SPECTRAL EVOLUTION OF A ONE PARAMETER EXTENSION OF A REAL SYMMETRIC TOEPLITZ MATRIX* William F. Trench. SIAM J. Matrix Anal. Appl. 11 (1990),

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1 SPECTRAL EVOLUTION OF A ONE PARAMETER EXTENSION OF A REAL SYMMETRIC TOEPLITZ MATRIX* William F Trench SIAM J Marix Anal Appl 11 (1990), Absrac Le T n = ( i j ) n i,j=1 (n 3) be a real symmeric Toepliz marix such ha T n 1 T n 2 have no eigenvalues in common We consider he evoluion of he specrum of T n as he parameer = n 1 varies over (, ) I is shown ha he eigenvalues of T n associaed wih symmeric (reciprocal) eigenvecors are sricly increasing funcions of, while hose associaed wih he skew symmeric (ani reciprocal) eigenvecors are sricly decreasing Resuls are obained on he asympoic behavior of he eigenvalues eigenvecors as ±, on he possible orderings of eigenvalues associaed wih symmeric skew symmeric eigenvecors *This work was parially suppored by Naional Science Foundaion Gran No DMS Suggesed running head: SYMMETRIC TOEPLITZ SPECTRAL EVOLUTION 1980 Mahemaics Subjec Classificaions: 15A18, 15A42 Key words phrases: Toepliz, exension, symmeric, eigenvalue, eigenvecor Mailing address of auhor: Deparmen of Mahemaics, Triniy Universiy, 715 Sadium Drive, San Anonio, Texas 78284, USA 1

2 SPECTRAL EVOLUTION OF A ONE PARAMETER EXTENSION OF A REAL SYMMETRIC TOEPLITZ MATRIX William F Trench Following Andrew [1], we will say ha an m vecor X = [x 1,x 2,,x m ] is symmeric if x j = x m j+1, 1 j m, or skew symmeric if x j = x m j+1, 1 j m (Some auhors call such vecors reciprocal ani-reciprocal) Canoni Buler [2] have shown ha if T m = ( i j ) m i,j=1 is a real symmeric Toepliz marix of order m, hen R m has an orhonormal basis consising of m [m/2] symmeric [m/2] skew symmeric eigenvecors of T m, where [x] is he ineger par of x A relaed resul of Delsare Genin [4, Thm 8] is ha if λ is an eigenvalue of T m wih mulipliciy greaer han one, hen he λ eigenspace of T m has an orhonormal basis which splis as evenly as possible beween symmeric skew symmeric λ eigenvecors of T m For convenience here, we will say ha an eigenvalue λ of T m is even (odd) if T m has a symmeric (skew symmeric) λ eigenvecor The collecion S + (T m )(S (T m )) of even (odd) eigenvalues will be called he even (odd) specrum of T m From he resul of Delsare Genin, a muliple eigenvalue is in boh he even odd specra of T m 2

3 This paper is moivaed by consideraions ha arose in connecion wih he inverse eigenvalue problem for real symmeric Toepliz marices Alhough we do no claim ha our resuls provide much insigh ino his problem, hey may neverheless be of some ineres in heir own righ The inverse eigenvalue problem for real symmeric Toepliz marices is usually saed as follows: Find a real symmeric Toepliz marix T m wih given specrum S(T m ) = {λ 1 λ 2 λ m } For our purposes i is convenien o impose an addiional condiion; namely, ha T m have even odd specra S + (T m ) S (T m ), conaining, respecively, m [m/2] [m/2] given elemens (couning repeaed eigenvalues according o heir mulipliciies) of S We will say ha S + (T m ) S (T m ) are inerlaced if whenever λ k λ l are in S + (T m ) (S (T m )) k < l, here is an elemen λ i in S (T m ) (S + (T m )) such ha λ k λ i λ l Delsare Genin [4] showed ha if m 4 hen he inverse eigenvalue problem always has a soluion (regardless of he numerical values of λ 1, λ 2, λ 3, λ 4 ) if S + (T m ) S (T m ) are inerlaced; however, if hey are no, hen he exisence or nonexisence of a soluion depends on he specific numerical values of he λ i s They also argue ha his negaive consequence of non inerlacemen of S + (T m ) S (T m ) holds for all m > 4; ha is, if he wo desired specra are no inerlaced, hen he inverse eigenvalue problem fails o have a soluion for some choices of desired eigenvalues Delsare Genin [4] formulaed he (sill open) conjecure ha he inverse eigenvalue problem always has a soluion (for arbirary m) provided ha he desired even odd specra are inerlaced (This was apparenly misinerpreed by Laurie [6], who cied a real symmeric Toepliz marix for which S + (T m ) S (T m ) are no inerlaced as a counerexample o he conjecure of Delsare Genin ha he eigenvecors of a symmeric Toepliz marix, corresponding o eigenval- 3

4 ues arranged in decreasing order, alernae beween reciprocal ani-reciprocal vecors ) In numerical experimens repored in [7] we compued he eigenvalues of hundreds of romly generaed real symmeric Toepliz marices wih orders up o 1000 (Since hen we have considered marices of order 2000) The even odd specra of hese marices are cerainly no necessarily inerlaced, bu hey seem o be almos inerlaced, in ha we seldom saw more han wo or hree successive even (or odd) eigenvalues In unsuccessfully rying o formulae a definiion of a measure of inerlacemen ha would be useful in connecion wih he inverse eigenvalue problem, we were led o sudy he problem considered here; namely, if 0 1 n n 3 T n 1 =, n 2 n 3 0 is a given real symmeric Toepliz marix of order n 1, hen how does he specrum of he n-h order marix (1) T n () = 0 1 n n 3 n 2 n 2 n n evolve as varies over (, )? We impose he following assumpion hroughou 4

5 Assumpion A: n 3 T n 2 T n 1 have no eigenvalues in common Assumpion A Cauchy s inerlace heorem imply ha T n 2 T n 1 have no repeaed eigenvalues Le α 1 < α 2 < < α n 1 be he eigenvalues of T n 1 le λ 1 () λ 2 () λ n () be he eigenvalues of T n () The Cauchy inerlace heorem implies ha λ i () α i λ i+1 (), 1 i n 1, < < I is also convenien o inroduce disinc names for he even odd eigenvalues of T n 2 T n () Define r = n [n/2] s = [n/2]; hus r = s if n is even r = s+1 if n is odd Denoe he even odd eigenvalues of T n 2 by β 1 < β 2 < < β r 1 γ 1 < γ 2 < < γ s 1, respecively, le (2) µ 1 () µ 2 () µ r () (3) ν 1 () ν 2 () ν s () 5

6 be he even odd eigenvalues, respecively, of T n () Now define p j (λ) = de(t j λi j ), 1 j n 1, p n (λ,) = de(t n () λi n ) As observed by Delsare Genin [4], a resul of Canoni Buler [2] implies ha p n (λ,) can be facored in he form p n (λ,) = p + n (λ,)p n (λ,), where p + n p n are of degrees r s respecively in λ, (4) p + n (µ i (),) = 0, 1 i r, < <, (5) p n (ν j (),) = 0, 1 j s, < < Moreover, an argumen of Delsare Genin [4, p 203, 208] implies ha he even (odd) eigenvalues of T n 2 separae he even (odd) eigenvalues of T n (); i e, (6) µ i () β i µ i+1 (), 1 i r 1, (7) ν i () γ i ν i+1 (), 1 i s 1 I now follows ha (2) (3) can be replaced by he sronger inequaliies µ 1 () < µ 2 () < < µ r () ν 1 () < ν 2 () < < ν s () 6

7 To see his, suppose for example ha µ i (ˆ) = µ i+1 (ˆ) for some i ˆ Then (6) implies ha β i, an eigenvalue of T n 2, is a repeaed eigenvalue of T n (ˆ) Cauchy s heorem hen implies ha β i is also an eigenvalue of T n 1, which violaes Assumpion A Since p + n (λ,) p n (λ,) have disinc roos for all, (4), (5), (6), (7) define µ 1 (),,µ r () ν 1 (), ν s () as coninuously differeniable funcions on (, ) However, (4) (5) do no provide convenien represenaions for he derivaives of hese funcions The nex wo lemmas will enable us o find such represenaions Lemma 1 Suppose ha Assumpion A holds, le α i (1 i n 1) be an eigenvalue of T n 1 Then here is exacly one value τ i of such ha α i is an eigenvalue of T n (τ i ) Moreover, α i is in fac an eigenvalue of T n (τ i ) wih mulipliciy wo, τ 1,,τ n 1 are he only values of for which T n () has repeaed eigenvalues Proof By an argumen of Iohvidov [5, p 98], based on Sylveser s ideniy, i can be shown ha n 2 0 λ 1 2 n 3 n 2 (8) p n (λ,)p n 2 (λ) = p 2 n 1(λ) n 3 n 4 n 5 0 λ 1 2 for all λ Exping he deerminan on he righ in cofacors of is firs row shows ha (8) can be rewrien as (9) p n (λ,)p n 2 (λ) = p 2 n 1(λ) [( 1) n+1 p n 2 (λ) + k n 2 (λ)] 2, 7

8 where k n 2 (λ) is independen of Therefore, p n (α i,τ i ) = 0 if only if τ i = ( 1)n k n 2 (α i ) p n 2 (α i ) Obviously, α i is a repeaed zero of he polynomial obained by seing = τ i on he righ of (9), herefore α i is an eigenvalue of T n (τ i ) wih mulipliciy m > 1 To see ha m = 2, suppose o he conrary ha m > 2 Then eiher µ l (τ i ) = µ l+1 (τ i ) = α i or ν l (τ i ) = ν l+1 (τ i ) = α i for some l Bu hen (6) (7) imply ha α i = β l or α i = γ l for some l, which conradics Assumpion A; hence, m = 2 To conclude he proof, we simply observe ha a repeaed eigenvalue of T n () mus be an eigenvalue of T n 1 This lemma is relaed o Theorem 3 of Cybenko [3], who also considered quesions conneced wih he eigensrucure of T n () regarded as an exension of T n 1 Now define q n (λ,) = p n(λ,) p n 1 (λ) The nex lemma can be proved by pariioning T n () λi n in he form [ 0 λ U T n () λi n = n 1() T U n 1 () T n 1 λi n 1 where U n 1 () is defined in (11), below (For deails, see he proof of Theorem 1 of [7]) ], Lemma 2 If λ is no an eigenvalue of T n 1, le X n 1 (λ,) = x 1,n 1 (λ,) x 2,n 1 (λ,) x n 1,n 1 (λ,) 8

9 be he soluion of he sysem (10) (T n 1 λi n 1 )X n 1 (λ,) = U n 1 (), where (11) U n 1 () = 1 n 2 Then q n (λ,) = 0 λ U T n 1()X n 1 (λ,); moreover, if q n (λ,) = 0, hen he vecor is a λ eigenvecor of T n (); hence [ ] 1 Y n (λ,) = X n 1 (λ,) (12) x n 1,n 1 (λ,) = ( 1) q+1, where { 0 if λ is an even eigenvalue of Tn 1 (), q = 1 if λ is an odd eigenvalue of T n 1 () We will call q he pariy of he eigenvalue λ Now suppose ha λ() is one of he funcions µ 1 (),,µ r () or ν 1 (),,ν s () Lemmas 1 2 imply ha 0 λ() U T n 1()X n 1 (λ(),) = 0, J, 9

10 where J is any inerval which does no conain any of he excepional poins τ 1,,τ n 1 defined in Lemma 1 Differeniaing his yields (13) ( 1 + U T n 1 () λ X n 1(λ(),) ) λ () + UT n 1() X n 1 (λ(),) + U T n 1() X n 1(λ(),) = 0 However, if λ is any number which is no an eigenvalue of T n 1, hen (14) U T n 1() = X T n 1(λ,)(T n 1 λi n 1 ) (see (10)), differeniaing (10) yields (15) (T n 1 λi n 1 ) λ X n 1(λ,) = X n 1 (λ,) 0 (16) (T n 1 λi n 1 ) X n 1(λ,) = U n 1() = 0 1 (see (11)) Seing λ = λ() in (14), (15), (16) subsiuing he resuls ino (13) yields ( 1 + Xn 1 (λ(),) 2) λ () + 2x n 1,n 1 (λ(),) = 0 (Euclidean norm); herefore, from (12), (17) λ () = ( 1) q X n 1 (λ(),) 2 10

11 Because of (12) we can wrie (18) X n 1 (λ(),) = [ ] ˆX n 2 (λ(),) ( 1) q+1, where q is he pariy of λ() ˆX n 2 (λ(),) is symmeric if q = 0 or skew symmeric if q = 1; hen (17) becomes (19) λ () = ( 1) q ˆX n 2 (λ(),) 2, which is valid for τ i, 1 i n 1 This formula does no ye apply a hese excepional poins, simply because he vecors ˆX n 2 (λ(τ i ),τ i ) = ˆX n 2 (α i,τ i ), 1 i n 1 are as ye undefined This is easily remedied; by Lemma 2, (20) ( ) Tn () λ()i n 1 ˆX n 2 (λ(),) = 0 ( 1) q+1 for all τ i, 1 i n 1 This (1) imply ha (21) ( Tn 2 λ()i n 2 ) ˆXn 2 (λ(),) = 1 2 n 2 + ( 1)q n 2 n 3 1 for all τ i, 1 i n 1 However, his sysem has a unique soluion when = τ i, since he marix T n 2 λ(τ i )I n 2 = T n 2 α i I n 2 is nonsingular, by Assumpion A Defining his soluion o be ˆX n 2 (λ(τ i ),τ i ) exends ˆX n 2 (λ(),) so as o make i coninuous on (, ) Since λ () is also coninuous for all, (19) mus hold for all 11

12 For fuure reference, noice from (12) he coninuiy of x n 1,n 1 (λ i (),) ha he pariy q i () of λ i () is consan on any inerval J which does no conain any of he excepional poins τ 1,,τ n 1 Theorem 1 The even eigenvalues µ 1 (),,µ r () are sricly increasing on (, ) he inequaliies (6) can be replaced by he sric inequaliies (22) µ i () < β i < µ i+1 (), < < ; 1 i r 1; moreover, { βi, 1 i r 1, (23) lim µ i () =, i = r, { (24) lim µ βi 1, 2 i r, i() =, i = 1 Proof Seing q = 0 in (17) shows ha µ 1 (), µ r () are sricly increasing for all ; herefore, (6) implies (22) For convenience, define β r = suppose ha (25) lim µ i () = ζ i < β i for some i in {1,,r} Since β i 1 < ζ i < β i, he sysem ( Tn 2 ζ i I n 2 ) Xi = n 2 + ( 1)q n 2 n 3 1

13 has a unique soluion,, from (21) wih λ() = µ i (), lim Consequenly, (19) implies ha lim µ i() = ˆX n 2 (µ i (),) = X i X i 2 > 0, herefore lim µ i () =, which conradics (25) This implies (23) A similar argumen implies (24) The proof of he nex heorem is similar o his Theorem 2 The odd eigenvalues ν 1 (),,ν s () are sricly decreasing on (, ) he inequaliies (7) can be replaced by he sric inequaliies moreover, ν i () < γ i < ν i+1 (), < < ; 1 i s 1; (26) lim ν i () = (27) lim ν i() = { γi 1, 2 i s,, i = 1, { γi, 1 i s 1,, i = s The remaining heorems deal wih he asympoic behavior of he vecors ˆX n 2 (λ(),) (see (18)) wih he orders of convergence in (23), (24), (26), (27) Theorem 3 Le A i = 13 a (i) 1 a (i) 2 a (i) n 2

14 be he β i eigenvecor of T n 2 which is normalized so ha Then n 2 a (i) 1 + n 3a (i) a (i) n 2 = 1 (28) lim ˆXn 2 (µ i (),) = A i, 1 i r 1, (29) µ i () = β i Also, 2(1 + o(1)),, 1 i r 1 A i 2 (30) lim ˆX n 2 (µ i (),) = A i 1, 2 i r, (31) µ i () = β i 1 2(1 + o(1)),, 2 i r A i 1 2 Proof I is easy o verify ha he vecor [ Ai 0 is he las column of (T n 1 β i I n 1 ) 1 Seing λ = µ i () in (10) shows ha ] X n 1 (µ i (),) = (T n 1 µ i ()I n 1 ) 1 U n 1 () for sufficienly large Therefore, (11), (23), (24) imply (28) (30) From (19) wih λ() = µ i () (28) (32) µ i() = 2(1 + o(1)) A i 2 2,, 1 i r 1 Similarly, (19) (30) imply ha (33) µ 2(1 + o(1)) i () =,, 2 i r A i

15 Since (32) (33) imply (29) (31), he proof is complee A similar argumen yields he following heorem Theorem 4 Le B i = b (i) 1 b (i) 2 b (i) n 2 be he γ i eigenvecor of T n 2 which is normalized so ha n 2 b (i) 1 + n 3b (i) b (i) n 2 = 1 Then lim ˆX n 2 (ν i (),) = B i 1, 2 i s, ν i () = γ i 1 + 2(1 + o(1)),, 2 i s B i 1 2 Also, lim ˆX n 2 (ν i (),) = B i, 1 i s 1, ν i () = γ i + 2(1 + o(1)),, 1 i s 1 B i 2 Theorems 3 4 provide no informaion on he asympoic behavior of he eigenvalues which end o infinie limis as ± The nex heorem fills his gap 15

16 Theorem 5 Le (34) Γ q = 1 2 n 2 + ( 1)q n 2 n 3 1 Then (35) µ r () = Γ o ( 1) ( ); (36) µ 1 () = Γ o ( 1) ( ); (37) lim ˆX n 2 (µ r (),) = lim ˆX n 2 (µ 1 (),) = Γ 0 ; (38) ν 1 () = + 0 Γ o ( 1) ( ); (39) ν s () = + 0 Γ o ( 1) ( ); (40) lim ˆX n 2 (ν 1 (),) = lim ˆX n 2 (ν s (),) = Γ 1 Proof We will prove (35) (38) verify he firs limis in (37) (40); he proof of (36) (39) he verificaion of he second limis in (37) 16

17 (40) are similar Le λ() = µ r () q = 0 or λ() = ν 1 () q = 1 We know from Theorems 1 2 ha (41) lim λ() = From (21) (34), ( λ() Tn 2 ) ˆX n 2 (λ(),) Γ q ; herefore, (41) implies ha From his (19), herefore lim ˆX n 2 (λ(),) = 0 lim λ () = ( 1) q, λ() lim by L Hôpial s rule Now (21) implies ha = ( 1) q, (42) lim ˆX n 2 (λ(),) = ( 1) q+1 Γ q, wih Γ q as in (34) This verifies he firs limis in (37) (40) Since he firs componen of he vecor on he lef of (20) is idenically zero, (43) λ() 0 + [ 1, 2,, n 2 ] ˆX n 2 (λ(),) + ( 1) q+1 = 0 From (35) (42), [ 1, 2,, n 2 ] ˆX n 2 (λ(),) = ( 1) q+1 [ 1, 2,, n 2 ]Γ q + o ( 1 ) = ( 1) q+1 Γ q o ( 1) 17

18 Subsiuing his ino (43) solving for λ() yields λ() = 0 + ( 1) q[ + Γ q o ( 1)], which proves (35) (38) We conclude wih a commen on he possible orderings of even odd eigenvalues of T n () Le he eigenvalues of T n 2 be ω 1 < ω 2 < < ω n 2 ; i e, {ω 1,,ω n 2 } = {β 1,,β r 1 } {γ 1,,γ s 1 } Define Q 0 = [q(ω 1 ),q(ω 2 ),,q(ω n 2 )], where q(ω i ) is he pariy of ω i Suppose ha he elemens τ 1,,τ n 1 of he excepional se discussed in Lemma 1 are disinc, ordered so ha (44) τ i1 < τ i2 < < τ in 1 Le (,τ i1 ), l = 1, J l = (τ il 1,τ il ), 2 l n 1, (τ il, ), l = n The eigenvalues of T n () saisfy he sric inequaliies λ 1 () < λ 2 () < < λ n () 18

19 for all in each inerval J 1,,J n 1 Recalling ha he pariies of λ 1 (),,λ n () are consan on each J l, we can define he n-vecors Q l = [q l1,q l2,,q ln ], 1 l n, where q lj is he pariy of λ j () on J l Since λ i () α i λ i+1 () for all α i is an eigenvalue wih mulipliciy wo of T n (τ i ), we mus have λ i (τ i ) = λ i+1 (τ i ) = α i Therefore, α i is in boh he even odd specrum of T n (τ i ) From he monooniciy properies of he even odd eigenvalues of T n (), i follows ha λ i () changes from even o odd λ i+1 () changes from odd o even as increases hrough τ i This (23), (24), (26), (27) imply he following heorem Theorem 6 If τ 1,,τ n 1 saisfy (44), hen Q 1 = [0,Q 0,1], Q n = [1,Q 0,0],, for 1 l n 1, (45) q i,l+1 = q i,l if i i l i i l+1, (46) q il,l = 0, q il +1,l = 1, q il,l+1 = 1, q il +1,l+1 = 0 From (45) (46), Q l+1 is obained by inerchanging he zero one which mus be in columns i l i l+1, respecively, of Q l The assumpion ha τ 1,,τ n 1 are disinc was imposed for simpliciy Theorem 6 can easily be modified o cover he excepional case where {τ 1,,τ n 1 } conains fewer han n 1 disinc elemens 19

20 References [1] AL Andrew, Eigenvecors of cerain marices, Lin Alg Appl, 7 (1973), [2] A Canoni F Buler, Eigenvalues eigenvecors of symmeric cenrosymmeric marices, Lin Alg Appl, 13 (1976), pp [3] G Cybenko, On he eigensrucure of Toepliz marices, IEEE Trans Acousics, Speech, Sig Proc, Vol ASSP-32 (1984), pp [4] P Delsare Y Genin, Specral properies of finie Toepliz marices, in Mahemaical Theory of Neworks Sysems, Proc MTNS 83 Inernaional Symposium, Beer Sheva, Israel (1983), pp [5] I S Iohvidov, Hankel Toepliz Marices Forms, Birkhauser, Boson, Basel, Sugar, 1982 [6] D P Laurie, A numerical approach o he inverse Toepliz eigenproblem, SIAM J Sci Sa Compu, 9 (1988), [7] W F Trench, Numerical soluion of he eigenvalue problem for Hermiian Toepliz marices, SIAM J Marix Th Appl 10 (1989),