d 1 sin n 1)θ ] k, when j =l when α = β = d 4 Case when d 1 d =0 In this case, we have proved in S Kouachi 6], the following Proposition 41 When d 1 d =0,the eigenvalues λ k σ), k =1,, n of the class of matrices A n σ) on the form 11) are independent of the entries a i,c i,i=

17 Volume 15, pp , April S Kouachi 1,, n 1) andofσ provided that condition ) is satisfied and their characteristic polynomials are given by d) n 1 1 α) ) β d, when n is odd; n = d) n 1 α + β) + αβ ) 19a), when n is even, when d 1 =0and n = d1) n 1 1 β) ) α d 1, when n is odd; 19b) d1) n ) α d 1 ) β d 1, when n is even, when d =0,where = Y is given by 3) An immediate consequence of this proposition is Proposition 4 If d 1 d =0,the eigenvalues λ k σ),k=1,, n of the class of matrices A n σ) on the form 11)are independent of the entries a i,c i,i=1,, n 1) and of σ provided that condition ) is satisfied: 1) When α = β =0, they are reduced to three eigenvalues when d 1 =0or when n is odd and d =0 and only two eigenvalues { b ± d, b} { b ± d 1, b} { b ± d 1 } when n is even and d =0 ) When α 0or β 0, they are reduced to five eigenvalues { : b ± d, b α, b 1 β ± 1 } β +4d when n is odd and d 1 =0,fivealso { b ± d 1,b β, b 1 α ± 1 } α +4d 1 when n is odd and d =0,four {b ± d,b α, b β} when n is even and d 1 =0and six eigenvalues { b ± d 1,b 1 α ± 1 α +4d 1, b 1 β ± 1 } β +4d 1

18 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 131 when n is even and d =0 3) All the corresponding eigenvectors u k) σ) = are are simple and given by 31)When λ k is simple u k) j σ) =ν j σ) when d 1 =0and u k) j σ) =ν j σ) ] n j b λ k ) d,whenj is odd, u k) 1 σ),, u k) n σ)) t, k =1,, n ] n j 1,n is odd, b λ k ) d b λ k ),whenj is even, ] 1+n j b λ k ) d,whenj is odd, ] n j b λ k ) d b λ k ),whenj is even,,n is even,, 0a) ] n j b λ k ) b λ k ) d 1,whenj is odd,,n is odd, ] n j 1 b λ k ) d +1 1,whenj is even, ] n 1 j b λ k ) b λ k ) d 1,whenj is odd,,n is even, ] n j b λ k ) d 1,whenj is even, 0b) when d =0,j=,, n and k =1,, n,where ν j σ) = 1) n j a σ1 a σj 1,j =,, n 3) When λ k is multiple, then all the components are zero except the last four ones at most and which we calculate directly Proof The expressions of the eigenvalues are trivial by annulling the corresponding characteristic determinants Following the same reasoning as the case d 1 d 0, by solving system 7), we get the expressions of the eigenvectors by formulas 13) ]] k) u k) j σ) = 1) n 1 ν j σ) u k) n j 1,j=,, n and k =1,, n, k) n 1 where ]] denote the reduced fraction ) k d m ) ) k) n 1 = k d k β k d, when n =m + 1is odd, ) k d m k β), k ) d, when n =m is even,

19 Volume 15, pp , April S Kouachi when d 1 =0and k) n 1 = ) k d m 1 1 k β ), when n =m + 1is odd, k ) k d m ) 1 k k β k d 1, when n =m is even, when d =0 k) n j = when d 1 =0and k) n j = k d) n j ) ) k d k β k d,whenj is odd, n is odd, k d) n j 1 1 ) k k β k d,whenj is even k d) n j 1 1 k β) k ) d,whenj is odd, n is even, k d) n j 1 ) k β k,whenj is even, k d1) n j 1 ) k β k,whenj is odd, k d1) n j 1 1 k β) n is odd; ) k d 1,whenj is even k d1) n j 1 1 ) k k β k d 1,whenj is odd, n is even, k d1) n j ) ) k d 1 k β k d 1,whenj is even when d =0 Then, when d 1 =0,wehave k d) 1 j u k) j σ) = 1) n 1 ν j σ),whenj is odd, k d ) j k,whenj is even k d ) 1 j,whenj is odd, k d ) j k,whenj is even, n is odd, n is even, j =,, n and k =1,, n By putting j = n, calculatingu k) 1 according to u k) n σ) and choosing u k) n σ) =a σ1 a σn 1,

20 Volume 15, pp , April Eigenvalues and Eigenvectors of Tridiagonal Matrices 133 we get 0a) Following the same reasoning as in the case when d 1 = 0, we deduce 0b) Acknowledgment The author thanks the superior education ministry and the national agency of development and scientific research ANDRU) for their financial support REFERENCES 1] Wen-Chyuan Yueh Eigenvalues of several tridiagonal matrices Applied Mathematics E-Notes, 5:66 74, 005 ] J F Elliott The characteristic roots of certain real symmetric matrices Master s thesis, Univ of Tennessee, ] R T Gregory and D Carney A Collection of Matrices for Testing Computational Algorithm Wiley-Interscience, New York, ] U Grenander and G Szego Toeplitz Forms and Their Applications University of California Press, Berkeley and Los Angeles, ] S Kouachi Eigenvalues and eigenvectors of several tridiagonal matrices Submitted to the Bulletin of the Belgian Mathematical Society 6] S Kouachi Eigenvalues and eigenvectors of tridiagonal matrices with non equal diagonal entries In preparation 7] S Kouachi Explicit Eigenvalues of tridiagonal matrices To appear in Applied Mathematics E-Notes