Orthogonal Symmetric Toeplitz Matrices

Transcription

1 Orthogonal Symmetric Toeplitz Matrices Albrecht Böttcher In Memory of Georgii Litvinchuk ( Abstract We show that the number of orthogonal and symmetric Toeplitz matrices of a given order is finite and determine all these matrices In this way we also obtain a description of the set of all symmetric Toeplitz matrices whose spectrum is a prescribed doubleton Mathematics Subject Classification (2000 Primary 47B35; Secondary 15A18 15A47 Keywords Toeplitz matrix matrix power orthogonal matrix symmetric matrix circulant skew circulant inverse eigenvalue problem 1 Introduction and main result Georgii Semyonovich Litvinchuk s favorites included singular integral equations with Carleman shifts that is with diffeomorphisms α for which the mth iterate α m is the identity map My favorites are Toeplitz matrices and hence when receiving the invitation to contribute to this volume I thought it might perhaps be an interesting problem to look for all Toeplitz matrices A whose mth power A m is equal to the identity matrix I The spectrum of an infinite Toeplitz matrix A that generates a bounded operator on l 2 is always connected [6] Consequently such a matrix satisfies A m = I if and only if A = e 2πij/m I for some j {0 1 m 1} and thus the question is not interesting for infinite matrices So let us consider finite matrices We denote by R n n the set of all real n n matrices and define O n S n T n as the subsets of R n n constituted by the orthogonal symmetric and Toeplitz matrices respectively We also put OS n = O n S n OT n = O n T n ST n = S n T n OST n = O n S n T n

2 2 A Böttcher The number of elements of a finite set E will be denoted by E Throughout the paper we assume that n 2 The following three simple propositions will be proved in Section 2 Proposition 11 For m 2 there are uncountably many matrices A T n such that A m = I and A k I for 1 k m 1 The infinite set we encounter in Proposition 11 seems to have no nice description Well letting ω m := e 2πi/m and T m := {1 ω m ωm m 1 } one could describe the set as the set of all diagonalizable Toeplitz matrices with eigenvalues in T m \ 1 k m 1 T k but this is not what I understand by nice However the problem becomes charming when restricting the search to symmetric matrices Proposition 12 (a For m 3 there is no A S n such that A m = I and A k I for 1 k m 1 (b The set of all A S n for which A 2 = I but A I coincides with OS n \ {I} This result shows that if A ST n then A m = I can only happen if A is in OST n in which case A 2 = I We are thus led to the set OST n = T n OS n = ST n OT n Proposition 13 The sets T n OS n ST n OT n are all uncountably infinite In spite of this proposition the following fact which is our main result is quite remarkable Theorem 14 The set OST n is finite and { 3 2 OST n = n/2 2 if n is even n/2 2 if n is odd Our proof of Theorem 14 is constructive and will yield all matrices in OST n Here is a simple consequence of Theorem 14 that concerns the inverse eigenvalue problem for Toeplitz matrices Corollary 15 Let α and β be two prescribed distinct real numbers The number of all matrices in ST n which have both α and β as eigenvalues and no other eigenvalues is 3 2 n/2 4 if n is even and n/2 4 if n is odd The paper is organized as follows Section 2 contains the proofs of Propositions 11 to 13 In Section 3 we show that every matrix in OST n is a circulant or a skew circulant This is enough to conclude that OST n is finite Theorem 14 is proved in Section 4 examples revealing the structure of the matrices in OST n are given in Section 5 and Corollary 15 is the subject of Section 6

3 Orthogonal Symmetric Toeplitz Matrices 3 2 Proofs of the surrounding results We denote by circ (a 1 a n the n n circulant matrix whose first column is ( a 1 a n and by F n the n n Fourier matrix ω n ωn 2 ωn n 1 F n = (ω (j 1(k 1 n n jk=1 = 1 ω n 1 n ωn 2(n 1 ω n (n 1(n 1 where ω n := e 2πi/n It is well known [2] that the eigenvalues of the circulant circ (a 1 a n are a(1 a(ω n a(ωn n 1 where a(z := a 1 +a 2 z + +a n z n 1 The equations a(ωn j 1 = µ j (j = 1 n may be written in the form or equivalently F n n a 1 a 1 a n = = Fn µ 1 µ n µ 1 (1 (2 a n µ n Lemma 21 Let α β R and λ 1 λ k C If n = 2k+2 there exists a circulant C R n n having the eigenvalues α β λ 1 λ k λ 1 λ k and if n = 2k + 1 there exists a circulant C R n n whose eigenvalues are α λ 1 λ k λ 1 λ k Proof Let first n = 2k + 2 We define a 1 a n by (2 with (µ 1 µ n = (α λ 1 λ k β λ 1 λ k It remains to show that a 1 a n are real But for j = 0 1 n 1 na j+1 = α + ω j(k+1 n β + ( ω jp n λ p + ω j(n p n λ p = α β + and this is a real number If n = 2k + 1 we take (µ 1 µ n = (α λ 1 λ k λ 1 λ k 2 Re(ω jp n λ p and define a 1 a n by (2 The same argument as before shows that these are real numbers Proof of Proposition 11 If n 3 Lemma 21 yields the existence of a circulant C R n n with the eigenvalues ω m ω m 1 1 It follows that C m = I and C k I for 1 k m 1 For µ R \ {0} put D µ = diag (1 µ µ n 1 Then A = D µ CD 1 µ is a matrix in T n for which A m = I and A k I for 1 k m 1

4 4 A Böttcher As different µ s produce different A s we get the assertion We are left with the case n = 2 The eigenvalues of ( a b A = T c a 2 are a ± bc Letting a b c be any real numbers such that a = cos(2π/m and bc = sin 2 (2π/m we obtain uncountably many matrices A with the eigenvalues ω m and ω m and thus with the desired property that A m = I and A k I for 1 k m 1 Proof of Proposition 12 (a Let m 3 and A S n Then A = U DU with an orthogonal matrix U and a diagonal matrix D which contains all eigenvalues of A If A m = I then sp A the set of the eigenvalues of A is contained in T m R Since T m R = {1} for odd m we see that A = I in this case which contradicts the requirement that A k I for 1 k m 1 If m is even we have T m R = { 1 1} and hence A 2 = I As necessarily m 4 this is again a contradiction to the requirement that A k I for 1 k m 1 (b Let A S n \ {I} and A 2 = I Write A = U DU as above The entries of D are all 1 or 1 and hence D is orthogonal This implies that A is orthogonal too Conversely if A OS n \ {I} then sp A T R = { 1 1} which shows that each diagonal entry of D is 1 or 1 It follows that A 2 = I Proof of Proposition 13 The assertion is trivial for T n and ST n Since each matrix of the form U DU with U O n and a diagonal matrix with sp D { 1 1} is in OS n we see that OS n is uncountably infinite Lemma 21 tells us that if τ T is any given number then there is a circulant C R n n (n 3 having the eigenvalues τ τ 1 1 Since C OT n we conclude that OT n is uncountably infinite for n 3 Finally the matrix ( a b T b a 2 is orthogonal whenever a 2 + b 2 = 1 This shows that OT 2 is an uncountable set Remark 22 Let UHT n be the set of all unitary Hermitian Toeplitz matrices in C n n The symmetric Toeplitz matrix B with the top row ( n 2 n 2 n 2 n belongs to OST n Let D µ = diag (1 µ µ n 1 If µ T then D µ is unitary and hence D µ BDµ 1 is in UHT n This shows that UHT n is uncountably infinite The matrices in OST 2 are ( ( ( ( and it is easily seen that UHT 2 equals {( } {( 0 µ a i 1 a : µ T 2 µ 0 i 1 a 2 a } : a [ 1 1] \ {0}

5 Orthogonal Symmetric Toeplitz Matrices 5 This reveals that not every matrix in UHT 2 is of the form D µ AD 1 µ with A OST n and µ T 3 The set OST n is finite We denote by T (a 1 a n the symmetric Toeplitz matrix whose top row is ( a 1 a n Let A = T (a 1 a n be a matrix in OST n Suppose first that n = 2k + 1 is odd Consider for example T (a b c d e = a b c d e b a b c d c b a b c d c b a b e d c b a Since each row has l 2 norm 1 it follows that e 2 = b 2 and d 2 = c 2 Thus d = δc and e = εb with ε δ { 1 1} and therefore T (a b c d e = T (a b c δc εb In the general case we see in this way that T (a 1 a n must be of the form A = T (a 1 a n = T (a b 1 b k ε k b k ε 1 b 1 (3 with ε j { 1 1} If ε 1 = = ε k = 1 then (3 is a circulant For instance a b c c b b a b c c T (a b c c b = c b a b c c c b a b b c c b a is a circulant If ε j = 1 and b j = 0 we change ε j to 1 This does not change A Thus we may assume that b j 0 whenever ε j = 1 Lemma 31 If 1 {ε 1 ε k } then b j = 0 whenever ε j = 1 Proof We denote the rows of A by r 1 r 2k+1 Since A is orthogonal the scalar product (r i r j is zero for i j Let ε i = 1 and thus b i 0 for some i Step 1 We show that if i 2 and ε i 1 = 1 then b i 1 = 0 Let us first consider the example A = T (a b c d e f f e d c b (4 in which ε 1 = 1 ε 2 = 1 ε 3 = 1 ε 4 = 1 ε 5 = 1 f 0 b 0 We have i = 5 and we want to show that e = 0 The rows 4 to 6 of (4 are r 4 r 5 = d c b a b c d e f f e e d c b a b c d e f f r 6 f e d c b a b c d e f

6 6 A Böttcher Thus 0 = (r 4 r 5 = (de + + ( ff + e( f 0 = (r 5 r 6 = (de + + ( ff + ef which gives e = 0 as desired because f 0 In the general case we have 0 = (r i 1 r i = Σ + b i 1 ( b i 0 = (r i r i+1 = Σ + b i 1 b i which implies that b i 1 = 0 Analogously one can show that b i+1 = 0 if i k 2 and ε i+1 = 1 Step 2 Suppose i 3 and consider b i 2 If ε i 2 = 1 we have nothing to prove So assume that ε i 2 = 1 We want to show that b i 2 = 0 If ε i 1 = 1 then b i 2 = 0 by Step 1 Thus let ε i 1 = 1 From Step 1 we know that b i 1 = 0 Example (4 for i = 5 illustrates just this situation We have (with e = 0 and hence r 3 r 4 r 5 r 6 = c b a b c d 0 f f 0 d d c b a b c d 0 f f 0 0 d c b a b c d 0 f f f 0 d c b a b c d 0 f 0 = (r 3 r 5 = (bd + + fd + d( f 0 = (r 4 r 6 = (bd + + fd + df which yields d = 0 as desired In the general case 0 = (r i 2 r i = Σ + b i 2 ( b i 0 = (r i 1 r i+1 = Σ + b i 2 b i and thus b i 2 = 0 Similarly one gets b i+2 = 0 if i k 3 and ε i+2 = 1 Step 3 Continuing as above we see that b i±l = 0 whenever i ± l {1 k} and ε i±l = 1 Lemma 31 implies that if at least one of the numbers ε 1 ε k is 1 then A is a skew circulant that is a matrix that results from a circulant by multiplying all entries below the main diagonal by 1 For example T (a 0 c c 0 = a 0 c c 0 0 a 0 c c c 0 a 0 c c c 0 a 0 0 c c 0 a is a skew circulant Let scirc (a 1 a n be the skew circulant whose first column is ( a 1 a n Thus scirc (a 1 a 2 a 3 = a 1 a 3 a 2 a 2 a 1 a 3 a 3 a 2 a 1

7 Orthogonal Symmetric Toeplitz Matrices 7 The eigenvalues of scirc (a 1 a n are a(σ n a(σ n ω n a(σ n ωn n 1 where σ n = e πi/n and ω n and a(z are as above (see [2] The equations a(σ n ωn j 1 = µ j (j = 1 n now take the form which is equivalent to F n a 1 µ 1 σ n a 2 = µ 2 (5 σn n 1 a n µ n a 1 a 2 n = F σ n µ 2 n (6 a n σn n 1 µ n At this point we are in a position to prove the following weakened version of Theorem 14 Proposition 32 If n is odd then OST n is a finite set and OST n 2 n+1 2 Proof We have seen that a matrix A in OST n is a circulant or a skew circulant The eigenvalues of A belong to { 1 1} Thus if A = circ (a 1 a n then a 1 a n are given by (2 with each µ j being 1 or 1 and if A = scirc (a 1 a n then a 1 a n are determined by (6 with each µ j in { 1 1} The two matrices I and I are counted both as a circulant and as a skew circulant This gives at most 2 2 n 2 solutions (That OST n will actually turn out to be much smaller than 2 n+1 2 is due to the circumstance that not every right-hand side of (2 and (6 with µ j { 1 1} gives a left-hand side with real numbers Now suppose n = 2k is even The case k = 0 was disposed of in Remark 22 where we observed that OST 2 consists of exactly four matrices So let k 1 We denote the rows of A by r 1 r 2k+2 Consideration of the scalar products (r j r j = 1 shows that A is of the form A = T (a 1 a n = T (a b 1 b k c b k b 1 with ε j { 1 1} Again we will assume without loss of generality that ε j = 1 if b j = 0 If all ε j are 1 then A is a circulant For example a b d c d b b a b d c d T (a b d c d b = d b a b d c c d b a b d d c d b a b b d c d b a is a circulant Thus assume there is a 1 among ε 1 ε k Lemma 33 Let 1 {ε 1 ε k } Then c = 0 and b j = 0 whenever ε j = 1 µ 1

8 8 A Böttcher Proof It can be shown as in the proof of Lemma 31 that b j = 0 if ε j = 1 Let us prove that c = 0 Again we first do an example If a 0 d c d 0 0 a 0 d c d A = T (a 0 d c d 0 = d 0 a 0 d c c d 0 a 0 d d c d 0 a 0 b d c d 0 a with d 0 then 0 = (r 3 r 4 = 2cd = 0 gives c = 0 In the general case the argument is as follows If ε k = 1 and hence b k 0 then (r k r k+1 = (r k+1 r k+2 2cb k and since (r k r k+1 = (r k+1 r k+2 = 0 it follows that c = 0 So let ε k = 1 and thus b k = 0 If ε k 1 = 1 then b k 1 0 and (r k 1 r k+1 = (r k r k+2 2cb k 1 which gives c = 0 as before If ε k = 1 b k = 0 ε k 1 = 1 b k 1 = 0 ε k 2 = 1 b k 2 0 we have (r k 2 r k+1 = (r k 1 r k+2 2cb k 2 and so on Eventually we get c = 0 Lemma 33 reveals that if 1 {ε 1 ε k } then A is a skew circulant For example a b d 0 d b b a b d 0 d T (a b d 0 d b = d b a b d 0 0 d b a b d d 0 d b a b b d 0 d b a is a skew circulant Proposition 34 If n is even then OST n is finite and OST n 2 n+1 2 Proof Proceed as in the proof of Proposition 32 4 The matrices in OST n In this section we prove Theorem 14 Let first n = 2k be odd By the results of Section 3 a matrix A is in OST n if and only if sp A { 1 1} and A is a circulant of the form A = circ (a 1 a n = circ (a b 1 b k b k b 1 (7 or a skew circulant of the form A = scirc (a 1 a n = scirc (a b 1 b k b k b 1 (8

9 Orthogonal Symmetric Toeplitz Matrices 9 Suppose A is the circulant (7 We denote the right-hand side of (1 by ( γ ε 1 ε k δ k δ 1 The numbers γ ε j δ j are all 1 or 1 Abbreviating ω n to ω we then get from (1 that ε j = a + b p (ω pj + ω (n pj p ( = a + b p ω pj + ω pj δ j = a + b p (ω p(n j + ω (n p(n j p ( = a + b p ω pj + ω pj k=1 k=1 whence ε j = δ j for 1 j k Conversely choose γ ε 1 ε k in { 1 1} insert (µ 1 µ n = (γ ε 1 ε k ε k ε 1 in (2 denote the column on the left of (2 by ( a b 1 b k c k c 1 and pass to complex conjugates Clearly a = γ + 2 ε j is real Furthermore nb j = γ + ε p (ω pj + ω j(n p ( = γ + ε p ω pj + ω pj nc j = γ + ε p (ω p(n j + ω (n j(n p ( = γ + ε p ω pj + ω pj It follows that b j = c j for 1 j k and since ω pj + ω pj is real the numbers b j are also real As there are exactly 2 k+1 choices of γ ε 1 ε k in { 1 1} k and as different right-hand sides of (1 give different (a 1 a n we conclude that OST n contains exactly 2 k+1 matrices of the form (7 Now suppose A is of the form (8 Put σ := σ n and notice that (1 σ σ n = (1 σ σ k σ k+1 σ 2k We denote the columns on the left and right of (5 by ( a σb 1 σ k b k σ k+1 b k σ 2k b 1 and ( ε 1 ε k γ δ k δ 1 respectively The numbers γ ε j δ j are all in { 1 1} From (5 we infer that ε j = a + δ j = a + b p (ω (j 1p σ p ω (j 1(n p σ 2k+1 p b p (ω (n jp σ p ω (n j(n p σ 2k+1 p

10 10 A Böttcher Since ω n = 1 σ 2k+1 = 1 ω = σ 2 we have ω (j 1p σ p = σ 2jp p ω (j 1(n p σ 2k+1 p = ω p(j 1 σ p = σ p 2jp ω (n jp σ p = ω jp σ p = σ p 2jp ω (n j(n p σ 2k+1 p = ω jp σ p = σ 2jp p which shows that ε j = δ j for 1 j k Conversely choose γ ε 1 ε k in { 1 1} insert (µ 1 µ n = (ε 1 ε k γ ε k ε 1 in (6 write (a 1 a n as (a b 1 b k c k c 1 and pass to complex conjugates It follows that a = γ + 2 ε k is real and that nb j = σ j ω jk γ + ε p (σ j ω j(p 1 + σ j ω j(n p nc j = σ n j ω (n jk γ + ε p (σ n j ω (n j(p 1 + σ n j ω (n j(n p Taking into account that ω n = 1 σ 2k+1 = 1 ω = σ 2 we see as above that c j = b j for 1 j k and that b 1 b k are real Thus what we obtained is a real matrix of the form (8 This proves that there are exactly 2 k+1 matrices of the form (8 in OST n The matrix A is both of the form (7 and of the form (8 if and only if b 1 = = b k = 0 that is if and only if A = I or A = I Thus in summary the number of matrices in OST n is 2 2 k+1 2 = 2 2 (n+1/2 2 = n/2 2 In addition we have proved the following Theorem 41 Let n = 2k The set OST n consists of the 2 k+1 circulants T (a 1 a n which are given by (2 with (µ 1 µ n = (γ ε 1 ε k ε k ε 1 and (γ ε 1 ε k { 1 1} k+1 and the 2 k+1 2 skew circulants T (a 1 a n which result from (6 with the choice (µ 1 µ n = (ε 1 ε k γ ε k ε 1 and (ε 1 ε k γ { 1 1} k+1 \ {(1 1 1 ( 1 1 1} Now suppose n = 2k is even We know from Section 3 that A belongs to OST n if and only if sp A { 1 1} and A is a circulant of the form A = circ (a 1 a n = circ (a b 1 b k c b k b 1 (9 or a skew circulant of the form A = scirc (a 1 a n = scirc (a b 1 b k 0 b k b 1 (10 Proceeding as in the case where n is odd we see that if A is given by (9 then the right-hand side of (1 is (µ 1 µ n = (γ ε 1 ε k β ε k ε 1

11 Orthogonal Symmetric Toeplitz Matrices 11 with γ β ε j { 1 1} and that different such right-hand sides produce different real matrices of the form (9 via (2 Analogously a matrix of the form (10 yields (µ 1 µ n = (ε 1 ε k γ β ε k ε 1 (11 with γ β ε j { 1 1} in (5 If we insert such a right-hand side in (6 we obtain a left-hand side with (a 1 a n = (a b 1 b k d c k c 1 The numbers b 1 b k and c 1 c k are real if and only if γ = β (This is not obvious but requires a computation which however is straightforward If γ = β then c j = b j for all j and d = 0 that is we get indeed a real matrix of the form (10 Different choices of (11 with γ = β lead to different left-hand sides of (9 and A = I and A = I are the only matrices that are both of the form (9 and the form (10 Thus we arrive at the conclusion that the number of matrices in OST n is 2 k k+1 2 = 3 2 k+1 2 = 3 2 n/2 2 and that moreover OST n can be described as follows Theorem 42 Let n = 2k The set OST n consists of the 2 k+2 circulants T (a 1 a n which result from (2 with (µ 1 µ n = (γ ε 1 ε k β ε k ε 1 and (γ ε 1 ε k β { 1 1} k+2 and the 2 k+1 2 skew circulants T (a 1 a n which are obtained from (6 with (µ 1 µ n = (ε 1 ε k β β ε k ε 1 and (ε 1 ε k β { 1 1} k+1 \ {(1 1 1 ( 1 1 1} 5 Examples OST 7 By Theorem 14 OST 7 = 30 Theorems 41 and 42 in conjunction with some obvious simplifications provide us with all matrices in OST 7 Formula (2 with (µ 1 µ 7 = (γ ε 1 ε 2 ε 3 ε 3 ε 2 ε 1 and (γ ε 1 ε 2 ε 3 { 1 1} 4 yields the 16 matrices with circ (a b 1 b 2 b 3 b 3 b 2 b 1 = T (a b 1 b 2 b 3 b 3 b 2 b 1 7a = γ + 2ε 1 + 2ε 2 + 2ε 3 7b 1 = γ + 2ε 1 cos 2π 7 + 2ε 2 cos 4π 7 + 2ε 3 cos 6π 7 7b 2 = γ + 2ε 1 cos 4π 7 + 2ε 2 cos 6π 7 + 2ε 3 cos 2π 7 7b 3 = γ + 2ε 1 cos 6π 7 + 2ε 2 cos 2π 7 + 2ε 3 cos 4π 7

12 12 A Böttcher The remaining 14 matrices are delivered by (6 with (µ 1 µ 7 = (ε 1 ε 2 ε 3 γ ε 3 ε 2 ε 1 and (ε 1 ε 2 ε 3 γ { 1 1} 4 \ {( ( } These matrices are scirc (a b 1 b 2 b 3 b 3 b 2 b 1 = T (a b 1 b 2 b 3 b 3 b 2 b 1 with 7a = γ + 2ε 1 + 2ε 2 + 2ε 3 7b 1 = γ + 2ε 1 cos π 7 + 2ε 2 cos 3π 7 2ε 3 cos 2π 7 7b 2 = γ + 2ε 1 cos 2π 7 2ε 2 cos π 7 2ε 3 cos 3π 7 7b 3 = γ + 2ε 1 cos 3π 7 2ε 2 cos 2π 7 + 2ε 3 cos π 7 This parametrization of OST 7 indicates that our approach perfectly fits with the problem and that any attempt to tackle the problem straightforwardly seems to be a hopeless venture For instance a matrix A of the form T (a b c d d c b is orthogonal if and only if a 2 + 2b 2 + 2c 2 + 2d 2 = 1 2ab + 2bc + 2cd + d 2 = 0 2ac + b 2 + 2bd + 2cd = 0 2ad + 2bc + 2bd + c 2 = 0 (Incidentally the same system is produced when applying the results of Gu and Patton [4] to the equation A A I I = 0 Thus we arrived at four equations with four unknowns This makes us expect that the set of solutions is finite However proving that the system has indeed only finitely many solution and even finding the solutions is another story The skeptical reader is invited to try and solve this system directly and to find the 16 solutions listed above OST 2 We have OST 2 = 4 and OST 2 = {T (ε 0 T (0 ε : ε { 1 1}} OST 3 The 6 matrices in the set OST 3 ε δ { 1 1} are T (ε 0 0 and T ( ε 3 2δ 3 2ε 3 with OST 4 The set OST 4 consists of the 10 matrices ( ε T (ε T (0 0 ε 0 T 2 δ 2 ε 2 δ 2 with ε δ { 1 1} ( T 0 ε 0 ε 2 2

13 Orthogonal Symmetric Toeplitz Matrices 13 OST 5 A list of the 14 matrices in OST 5 is with ε δ γ { 1 1} T (ε T ( 1 T 5 ε 1 + γ 5 5 ( 3ε 5 2δ 5 2ε 5 2δ 5 2ε 5 δ 1 γ 5 5 ε 1 γ 5 5 OST 6 The set OST 6 is constituted by the 22 matrices with ε δ { 1 1} T (ε T (0 0 0 ε 0 0 ( T 0 2ε 3 0 ε ( ε 0 2ε T ( ε T 3 δ 3 ε 3 2δ 3 ε 3 δ T 3 ( ε T 3 δ 3 ε 3 0 ε 3 δ 3 δ 1 + γ 5 ε 5 2δ 0 0 2ε ( 2ε 3 δ 3 ε 3 δ 3 ε 3 δ 3 6 An inverse eigenvalue problem The inverse eigenvalue problem for symmetric Toeplitz matrices consists in finding all A ST n for which sp A is a prescribed set of at most n points on R The pioneering work on this problem is due to Delsarte and Genin [3] who in particular solved the problem for n 4 Henry Landau [5] proved that if E R is any set with E n then there exists a matrix A ST n with sp A = E In connection with the problem studied in the present paper we mention the paper [1] by Chu and Erbrecht which deals with the question of finding all matrices in ST n that have two prescribed pairs of eigenvalues and eigenvectors The following result in conjunction with Theorems 41 and 42 provides us with all matrices A ST n for which sp A is a given doubleton Proposition Let α < β be two real numbers The set of all matrices A ST n with sp A = {α β} is α + β + β α OST n 2 2 where OST n := OST n \ {I I} Proof Let f(λ := α+β 2 + β α 2 λ and f 1 (λ := 2 β α λ α+β 2 sp A = {α β} then f 1 (A ST n and sp f 1 (A = {f 1 (α f 1 (β} = { 1 1} ( If A ST n and It follows that f 1 (A OST n and hence A f(ost n Conversely if A is in f(ost n then A is in ST n and sp A = {f( 1 f(1} = {α β} Clearly this proposition and Theorem 14 yield Corollary 15

14 14 A Böttcher References [1] M T Chu and M A Erbrecht Symmetric Toeplitz matrices with two prescribed eigenpairs SIAM J Matrix Anal Appl 15 ( [2] P J Davis Circulant Matrices John Wiley & Sons New York Chichester Brisbane 1979 [3] P Delsarte and Y Genin Spectral properties of finite Toeplitz matrices In: Mathematical Theory of Networks and Systems (Beer Sheva 1983 Lecture Notes in Control and Inform Sci 58 Springer London 1984 pp [4] C Gu and L Patton Commutation relations for Toeplitz and Hankel matrices SIAM J Matrix Anal Appl 24 ( [5] H J Landau The inverse eigenvalue problem for real symmetric Toeplitz matrices J Amer Math Soc 7 ( [6] H Widom On the spectrum of a Toeplitz operator Pacific J Math 14 ( Albrecht Böttcher Fakultät für Mathematik TU Chemnitz D Chemnitz Germany aboettch@mathematiktu-chemnitzde