Solution of a tridiagonal operator equation

Transcription

1 Liear Algebra ad its Applicatios 44 (006) Solutio of a tridiagoal operator equatio R. Balasubramaia a, S.H. Kulkari b,, R. Radha b a The Istitute of Mathematical Scieces, C.I.T. Campus, Tharamai, Cheai 600 3, Idia b Departmet of Mathematics, Idia Istitute of Techology, Cheai , Idia Received 6 August 004; accepted 8 October 005 Available olie 0 December 005 Submitted by R. Bhatia Abstract Let H be a separable Hilbert space with a orthoormal basis {e / N}, T be a bouded tridiagoal operator o H ad T be its trucatio o spa ({e,e,...,e }). We study the operator equatio Tx = y through its fiite dimesioal trucatios T x = y. It is show that if { T e } ad { T e } are bouded, the T is ivertible ad the solutio of Tx = y ca be obtaied as a limit i the orm topology of the solutios of its fiite dimesioal trucatios. This leads to uiform boudedess of the sequece {T }. We also give sufficiet coditios for the boudedess of { T e } ad { T e } i terms of the etries of the matrix of T. 005 Elsevier Ic. All rights reserved. AMS classificatio: Primary: 47B37; Secodary: 5A60, 65F99 Keywords: Diagoal domiace; Determiat; Gerschgori disc; Tridiagoal matrix; Tridiagoal operator. Itroductio The umerical solutio of ordiary ad partial differetial equatios frequetly leads to liear systems of equatios ivolvig matrices whose elemets are zero except i a bad sorroudig the mai diagoal. For details regardig liear system arisig from partial differetial equatios, we refer to [3]. Oe of the special type is a tridiagoal matrix. Applicatio of Fiite Differeces or Fiite Elemet methods to solve boudary value problems i oe variable results i systems Correspodig author. addresses: balu@imsc.eret.i (R. Balasubramaia), shk@iitm.ac.i (S.H. Kulkari), radharam@iitm.ac.i (R. Radha) /$ - see frot matter ( 005 Elsevier Ic. All rights reserved. doi:0.06/j.laa

2 390 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) of equatios whose matrices are baded ad i case of some importat examples these matrices tur out to be tridiagoal. There exist may well developed methods ad efficiet algorithms i the literature for solvig these matrix equatios or fidig eige values of these matrices. We use the followig otatios throughout the paper. Let H deote a separable Hilbert space with a orthoormal basis {e / N}. Let H deote the liear spa of {e,e,...,e }, P, the orthogoal projectio of H oto H.ForT B(H ), the class of bouded liear operators o H. T = P T H ad Sp(T ) will deote the spectrum of T. The operators {T } are kow as fiite sectios or Galerki approximatios of T. Matrix of {T } (with respect to the basis {e j /j N}) cosists of first rows ad colums of the matrix T.Forx = j α j e j, x will deote P (x). Thus x = j= α j e j. Our iterest is to study the solutio of the operator equatio Tx = y where T is a ifiite tridiagoal matrix which ca be regarded as a bouded operator o H. I particular, we wish to first aswer the questio of ivertibility of a tridiagoal operator ad the obtai the solutio of the operator equatio Tx = y. For this, we cosider the fiite dimesioal approximatios T of T. By assumig that each T is ivertible alog with certai other coditios, we show that T is ivertible. The result is cotaied i Theorem 5.. I the ext step, we try to fid certai coditios so that the hypothesis of Theorem 5. could be verified for a give operator. It turs out that the required coditios ca be stated i terms of the etries of the matrices T. We orgaize the paper i the followig way. I Sectio, we provide some mathematical ad historical backgroud. I Sectio 3, we record certai ice properties of the tridiagoal operator. I Sectio 4, we obtai some cosequeces of the boudedess of { T (e ) }. These are used i Sectio 5 to prove the mai result (Theorem 5.). I Sectio 6, we discuss some verifiable criterios ad illustrate these with some examples. Ulike i the case of oe variable, the applicatio of Fiite Differeces or Fiite Elemet methods to partial differetial equatios lead to matrices that are o loger baded. But these matrices are baded i a differet sese. For example, whe these methods are applied to two dimesioal Laplace equatio, the resultig matrices ca be viewed as tridiagoal matrices whose etries are tridiagoal matrices (see [9] for details). At preset it is ot clear whether our methods ca be applied to this class of problems. This is a iterestig questio to be ivestigated i future.. Backgroud Though it may seem atural to expect that the behaviour of T ca be predicted from the behaviour of T for large values of, it is well kow that this expectatio is false uless some additioal assumptios are made about T ad/or T. For example, i geeral the ivertibility of T for all does ot imply the ivertibility of T. Cosider for example T : l l defied by ( Tx = α, α ),...,α,..., (α,α,...) l. The each T is ivertible. I fact, T (α,α,...,α ) = (α, α,...,α ). But T is ot ivertible as Sp(T ) ={0} {/, N}. Notice that T = = T e. Similarly diagoal domiace property of a fiite matrix implies its ivertibility whereas the above example illustrates that this is ot the case with a ifiite matrix. For the study of diagoal domiace property ad fiite dimesioal matrices we refer to [7]. However, i the case of a ifiite matrix it ca be proved that if T B(H ) has a strict row ad colum domiace property, that is if for some ɛ>0, α jj > i/=j α ij +ɛfor all j ad α ii > j/=i α ij +ɛfor all i, the

3 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) T is ivertible. We ca deduce from this a ifiite dimesioal versio of Gerschgori theorem. Let T = (α ij ), r i = j/=i α ij,r i = k/=i α ki. IfD i = B(α ii,r i ) ad D i = B(α ii,r i ), the Sp(T ) ( i D i D i). Further it is also kow that if there exists a 0 such that T is ivertible for all 0, the the solutios x of T x = y lead to the solutio x of Tx = y where x = lim x if ad oly if T is ivertible. The ivertibility of a operator T ad the ivertibility of its fiite dimesioal trucatios T are discussed i detail for Toeplitz operators i [6]. The problem of computig spectrum Sp(T ) through its fiite dimesioal trucatios uder certai assumptios are discussed i [,]. Fially, it may be oted that several of tridiagoal operators are ot ivertible. Promiet amog these are certai class of almost Mathieu operators ad, by cosistecy, ay discretizatio by Fiite Differeces of differetial equatios (see [4,8] for details). I this paper, we try to aswer the questio of fidig coditios uder which a tridiagoal operator equatio Tx = y has a solutio, whe each of the correspodig fiite dimesioal trucatios T x = y has a solutio. Further such coditios prove that {T } is uiformly bouded. I practice, to apply these sort of theorems to cocrete cases, we must have easily verifiable coditios. For example, if the coditios ca be stated i terms of the etries of the matrices, the it serves the purpose. The best kow illustratio of such a coditio is the diagoal domiace property metioed above. Suppose the tridiagoal operator is such that its off-diagoal elemets are ad product of k diagoal elemets i absolute value is greater tha k (istead of assumig that absolute value of each diagoal elemet to be greater tha ), the the operator T is ivertible (see [5]). This essetially tells us that eve if oe of the diagoal elemets is very small but the product is large eough, the we obtai the ivertibility of T. I this paper we prove a still weaker coditio. For example, whe k =, ad agai if we assume that T to be a tridiagoal operator with off-diagoal elemets, the we prove that the coditio ( d i d i+ )( d i d i ) 4 + ɛ gives the sufficiet coditios for the hypothesis of Theorem 5. to be satisfied. Similarly for k = 3, if the sequece {d i } satisfies the coditio d i ( d i+ d i+ )( d i d i ) ( d i d i ) d i+ +( d i+ d i+ ) d i +( d i+ d i+ ) + ( + ɛ)( d i d i ) ad ( d i d i+ ) η for all i where η>0 with det T 3 ( + ɛ), the { T e } ad { T e } are bouded. 3. Properties of a tridiagoal operator Let T be the tridiagoal operator defied by Te = d e + u e ad Te = c e + d e + u + e + for, where {c }, {d } ad {u } are bouded sequeces of complex umbers. For the sake of otatioal coveiece, we give the proof whe c,u,d are all real. We ca modify this i a obvious maer if c, d, u are complex umbers. Oe of the iterestig characteristics of the tridiagoal operator T is the recurrece relatio cocerig the determiats of its fiite sectios T, amely

4 39 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) det T = d det T u c det T. () Relatio () ca be easily verified by expadig the matrix of T alog the last row. The tridiagoal ature of T helps us to expad T e ad T (e ) i terms of determiats of T,T,...,T. I fact, if we cosider the matrix form of T, the T (e )=(det T ) [k e + k e + +k e ] where k,k,...,k are the cofactors of the elemets i the last row respectively. The later equality follows from the fact that for a give matrix, A = (a ij ), a ij A ri = δ jr (det A), where A r,i deotes the cofactor of a r,i. The tridiagoal form of T simplifies our task of computig the cofactors of the elemets of the last row. k m is the cofactor of the elemet i the place (, m) of T where (, m) deotes the th row ad mth colum. It is othig but ( ) +m multiplied by the determiat of T with th row ad mth colum deletio. If we ow expad this determiat alog the last colum, there is oly oe o-zero elemet i the last colum amely u, the expadig alog the last colum the oly o-zero elemet i the last colum uder cosideratio is u. Repeatig this process upto (m + )th colum (i the origial determiat), we get that u u u m+. Because of mth colum deletio, the left out determiat will be precisely det Tm = det T m. Thus k m = ( ) m+ u m+ u m+ u u (det T m ). The above techique ca be uderstood by the followig simple example: d c 0 0 T 4 = u d c 0 0 u 3 d 3 c 3, 0 0 u 4 d 4 d u 0 0 T4 = c d u c d 3 u 4, 0 0 c 3 d 4 T 4 (e 4 ) = (det T 4 ) [ (cofactor 0 (4, )) e + (cofactor 0 (4, )) e +(cofactor c 3 )e 3 + (cofactor d 4 )e 4 ]. Hece u 0 0 (det T 4 )T4 (e 4 ) = d u 3 0 c d 3 u 4 e d c u d 3 u 4 e d u 0 c d 0 0 c u 4 e d u c d u 3 0 c d 3 e 4 = u u 3 u 4 e + u 3 u 4 (det T )e u 4 (det T )e 3 + (det T 3 )e 4. Thus we get T (e ) = ( ) + (det T ) [ u u 3 u e u 3 u 4 u (det T )e + +( ) u (det T )e + ( ) (det T )e ]. ()

5 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Similarly it ca be show that If T (e ) = ( ) + (det T ) [ c c c e c c (det T )e + +( ) (det T )e ]. (3) x = α () e + α () e + +α () e, the T(x ) ad T (x ) differs oly i the last compoet, amely T(x ) = T (x ) + α () u +e +. (4) Further if x = α i e i, the it ca be easily show that P T(x)= T (x ) + α + c e. 4. Boudedess of { T e } Assume that { T e } is bouded by a umber say L. Ifwelet x = α () e + α () e + +α () e, the we will show that {u + α () } l. The coditio T e L is equivalet to u u 3 u + u 3 u 4 u (det T ) + +u (det T ) + (det T ) L (det T ). (5) Defie a i = det T i. u u i+ The (3) ca be rewritte as + a + +a L u + a, (6) or usig the boudedess of {u }, we ca write + a + a + +a B a. (7) But α () = x,e = T (y ), e = β i T e i,e = β i e i,t e. = (det T ) ( ) i+ β i u i+ u det T i

6 394 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Now, i order to show that {u + α () = (det T ) ( ) i+ β i u u a i = (det T ) u u ( ) i+ β i a i. Lemma 4. r + r + + r r+ ( + )r+ r + r+ (r + ). } l, we make use of certai estimates. Proof j r+ (j ) r+ j j = u r du j r du = j r. r + j j Summig for j =,,...,, we obtai the result. Lemma 4.. For ay o-egative iteger r ad, we have a k+ ( + a + +a k ) Here B is defied by Eq. (7). r r(r+) r!b (r+). Proof. Whe r = 0, the result will follow immediately from (7). Assume that the result is true for r. The, by summig over =,,...,(N ),weget Thus ak+ + +a k+n ( + a + +a k ) r(r+) r!b (r+) N = r ( + a + +a k ) N r+ r(r+) r!b (r+) r+ (r + ) )N r+ = ( + a + +a k. (r + )!B (r+) (r+)(r+) a k+n B [ ( + a + +a k ) + (a k+ + +a k+n ) ] ( + a + +a k (r+)(r+) )N r+ (r + )!B (r+) (by Lemma 4.) Hece the result is true for r +. Now we are i a positio to prove the required result. Propositio 4.3. Let T be the tridiagoal operator defied by Te = c e + d e + u + e +,

7 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) where {c }, {d } ad {u } are bouded sequeces of complex umbers. Suppose T is ivertible for all ad there exists a costat L>0such that T (e ) L for all, the {u + α () } l. Proof. Recall α () Therefore, = (det T ) u u = a u + u + α () = a ( ) i+ β i a i ( ) i+ β i a i. u + α () a ( ) i+ β i a i β i a i By Lemma 4., we get β i a i c a β i ( i + ) 4 by takig r = 8. We break the sum for i as i ad i>. Whe i, we get the R.H.S. to O ( 3 ), which gives a l sequece. Now, let M< N. The Cosider Now, u + α () c i>m/ i M < N i,i, β i ( + i) 4 i> M β i ( i + ) 4 + a l sequece. = M < i,i i > M i i i,i i,i β i ( + i ) 4 ( + i ) 4 β i β i ( + i ) 4 ( + i ) 4 β i + β i ( + i i ) 4 ( + i ) 4. N i > M β i <ɛ as M,N. (i + i ) 4 i i ( + i ) 4 i

8 396 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Similarly, we ca show that β i ( + i ) 4 <ɛ as M,N, ( + i ) 4 i,i, thus provig our assertio. 5. The mai result I this sectio we prove our mai result. Theorem 5.. Let T be the tridiagoal operator defied by Te = c e + d e + u + e + where {c }, {d } ad {u } are bouded sequeces of complex umbers. Suppose T is ivertible for all ad there exists a costat K such that 0 <K< ad T e K for all. If the operator equatio Tx = y has a solutio (i.e. if y R(T ), with R(T ) beig the rage of T),the this solutio ca be obtaied as the limit of the solutios x of the operator equatio T x = y H i the orm topology. I other words, T (y ) x. I particular, Tis. I additio if c /= 0,u /= 0 for all ad there exists L>0 such that T (e ) L for all, the T is oto ad hece ivertible. Proof. Let y R(T ), x = α i e i, ad T(x)= y. The T (x ), e =u α + d α ad T(x),e =u α + d α + c α +. Thus T (x ) + α + c e = y which implies that T (y ) = x + c α + T (e ). As 0, x x,α 0 ad { T (e ) } is bouded, we see that T (y ) x. It follows that T is (by takig y = 0). I order to prove T is oto, let y H. The y = β i e i, β i <. I fact, i β i = y. Put y = β i e i. Sice each T is oto x H such that T x = y. We ca write x = α () e + α () e + +α () e. The it follows from Propositio 4.3 that α () u + 0as. Further by (4), we have, T (x ) = T(x ) α () u +e +. (8) Now, if we show that {x } is a Cauchy sequece i H the there will exist a x H such that x x i H, the by cotiuity T(x ) T(x)ad i the limit T(x ) ad T (x ) coicide by (). But y = lim y = lim T (x ) = lim T(x ) = Tx, thus showig that T is oto. Further this argumet clearly shows that the operator equatio Tx = y could be solved by restrictig y to each H, for, the ivertibility of each T provides the solutio x for T x = y ad the resultig x which is obtaied as a limit (i the orm) of x turs out to

9 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) be the solutio of Tx = y. Thus it remais to show that {x } is a Cauchy sequece i H, which is proved i Lemma 5.. Lemma 5.. {x } is a Cauchy sequece i H. Proof. Cosider x + x = T + (y +) T + T +(x ) = T + (y +) T + (y ) α () = β + T + (e +) α () The for M>N, x M x N = By (3), we have N <M u +T + (e +) u +T + (e +). (β + α () u +)T + (e +). T (e ) = ( ) + (det T ) [ c c c e c c (det T )e + +( ) ] (det T )e. Let b i = det T i c c c i. The T + (e + +) = (c + ) (b + ) ( ) i+ b i e i. Let ν = β + α () u +. The by Propositio 4.3, ν l. The x M x N = ν c + (b +) b i N <M + i [ ] = ν ν c + b + c + b +,, ν ν c + c + b + b + i + x M x N c ν ν c + b + c, + c + b + b + = c ν ν c + b + c + b + c ν + c ν ν c + b + b < c ν + c ν ν b + b < b i b i

10 398 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) as {c } is bouded. Cosider ν ν b + ν b + + ν b + b < b < b < ν + ν > + > + ν b +. b ν b + b Here, {ν } l. Now, applyig Lemma 4. for {b i } istead of {a i }, ad followig the same techique i the proof of Propositio 4.3, we ca show that x M x N <ɛ as N, M. Corollary 5.3. If T satisfies the hypothesis of Theorem 5., the { T } is uiformly bouded. Proof. Let y H. The P y H ad by Theorem 5., {T (P y)} coverges, which shows that there exists M y such that T P y M y. The by uiform boudedess priciple { T P } is uiformly bouded. But it ca be easily show that T P = T, thus provig our assertio. 6. A verifiable criterio As metioed earlier, we obtai some sufficiet coditios i terms of the etries of the fiite sectios of the tridiagoal operator T so that { T e } ad { T e } are bouded. I order to state the mai theorem, we eed the followig otatios. Let k be a positive iteger. Let A j, deote the j j matrix d k u k c k d k u k c k (j ) d k (j ) ad B j, deote the j j matrix B j, = k (j+) k k (j+) c k l k 3 u k l d k (k ) c k (k ) 0 u k (k 3) d k (k 3) c k (k 3) u k (k (j+)) d k (k (j+)) Let A () j,, B() j, deote similar matrices where d i, u i, c i are replaced by d i+, u i+, c i+ (with the correspodig chages i other etries) respectively. I fact, let A (r) j,, B(r) j, deote the matrix where the etries of A j,, B j, are shifted to the right by r respectively. Now we are i a positio to state the required result..

11 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Theorem 6.. Let T be a tridiagoal operator Te = c e + d e + u + e + where {d }, {u }, {c } are complex sequeces whose absolute values are both bouded above ad below by a positive costat. Suppose there exists a positive iteger k ad positive real umbers η, η such that for some ɛ>0, the followig coditios are satisfied for 0 r k ad N: (i) det T k ( + ɛ) c c c k, det T k ( + ɛ) u u k+, (ii) d k (k )+r det A (r) k, det B(r) k, c k( )++r c k( )+r det B (r) k, det A(r) k, u k( )+r c k( )++r c k( )+r + det B (r) k, det A(r) k, c k( )++r c k( )+r + det A (r) k, u(r) k( )++r +( + ɛ) c k( )++r c k( )+r c k( )++r c k+r det B (r) k,, ad (iii) d k (k )+r det A (r) k, det B(r) k, u k( )+3+r u k( )++r det B (r) k, det A(r) k, c k( )++r u k( )+3+r u k( )++r u k( )++r + det B (r) k, det A(r) k, u k( )+3+r u k( )++r + det A (r) k, c k( )++r +( + ɛ) det B k u k( )+3+r u k( )++r u k( )++r u k++r, (iv) det A (r) k, >η, det B(r) k, >η. The { T e } ad { T e } are bouded. Here, we observe that if the give operator T is self adjoit, viz, whe u j = c j the coditios (ii) ad (iii) are the same. Also there is oly oe iequality i (i). Proof. First we obtai a recurrece relatio betwee det T k, det T k( ) ad det T k( ). For this purpose we cosider the followig k equatios: () : det T k d k det T k + u k c k det T k = 0. () : det T k d k det T k + u k c k det T k 3 = 0. (k ) : det T k (k ) d k (k ) det T k (k ) + u k (k ) c k (k ) det T k k = 0. (k) : det T k (k ) d k (k ) det T k k + u k (k ) c k k det T k (k+) = 0. (k + ) : det T k k d k k det T k (k+) + u k k c k (k+) det T k (k+) = 0. (k ) : det T k (k 3) d k (k 3) det T k (k ) + u k (k 3) c k (k ) det T k (k ) = 0. (k ) : det T k (k ) d k (k ) det T k (k ) + u k (k ) c k (k ) det T k (k) = 0.

12 400 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Multiply first (k ) equatios by det A j det B k for j = 0,,,...,k (with det A 0 = ) respectively. Multiply the last set of (k ) equatios by det A k det B j for j = k,k 3,...,0 (with det B 0 = ) respectively. Multiply the kth equatio by det A k det B k ad add all the resultig equatios. The compute the coefficiets of each det T k j. Calculatig the coefficiet of each det T k j is easy because, the o-zero coefficiets occur at the most oly i three equatios. With the straight forward calculatios, oe ca see that all the coefficiets except of det T k, det T k( ), det T k( ) vaish. As a cosequece, we obtai a recurrece relatio i terms of these three determiats, amely, det B k, (det T k ) +[u k (k ) c k (k ) det B k, det A k, d k (k ) det A k, det B k, + det B k, det A k, ](det T k( ) ) + det A k, u k (k ) c k (k ) (det T k( ) ) = 0. I the same way, we ca obtai a recurrece relatio for det T k+r, det T k( )+r, det T k( )+r, r k, where A k j,, B k j, are replaced by A (r) k j,, B(r) k j,, ad u m, c m, d m are replaced by u m+r, c m+r, d m+r respectively. Next, we claim that for =,, 3,..., det T k ( + ɛ) c k( )+ c k det T k( ), where det T 0 =. The result is true for = by coditio (i) of the hypothesis. Assume the result for. Thus, we assume det T k( ) ( + ɛ) c k( )+ c k( ). Now, usig the recurrece relatio betwee det T k, det T k( ), det T k( ),weget det B k, det T k ( d k (k ) det A k, det B k, u k (k ) c k (k ) det B k, det A k, det B k, det A k, ) det T k( ) det A k, u k (k ) c k (k ) det T k( ) ( ) det T k( ) det A k, u k (k ) + ɛ c k( )+ c k( ) det T k( ) (by usig iductio hypothesis) ( ) det T k( ) det A k, u k (k ) c k( )+ c k( ) det T k( ) = d k (k ) det A k, det B k, det A k, det B k, c k( )+ c k( ) u k (k ) det B k, det A k, c k( )+ c k( )+ det B k, det A k, c k( )+ c k( ) det A k, u k (k ) c k( )+ c k( k) det T k( ) ( + ɛ) c k( )+ c k( )+ c k det B k, det T k( )

13 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) (by coditio (ii) with r = 0). Thus, it follows that det T k ( + ɛ) c k( )+ c k( )+ c k det T k( ) (9) as det B k /= 0. Similarly, we ca show that for r =,,...,k, det T k+r ( + ɛ) c k( )++r c k( )+(+r) c k+r det T k( )+r. (0) I a similar way, by usig coditio (iii), we ca show that det T k+r ( + ɛ) u k( )++r u k++r det T k( )+r () for r = 0,,,...,k. Next, we claim that det T k det T, det T k k det T,..., det T k (k ) k det T k are bouded. I this case, we agai eed to cosider k equatios, where each equatio is the recurrece relatio ivolvig three determiats amely det T k i, det T k (i+), det T k (i+) for i = 0,,,...,k. We may have to multiply each equatio by appropriate coefficiets (as it was doe earlier) i such a way that the required boud, amely, det T k j det T k is expressed i terms of det T k k. det T k Sice by (9), det T k k det T k is bouded by where α = if c (+ɛ)α k i we obtai the required boud for det T k j. det T k To avoid the computatioal complexity i a geeral k case, we illustrate the situatio i the case k = 3. We have the followig recurrece relatios: det T 3 d 3 det T 3 + u 3 c 3 det T 3 = 0, () det T 3 d 3 det T 3 + u 3 c 3 det T 3 3 = 0. (3) Multiply (3) byd 3 ad addig with (), we get det T 3 + (u 3 c 3 d 3 d 3 ) det T 3 + d 3 u 3 c 3 det T 3 3 = 0. Thus, (d 3d 3 u 3 c 3 ) det T 3 det T = ( + d 3 u 3 c 3 ) det T det T 3 + d 3 u 3 c 3 det T 3 3 det T 3 + k k k 3 ( + ɛ)α k, where k = sup i d i,k = sup i u i,k 3 = sup i c i. Thus ( det T 3 det T + k ) k k 3 3 d 3 d 3 u 3 c 3 ( + ɛ)α k. But Sice { det A k } is bouded below, we get the upper boud for det T 3 det T 3. Agai, multiplyig () byd 3 ad (3)byu 3 c 3,weget d 3 det T 3 + (u 3 c 3 d 3 d 3 ) det T 3 + u 3 u 3 c 3 c 3 det T 3 3 = 0.

14 40 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Thus, we have (d 3 d 3 u 3 c 3 ) det T 3 det T 3 = d 3 + u 3 u 3 c 3 c 3 det T 3 det T 3 from which it follows that det T 3 det T d 3 +k k det T det T 3 which is bouded above. Now cosider Tk e k =[c c c k + c c k det T + ck det T k + det T k ](det T ) {( ) (det T k ) ( + ɛ) ( ) + ( + ɛ) ( ) + + ( + ɛ) + } (det Tk + det T k + +det T k( ) ) (by usig (9) ad (0)). Sice det T k j det T k is bouded for each j=,,...,k, it follows that { Tk e k } is bouded. Similarly usig coditio (0), we ca show that { Tk e k } to be bouded. The coditios i Theorem 6. appear quite clumsy at first sight. However, for small values of k, these coditios take simple forms. We illustrate this for k = ad 3. These may be compared with the coditios give i Theorem.3, Remark.4 ad Corollary.6 of [5]. Corollary 6.. Suppose Te = c e + d e + u e + where {c }, {u }, {d } are sequeces of umbers such that their absolute values are bouded above ad below by a costat ad satisfy the followig coditios: There exists ɛ>0 such that (i) d i d i c i ( u i + c i ) ad d i d i u i ( c i + u i ), (ii) d i d i+ u i+ ( c i +( + ɛ) u i+ ) ad d i d i+ c i ( u i+ +( + ɛ) c i+ ) for all i, the { T e } ad { T e } are bouded. Proof. We show that these coditios (i) ad (ii) imply the coditios (i) (iv) of Theorem 6. with k =. For this purpose, ote that det B, = d, det A, = d, det A 0, =, ad det B 0, =. c u If we take r = 0 i coditio (ii) of Theorem 6., after simplificatio it turs out to be d d d d u ( c + u ) + d c ( u +( + ɛ) c ). Similarly coditio (iii) leads to d d d d c ( u + c 3 ) + d u ( c + u + ( + ɛ)).

15 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) We obtai similar coditios for r =. The the coditios (ii) ad (iii) ca be rewritte as d i d i d i+ d i+ u i ( c i + u i ) + d i c i ( u i+ +( + ɛ) c i+ ) ad d i d i d i+ d i+ c i ( u i + c i ) + d i u i+ ( c i + u i+ ( + ɛ)) (4) for all i. But we kow that Axy Bx + Cy iff (Ax C)(Ay B) BC wheever A>0. These are satisfied if the followig coditios hold: d i d i c i ( u i + c i ), d i d i+ u i+ ( c i +( + ɛ) u i+ ), (5) d i d i u i ( c i + u i ), d i d i+ c i ( u i+ +( + ɛ) c i+ ). By simple calculatios, oe ca verify that these coditios imply that det T ( + ɛ) c c ad det T ( + ɛ) u u 3. Thus coditio (i) of Theorem 6. is satisfied. Also i this case, { det A (r) } ad { det B(r) } are bouded below. Hece coditio (iv) of Theorem 6. is satisfied, which completes the proof. Next, we cosider a special case of this corollary whe u i = c i = i. Corollary 6.3. Let Te = e + d e + e +. If the sequece {d i } satisfy ( d i d i+ )( d i d i ) 4 + ɛ, for each i, the { T e } ad { T e } are bouded. Proof. The result follows from Eqs. (5) with u i = c i =. Example 6.4. Let u i = c i = for all i. Defie d = d = 3, d 3 = 4 ad for i =, 3,..., d 3i = 3 + 3i, d 3i = 9/4 d 3i (say), d 3i = d 3i. First ote that for each i, d 3i 4. ( This ca be ) proved by iductio. Next, d 3i d 3i = 9/4, d 3i d 3i = ad d 3i+ d 3i = d 3 3i + 3i+ 4 3/ =. Thus the coditios of Corollary 6.3 are satisfied with ɛ = 3/4. Hece the correspodig operator T is ivertible. Corollary 6.5. Let Te = e + d e + e +. If the sequece {d i } satisfy d i ( d i+ d i+ )( d i d i ) ( d i d i ) d i+ +( d i+ d i+ ) d i +( d i+ d i+ ) +( + ɛ)( d i d i ) ad ( d i d i+ ) η for all i where η>0 with det T 3 ( + ɛ), the { T e } ad { T e } are bouded.

16 404 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) Proof. The result follows from the theorem 6. with k = 3. Example 6.6. Take u i = c i = for all i. Defie d = d = 3, d 3 = 9 ad for i =, 3,..., d 3i = 3 + 3i, d 3i = 9/4 d 3i =, d 3i = d 9 3i. Note that for ay i, ( ) 9 (d 3i d 3i )(d 3i d 3i ) = 4 (9 ) = 7 4. Thus the coditios i Corollary 6.3 are ot satisfied. O the other had, for large values of i, d 3i 3/ d 3i ad d 3i = 6. Thus it is easy to verify the coditios i Corollary 6.5 are satisfied. Remark 6.7. Theorem 6. gives a set of sufficiet coditios for each positive iteger k. If for ay such k, the correspodig coditios are satisfied, the the theorem implies that the sequeces { T e } ad { T e } are bouded ad hece T is ivertible. I a earlier versio of this paper, oly the case k = was discussed after provig Theorem 6.. That versio did ot cotai Corollary 6.5 ad Example 6.6 which pertai to case k = 3. The referee s report o that versio cotaied a followig very importat observatio: If the tridiagoal matrix T satisfies the assumptios i Corollary 6.3., the T ca be factorized as T = T where is the diagoal matrix with jth diagoal etry d / j ad T is the tridiagoal matrix whose jth row is give by 0, 0,...,0,(d j d j ) /,,(d j d j+ ) /, 0, 0,...,0. Notig that { } is bouded, it follows by applyig the classical Gerschgori theorem to Tˆ that the coditio d j d j / + d j d j+ / ɛ (6) for all j ad some ɛ>0is sufficiet to imply the boudedess of { T e } ad { T e }.It turs out that the coditio (6) is weaker tha the coditio ( d j d j )( d j d j+ ) 4 + ɛ for all j ad some ɛ>0 give i Corollary 6.3. I view of this commet, we were led to ivestigate the case k = 3 i Corollary 6.5 ad Example 6.6. Note that for the operator i Example 6.6 takig j = 3i, d j d j = 9 4 ad d j d j+ = 9. Thus d j d j / + d j d j+ / = =. Hece the coditio (6) is ot satisfied. O the other had, we have show above that this operator satisfies assumptios i Corollary 6.5 ad is hece ivertible. Ackowledgemets The authors are highly grateful to the referee for his valuable commets ad suggestios i improvig the earlier versio of the paper ad drawig our attetio to the refereces of [3,8,9]

17 R. Balasubramaia et al. / Liear Algebra ad its Applicatios 44 (006) (see also Remark 6.7). The authors S.H.K ad R.R thak Natioal Board for Higher Mathematics, Departmet of Atomic Eergy for the project fud for the project No: 48/4/004/R&DII/9. Refereces [] W. Arveso, C -algebras ad umerical liear algebra, J. Fuct. Aal. () (994) [] W. Arveso, The role of C -algebras i ifiite dimesioal umerical liear algebra, Cotemp. Math. 67 (994) 5 9. [3] O. Axelsso, V. Barker, Fiite Elemet Solutio of Boudary Value Problems, Theory ad Computatio, Academic Press, New York, 984. [4] R. Balasubramaia, S.H. Kulkari, R. Radha, No-ivertibility of certai almost Mathieu operators, Proc. Amer. Math. Soc. 9 (7) (000) [5] R. Balasubramaia, S.H. Kulkari, R. Radha, Fiite dimesioal approximatios of the operator equatios, i: Proceedigs of the Iteratioal Workshop o Liear Algebra, Numerical Fuctioal Aalysis ad Wavelet Aalysis, Allied Publishers, 003, pp [6] A. Bottcher, B. Silberma, Itroductio to Large Trucated Toeplitz Matrices, Spriger-Verlag, New York, 998. [7] G. Hammerli, K. Hoffma, Numerical Mathematics, Readigs i Mathematics, Spriger-Verlag, New York, 99. [8] S. Serra Capizzao, C. Tablio Possio, Spectral ad structural aalysis of high precisio Fiite Differece matrices for elliptic operators, Liear Algebra Appl. 93 (999) [9] J.C. Strikewerda, Fiite Differece schemes ad partial differetial equatios, Chapma ad Hall, Iteratioal Thompso Publ., New York, 989.