Properties of Some Generalizations of Kac-Murdock-Szegö Matrices

Appeared In Structured Matrices in Mathematics, Computer Science and Engineering II AMS Contemporary Mathematics Series 281 (2001) 233 245 Properties of Some Generalizations of Kac-Murdock-Szegö Matrices William F Trench 95 Pine Lane Woodland Park, CO 80863, USA Abstract We consider generalizations of the Kac-Murdock-Szegö matrices of the forms L n = (ρ r s c min(r,s) ) n r,s=1 and U n = (ρ r s c max(r,s) ) n r,s=1, where ρ and c 1, c 2,, c n are real numbers We obtain explicit expressions for the determinants and inverses of L n and U n, determine their inertias, and diagonalize their quadratic forms We also consider the spectral distributions of two special cases Key words: Kac-Murdock-Szegö matrices; Toeplitz matrices; determinant; inverse; inertia; diagonalization; spectrum 1 Introduction The Kac-Murdock-Szegö (KMS) matrices [4] are the symmetric Toeplitz matrices K n (ρ) = (ρ r s ) n, n = 1, 2,, r,s=1 where ρ is real It is known [3, Section 72, Problems 12-13] that and, if ρ ±1, then K 1 n (ρ) = 1 1 ρ 2 det(k n (ρ)) = (1 ρ 2 ) n 1 (1) 1 ρ 0 0 0 0 ρ 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 0 0 0 0 0 0 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 ρ 0 0 0 0 ρ 1 ; 1

2 William F Trench thus, except for its first and last rows, K 1 n (ρ) is a tridiagonal Toeplitz matrix In this paper we consider generalizations of the KMS matrices of the form L n = (ρ r s c min(r,s) ) n r,s=1 and U n = (ρ r s c max(r,s) ) n r,s=1, where ρ and c 1, c 2,, c n are real numbers; thus, c 1 ρc 1 ρ 2 c 1 ρ n 3 c 1 ρ n 2 c 1 ρ n 1 c 1 ρc 1 c 2 ρc 2 ρ n 4 c 2 ρ n 3 c 2 ρ n 2 c 2 ρ 2 c 1 ρc 2 c 3 ρ n 5 c 3 ρ n 4 c 3 ρ n 3 c 3 L n = ρ n 3 c 1 ρ n 4 c 2 ρ n 5 c 3 c n 2 ρc n 2 ρ 2 c n 2 ρ n 2 c 1 ρ n 3 c 2 ρ n 4 c 3 ρc n 2 c n 1 ρc n 1 ρ n 1 c 1 ρ n 2 c 2 ρ n 3 c 3 ρ 2 c n 2 ρc n 1 c n (2) and U n = c 1 ρc 2 ρ 2 c 2 ρ n 3 c n 2 ρ n 2 c n 1 ρ n 1 c n ρc 2 c 2 ρc 3 ρ n 4 c n 2 ρ n 3 c n 1 ρ n 2 c n ρ 2 c 3 ρc 3 c 3 ρ n 5 c n 2 ρ n 4 c n 1 ρ n 3 c n ρ n 3 c n 2 ρ n 4 c n 2 ρ n 5 c n 2 c n 2 ρc n 1 ρ 2 c n ρ n 2 c n 1 ρ n 3 c n 1 ρ n 4 c n 1 ρc n 1 c n 1 ρc n ρ n 1 c n ρ n 2 c n ρ n 3 c n ρ 2 c n ρc n c n Although we do not know of any practical applications in which these matrices occur, we believe that they have interesting properties In particular, we hope to discover conditions on sequences {c n } n=1 which guarantee that the spectra of the family {L n } n=1 and/or the family {U n } n=1 have predictable distributions as n Theorems 5-8 provide a modest start in this direction In Section 2 we obtain explicit expressions for the determinants and inverses of L n and U n We also determine their inertias and diagonalize their quadratic forms In Section 3 we discuss the distribution of the eigenvalues of the matrices ( ) n K n (ρ, γ) = ρ r s + γρ r+s ) r,s=1 (which is of the form (2) with c r = 1+γρ 2r ), where 0 < ρ < 1 and γ is an arbitrary real number In Section 4 we discuss the distribution of the eigenvalues of L n = (min(r, s) γ) n r,s=1 (which is of the form (2) with c r = r γ), where γ 1/2

Generalized KMS Matrices 3 2 Properties of L n and U n Let A n be the n n matrix with 1 s on the diagonal, ρ s on the super diagonal, and zeros elsewhere; thus, 1 ρ 0 0 0 0 0 1 ρ 0 0 0 0 0 1 0 0 0 A n = 0 0 0 1 ρ 0 0 0 0 0 1 ρ 0 0 0 0 0 1 It is straightforward to verify that c 1 0 0 0 0 0 ρc 1 α 1 0 0 0 0 ρ 2 c 1 ρα 1 α 2 0 0 0 L n A n =, ρ n 3 c 1 ρ n 4 α 1 ρ n 5 α 2 α n 3 0 0 ρ n 2 c 1 ρ n 3 α 1 ρ n 4 α 2 ρα n 3 α n 2 0 ρ n 1 c 1 ρ n 2 α 1 ρ n 1 α 2 ρ 2 α n 3 ρα n 2 α n 1 where and that α i = c i+1 ρ 2 c i, i = 1,, n 1, A T nl n A n = diag(c 1, α 1, α 2,, α n 1 ) = diag(c 1, c 2 ρ 2 c 1, c 3 ρ 2 c 2,, c n ρ 2 c n 1 ) (3) It is also straightforward to verify that β 1 0 0 0 0 0 ρβ 2 β 2 0 0 0 0 ρ 2 β 3 ρβ 3 β 3 0 0 0 A n U n =, ρ n 3 β n 2 ρ n 4 β n 2 ρ n 5 β n 2 β n 2 0 0 ρ n 2 β n 1 ρ n 3 β n 1 ρ n 4 β n 1 ρβ n 1 β n 1 0 ρ n 1 c n ρ n 2 c n ρ n 1 c n ρ 2 c n ρc n c n where β i = c i ρ 2 c i+1, i = 1,, n 1,

4 William F Trench and that A n U n A T n = diag(β 1, β 2,, β n 1, c n ) = diag(c 1 ρ 2 c 2, c 2 ρ 2 c 3,, c n 1 ρ 2 c n, c n ) (4) and Since det(a n ) = 1, (3) and (4) imply that n 1 det(l n ) = c 1 (c i+1 ρ 2 c i ) (5) i=1 n 1 det(u n ) = c n (c i ρ 2 c i+1 ) (6) Note that (5) and (6) both reduce to (1) when c 1 = c 2 = = c n = 1 i=1 We will prove the following two theorems together Theorem 1 The inertia of L n is (m, z, p), where m, z, and p are the numbers of negative, zero, and positive elements in the set Moreover, r,s=1 {c 1, c 2 ρ 2 c 1, c 3 ρ 2 c 2,, c n ρ 2 c n 1 } 2 ρ r s c min(r,s) x r x s = c 1 ρ j 1 x j + (c i+1 ρ 2 c i ) i=2 j=i 2 ρ j i x j (7) Theorem 2 The inertia of U n is (m, z, p), where m, z, and p are the numbers of negative, zero, and positive elements in the set Moreover, r,s=1 {c 1 ρ 2 c 2, c 2 ρ 2 c 3,, c n 1 ρ 2 c n, c n } 2 n 1 i ρ r s c max(r,s) x r x s = (c i ρ 2 c i+1 ) ρ i j x j + c n i=1 2 ρ n j x j Proof: By Sylvester s theorem, (3) and (4) imply the statements concerning inertia From (3), L n = (A 1 n ) T diag(c 1, c 2 ρ 2 c 1, c 3 ρ 2 c 2,, c n ρ 2 c n 1 )A 1 n (9) (8)

Generalized KMS Matrices 5 From (4), U n = A 1 n diag(c 1 ρ 2 c 2, c 2 ρ 2 c 3,, c n 1 ρ 2 c n, c n )(A 1 n )T (10) Since A 1 n = (9) implies (7) and (10) implies (8) 1 ρ ρ 2 ρ n 3 ρ n 2 ρ n 1 0 1 ρ ρ n 4 ρ n 3 ρ n 2 0 0 1 ρ n 5 ρ n 4 ρ n 3 0 0 0 1 ρ ρ 2 0 0 0 0 1 ρ 0 0 0 0 0 1 Example 1 With ρ = 1 and and c r = r, (7) and (8) reduce to min(r, s)x r x s = r,s=1 i=1 j=i 2 x j, (11) and n 1 i max(r, s)x r x s = r,s=1 i=1 x j 2 + n These diagonalizations have recently been obtained by T Y Lam [5], who observed that (11) was previously stated in [2] Example 2 With c r = 1, (7) and (8) provide distinct diagonalizations of the quadratic form associated with K n (ρ): ρ r s x r x s = ρ j 1 x j r,s=1 r,s=1 i=1 2 i=2 j=i x j 2 + (1 ρ 2 ) ρ j i x j, 2 2 n 1 i ρ r s x r x s = (1 ρ 2 ) ρ i j x j + ρ n j x j Theorem 3 If det(l n ) 0 define σ i = 1 c i+1 ρ 2 c i, i = 1,, n 1 2

6 William F Trench Then L 1 n = (u rs ) n r,s=1 is the symmetric tridiagonal matrix with u 11 = 1/c 1 + ρ 2 σ 1, u nn = σ n 1, and For example, L 1 5 = u rr = σ r 1 + ρ 2 σ r, r = 2,, n 1, u r+1,r = u r,r+1 = ρσ r, r = 1,, n 1 1/c 1 + ρ 2 σ 1 ρσ 1 0 0 0 ρσ 1 σ 1 + ρ 2 σ 2 ρσ 2 0 0 0 ρσ 2 σ 2 + ρ 2 σ 3 ρσ 3 0 0 0 ρσ 3 σ 3 + ρ 2 σ 4 ρσ 4 0 0 0 ρσ 4 σ 4 Proof: From (9), L 1 n = A n diag(1/c 1, σ 1, σ 2,, σ n 1 )A T n, and routine manipulations verify the stated result Example 3 Let c r = 1 + γρ 2r, where γ is an arbitrary real number Then ρ r s c min(r,s) = ρ r s + γρ r+s We denote L n by K n (ρ, γ), since we will return to this matrix in Section 3; thus K n (ρ, γ) = (ρ r s + γρ r+s) n In this case c i+1 ρ 2 c i = 1 ρ 2, so (5) implies that Since Theorem 3 implies that K 1 n (ρ, γ) = 1 1 ρ 2 r,s=1 det(k n (ρ, γ)) = (1 + γρ 2 )(1 ρ 2 ) n 1 (12) σ i = 1,, i = 1,, n 1, 1 ρ2 1 + γρ 4 1 + γρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 0 0 0 0 0 0 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 ρ 0 0 0 0 ρ 1

Generalized KMS Matrices 7 if ρ ±1 and γρ 2 1 Example 4 If ρ = 1 and c r = r γ then L n = (min(r, s) γ) n r,s=1 Since σ i = 1, i = 1,, n 1, Theorem 3 implies that 2 γ 1 0 0 0 0 0 0 1 γ 1 2 1 0 0 0 0 0 0 1 2 1 0 0 0 0 L 1 n = 0 0 0 0 1 2 1 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 1 (13) if γ 1 Theorem 4 If det(u n ) 0 define Then U 1 n and For example, τ i = 1 c i ρ 2 c i+1, i = 1,, n 1 = (v rs ) n r,s=1 is the symmetric tridiagonal matrix with U 1 5 = Proof: From (10), v 11 = τ 1, v nn = ρ 2 τ n 1 + 1/c n, v rr = ρ 2 τ r 1 + τ r, r = 2,, n 1, v r+1,r = v r,r+1 = ρτ r, r = 1,, n 1 τ 1 ρτ 1 0 0 0 ρτ 1 ρ 2 τ 1 + τ 2 ρτ 2 0 0 0 ρτ 2 ρ 2 τ 2 + τ 3 ρτ 3 0 0 0 ρτ 3 ρ 2 τ 3 + τ 4 ρτ 4 0 0 0 ρτ 4 ρ 2 τ 4 + 1/c 5 U 1 n = A T n diag(τ 1, τ 2, τ 3,, τ n 1, 1/c n )A n,

8 William F Trench and routine manipulations verify the stated result Example 5 Let c r = 1 + γρ 2r, where γ is real Then U n = (ρ r s + γρ r s) n r,s=1 In this case c i ρ 2 c i+1 = 1 ρ 2, so (6) implies that det(u n ) = (1 + γρ 2n )(1 ρ 2 ) n 1 Since τ i = 1,, i = 1,, n 1, 1 ρ2 Theorem 4 implies that 1 ρ 0 0 0 0 ρ 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 0 0 0 Un 1 = 1 1 ρ 2 0 0 0 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 ρ ρ 2n + γρ 2 0 0 0 0 ρ ρ 2n + γ if ρ ±1 and γ ρ 2n Example 6 If ρ = 1 and c r = r γ then U n = (max(r, s) γ) n r,s=1 Since τ i = 1, i = 1,, n 1, Theorem 4 implies that U 1 n = 1 1 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 1 n + 1 + γ n γ if γ n

Generalized KMS Matrices 9 3 Spectral Properties of K n (ρ, γ) If 0 < ρ < 1 then n= ρ n e inθ = F(θ) = 1 ρ 2 1 2ρcosθ + ρ 2, (14) and it is known that the eigenvalues λ 1n < λ 2n < < λ nn of K n (ρ) = (ρ r s ) n r,s=1 are given by λ jn = F(φ n j+1,n ), where (j 1)π n + 1 < φ jn < jπ n + 1, j = 1, 2, n (See [6] for more on this) This illustrates a theorem of Szegö [1, Chapter 5] which implies that if {c r } r= are the Fourier coefficients of a bounded real-valued even function f L[ π, π] then the spectra of the symmetric Toeplitz matrices T n = (c r s ) n r,s=1, n = 1, 2,, are equally distributed in the sense of H Weyl [1, p 62] with values of f at n equally spaced points in [0, π], as n We will now obtain related results on the spectrum of K n (ρ, γ) as n, assuming that 0 < ρ < 1 We also assume temporarily that γρ 2 1, so K n (ρ, γ) is invertible We begin by considering the spectrum of 1 + γρ 4 ρ 0 0 0 0 1 + γρ 2 ρ 1 + ρ 2 ρ 0 0 0 V n = (1 ρ 2 )Kn 1 0 ρ 1 + ρ 2 0 0 0 (ρ, γ) = 0 0 0 1 + ρ 2 ρ 0 0 0 0 ρ 1 + ρ 2 ρ 0 0 0 0 ρ 1 It is straightforward to verify that if x 0, x 1,, x n, x n+1 (not all zero) satisfy and the boundary conditions ρx r 1 + [1 + ρ 2 µ]x r ρx r+1 = 0, 1 r n, (15) (1 + γρ 2 )x 0 = ρ(1 + γ)x 1 and x n+1 = ρx n, (16) then x = [x 1 x 2 x n ] T is a µ-eigenvector of V n The solutions of (15) are of the form x r = c 1 ζ r + c 2 ζ r, (17)

10 William F Trench where ζ and 1/ζ are the zeros of the reciprocal polynomial The boundary conditions (16) require that P(z) = ρz 2 + (1 + ρ 2 µ)z ρ (18) (1 + γρ 2 )(c 1 + c 2 ) = ρ(1 + γ)(c 1 ζ + c 2 /ζ) c 1 ζ n+1 + c 2 ζ n 1 = ρ(c 1 ζ n + c 2 ζ n ) (19) The determinant of this system is D n (ζ) = 1 + γρ2 ρ(1 + γ)ζ 1 + γρ 2 ρ(1 + γ)/ζ ζ n+1 (1 ρ/ζ) ζ n 1 (1 ρζ) = (1 + γρ 2 )(ζ n 1 ζ n+1 ) ρ(2 + γ(1 + ρ 2 ))(ζ n ζ n ) +ρ 2 (1 + γ)(ζ n+1 ζ n 1 ) With ζ = ±1, (19) has the nontrivial solution (1, 1), but (17) yields x r = 0 for all r Therefore the zeros ±1 of D n are not associated with eigenvalues of V n The remaining 2n zeros of D n occur in reciprocal pairs (ζ, 1/ζ) Corresponding to a given pair, x as defined in (17) is an eigenvector of V n, and therefore of K n (ρ, γ) To determine the eigenvalue µ of V n with which it is associated, we note that since (18) implies that Therefore P(z) = ρ(z ζ)(z 1/ζ) = ρ(z 2 (ζ + 1/ζ)z + 1), µ = 1 ρ(ζ + 1/ζ) + ρ 2 λ = G(ζ) = 1 ρ 2 1 ρ(ζ + 1/ζ) + ρ 2 is an eigenvalue of K n (ρ, γ) In particular, if ζ = e iθ then F(θ) (see (14)) is an eigenvalue of K n (ρ, γ) Theorem 5 Let ρ and γ be real numbers, with 0 < ρ < 1 Then: (a) K n (ρ, γ) has eigenvalues of the form F(θ jn ), j = 2,, n 1, where (j 1)π n (20) < θ jn < jπ, j = 2,, n 1 (21) n (b) If γ 1/ρ then K n (ρ, γ) has an eigenvalue of the form F(θ 1n ), where 0 < θ 1n < π n (c) If γ 1/ρ then K n (ρ, γ) has an eigenvalue of the form F(θ nn ), where (n 1)π n < θ nn < π

Generalized KMS Matrices 11 Proof: It suffices to prove (a) and (b) under the additional assumption that γ 1/ρ 2 (so that V n is defined), since the conclusions will then follow in the case where γ = 1/ρ 2 by a continuity argument We first isolate the zeros ζ = e iθ of D n with 0 < θ < π Define S n (θ) = (1 + γρ 2 sin(n + 1)θ ) ρ(2 + γ(1 + ρ 2 sin nθ )) sin θ sin θ + sin(n 1)θ ρ2 (1 + γ) sinθ on [0, π], where the definition at the endpoints is by continuity; then D n (e iθ ) = D n (e inθ ) = 0 if and only if S n (θ) = 0 and It is routine to verify that S n (0) = (1 ρ)[1 + ρ + n(1 ρ)(1 γρ)], (22) ( ) jπ S n = ( 1) j (1 ρ 2 ), j = 1,, n 1, (23) n S n (π) = ( 1) n (1 ρ 2 + n(1 + ρ) 2 (1 + γρ)) (24) From (23), S n changes sign on ((j 1)π/n, jπ/n), j = 2,, n 1 This implies (a) If γ 1/ρ then (22), and (23) with j = 1 imply that S n changes sign on (0, π/n) This implies (b) If γ 1/ρ then (23) with j = n 1 and (24) imply that S n changes sign on ((n 1)π/n, π) This implies (c) Now let λ 1n < λ 2n < < λ nn be the eigenvalues of K n (ρ, γ) Let α = 1 ρ 1 + ρ = min 1 + ρ F(θ) and β = 0 θ π 1 ρ = max F(θ), 0 θ π and define ( ) (2n 2j + 1)π χ jn = F 2n j = 1,, n (25) Theorem 6 Suppose that 0 < ρ < 1, γ 1/ρ, and H is continuous on [α, β] Then 1 lim H(λ jn ) H(χ jn ) = 0 (26) n n According to a definition given in [8], the sets {λ jn } n and {χ jn} n are absolutely equally distributed as n This is stronger than Weyl s definition of equally distributed as n, which does not require the absolute value signs in (26) The proof that we are about to give is similar to the proof of Theorem 4 in [7] We repeat the proof here because there were minor but potentially confusing errors in the enumeration of {λ jn } n and {χ jn} n in [7]

12 William F Trench Proof: Since F is decreasing, Theorem 5 implies that λ jn = F(θ n j+1,n ), j = 1,, n Therefore (21), (25), and the mean value theorem imply that where K = max 0 θ π F (θ) Let W n (H) = λ kn χ kn Kπ 2n, (27) H(λ kn ) H(χ kn ) k=1 If H is constant then W n (H) = 0 If N is a positive integer then (27) and the mean value theorem imply that λ N kn χ N kn Nβ N 1 λ kn χ kn NβN 1 Kπ, 2n so (26) holds if H is a polynomial Now suppose H is an arbitrary continuous function on [α, β] and let ɛ > 0 be given From the Weierstrass approximation theorem, there is a polynomial P such that H(u) P(u) < ɛ for all u in [α, β] Therefore W n (H) < W n (P) + 2nɛ, and lim sup n W n (H) n W n (P) lim + 2ɛ = 2ɛ n n Now let ɛ 0 to conclude that lim n W n (H)/n = 0 Theorem 7 Suppose that 0 < ρ < 1 and H is continuous on [α, β] Then: and and (a) If γ > 1/ρ then (b) If γ < 1/ρ then lim λ nn = (1 + γ)(1 + γρ2 ) n γ(1 ρ 2 ) (28) n 1 1 lim H(λ jn ) H(χ jn ) = 0 (29) n n lim λ 1n = (1 + γ)(1 + γρ2 ) n γ(1 ρ 2 ) 1 lim n n (30) H(λ jn ) H(χ jn ) = 0 (31) j=2

Generalized KMS Matrices 13 Proof: If γ = 1/ρ 2 then (12) implies that λ 1n = 0, which verifies (30) in this case Henceforth we assume that γ > 1/ρ, but γ 1/ρ 2 For all values of n and γ, Theorem 5 implies that at least n 1 eigenvalues of K n (ρ, γ) are values of F(θ) and therefore in (α, β) This and the fact that D n (1) = D n ( 1) = 0 account for at least 2n zeros of D n If γ > 1/ρ then S n (0) and S n (π/n) are both negative for n sufficiently large, while if γ < 1/ρ 2 then S n (π) and S n ((n 1)π/n) and S n (π) have the same sign for n sufficiently large Therefore, there is an N such that if n N then D n has exactly one pair (ζ n, 1/ζ n ) of zeros which are not on the unit circle Hence, ζ n is real, and we may assume without loss of generality that ζ n > 1 We denote the eigenvalue corresponding to ζ n by ν n ; thus, ν n = G(ζ n ) = 1 ρ 2 1 ρ(ζ n + 1/ζ n ) + ρ 2 (32) Since ζ n is not on the unit circle, ν n / [α, β] Therefore the Cauchy interlacement theorem implies that ν n = λ nn for all n N or ν n = λ 1n for every n N, and that ν n+1 > ν n Therefore (32) implies that ζ n+1 > ζ n Now it is convenient to rewrite (20) as D n (ζ) = ζ n+1 H(1/ζ) ζ n 1 H(ζ), (33) with where H(ζ) = (1 + γρ 2 )ζ 2 ρ(2 + γ(1 + ρ 2 )ζ + ρ 2 (1 + γ) = (1 + γρ 2 )(ζ ρ)(ζ ζ ), ζ = Since D n (ζ n ) = 0, (33) and (34) imply that ρ(1 + γ) 1 + γρ 2 (34) ζ n ζ = ζ 2n+2 n H(1/ζ n ) (1 + γρ 2 )(ζ n ρ) Since ζ n is increasing and greater than 1, this implies that lim n ζ n = ζ Therefore lim ν n = G(ζ ) = (1 + γ)(1 + γρ2 ) n γ(1 ρ 2 ) Since the quantity on the right is greater than β if γ > 1/ρ, or less than α if γ < 1/ρ, this implies (28) if γ > 1/ρ, or (30) if γ < 1/ρ Now Theorem 5 implies that if γ > 1/ρ then λ jn = F(θ n j+1,n ), j = 1,, n 1, while if γ < 1/ρ then λ jn = F(θ n j+1,n ), j = 2,, n, and arguments similar to the proof of Theorem 6 yield (29) and (31)

14 William F Trench 4 Spectral Properties of L n = (min(r, s) γ) n r,s=1 We now consider the spectrum of L n = (min(r, s) γ) n r,s=1 in the case where γ 1/2 We begin by considering the spectrum of L 1 n (see (13)) It is straightforward to verify that if x 0, x 1,, x n, x n+1 (not all zero) satisfy the difference equation and the boundary conditions x r 1 (2 µ)x r + x r+1 = 0, 1 r n, (35) (1 γ)x 0 + γx 1 = 0 and x n x n+1 = 0, (36) then x = [x 1 x 2 x n ] T satisfies L 1 n x = µx; therefore, µ is an eigenvalue of L 1 n if and only if (35) has a nontrivial solution satisfying (36), in which case x is µ- eigenvector of L 1 n The solutions of (35) are of the form x r = c 1 ζ r + c 2 ζ r, (37) where ζ and 1/ζ are the zeros of the reciprocal polynomial The boundary conditions (36) require that P(z) = z 2 (2 µ)z + 1 (38) (1 γ)(c 1 + c 2 ) + γ(c 1 ζ + c 2 /ζ) = 0 (c 1 ζ n + c 2 ζ n ) (c 1 ζ n+1 + c 2 ζ n 1 ) = 0 The determinant of this system is D n (ζ) = 1 γ + γζ 1 γ + γ/ζ ζ n ζ n+1 1/ζ n 1/ζ n+1 = ζ n 1 (ζ 1)[(1 γ)(ζ 2n+1 + 1) + γ(ζ 2n + ζ)] With ζ = 1, (39) has the nontrivial solution (1, 1), but (37) yields x r = 0 for all r Therefore ζ = 1 is not associated with an eigenvalue of L 1 n The remaining 2n zeros of D n occur in reciprocal pairs (ζ, 1/ζ) Corresponding to a given pair, x as defined in (37) is an eigenvector of L 1 n (and therefore of L n) To determine the eigenvalue µ of L 1 n with which it is associated, we note that since (38) implies that P(z) = (z ζ)(z 1/ζ) = z 2 (ζ + 1/ζ)z + 1, µ = ( 2 ζ 1 ) ζ (39) Therefore is an eigenvalue of L n λ = 1 2 ζ 1/ζ (40)

Generalized KMS Matrices 15 Theorem 8 If γ 1/2 then the eigenvalues λ 1n < λ 2n < < λ nn of L n = (min(r, s) γ) n r,s=1 are of the form where λ jn = 1 θ 4 csc2 n j+1,n, 2 2(j 1)π 2n + 1 < θ jn < 2jπ 2n + 1 Proof: It suffices to isolate the zeros ζ = e iθ of D n with 0 < θ < π Define C n (θ) = (1 γ)cos(n + 1/2)θ + γ cos(n 1/2)θ Then D n (e inθ ) = D n (e inθ ) = 0 if C n (θ) = 0 If γ 1/2 then S n changes sign on each interval ( ) 2(j 1)π I jn = 2n + 1, 2jπ, j = 1,, n 2n + 1 This implies that S n (θ jn ) = 0 for some θ jn in I jn From (40), (1/4)csc 2 (θ jn /2) is an eigenvalue of L n Since csc 2 (θ/2) is decreasing on (0, π), the conclusion follows References [1] U Grenander and G Szëgo, Toeplitz Forms and Their Applications, Univ of California Press, Berkeley and Los Angeles, 1958 [2] T A Hannula, T G Ralley, and I Reiner, Modular representation algebras, Bull Amer Math Soc 73 (1967), 100-101 [3] R A Horn and C R Johnson, Matrix Analysis, Cambridge University Press, 1991 [4] M Kac, W L Murdock, and G Szegö, On the eigenvalues of certain Hermitian forms, J Rat Mech and Anal 2 (1953), 787-800 [5] T Y Lam, On the diagonalization of quadratic forms, Math Mag 72 (1999), 231-235 [6] W F Trench, Numerical solution of the eigenvalue problem for symmetric rationally generated Toeplitz matrices, SIAM J Matrix Anal Appl 9 (1988), 291-303

16 William F Trench [7] W F Trench, Asymptotic distribution of the spectra of a class of generalized Kac-Murdock-Szegö matrices, Lin Alg Appl 294 (1999), 181-192 [8] W F Trench, Asymptotic distribution of the even and odd spectra of real symmetric Toeplitz matrices, Lin Alg Appl 302-303 (1999), 155-162