Gene H. Golub y. Computer Science Department, Stanford University. Stanford CA 94305, USA. Gerard Meurant

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1 Mtrices, moments nd qudrture Gene H. Golub y omputer Science Deprtment, Stnford University Stnford A 9435, USA golubsccm.stnford.edu Gerrd Meurnt EA, entre d'etudes de Limeil-Vlenton, 9495 Villeneuve St Georges cedex, Frnce meurntetc.fr September 29, 994 Presented s the rst A.R. Mitchell lecture in Dundee, Scotlnd, July 993. We dedicte this pper to Ron Mitchell who hs given intellectul ledership nd generous support to ll. ythe work of this uthor ws supported in prt by the Ntionl Science Foundtion under Grnt NSF R-88278

2 Abstrct In this pper we study methods to obtin bounds or pproximtions of elements of mtrix f(a) where A is symmetric positive denite mtrix nd f is smooth function. These methods re bsed on the use of qudrture rules nd the Lnczos lgorithm for digonl elements nd the block Lnczos or the non{symmetric Lnczos lgorithms for the non digonl elements. We give some theoreticl results on the behvior of these methods bsed on results for orthogonl polynomils s well s nlyticl bounds nd numericl experiments on set of mtrices for severl functions f. Denition of the problem Let A be rel symmetric positive denite mtrix of order n. We wnt to nd upper nd lower bounds (or pproximtions, if bounds re not vilble) for the entries of function of mtrix. We shll exmine nlyticl expressions s well s numericl itertive methods which produce good pproximtions in few steps. This problem leds us to consider u T f(a)v (.) where u nd v re given vectors nd f is some smooth (possibly ) function on given intervl of the rel line. As n exmple, if f(x) = nd x ut = e T i = ( ), the non zero element being in the i{th position nd v = e j, we will obtin bounds on the elements of the inverse A. We shll lso consider W T f(a)w where W is n n m mtrix. For specicity, we shll most often consider m = 2. Some of the techniques presented in this pper hve been used (without ny mthemticl justiction) to solve problems in solid stte physics, prticulrly to compute elements of the resolvnt of Hmiltonin modeling the interction of toms in solid, see [2], [4], [5]. In these studies the function f is the inverse of its rgument. Anlytic bounds for elements of inverses of mtrices using dierent techniques hve been recently obtined in [7]. The outline of the pper is s follows. Section 2 considers the problem of chrcterizing the elements of function of mtrix. The theory is developed in Section 3 nd Section 4 dels with the construction of the orthogonl polynomils tht re needed to obtin numericl method for computing bounds. The Lnczos, non{symmetric Lnczos nd block Lnczos methods used for the computtion of the polynomils re presented there. Applictions to the computtion of elements of the inverse of mtrix re described in Section 5 where very simple itertive lgorithms re given to compute bounds. Some numericl exmples re given in Section 6, for dierent mtrices nd functions f. 2 Elements of function of mtrix Since A = A T, we write A s A = QQ T 2

3 where Q is the orthonorml mtrix whose columns re the normlized eigenvectors of A nd is digonl mtrix whose digonl elements re the eigenvlues i which we order s 2 n y denition, we hve Therefore, f(a) = Qf()Q T u T f(a)v = u T Qf()Q T v = T f() = nx i= f( i ) i i This lst sum cn be considered s Riemnn{Stieltjes integrl I[f] = u T f(a)v = where the mesure is piecewise constnt nd dened by () = 8 >< > if < = P i j= j j if i < P i+ n j= j j if b = n f() d() (2.) When u = v, we note tht is n incresing positive function. The block generliztion is obtined in the following wy. Let W be n n 2 mtrix, W = (w w 2 ), then W T f(a)w = W T Qf()Q T W = f() T where, of course, is 2 n mtrix such tht = ( n ) nd i is vector with two components. With these nottions, we hve W T f(a)w = nx i= f( i ) i T i This cn be written s mtrix Riemnn{Stieltjes integrl I [f] = W T f(a)w = f() d() I [f] is 2 2 mtrix where the entries of the (mtrix) mesure re piecewise constnt nd dened by () = lx k= k T k l < l+ 3

4 In this pper, we re looking for methods to obtin upper nd lower bounds L nd U for I[f] nd I [f], L I[f] U L I [f] U In the next section, we review nd describe some bsic results from Guss qudrture theory s this plys fundmentl role in estimting the integrls nd computing bounds. 3 ounds on mtrix functions s integrls A wy to obtin bounds for the Stieltjes integrls is to use Guss, Guss{Rdu nd Guss{Lobtto qudrture formuls, see [3],[8],[9]. For., the generl formul we will use is f() d() = NX j= w j f(t j ) + MX k= v k f(z k ) + R[f] (3.) where the weights [w j ] N j= [v k] M k= nd the nodes [t j] N j= re unknowns nd the nodes [z k] M k= re prescribed, see [4],[5],[6],[7]. 3. The cse u = v When u = v, the mesure is positive incresing function nd it is known (see for instnce [8]) tht R[f] = f (2N +M ) () (2N + M)! MY k= ( z k ) 2 Y 4 N j= ( t j ) d() < < b (3.2) If M =, this leds to the Guss rule with no prescribed nodes. If M = nd z = or z = b we hve the Guss{Rdu formul. If M = 2 nd z = z 2 = b, this is the Guss{Lobtto formul. Let us recll briey how the nodes nd weights re obtined in the Guss, Guss{ Rdu nd Guss{Lobtto rules. For the mesure, it is possible to dene sequence of polynomils p () p () tht re orthonorml with respect to if i = j p i ()p j () d() = otherwise nd p k is of exct degree k. Moreover, the roots of p k re distinct, rel nd lie in the intervl [ b]. We will see how to compute these polynomils in the next Section. This set of orthonorml polynomils stises three term recurrence reltionship (see [2]) j p j () = (! j )p j () j p j 2 () j = 2 N (3.3) if R d =. p () p () 4

5 where In mtrix form, this cn be written s J N = p() = J N p() + N p N ()e N p() T = [p () p () p N ()] e T N = ( )!! N 2! N N N! N A (3.4) The eigenvlues of J N (which re the zeroes of p N ) re the nodes of the Guss qudrture rule (i. e. M = ). The weights re the squres of the rst elements of the normlized eigenvectors of J N, cf. [7]. We note tht ll the eigenvlues of J N re rel nd simple. For the Guss qudrture rule (renming the weights nd nodes w G j nd t G j ) we hve with nd the next theorem follows. f() d() = R G [f] = f (2N ) () (2N)! NX j= w G j f(t G j ) + R G [f] 2 Y 4 N j= ( t G j ) d() Theorem 3. Suppose u = v in 2. nd f is such tht f (2n) () > 8n 8 < < b nd let Then, 8N, 9 2 [ b] such tht L G [f] = NX j= w G j f(tg j ) L G [f] I[f] I[f] L G [f] = f (2N ) () (2N)! Proof See [8]. The min ide of the proof is to use Hermite interpoltory polynomil of degree 2N on the N nodes which llows us to express the reminder s n integrl of the dierence between the function nd its interpoltory polynomil nd to pply the men vlue theorem (s the mesure is positive nd incresing). As we know the sign of the reminder, we esily obtin bounds. To obtin the Guss{Rdu rule (M = in 3.{3.2), we should extend the mtrix J N in 3.4 in such wy tht it hs one prescribed eigenvlue, see [8]. 5

6 Assume z =, we wish to construct p N + such tht p N + () =. From the recurrence reltion 3.3, we hve This gives We hve lso = N + p N + () = (! N + )p N () N p N ()! N + = N p N () p N () (J N I)p() = N p N ()e N Let us denote () = [ () N ()] T with This gives! N + = + N () nd l () = N p l () p N () l = N (J N I)() = 2 N e N (3.5) From these reltions we hve the solution of the problem s ) we generte N by the Lnczos process (see Section 4 for the denition), 2) we solve the tridigonl system 3.5 for () nd 3) we compute! N +. Then the tridigonl mtrix ^J N + dened s JN ^J N + = N e N N e T N! N + will hve s n eigenvlue nd gives the weights nd the nodes of the corresponding qudrture rule. Therefore, the recipe is to compute s for the Guss qudrture rule nd then to modify the lst step to obtin the prescribed node. For Guss{Rdu the reminder R GR is R GR [f] = f (2N +) () (2N + )! ( z ) 2 Y 4 N j= ( t j ) d() Agin, this is proved by constructing n interpoltory polynomil for the function nd its derivtive on the t j s nd for the function on z. Therefore, if we know the sign of the derivtives of f, we cn bound the reminder. This is stted in the following theorem. Theorem 3.2 Suppose u = v nd f is such tht f (2n+) () < 8n 8 < < b. Let U GR be dened s w j v t j U GR [f] = NX j= w j f(t j ) + v f() being the weights nd nodes computed with z = nd let L GR be dened s L GR [f] = NX j= w b j f(tb j ) + vb f(b) 6

7 w b j v b t b j being the weights nd nodes computed with z = b. Then, 8N we hve L GR [f] I[f] U GR [f] nd I[f] U GR [f] = f (2N +) () (2N + )! I[f] L GR [f] = f (2N +) () (2N + )! ( ) ( b) 2 Y 4 N j= j= ( t j ) d() 2 3 Y 4 N 2 ( t b ) 5 j d() Proof With our hypothesis the sign of the reminder is esily obtined. It is negtive if we choose z =, positive if we choose z = b. Remrks i) if the sign of the f derivtives is positive, the bounds re reversed. ii) it is enough to suppose tht there exists n n such tht f (2n +) () < but, then N = n is xed. Now, consider the Guss{Lobtto rule (M = 2 in 3.{3.2), with z = nd z 2 = b s prescribed nodes. Agin, we should modify the mtrix of the Guss qudrture rule, see [8]. Here, we would like to hve p N + () = p N + (b) = Using the recurrence reltion 3.3 for the polynomils, this leds to liner system of order 2 for the unknowns! N + nd N pn () p N ()!N + = p N (b) p N (b) Let nd be dened s vectors with components N pn () (3.6) b p N (b) l = p l () N p N () l = p l (b) N p N (b) then (J N I) = e N (J N bi) = e N nd the liner system 3.6 cn be written N!N + = N 2 N b giving the unknowns tht we need. The tridigonl mtrix ^JN + is then dened s in the Guss{Rdu rule. Hving computed the nodes nd weights, we hve f()d() = NX j= w GL j f(t GL j ) + v f() + v 2 f(b) + R GL [f] 7

8 where R GL [f] = f (2N +2) () (2N + 2)! Then, we hve the following obvious result. ( )( b) 2 Y 4 N j= ( t j ) d() Theorem 3.3 Suppose u = v nd f is such tht f (2n) () > 8n 8 < < b nd let Then, 8N U GL [f] = NX j= I[f] U GL [f] = f (2N +2) () (2N + 2)! w GL j f(t GL j ) + v f() + v 2 f(b) I[f] U GL [f] ( )( b) 2 Y 4 N j= ( t GL j ) d() We remrk tht we need not lwys compute the eigenvlues nd eigenvectors of the tridigonl mtrix. Let Y N be the mtrix of the eigenvectors of J N (or ^JN ) whose columns we denote by y i nd T N be the digonl mtrix of the eigenvlues t i which give the nodes of the Guss qudrture rule. It is well known tht the weights w i re given by (cf. [2]) It cn be esily shown tht where y i is the rst component of y i. ut, since p (), we hve, Theorem 3.4 Proof NX l= X N = p 2 l (t i ) w i w i = l=! y 2 i p (t i ) w i = (y i ) 2 = (e T y i ) 2 NX l= w l f(t l ) = e T f(j N )e NX w l f(t l ) = e T y lf(t l )y T e l l=! NX = e T y l f(t l )y T l e l= = e T Y Nf(T N )Y T N e = e T f(j N )e The sme sttement is true for the Guss{Rdu nd Guss{Lobtto rules. Therefore, in some cses where f(j N ) (or the equivlent) is esily computble (for instnce, if f() = =, see Section 5), we do not need to compute the eigenvlues nd eigenvectors of J N. 8

9 3.2 The cse u 6= v We hve seen tht the mesure in 2. is piecewise constnt nd dened by () = lx k= k k l < l+ For vrible signed weight functions, see [9]. We will see lter tht for our ppliction, u nd v cn lwys be chosen such tht k k. Therefore, in this cse will be positive incresing function. In the next Section, we will show tht there exists two sequences of polynomils p nd q such tht j p j () = (! j )p j () j p j 2 () p () p () j q j () = (! j )q j () j q j 2 () q () q () nd Let Then, we cn write J N = p() T = [p () p () p N ()] q() T = [q () q () q N ()]!! N 2! N N N! N A p() = J N p() + N p N ()e N q() = J T Nq() + N q N ()e N Theorem 3.5 p j () = j j q j () Proof The theorem is proved by induction. We hve p () =! therefore Now, suppose tht q () =! p () = q () p j () = j j q j () 9

10 We hve j p j () = (! j )p j () j p j 2 () = (! j ) j j q j () j j 2 j 2 q j 2 () Multiplying by j j we obtin the result. Hence q N is multiple of p N nd the polynomils hve the sme roots which re lso the common eigenvlues of J N nd JN. T We will see tht it is possible to choose j nd j such tht with, for instnce, j. Then, we hve We dene the qudrture rule s f() d() = j = j p j () = q j () NX j= f( j )s j t j + error (3.7) where j is n eigenvlue of J N, s j is the rst component of the eigenvector u j of J N corresponding to j nd t j is the rst component of the eigenvector v j of J T N corresponding to the sme eigenvlue, normlized such tht v T j u j =. We hve the following results Proposition 3. Suppose tht j j 6=, then the (non{symmetric) Guss qudrture rule 3.7 is exct for polynomils of degree less thn or equl to N. Proof The function f cn be written s f() = N X k= nd becuse of the orthonormlity properties For the qudrture rule, we hve NX j= f( j )s j t j q l ( j ) = c k p k () f() d() = c = NX N X j= k= X N k= c k p k ( j )s j t j q l ( j ) c k N X j= p k ( j )s j t j q l ( j )

11 ut p k ( j )s j nd q l ( j )t j re respectively the components of the eigenvectors of J N nd J T N corresponding to j. Therefore they re orthonorml with the normliztion tht we chose. Hence, nd consequently which proves the result. NX j= f( j )s j t j q l ( j ) = c l NX j= f( j )s j t j = c Now, s in [4], we extend the result to polynomils of higher degree. Theorem 3.6 Suppose tht j j 6=, then the (non{symmetric) Guss qudrture rule 3.7 is exct for polynomils of degree less thn or equl to 2N. Proof Suppose f is polynomil of degree 2N. Then, f cn be written s f() = p N ()s() + r() where s nd r re polynomils of degree less or equl to N. Then, f() d() = p N ()s() d() + r() d() = r() d() since p N is orthogonl to ny polynomil of degree less or equl to N becuse of the orthogonlity property of the p nd q's. For the qudrture rule, we hve NX j= p N ( j )s( j )s j t j + ut, s j is n eigenvlue of J N, it is root of p N NX j= NX j= r( j )s j t j nd p N ( j )s( j )s j t j = As the qudrture rule hs been proven to be exct for polynomils of degree less thn N, r() d() = NX j= r( j )s j t j which proves the Theorem. We will see in the next Section how to obtin bounds on the integrl 2.. Now, we extend the Guss{Rdu nd Guss{Lobtto rules to the non{symmetric cse. This is lmost identicl (up to technicl detils) to the symmetric cse.

12 For Guss{Rdu, ssume tht the prescribed node is, then, we would like to hve p N + () = q N + () =. This gives If we denote () = [ () N ()] T, with we hve where (! N + )p N () N p N () = l () = N p l () p N ()! N + = + N () (J N I)() = N N e N Therefore, the lgorithm is essentilly the sme s previously discussed. For Guss{Lobtto, the lgorithm is lso lmost the sme s for the symmetric cse. We would like to compute p N + nd q N + such tht p N + () = p N + (b) = q N + () = q N + (b) = This leds to solving the liner system pn () p N () p N (b) p N (b)!n + N = pn () bp N (b) The liner system for the q's whose solution is (! N + N ) T cn be shown to hve the sme solution for! N + nd N = N depending on the signs reltions between the p's nd the q's. Let () nd (b) be the solutions of (J N I)() = e N (J N bi)(b) = e N Then, we hve ()N!N + = (b) N N N b When we hve the solution of this system, we choose N = N nd N. The question of estblishing bounds on the integrl will be studied in the next Section. As for the cse u = v, we do not lwys need compute the eigenvlues nd eigenvectors of J N but only the ( ) element of f(j N ). 3.3 The block cse Now, R we consider the block cse. The problem is to nd qudrture rule. The integrl b f()d() is 2 2 symmetric mtrix. The most generl qudrture formul is of the form f()d() = NX j= W j f(t j )W j + error 2

13 where W j nd T j re symmetric 2 2 mtrices. In this sum, we hve 6N unknowns. This qudrture rule cn be simplied, since T j = Q j j Q T j where Q j is the orthonorml mtrix of the eigenvectors, nd j, the digonl mtrix of the eigenvlues of T j. This gives NX j= ut W j Q j f( j )Q T j W j cn be written s W j Q j f( j )Q T j W j f( )z z T + f( 2)z 2 z T 2 where the vector z i is 2. Therefore, the qudrture rule cn be written s 2NX j= f(t j )w j w T j where t j is sclr nd w j is vector with 2 components. In this qudrture rule, there re lso 6N unknowns. In the next Section, we will show tht there exists orthogonl mtrix polynomils such tht This cn be written s where p j () = p j () j + p j () j + p j 2 () T j p () I 2 p () [p () p N ()] = [p () p N ()]J N + [ p N () N ] J N = T 2 T N 2 N T N N N A (3.8) is block tridigonl mtrix of order 2N nd bnded mtrix whose hlf bndwidth is 2 (we hve t most 5 non zero elements in row). If we denote P () = [p () p N ()] T we hve s J N is symmetric J N P () = P () [ p N () N ] T We note tht if is n eigenvlue, sy r, of J N nd if we choose u = u r to be two element vector whose components re the rst two components of n eigenvector corresponding to r, then P ( r )u is this eigenvector (becuse of the reltions tht re stised) nd if 3

14 N is non singulr, p T N( r )u =. The dierence with the sclr cse is tht lthough the eigenvlues re rel, it might be tht they re of multiplicity greter thn (lthough this is unlikely except in the cse of the Guss-Rdu nd Guss-Lobtto rule where this condition is enforced). We dene the qudrture rule s f() d() = 2NX i= f( i )u i u T i + error (3.9) where 2N is the order of J N, the eigenvlues i re those of J N nd u i is the vector consisting of the two rst components of the corresponding eigenvector, normlized s before. In fct, if there re multiple eigenvlues, the qudrture rule should be written s follows. Let i i = l be the set of distinct eigenvlues nd q i their multiplicities. The qudrture rule is then lx q i X (w j i )(w j i ) T A f(i ) (3.) i= j= We will show in the next Section tht the Guss qudrture rule is exct for polynomils of degree 2N nd how to obtin estimtes of the error. We extend the process described for sclr polynomils to the mtrix nlog of the Guss{Rdu qudrture rule. Let be n extreme eigenvlue of A. We would like to be double eigenvlue of J N +. We hve J N + P () = P () [ p N + () N + ] T Then, we need to require p N + (). From the recurrence reltion this trnsltes into Therefore, if p N () is non singulr, we hve p N () p N () N + p N () T N = N + = I 2 p N () p N () T N We must compute the right hnd side. This cn be done by noting tht J N p () T. A = p N () T p () T. p N () T A. T Np N () T Multiplying on the right by p N () T, we get the mtrix eqution Thus, we solve (J N I) (J N I) p () T p N () T. p N () T p N () T (). A = N () 4 A =. T N A. T N A A

15 nd hence N + = I 2 + N () T T N The generliztion of Guss{Lobtto to the block cse is little more tricky. We would like to hve nd b s double eigenvlues of the mtrix J N +. This leds to stisfying the following two mtrix equtions This cn be written s I2 p N () p N () N + p N () T N = bp N (b) p N (b) N + p N (b) T N = I 2 p p N ()p N () N (b)p N (b) N + T N = I2 bi 2 We now consider the problem of computing (or void computing) p N ()p N (). Let () be the solution of (J N I)() = ( I 2 ) T Then, s before N () = p N () T p N () T T N We cn esily show tht N () is symmetric. We consider solving 2 2 block liner system I X U I = I Y V bi onsider the block fctoriztion I I thus W Z = Y X. The solution of the system gives X Y I I W I = I W U = V I W V = (b )I The next step is I X U Z V nd we get or Therefore X Z I bi = U V ZV = V = W (b )I (W Z)V = (b )I V = (b )(Y X) 5

16 Hence, we hve Y X = p N (b)p N (b) p N ()p N () = N ( N () N (b)) This mens tht or T N = (b )( N () N (b)) N T N N = (b )( N () N (b)) Then, N is given s holesky decomposition of the right hnd side mtrix. The right hnd side is positive denite becuse N () is digonl block of the inverse of (J N I) which is positive denite becuse the eigenvlues of J N re lrger tht nd N (b) is the negtive of digonl block of (J N bi) which is positive denite becuse the eigenvlues of J N re smller tht b. From N, we cn compute N + N + = I 2 + N N () T N As for the sclr cse, it is not lwys needed to compute the weights nd the nodes for the qudrture rules. Theorem 3.7 We hve where e T = (I 2 ). 2NX i= f( i )u i u T i = e T f(j N )e Proof The qudrture rule is If y i re the eigenvectors of J N 2NX i= u i f( i )u T i then u i = e T y i nd 2NX u i f( i )u T i = 2NX i= i= e T y i f( i )y T i e = e T 2N X i= y i f( i )y T i! e = e T Y N f(t N )Y T N e = e T f(j N )e where Y N is the mtrix of the eigenvectors nd T N the digonl mtrix of the eigenvlues of J N. Note tht bounds for non digonl elements cn lso be obtined by considering e T i f(a)e i, e T j f(a)e j nd 2 (e i + e j ) T f(a)(e i + e j ). 6

17 4 onstruction of the orthogonl polynomils In this section we consider the problem of computing the orthonorml polynomils or equivlently the tridigonl mtrices tht we need. A very nturl nd elegnt wy to do this is to use Lnczos lgorithms. 4. The cse u = v When u = v, we use the clssicl Lnczos lgorithm. Let x = nd x be given such tht kx k =. The Lnczos lgorithm is dened by the following reltions, j x j = r j = (A! j I)x j j x j 2 j =! j = x T j Ax j j = kr j k The sequence fx j g l j= is n orthonorml bsis of the Krylov spce spnfx Ax A l x g Proposition 4. The vector x j is given by x j = p j (A)x where p j is polynomil of degree j dened by the three term recurrence (identicl to 3.3) Proof j p j () = (! j )p j j() j p j 2 () p () p () x = (A! I)x is rst order polynomil in A. Therefore, the Proposition is esily obtined by induction. Theorem 4. If x = u, we hve x T k x l = Proof As the x j 's re orthonorml, we hve where ^x = Q T x. p k ()p l ()d() x T k x l = x T P k (A) T P l (A)x = x T QP k()q T QP l ()Q T x = x T QP k ()P l ()Q T x = nx j= p k ( j )p l ( j )^x 2 j Therefore, the p j 's re the orthonorml polynomils relted to tht we were referring to in

18 4.2 The cse u 6= v We pply the non{symmetric Lnczos lgorithm to symmetric mtrix A. Let x = ^x = nd x ^x be given with x 6= ^x nd x T ^x =. Then we dene the itertes for j = by j x j = r j = (A! j I)x j j x j 2 (4.) j ^x j = ^r j = (A! j I)^x j j ^x j 2 (4.2)! j = ^x T j Ax j j j = ^r T j r j This lgorithm genertes two sequences of mutully orthogonl vectors s we hve x T l ^x k = kl We hve bsiclly the sme properties s for the Lnczos lgorithm. Proposition 4.2 x j = p j (A)x ^x j = q j (A)^x where p j nd q j re polynomils of degree j dened by the three term recurrences j p j () = (! j )p j () j p j 2 () p () p () j q j () = (! j )q j () j q j 2 () q () q () Proof The Proposition is esily obtined by induction. Theorem 4.2 If x = u nd ^x = v, then x T k ^x l = p k ()q l ()d() = kl Proof As the x j 's nd ^x j 's re orthonorml the proof is identicl to the proof of Theorem 4.. We hve seen in the previous Section the reltionship between the p nd q's. The polynomils q re multiples of the polynomils p. In this prticulr ppliction of the non{symmetric Lnczos lgorithm it is possible to choose j nd j such tht q j = j = j^r j T r j j with, for instnce, j nd j = sgn(^r T j r j) j. Then, we hve p j () = q j () 8

19 The min dierence with the cse of symmetric Lnczos is tht this lgorithm my brek down, e.g. we cn hve j j = t some step. We would like to use the non{symmetric Lnczos lgorithm with x = e i nd ^x = e j to get estimtes of f(a) ij. Unfortuntely, this is not possible s this implies x T ^x =. A wy to get round this problem is to set x = e i = nd ^x = e i + e j. This will give n estimte of f(a) ij = + f(a) ii nd we cn use the bounds we get for the digonl elements (using for instnce symmetric Lnczos) to obtin bounds for the non digonl entry. An dded dntge is tht we re ble to choose so tht j j > nd therefore p j () = q j (). This cn be done by strting with = nd restrting the lgorithm with lrger vlue of s soon s we nd vlue of j for which j j. Regrding expressions for the reminder, we cn do exctly the sme s for symmetric Lnczos. We cn write R(f) = f (2N ) () (2N)! However, we know tht p N () = q N () nd p N () 2 d() p N ()q N () d() = This shows tht the sign of the integrl in the reminder cn be computed using the lgorithm nd we hve the following result. Theorem 4.3 Suppose f is such tht f (2n) () > 8n 8 < < b Then, the qudrture rule 3.7 gives lower bound if NY j= sgn(^r T j r j) = nd n upper bound otherwise. In both cses, we hve jr(f)j = f (2N ) () (2N)! For Guss{Rdu nd Guss{Lobtto we cnnot do the sme thing. ounds cn be obtined if we choose the initil vectors (e.g. ) such tht the mesure is positive nd incresing. In this cse we re in exctly the sme frmework s for the symmetric cse nd the sme results re obtined. Note however tht it is not esy to mke this choice priori. Some exmples re given in Section 6. A wy to proceed is to strt with = nd to restrt the lgorithm 4.2 with lrger vlue of whenever we hve j j. 4.3 The block cse Now, we consider the block Lnczos lgorithm, see [],[6]. Let X be n n 2 given mtrix, such tht X T X = I 2 ( chosen s U dened before). Let X = be n n 2 mtrix. Then j = Xj AX T j 9

20 R j = AX j X j j X j 2 T j X j j = R j The lst step is the QR decomposition of R j such tht X j is n 2 with X T X j j = I 2 nd j is 2 2. The mtrix j is 2 2 nd j is upper tringulr. It my hppen tht R j is rnk decient nd in tht cse j is singulr. The solution of this problem is given in []. One of the columns of X j cn be chosen rbitrrily. To complete the lgorithm, we choose this column to be orthogonl with the previous block vectors X k. We cn for instnce choose nother vector (rndomly) nd orthogonlize it ginst the previous ones. This lgorithm genertes sequence such tht where I 2 is the 2 2 identity mtrix. Proposition 4.3 where (i) k re 2 2 mtrices. X T j X i = ij I 2 X i = Proof The proof is given by induction. ix k= A k X (i) k We dene mtrix polynomil p i (), 2 2 mtrix, s Thus, we hve the following result. Theorem 4.4 X T i X j = p i () = ix k= k (i) k p i () T d()p j () = ij I 2 Proof Using the orthogonlity of the X i s, we cn write ij I 2 = X T i X j = = X kl = X kl = X kl = ix k= nx m=! X k ) T X T A k j ( (i) l= A l X (j) ( (i) k )T X T Q k+l Q T X (j) l ( (i) k )T k+l T (j) l ( (i) k )T X k nx m= k+l m m T m! k m( (i) k ) T! m T m 2 l (j) l X l A l m (j) l!

21 The p j s cn be considered s mtrix orthogonl polynomils for the (mtrix) mesure. To compute the polynomils, we need to show tht the following recurrence reltion holds. Theorem 4.5 The mtrix vlued polynomils p j stisfy p j () j = p j () p j () j p j 2 () T j p () p () I 2 where is sclr. Proof From the previous denition, it is esily shown by induction tht p j cn be generted by the given (mtrix) recursion. As we hve seen before this cn be written s [p () p N ()] = [p () p N ()]J N + [ p N () N ] nd s P () = [p () p N ()] T J N P () = P () [ p N () N ] T with J N dened by 3.8. Most of the following results on the properties of the mtrix polynomils re derived from []. Proposition 4.4 The eigenvlues of J N re the zeroes of det[p N ()]. Proof Let be zero of det[p N ()]. As the rows of p N () re linerly dependent, there exists vector v with two components such tht This implies tht v T p N () = [v T p () v T p N ()] = [v T p () v T p N ()]J N Therefore is n eigenvlue of J N. det[p N ()] is polynomil of degree 2N in. Hence, there exists 2N zeroes of the determinnt nd therefore ll eigenvlues re zeroes of det[p N ()]. Proposition 4.5 For nd rel, we hve the nlog of the hristoel{drboux identity (see [2]) X N p N () T Np T N() p N () N p T N () = ( ) m= p m ()p T m() (4.3) 2

22 Proof From the previous results, we hve T j+ pt j+ () = pt j () j+p T j () jp T j () p j+ () j+ = p j () p j () j+ p j () T j Multiplying the rst reltion by p j () on the left nd the second one by p T j () on the right gives p j () T j+p T j+() p j+ () j+ p T j () = = ( )p j ()p T j () p j () j p T j () + p j () T j p T j () Summing these equlities, some terms cncel nd we get the desired result. In prticulr, if we choose = in 4.3, we hve tht p N () N p T N () is symmetric. Proposition 4.6 N X m= u T s p m( s )p T m ( r)u r = rs Proof If we set = s nd = r nd multiply the hristoel{drboux identity 4.3 on the left by u T s nd on the right by u r, we hve p T N( r )u r = nd p T N( s )u s =, nd we get the result if s 6= r. Let K N ( ) = N X m= p m ()p T m() As p () = I 2, K N ( ) is symmetric positive denite mtrix nd therefore denes sclr product. If r is multiple eigenvlue, there exist linerly independent eigenvectors tht we could orthonormlize. If r is n eigenvlue of multiplicity q r, there exist q r linerly independent vectors v r vq r r with two components such tht the vectors P ( r )vr j j = q r re the eigenvectors ssocited with r. We cn certinly nd set of vectors (wr w q r r ) spnning the sme subspce s (vr v q r r ) nd such tht (w k r )T K N ( r r )w l r = kl This property is nothing else thn the orthogonlity reltion of the eigenvectors. Proposition 4.7 2NX r= p T i ( r )u r u T r p l ( r ) = il I 2 i = N l = N Proof Note tht the eigenvectors of J N re linerly independent. We tke set of N vectors with two elements fy y N g, nd write [y T yt N ]T = 2NX r= r [u T r p ( r ) u T r p N ( r )] T 22

23 or y l = 2NX r= p T l ( r )u r r l = N Multiplying on the left by u T s p l ( s ) nd summing Therefore, N X l= This gives the desired result. u T s p l ( s )y l = s s = 2N y i = 2NX N X r= l= p T i ( r )u r u T r p l ( r )y l To prove tht the block qudrture rule is exct for polynomils of degree up to 2N, we cnnot use the sme method s for the sclr cse where the given polynomil is fctored becuse of commuttivity problems. Therefore, we tke nother pproch tht hs been used in dierent setting in [2]. The following results re tken from [2]. We will consider ll the monomils k k = 2N. Let M k the moment mtrix, be dened s M k = k d() We write the (mtrix) orthonorml polynomils p j s p j () = jx k= p (j) k k p (j) k being mtrix of order 2. Then, we hve p T j () d() = jx (p (j) k k= ) T k d() = jx k= (p (j) k ) T M k nd more generlly p T j () q d() = jx k= (p (j) k ) T M k+q We write these equtions for j = N. Note tht becuse of the orthogonlity of the polynomils, we hve p T N () q d() = q = N 2 Let H N be the block Hnkel mtrix of order 2N, dened s H N = M M N.. M N M 2N 2 A 23

24 Then H N p (N ). p (N ) N 2 p (N ) N A =. R b p T N () N d() We introduce some dditionl nottion. Let L N be block upper tringulr mtrix of order 2N, p () p () p (N ) L N = p () p (N ).... A Let V N be 4N 2N mtrix dened in block form s where j is q j 2N mtrix, j = nd the j re the eigenvlues of A. Let K j i be 2q i 2q j mtrix nd Proposition 4.8 K j i = where j is 2q j 2N mtrix, j = V N = 2. l A p (N ) N I 2 j I 2 N j I I 2 j I 2 N j I 2 A A K N ( i j ) K N ( i j )... A K N ( i j ) K N ( i j ) K = K K 2 K l K 2 K l 2.. A K l K l l V N L N = 2. l A p ( j ) p ( j ) p N ( j ).... A p ( j ) p ( j ) p N ( j ) 24

25 Proof This is strightforwrd by the denition of the polynomils p j (). Proposition 4.9 L T NH N L N = I Proof the generic term of H N L N is (H N L N ) ij = Therefore the generic term of L T NH N L N (L T NH N L N ) ij = ix jx r= s= is jx s= M s+i 2 p (j ) s Splitting the power of we cn esily see tht this is (L T N H NL N ) ij = (p (i ) r ) T s+r 2 d()p (j ) s p T i () d() p j () Therefore becuse of the orthonormlity properties, we hve (L T I2 if i = j, NH N L N ) ij = otherwise. This result implies tht Proposition 4. H N = L N L T N V N L N (V N L N ) T = K Proof this is just using the denition of K j i. Now, we dene 2N 4N mtrix W T N whose only non zero components in row i re in position (i 2i ) nd (i 2i) nd re successively the two components of (w ) T (w q ) T (w 2) T (w q l l )T. Then becuse of the wy the w j i re constructed we hve Proposition 4. Proposition 4.2 W T N V N W T N KW N = I is non singulr 2N 2N mtrix. Proof This shows tht W T N V N W T N V NH N V T N W N = W T N V NL N L T N V T N W N = W T N KW N = I is non singulr. Then, we hve the min result 25

26 Theorem 4.6 The qudrture rule 3.9 or 3. is exct for polynomils of order less thn or equl to 2N. Proof From Proposition 4.2, we hve Therefore, H N = (W T N V N) (V T N W N) H N = (V T N W N)(W T N V N) y identiction of the entries of the two mtrices we hve, lx M k = q i X (w j i )(w j i ) T A k i k = 2N 2 i= j= It remins to prove tht the qudrture rule is exct for k = 2N. As we hve, H N + p (N ). p (N ) N p (N ) N A =. R b pt N () N Writing the (N )th block row of this equlity, we get We hve proved before tht y substitution, we get We use the fct tht M 2N p (N ) N M N +r = M 2N p (N ) N N X (w j i ) T r= N X = lx X i= N X = q i j= r= r= i= j= A d() M N +r p (N ) r w j i (w j i ) T A N +r i q lx X i w j i (w j i ) T N +r i p (N ) r r i p(n ) r = (w j i ) T p N ( i ) (w j i ) T N i p(n ) N nd This shows tht M 2N p (N ) (w j i ) T p N ( i ) = q lx X = i N i= j= (w j i )(w j i ) T 2N i p (N ) N 26

27 As p (N ) N is non singulr, we get the result. To obtin expressions for the reminder, we would like to use similr pproch s for the sclr cse. However there re some dierences, s the qudrture rule is exct for polynomils of order 2N nd we hve 2N nodes, we cnnot interpolte with n Hermite polynomil nd we hve to use Lgrnge polynomil. y Theorems 2... nd of [8], there exists polynomil q of degree 2N such tht q( j ) = f( j ) j = 2N nd where f(x) q(x) = s(x)f (2N ) ((x)) (2N)! s(x) = (x ) (x 2N ) If we cn pply the men vlue theorem, the reminder R(f) which is 2 2 mtrix cn be written s R(f) = f (2N ) () s() d() (2N)! Unfortuntely s does not hve constnt sign over the intervl [ b]. Therefore this representtion formul for the reminder is of little prcticl use for obtining bounds with the knowledge of the sign of the entries of the reminder. It is esy to understnd why we cnnot directly obtin bounds with this block pproch. We must use W = (e i e j ) nd the block Lnczos lgorithm with X = W. For the block Lnczos lgorithm we multiply successively A with the Lnczos vectors. If A is sprse, most of the components of these products re for the rst few itertes of the lgorithm. Therefore, it is likely tht t the beginning, the estimtes tht we will get for the non digonl entries will be. This explins why we cnnot directly obtin upper or lower bounds with Guss, Guss{Rdu or Guss{Lobtto. A wy to voiding this diculty is to use W = (e i + e j e j ) but this cnnot be done since X T X 6= I 2. However, we will see in the numericl experiments tht the estimtes we get re often quite good. 5 Appliction to the inverse of mtrix In this Section we consider obtining nlyticl bounds for the entries of the inverse of given mtrix nd simplifying the lgorithms to compute numericl bounds nd pproximtions. We consider f() = < < b nd hence nd f (2n+) () = (2n + )! (2n+2) f (2n) () = (2n)! (2n+) 27

28 Therefore, the even derivtives re positive on [ b] nd the odd derivtives re negtive which implies tht we cn pply Theorems 3., 3.2 nd 3.3. onsider dense non singulr mtrix A = ( ij ) ij=m. We choose u = x = e i nd we pply the Lnczos lgorithm. From results of the rst itertion we cn obtin nlyticl results. The rst step of the Lnczos lgorithm gives us! = e T i Ae i = ii Let s i be dened by nd let Then x = r = (A! I)e i X s 2 i = 2 ji j6=i d T i = ( i i i i+i mi ) = s i x = d i s i From this, we hve! 2 = s 2 i X X k6=i l6=i ki kl li From this dt, we compute the Guss rule nd get lower bound on the digonl element! J 2 =! 2 The lower bound is given by! 2!! 2 2 J 2 = =!! 2 2 ii Pk6=i P P k6=i!2! l6=i ki kl li P l6=i ki kl li P k6=i 2 ki Note tht this bound does not depend on the eigenvlues nd b. Now, we consider the Guss{Rdu rule. Then,! ~J 2 = x the eigenvlues re the roots of (! )(x ) 2 =, which gives the reltion x = + 2! To obtin n upper bound we set =. The solution is x = x = + 2! 28 2

29 For the Guss{Lobtto rule, we hve the sme problem except tht we wnt J ~ 2 to hve nd b s eigenvlues. This leds to solving the following liner system,!! = 2 b! b 2! b x 2 Solving this system nd computing the ( ) element of the inverse gives Hence we hve the following result. + b! b Theorem 5. We hve the following bounds ii Pk6=i P k6=i P l6=i ki kl li P l6=i ki kl li P k6=i 2 ki 2 (A ) ii ii b + s2 i b 2 ii iib + s 2 i (A ) ii ii + s 2 i 2 ii ii + s 2 i (A ) ii + b ii b It is not too esy to derive nlyticl bounds from the block Lnczos lgorithm s we hve to compute repeted inverses of 2 2 mtrices. It is much esier to use the non{symmetric Lnczos method with the Guss{Rdu rule. We re looking t the sum of the (i i) nd (i j) elements of the inverse. Let t i = = X k6=i ki ( ki + kj ) ij ( ij + ii ) Then, the computtions re essentilly the sme s for the digonl cse. Theorem 5.2 For (A ) ij + (A ) ii we hve the two following estimtes ii + ij + t i ( ii + ij ) 2 ( ii + ij ) + t i ii + ij b + t i b ( ii + ij ) 2 b( ii + ij ) + t i If t i, the rst expression with gives n upper bound nd the second one with b lower bound. Then, we hve to subtrct the bounds for the digonl term to get bounds on (A ) ij. The previous results cn be compred with those obtined by other methods in [7]. Results cn lso be obtined for sprse mtrices tking into ccount the sprsity structure. In the computtions using the Lnczos lgorithm for the Guss, Guss{Rdu nd Guss{Lobtto rules, we need to compute the ( ) element of the inverse of tridigonl mtrix. This my be done in mny dierent wys, see for instnce [3]. Here, we will show tht we cn compute this element of the inverse incrementlly s we go through the Lnczos lgorithm nd we obtin the estimtes for very few dditionl opertions. This is stted in the following theorem where b j stnds for the bounds for Lnczos itertion j nd the! j s nd the j s re generted by Lnczos. 29

30 Theorem 5.3 The following lgorithm yields lower bound b j of A ii by the Guss qudrture rule, lower bound b j nd n upper bound ^b j through the Guss{Rdu qudrture rule nd n upper bound b j through the Guss{Lobtto rule. Let x = nd x = e i,! = ii, = k(a! I)e i k, b =!, d =!, c =, ^d =!, d =! b, x = (A! I)e i =. Then for j = 2 we compute r j = (A! j I)x j j x j 2 b j = b j +! j = x T j Ax j j = kr j k x j = r j j 2 j c 2 j d j (! j d j 2 j ) d j =! j 2 j d j c j = c j j d j ^d j =! j 2 j ^d j d j =! j b 2 j d j ^bj = b j + bj = b j + ^! j = + 2 j ^d j! j = ^d j dj d j ^d j! j = b + 2 j d j 2 j c 2 j d j (^! j d j 2 j ) 2 j c2 j d j (! j d j 2 j )! b ^d j dj 2 j = ^d j dj d j ^d j (b ) bj = b j + 2 j c 2 j d j (! j d j j 2 ) 3

31 Proof We hve from 3.4 J N =!! N 2! N N N! N A Let x T N = ( N ), so tht J N + = JN x T N x N! N + Letting ~J = J N x Nx T N! N + the upper left block of J N + is ~ J. This cn be obtined through the use of the Shermn{ Morrison formul (see []), ~J = J N (JN x N )(x T + NJ! N + x T NJ N x N Let j N = J N e N be the lst column of the inverse of J N. With this nottion, we hve N ) ~J = J N + 2 N j Nj T N! N + 2 N (j N) N Therefore, it is cler tht we only need the rst nd lst elements of the lst column of the inverse of J N. This cn be obtined using the holesky decomposition of J N. It is esy to check tht if we dene d =! d i =! i 2 i d i i = 2 N then (j N ) = ( ) N N d d N (j N ) N = d N When we put ll these results together we get the proof of the Theorem. The lgorithm is essentilly the sme for the non{symmetric cse. lgorithm is the following, using f = The modied r j = (A! j I)x j j x j 2 ^r j = (A! j I)^x j j ^x j 2! j = ^x T Ax j j j j = ^r T j r j 3

32 b j = b j + x j = r j j ^x j = ^r j j j j c j f j d j (! j d j j j ) d j =! j 2 j d j c j = c j j d j f j = f j j d j ^d j =! j j j ^d j d j =! j b j j d j ^bj = b j + bj = b j + ^! j = + j j ^d j! j = b + j j d j! j = ^d j dj d j ^d j j j c j f j d j (^! j d j j j ) j j c j f j d j (! j d j j j )! b ^d j dj 2 j = ^d j dj d j ^d j (b ) bj = b j + j j c j f j d j (! j d j j j ) We hve the nlog for the block cse. For simplicity we only consider the Guss rule. Then, in 3.8 T T J N = N 3 N 2 T A N 2 N 2 N 32

33 nd with nd JN x J N + = N x T N! N + x T N = ( N ) ~J = J n x N N x T N Here, we use the Shermn{Morrison{Woodbury formul (see []) which is generliztion of the formul we used before. Then, ~J = J N + J N x N ( N x T NJ x N ) x T NJ In order to compute ll the elements, we need block holesky decomposition of J N. We obtin the following lgorithm which gives 2 2 mtrix i, the block element of the inverse tht we need. Theorem 5.4 Let = D = = I, for i = we compute i = i + i D i T i ( i i D i T i ) i D i T i N D i = i i D i T i i = i T i D i These recurrences for 2 2 mtrices cn be esily computed. Hence, the pproximtions cn be computed s we pply the block Lnczos lgorithm. Given these lgorithms to compute the estimtes, we see tht lmost ll of the opertions re result of the Lnczos lgorithm. omputing the estimte hs complexity independent of the problem size. To compute digonl entry, the Lnczos lgorithm needs per itertion the following opertions mtrix{vector product, 4N multiplies nd 4N dds. To compute two digonl entries nd non digonl one, the block Lnczos lgorithm needs per itertion 2 mtrix{vector products, 9N multiplies, 7N dds plus the QR decomposition which is 8N ops (see []). The non{symmetric Lnczos lgorithm requires per itertion mtrix{vector product, 6N multiplies nd 6N dds. Therefore, if we only wnt to estimte digonl elements it is best to use the Lnczos lgorithm. If we wnt to estimte non digonl element, it is best to use the block Lnczos lgorithm since we get three estimtes in one run while for the non{symmetric Lnczos method we need lso to hve n estimte of digonl element. The number of ops is the sme but for the block Lnczos we hve three estimtes insted of two with the non{symmetric Lnczos. On the other hnd, the non{symmetric Lnczos gives bounds but the block Lnczos yields only estimtes. As we notice before, we cn compute bounds for the non digonl elements by considering 2 (e i + e j ) T A (e i + e j ). For this, we need to run three Lnczos lgorithms tht is per itertion 3 mtrix{vector products, 2N multiplies nd 2N dds to get 3 estimtes. This no more opertions thn in the block Lnczos cse but here, we cn get bounds. 33 N

34 With the non{symmetric Lnczos, we hve 2 bounds with 2 mtrix{vector products, N multiplies nd N dds. One cn sk why in the cse of the inverse re we not solving the liner system Au = e i to obtin the i th column of the inverse t once. To our knowledge, it is not possible then to tell if the estimtes re upper or lower bounds. Moreover this cn be esily dded to the lgorithm of Theorem 5.3. Let Q N = [x x N ] be the mtrix of the Lnczos vectors. Then we hve the pproximte solution u N = Q N y N where y N is the solution of J N y N = Q T Ne i = e This tridigonl liner system cn be esily solved incrementlly from the LDL T decomposition, see []. This yields vrint of the conjugte grdient lgorithm. We give numericl exmples in Section 6 nd show tht our methods give better bounds. 6 Numericl exmples In this Section, we rst describe the exmples we use nd then we give numericl results for some specic functions f. 6. Description of the exmples First we look t exmples of smll dimension for which the inverses re known. Then, we will turn to lrger exmples rising from the discretiztion of prtil dierentil equtions. Most of the numericl computtions hve been done with Mtlb 3.5 on n Apple Mcintosh Powerbook 7 nd few ones on Sun worksttion. Exmple. First, we consider A = I + uu T u T = ( ) This mtrix hs two distinct eigenvlues nd n +. Therefore, the miniml polynomil is of degree 2 nd the inverse cn be written s A = 2 + n + n I + n A Exmple 2. The entries of the Hilbert mtrix re given by. We consider mtrix of dimension i+j 5 which is =2 =3 =4 =5 =2 =3 =4 =5 =6 A() = I 5 + =3 =4 =5 =6 =7 =4 =5 =6 =7 =8 A =5 =6 =7 =8 =9 34

35 The inverse of A() is A = nd the eigenvlues of A() re ( ) Exmple 3. We tke n exmple of dimension, A = It is esily seen (cf. [3]) tht the inverse is tridigonl mtrix A = The eigenvlues of A re therefore distinct nd given by ( ) A A A Exmple 4. We hve A = A 35

36 whose inverse is whose eigenvlues re A = A ( ) Exmple 5. We use mtrix of dimension constructed with the TOEPLITZ function of Mtlb, A = 2I + TOEPLITZ( ) This mtrix hs distinct eigenvlues but most of them re very close together ( ) Exmple 6. This exmple is the mtrix rising from the 5{point nite dierence of the Poisson eqution in unit squre. This gives liner system of order m 2, where A = ech block being of order m nd T = Ax = b T I I T I I T I I T For m = 6, the minimum nd mximum eigenvlues re 396 nd Exmple 7. This exmple rises from the 5{point nite dierence pproximtion of the following eqution in unit squre, div(ru)) = f with Dirichlet boundry conditions. (x y) is digonl mtrix with equl digonl elements. This element is equl to in squre ]=4 3=4[]=4 3=4[, otherwise. For m = 6, the minimum nd mximum eigenvlues re 4354 nd A A

37 6.2 Results for polynomil function To numericlly check some of the previous theorems, f ws chosen s polynomil of degree q, f() = qy i= ( i) We chose Exmple 6 with m = 6, tht is mtrix of order 36. ) We compute the (2 2) element of f(a) nd we vry the order of the polynomil. In the next tble, we give, s function of the degree q of the polynomil, the vlue of N to hve n \exct" result (4 digits in Mtlb) for the Guss rule. q N From these results, we cn conclude tht the mximum degree for which the results re exct is q = 2N, s predicted by the theory. For the Guss{Rdu rule, we get q N From this we deduce q = 2N s predicted. For the Guss{Lobtto rule, we hve q N This shows tht q = 2N + which is wht we expect. 2) If we consider the block cse to compute the (3 ) element of the polynomil, we get the sme results, therefore the block Guss rule is exct for the degree 2N, the block Guss{Rdu rule is exct for degree 2N nd the block Guss{Lobtto is exct for degree 2N +. 3) The sme is lso true for the non{symmetric Lnczos lgorithm if we wnt to compute the sum of the (3 ) nd (3 3) elements. 6.3 ounds for the inverse 6.3. Digonl elements Now, we turn to some numericl experiments using Mtlb on the exmples described bove. Usully the results will be given using the \short" formt of Mtlb. In the following results, N it denotes the number of itertions of the Lnczos lgorithm. This corresponds to N for the Guss nd Guss{Rdu rules nd N for the Guss{Lobtto rule. Exmple. 37

38 ecuse of the properties of the mtrix we should get the nswer in two steps. We n hve! = 2 nd = n, therefore, the lower bound from Guss{Rdu is nd n+ n the upper bound is, the exct result. If we look t the lower bound from the Guss n+ rule, we nd the sme vlue. This is lso true for the numericl experiments s well s for Guss{Lobtto. Exmple 2. Let us consider (A() ) 33 whose exct vlue is The Guss rule, s function of the degree of the qudrture, gives Results from Guss rule Nit= The Guss{Rdu rule gives upper nd lower bounds. For nd b, we use the computed vlues from the EIG function of Mtlb. Results from the Guss{Rdu rule Nit= lw bnd up bnd Results from Guss{Lobtto rule Nit= The results re not s good s expected. The exct results should hve been obtined for Nit = 3. The discrepncy comes from round o errors, prticulrly for the lower bound, becuse of the eigenvlue distribution of A nd poor convergence rte of the Lnczos lgorithm in this cse. To see how this is relted to the conditioning of A, let us vry. For simplicity we consider only the Guss rule. The following tbles give results for dierent vlues of nd the (3 3) element of the inverse. The exct vlues re ( = 73949), ( = 77793), ( = 954). Lower bound from Guss rule for = Nit= Lower bound from Guss rule for = Nit= Lower bound from Guss rule for = Nit=

39 We see tht when A is well conditioned, the numericl results follow the theory. The discrepncies probbly rise from the poor convergence of the smllest eigenvlues of J N towrds those of A. Exmple 3. We re looking for bounds for (A ) 55 whose exct vlue is, of course, 2. Lower bounds for (A ) 55 from the Guss rule Nit= Results for (A ) 55 from the Guss{Rdu rule Nit= b b Upper bounds for (A ) 55 from the Guss{Lobtto rule Nit= In this exmple 5 or 6 itertions should be sucient, so we re little o the theory. Exmple 4. We look t bounds for (A ) 55 whose exct vlue is 45 Lower bounds for (A ) 55 from the Guss rule Nit= Lower nd upper bounds for (A ) 55 from the Guss{Rdu rule Nit= lw bnd up bnd Upper bounds for (A ) 55 from the Guss{Lobtto rule Nit= Exmple 5. We get for (A ) 55, whose vlue is 595, Lower bounds for (A ) 55 from the Guss rule Nit= Lower nd upper bounds for (A ) 55 from the Guss{Rdu rule 39

40 Nit= lw bnd up bnd Upper bounds for (A ) 55 from the Guss{Lobtto rule Nit= ecuse some eigenvlues re very close together, we get the exct nswers little sooner thn it is predicted by theory. Exmple 6. onsider m = 6. Then we hve system of order 36 nd we look for bounds on (A ) 88 whose vlue is 355. There re 9 distinct eigenvlues, therefore we should get the exct nswer in bout itertions for Guss nd Guss{Rdu nd 9 itertions for Guss{Lobtto. Lower bounds for (A ) 88 from the Guss rule Nit= Lower nd upper bounds for (A ) 88 from the Guss{Rdu rule Nit= lw bnd up bnd Upper bounds for (A ) 88 from the Guss{Lobtto rule Nit= Now, we consider m = 6 which gives mtrix of order 256. We wnt to compute bounds for the (25 25) element whose vlue is 564. In this cse there re 29 distinct eigenvlues, so we should nd the exct nswer in bout 65 itertions t worst. These computtions for lrger problem hve been done on Sun Sprcsttion +. We nd the following results. Lower bounds for (A ) 2525 from the Guss rule Nit= Lower nd upper bounds for (A ) 2525 from the Guss{Rdu rule Nit= lw bnd up bnd