Linear Algebra and its Applications 306 (2000) 103–121 www.elsevier.com/locate/laa On almost regular tournament matrices Carolyn Eschenbach a , Frank Hall a , Rohan Hemasinha b , Stephen J. Kirkland c , Zhongshan Li a , Bryan L. Shader d , Jeffrey L. Stuart e , James R. Weaver b,∗ a Department of Mathematics and Statistics, Georgia State University, Atlanta, GA 30303, USA b Department of Mathematics and Statistics, University of West Florida, Pensacola, FL 32514-5751, USA c Department of Mathematics and Statistics, University of Regina, Regina, Saskatchewan, Canada S4S 0A2 d Department of Mathematics, University of Wyoming, Laramie, WY 82071-3036, USA e Department of Mathematics, University of Southern Mississippi, Hattiesburg, MI 39406-5045, USA Received 29 April 1999; accepted 22 October 1999 Submitted by R.A. Brualdi Abstract Spectral and determinantal properties of a special class Mn of 2n × 2n almost regular tournament matrices are studied. In particular, the maximum Perron value of the matrices in this class is determined and shown to be achieved by the Brualdi–Li matrix, which has been conjectured to have the largest Perron value among all tournament matrices of even order. We also establish some determinantal inequalities for matrices in Mn and describe the structure of their associated walk spaces. © 2000 Elsevier Science Inc. All rights reserved. AMS classification: 05C20; 05C50; 15A15; 15A18 Keywords: Almost regular tournament; Brualdi–Li conjecture; Determinant; Eigenvalues; Spectral radius; Tournament ∗ Corresponding author. E-mail addresses: ceschenbach@cs.gsu.edu (C. Eschenbach), fhall@cs.gsu.edu (F. Hall), rhemasin@uwf.edu (R. Hemasinha), kirkland@math.uregina.ca (S.J. Kirkland), zli@cs.gsu.edu (Z. Li), bshader@wyo.edu (B.L. Shader), 112358etc@msn.com (J.L. Stuart), jweaver@uwf.edu (J.R. Weaver). 0024-3795/00/$ - see front matter ( 2000 Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 9 9 ) 0 0 2 4 9 - 9 104 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 1. Introduction A tournament matrix T is a (0, 1)-matrix that satisfies T + T t = J − I , where J is the all ones matrix and I is the identity matrix. Thus T has zero diagonal and tij = 1 / j . The directed graph associated with a tournament if and only if tj i = 0 for each i = matrix is known as a tournament and this is a well-studied class of graphs (see [16]). There is a growing body of work on tournament matrices (see references). Given a tournament matrix T of order n, its score vector is given by Te where e is the all ones vector. The tournament matrix T is regular provided all of the entries in its score vector are the same, equivalently if each of the row sums of T is (n − 1)/2. (Observe that necessarily n must be odd.) It is known that for odd n, the regular tournament matrices maximize the Perron value over the class of n × n tournament matrices [1]. If n is even, an n × n tournament matrix is called almost regular provided half of its row sums are (n − 2)/2 and the other half are n/2. For even n, it is not known which n × n tournament matrices maximize the Perron value, but it is shown in [11] that for all sufficiently large n the maximizers are almost regular. In [3], Brualdi and Li conjecture that  t  Un/2 Un/2 , Bn = t +I Un/2 Un/2 where Uk is the k × k strictly upper triangular matrix with ones above the main diagonal, maximizes the Perron value over the class of tournament matrices of even order n. Inspired by this recent interest in almost regular tournament matrices we investigate some of their structural and algebraic properties in this paper. We focus particularly on the subclass Mn of almost regular tournament matrices of order 2n given by    T Tt : T is an n × n tournament matrix . Mn = Tt +I T Note that Mn contains the Brualdi–Li matrix B2n . For a tournament matrix T, we let   T Tt . MT = Tt +I T The following result summarizes some of the spectral results from [1,12] on tournament matrices employed in this paper. Proposition 1.1. Let T be an n × n tournament matrix with eigenvalue λ and corresponding eigenvector v. (i) −1/2 6 Re λ 6 (n − 1)/2; (ii) Re λ = −1/2 if and only if et v = 0, where e is the all ones vector. In this instance, T t v = λv and v is orthogonal to any eigenvector corresponding to another eigenvalue of T; C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 105 (iii) the walk space WT of T, defined by e, T e, . . . , T n−1 e , is the same as the subspace spanned by the eigenvectors and generalized eigenvectors of T corresponding to eigenvalues with real part strictly larger than −1/2; (iv) there is an orthogonal basis of eigenvectors of T that spans WT⊥ . Each eigenvector in WT⊥ corresponds to an eigenvalue whose real part is −1/2. Throughout the sequel we will refer to basic ideas and techniques from the theory of matrices. The reader is referred to [6,14] for further details. 2. Structure for almost regular tournament matrices Our first result provides a construction for almost regular tournament matrices with certain specified submatrices. Theorem 2.1. Let S and T be tournament matrices of order n. There is an almost regular tournament matrix of order 2n of the form   S X , A= J − Xt T where the first n row sums of A are equal to n − 1 and the last n row sums of A are equal to n. Proof. Let the score vectors of S and T be s and t, respectively. Evidently, we need only prove the existence of an n × n, (0, 1)-matrix X with row sum vector (n − 1)e − s and column sum vector t. According to, Corollary 6.2.3 of [4] such an X exists provided that for each pair of subsets α and β of {1, 2, . . . , n}, X X tj 6 |α| |β| . (n − 1 − si ) − i∈α j ∈β Since every  ℓ by ℓprincipal submatrix  of a tournament matrix has  X X |α| n − |β| and si > . tj > 2 2 i∈α Hence X i∈α ℓ
2 1’s, j ∈β (n − 1 − si ) − X j ∈β tj 6 |α|(n − 1) −  |α| 2  −  n − |β| 2  (|α| − n + |β|)2 + (|α| − n + |β|) . 2 Letting δ = |α| − n + |β|, which is an integer, we see δ 2 + δ > 0. Thus X X δ2 + δ 6 |α||β|.  tj 6 |α||β| − (n − 1 − si ) − 2 = |α||β| − i∈α j ∈β 106 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 The next two results give alternative descriptions of the walk space of an almost regular tournament matrix. We will use the notation n n z }| { z }| { u = [1, 1, . . . , 1| 0, 0, . . . , 0] t in our discussion. n n }| { z }| { z and ℓ = [0, 0, . . . , 0| 1, 1, . . . , 1] t Theorem 2.2. Let A be a 2n × 2n almost regular tournament matrix such that the first n rows of A have sum n − 1 and the last n rows have sum n. Then for each j > 0, e, ℓ, Aℓ, A2 l, . . . , Aj ℓ = e, Ae, A2 e, . . . , Aj +1 e . In particular, WA = e, ℓ, Aℓ, A2 ℓ, . . . , A2n−2 ℓ . Proof. Note that the case j = 0 follows upon observing that Ae = (n − 1) e + ℓ. For j > 1, Aj +2 e = (n − 1) Aj +1 e + Aj +1 ℓ, and hence Aj +2 e is a linear combination of Aj +1 e and Aj +1 ℓ, and Aj +1 ℓ is a linear combination of Aj +2 e and Aj +1 e. The result now follows by a simple inductive argument.  Corollary 2.3. If A is a 2n × 2n almost regular tournament matrix with n > 2, then WA = ℓ, Aℓ, A2 ℓ, . . . , A2n−1 ℓ . Proof. Let R = ℓ, Aℓ, A2 ℓ, . . . , A2n−1 ℓ . Then R is A-invariant by the Cayley– Hamilton theorem. Since WA is A-invariant, it follows from Theorem 2.2 that R ⊆ WA and that R = WA if e ∈ R. Suppose R = / WA . Since WA is spanned by a set of (generalized) eigenvectors of A, there must be a (generalized) eigenvector v in WA that is not in R. Among all such vectors, pick v to be of minimum height k. Let w = (A − λI )v. Then either w = 0 or else w is a (generalized) eigenvector of height k − 1. Thus w ∈ R. By Theorem 2.2, v = r + ae for some r ∈ R, and some nonzero scalar a. Premultiplying both sides of this equation by A − λI yields w = Ar − λr + aAe − aλe. Since Ae = (n − 1)e + ℓ, we have that w − Ar + λr − aℓ = a(n − 1 − λ)e. Because a = / 0, and λ < n − 1, we are led to the contradiction that e ∈ R. Therefore WA = R.  3. The Brualdi–Li conjecture and the class Mn In this section, we prove that the Brualdi–Li matrix, B2n , has the largest Perron value among the matrices in Mn . Theorem 3.1. Suppose T is an n × n tournament matrix with n > 2 and let x be a real number with 0 < x < 2/(n − 1). Then C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 107 (1 + x)n − 1 x with equality if and only if T is permutationally similar to Un . ent (I − xT )−1 en > Proof. Since the spectral radius of T is at most (n − 1)/2, and x < 2/(n − 1), (I − xT )−1 exists and is given by ∞ X xkT k. (I − xT )−1 = k=0 For k > 0, et T k e equals the number of walks of length k in the tournament associated with T. If k 6 n − 1, then since every tournament on k + 1 vertices has a path of length k [18], we find that   n . et T k e > k+1 It follows that   n−1 X n xk et (I − xT )−1 e > k+1 k=0 with equality only if T has no walks of length n or more. Simple algebra and the binomial theorem yield   n−1 X (x + 1)n − 1 n k = x . k+1 x k=0 Thus, we have established the inequality. Up to isomorphism, the only tournament on n vertices with no walks of length n
or more is the transitive one. Since the transitive n walks of length k for k 6 n − 1, equality holds tournament on n vertices has k+1 for this tournament. Hence equality holds if and only if T is permutationally similar to Un .  Theorem 3.2. Let T be an n by n tournament matrix. Let ρ be the Perron value of MT . Then !−1   ρ + 1 2n 2 +1 2ρ − 2 (n − 1) ρ − (n − 1) 6 ρ and equality holds if and only if T is permutationally similar to Un .   Proof. Let the Perron vector of MT be conformally partitioned as u v , where u and t t v are n × 1 vectors with e u + e v = 1. Then and T u + T t v = ρu (3.1) (T t + I )u + T v = ρv. (3.2) 108 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 Adding these equations, using et u + et v = 1, and then solving for ρu yields e − (ρ + 1)v = ρu. Pre-multiplying both sides of (3.3) by (3.3) et gives n − et v = ρ. (3.4) Next multiply both sides of (3.3) by ρ and use (3.4) to get ρ 2 v =(J − T )ρu + ρT v =(ρ(ρ − n + 1))e − T e + (2ρ + 1)T v. Thus, If ρ 2 (ρ 2 I − (2ρ + 1)T )v = ρ(ρ − n + 1)e − T e. (3.5) + 1)(n − 1)/2, then 2ρ 2 6 (2ρ − 2ρ(n − 1) − (n − 1) 6 0, and hence the desired inequality holds. Otherwise, ρ 2 I −
(2ρ + 1)T is a nonsingular M-matrix. Let w be the vector with ρ 2 I − (2ρ + 1)T w = e. By (3.5), v = ρ(ρ − n + 1)w − T w 1 e − ρ2 w = ρ(ρ − n + 1)w + 2ρ + 1 2ρ + 1 ρ = 2ρ + 1 (2ρ 2 − 2ρ(n − 1) − (n − 1))w + 2ρ 1+ 1 e. Pre-multiplying both sides of the above equation by et and using (3.4) implies that ρ n (2ρ 2 − 2ρ(n − 1) − (n − 1))et w + , n−ρ = 2ρ + 1 2ρ + 1 which upon simplifying becomes 2n − 2ρ − 1 = (2ρ 2 − 2ρ(n − 1) − (n − 1))etw. By Theorem 3.1, et w =et (ρ 2 I − (2ρ + 1)T )−1 e −1  2ρ + 1 1 t T e = 2e I − ρ ρ2 2n  ρ+1 −1 ρ . > 2ρ + 1 Now substitute in (3.6). Since 2ρ 2 − 2ρ(n − 1) − (n − 1) > 0 (2ρ + 1)(2n − 2ρ − 1) > (2ρ 2 − 2ρ(n − 1) − (n − 1))(x − 1), where x = ((ρ + 1)/ρ)2n . Simple algebra now shows that 1 > (2ρ 2 − 2ρ(n − 1) − (n − 1))(x + 1) (3.6) C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 109 Fig. 1. Graphs of f (r) and g(r). and the inequality follows. Note that equality implies equality in Theorem 3.1, and hence the T is permutationally similar to Un . It is shown in [9], that if T = Un , then equality holds.  Theorem 3.3. Let T be a tournament matrix of order n, let the Perron value of MT be ρ, and let ρ ∗ denote the Perron value of the Brualdi–Li matrix B2n . Then ρ 6 ρ ∗ , and equality holds if and only if MT is permutationally similar to B2n . Proof. From Theorem 3.2, we have 2 2ρ − 2 (n − 1) ρ − (n − 1) 6  ρ +1 ρ 2n +1 !−1 . If n = 2 or 3 the result is immediate, so suppose that n > 4. In [9] it is shown that !−1  ∗ 
ρ + 1 2n ∗ 2 ∗ − 2 (n − 1) ρ − (n − 1) = +1 . 2 ρ ρ∗ It remains to show that ρ 6 ρ ∗ . To see this, we will prove that !−1   r + 1 2n +1 r is concave down as a function of r for r > n − 1, from which it will follow that for such r, if !−1   r + 1 2n 2 +1 2r − 2 (n − 1) r − (n − 1) 6 r then necessarily r 6 ρ ∗ . Let f (r) = (((r + 1)/r)2n + 1)−1 and let g(r) = 2r 2 − 2 (n − 1) r − (n − 1). See Fig. 1. It can be shown that 110 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 # " 2n  4n−2 (r + 1)2n−2 2nr r + 1 − (2n + 1 + 2r) . f ′′ (r) =  3 (2n − 2r − 1) r r 2n + (r + 1)2n We are done if we can show   r + 1 2n < (2n + 1 + 2r) (2n − 2r − 1) r for r > n − 1. Observe that (2n − 2r − 1)((r + 1)/r)2n is decreasing in r, while (2n + 1 + 2r) is increasing. Now  2n   1 r + 1 2n < 1+ . (2n − 2r − 1) r n−1 Observe that (1 + 1/(n − 1))2n is decreasing in n. Since n > 4, we have (1 + 1/ (n − 1))2n 6 ( 43 )8 ⊜ 9.9887, while 2n + 1 + 2r > 4n − 1 > 15. Thus we see that f ′′ (r) < 0 for r > n − 1, so f (r) is concave down. We then can conclude that ρ 6 ρ ∗ . Further if ρ = ρ ∗ , then !−1   ρ + 1 2n 2 +1 , 2ρ − 2 (n − 1) ρ − (n − 1) = ρ so by Theorem 3.2, T is transitive. It now follows that MT is permutationally similar to B2n .  4. Other spectral properties for matrices in Mn In this section, we focus on the connection between the spectral properties of T and those of MT . Theorem 4.1. Let T be a regular tournament matrix of order n. Then (i) MT is diagonalizable, and (ii) if T has k distinct eigenvalues then MT has 2k distinct eigenvalues. Proof. (i) Observe that since T is regular, the walk space of MT has dimension 2. Thus by results in [8] MT has a positive eigenvalue ρ, a negative eigenvalue r, and 2n − 2 eigenvalues with real parts equal to −1/2. By Proposition 1.1 there is a collection of 2n − 2 orthogonal eigenvectors of MT corresponding to those eigenvalues with real part −1/2, it now follows that MT is diagonalizable. (ii) Since T is regular, it is normal and commutes with T t . Hence the characteris
tic polynomial of MT is given by det (zI − MT ) = det z2 I − (2z + 1) T . Consequently z is an eigenvalue of MT if and only if z2 = (2z + 1) λ for some eigenvalue λ of T. Suppose that z is an eigenvalue of MT corresponding to two eigenvalues λ1 and λ2 of T. Then z2 = (2z + 1) λ1 and z2 = (2z+1) λ2 , so that (2z+1) (λ1 −λ2 ) = 0. C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 111 Since the characteristic polynomial of MT is a monic polynomial with integer coefficients, the eigenvalues are algebraic integers. Since −1/2 is not an algebraic integer, z= / −1/2. So we deduce that λ1 = λ2 . Note that Re λ > −1/2 and since T is regu+λ= / 0. Thus each distinct eigenvalue λ of T corresponds to two lar, λ = / 0, so λ2 √ eigenvalues λ ± λ2 + λ of MT . Thus T has k distinct eigenvalues and MT has 2k distinct eigenvalues.  Theorem 4.2. Suppose T is a tournament matrix of order n. Then w is an eigenvector of MT corresponding to an eigenvalue γ with Re γ = −1/2 if and only if w has the form # "p λu √ , ± −λu where u is an eigenvector of T corresponding to the eigenvalue λ with Re λ = −1/2. p When this is the case, γ = λ ± −λλ. Proof. Suppose that T u = λu and that Re λ = −1/2. Then T t u = λu and it is readily verified that  √ λu √ ± −λu p is an eigenvector of MT corresponding to the eigenvalue λ ± −λλ. Conversely, suppose that w is an eigenvector of MT corresponding to an eigenvalue γ with Re γ = −1/2, and partition w conformally with MT as   w1 . w= w2 Then et w = et w1 + et w2 = 0. Now T w1 + T t w2 = γ w1 and (T t + I )w1 + T w2 = γ w2 , so that J w1 + (J − I ) w2 = γ (w1 + w2 ) . But J (w1 + w2 ) = 0 and Re γ = −1/2, so we find that γ w1 = γ w2 . In particular, both w1 and w2 are orthogonal to e. T w1 + (J − I − T ) w2 = γ w1 implies that w1 is an eigenvector of T, necessarily corresponding to an eigenvalue λ of T with Re λ = −1/2. Writing √ w1 = λu (where T u = λu), we find that w2 = xu for some scalar x. Using the eigenvalue–eigenvector relation we deduce that x 2 = −λ, and the results follows.  Corollary 4.3. Suppose T is a tournament matrix of order n. Then 2 dim WT⊥ = ⊥ , 2 dim W = dim W dim WM T MT and T WMT    u : u, v ∈ WT . = v 112 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 ⊥ follows from Theorem 4.2. We then Proof. The fact that 2 dim WT⊥ = dim WM T deduce that 2 dim WT = dim WMT and    u : u, v ∈ WT .  WMT = v 5. Determinants of tournament matrices in Mn We now investigate determinants for matrices in Mn . Theorem 5.1. Let T be a tournament matrix of order n.Then  n−1 n−1 J . det(MT ) = (−1) (2n − 1) det T − 2n − 1 (5.1) Proof. Note that      I O I I 2J − I J − I = . MT I I O I J T Taking determinants and using the Schur complement of the last matrix above with respect to 2J − I yields det(MT ) = det(2J − I ) det[T − J (2J − I )−1 (J − I )]. Observe that 1 n−1 J (J − I ) = J. J (2J − I )−1 (J − I ) = 2n − 1 2n − 1 Eq. (5.1) now follows upon observing that det(2J − I ) = (−1)n−1 (2n − 1).  Corollary 5.2. If T is an n × n tournament matrix with n > 1. Then det MT = (−1) (n − 1) det (T + I ) + (−1)n−1 n det(T ). Further det MT < 0 and hence |det MT | = (n − 1) det (T + I ) + (−1)n n det T . Proof. From Theorem 5.1, det MT = (−1) Note that n−1     n−1 J . (2n − 1) det T − 2n − 1  t11 − x  t21 − x det(T − xJ )=det   ...  tn1 − x t11 − x  t21 − t11 =det  ..  . tn1 − t11  t1n − x t2n − x  ..  .  t12 − x t22 − x .. . ··· ··· .. . tn2 − x · · · tnn − x  t1n − x t2n − t1n  . ..  . t12 − x t22 − t12 .. . ··· ··· .. . tn2 − t12 · · · tnn − t1n C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 113 Thus det (T − xJ ) is a linear function of x, say det(T − xJ ) = a + bx. Letting x = 0 yields a = det(T ), while letting x = 1 yields a + b = det(T − J ) = (−1)n det(T + I ). Solving for b, substituting in x = (n − 1)/(2n − 1), and simplifying now yields det MT = (−1) (n − 1) det (T + I ) + (−1)n−1 n det T . Lastly we claim that (n − 1) det (T + I ) > |n det(T )|, so that in fact |det MT | = (n − 1) det (T + I ) + (−1)n n det T . To see the claim, let the spectral radius of T be ρ, and suppose that its other eigenvalues are xj + iyj , 1 6 j 6 n − 1. Note that since xj > −1/2 for all j, any real eigenvalue of T + I is positive, so that det(T + I ) > 0. In order to prove strict inequality we first assume ρ = 0, then clearly, (n − 1) det(T + I ) > |n det(T )|. If ρ > 0, since xj > −1/2 for all j and ρ 6 (n − 1)/2 < n − 1, we have det(T + I )=(ρ + 1) n−1 Y j =1 |xj + iyj + 1| =(ρ + 1) n−1 Yq xj2 + yj2 + 2xj + 1 > (ρ + 1) n−1 Yq xj2 + yj2 j =1 j =1 n ρ +1 | det(T )| > | det(T )|, = ρ n−1 and the proof is complete.  The next result gives a general upper bound on |det MT | for MT ∈ Mn . Theorem 5.3. Let T be a tournament matrix of order n. Let s = T e be the score vector for T . Then |det(MT )| 6 1 (2n − 1)n−1 n Y [(n − 1)2 + n2 (n − 1) − (2n − 1)si ]1/2 (5.2) n Y [n(n − 1)2 + (2n − 1)si ]1/2 . (5.3) i=1 and |det(MT )| 6 1 (2n − 1)n−1 i=1 Proof. From Theorem 5.1,   n−1 |det(MT )| = (2n − 1) det T − J . 2n − 1 114 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 Observe that in the ith row of T − ((n − 1)/(2n − 1))J, there are si entries equal to n/(2n − 1) and n − si entries equal to (−n + 1)/(2n − 1). By Hadamard’s inequality we find that !1/2  Y  n n2 n−1 (n − 1)2 J 6 + (n − si ) si det T − 2n − 1 (2n − 1)2 (2n − 1)2 i=1 = n Y 1 [n(n − 1)2 + (2n − 1)si ]1/2 , (2n − 1)n i=1 which yields (5.3). Note that     n−1 n−1 J = det T t − J det T − 2n − 1 2n − 1 so arguing as above readily establishes (5.2) .  and T t e = (n − 1)e − s, Corollary 5.4. Let T be a tournament matrix of order n. Then  1  [(n − 1)(n2 − 21 )]n/2 (2n−1)n−1 | det(MT )| 6 1  [n6 − 2n5 + 2n3 − 72 n2 + n2 ]n/4 (2n−1)n−1 if n is odd, if n is even. Proof. Let s be the  vector of a tournament matrix of order n and consider the Q score quantity f (s) = ni=1 n (n − 1)2 + (2n − 1) si . We claim that if s has two entries which differ by 2 or more, then there is another score vector s̃ of an n × n tournament matrix such that f (s) < f (s̃) . To see the claim, suppose that we have si 6 sj + 2. / i, j, s˜i = si + 1, and s˜j = sj − 1. It follows Let s̃ be the vector with s˜l = sl for l = from Landau’s theorem [13] that s̃ is the score vector of some tournament matrix of order n. Further, since [n(n − 1)2 + (2n − 1)s̃i ][n(n − 1)2 + (2n − 1)s̃j ] −[n(n − 1)2 + (2n − 1)si ][n(n − 1)2 + (2n − 1)sj ] = (2n − 1)2 (sj − si − 1) > 0, it follows that f (s̃) > f (s) . Thus we see that f (s) is maximized over the class of score vectors of order n by a vector s for which any two entries differ by at most 1. Thus for n odd each entry in s must be (n − 1)/2, while for n even, half of the entries of s are n/2 and the rest are (n − 2)/2. The upper bounds on |det MT | are now established by computing f (s) for the cases that n is odd or even.  Remark 1. By examining the equality conditions for the Hadamard inequality it can be shown that the inequalities in Theorem 5.3 and Corollary 5.4 are strict. C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 115 The next few results deal with the determinant of a matrix in Mn arising from a regular tournament matrix T . Using part (ii) of Theorem 4.1 and setting z = 0 in the expression for det(zI − MT ) yields the following lemma. Lemma 5.5. Let T be a regular tournament matrix of odd order n > 1. Then det(MT ) = − det(T ). Theorem 5.6. Let T be a regular tournament matrix of odd order n > 1. Then |det(MT )| > |det(MC )| = (n − 1)/2, where C is the circulant tournament matrix of order n whose first row is n−1 2 n−1 2 z }| { z }| { [0, 1, . . . , 1, 0, . . . , 0] . Proof. Note that |det(MT )| = |det(T )|, and since T is a regular tournament of odd order n, its row and column sums are (n − 1)/2. Adding the last n − 1 columns of T to the first column of T, we have that  n−1 t12 2  n−1  2 |det(MT )| = det   .  .. t22 .. . tn2 n−1 2 = =   · · · t1n  .. . t2n   ..  .. . .  · · · tnn 1 t12  n−1 1 det  .  .. 2 1 t22 .. .  tn2 ··· .. . .. . t1n   t2n  ..  .  · · · tnn    n−1 n−1 |q| > 2 2 since q = / 0 is an integer. Using Lemma 5.5, |det(MC )| = |det(C)| . Let P be the circulant permutation matrix whose first row is [0 1 0 · · · 0]. Then C = P + P 2 + · · · + P ((n−1)/2) . Observe that (I + P (n−1)/2 )C = J − I. Since (I + P (n−1)/2 ) is permutationally similar to I + P it has determinant 2. Thus det[(I + P (n−1)/2 )C] = 2 det C = det[J − I ] = (−1)n−1 (n − 1). The result now follows.  Conjecture 5.7. We conjecture that for any matrix MT in the class Mn , |det MT | >   n . 2 116 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 Observe that for odd n, the matrix MC of Theorem 5.6 provides an example for which equality holds in this conjectured lower bound; for n even, it can be shown that   jnk Bnt B . = det t n Bn + I Bn 2 A tournament matrix H of order n is a Hadamard tournament matrix if it satisfies the equation H H t = ((n + 1)/4)I + ((n − 3)/4)J (necessarily n ≡ 3 mod 4). The existence of such matrices is a difficult unsolved problem since Hadamard tournaments of order n are coexistent with skew Hadamard matrices of order n + 1. Using the fact that the determinant of the n × n matrix aI + bJ is given by (a + nb) a n−1 , we obtain the following lemma. Lemma 5.8. If H is a Hadamard tournament matrix then det H = (n − 1) (n + 1)(n−1)/2 . 2n Proof. Note that  (n − 3) (n + 1) I+ J 4 4   n−1   (n − 1)2 (n + 1)n−1 n+1 n−3 (n + 1) +n . = = 4 4 4 22n (det H )2 = det(H H t) = det  The result follows.  Theorem 5.9. Suppose that T is an n × n regular tournament matrix with n > 3. Then |det MT | 6 (n − 1) (n + 1)(n−1)/2 , 2n with equality if and only if T is a Hadamard tournament matrix. Proof. By Lemma 5.5, det MT = −1 det T . Using [19, Theorem 5.2, Chapter 8],   n − 1 n − 1 n − 3 (n−1)/2 (n − 1) (n + 1)(n−1)/2 |det MT | = |det T | 6 − = . 2 2 4 2n Again appealing to [19, Theorem 5.2], equality holds if and only if T is an incidence matrix of a design, and hence T is a Hadamard tournament matrix.  C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 117 Remark 2. Let T be a tournament of order n > 1. Then since MT is a nonsingular (0, 1)-matrix, |det (MT )| > 1. The next two results deal with matrices in Mn whose associated T is singular. Theorem 5.10. Suppose that T is a singular n × n tournament matrix. Then |det MT | > n − 1, with equality if and only if MT is permutationally similar to the Brualdi–Li matrix. Proof. From Corollary 5.2, |det MT | = (n − 1) det (T + I ) . As argued in that Corollary, T + I is nonsingular with positive determinant. Further, since T + I is an integral matrix, we see that det(T + I ) > 1. This gives us the inequality on |det MT |. We deal with the case of equality by proving that if det(T + I ) = 1, then T is transitive (the converse being obvious) and MT is permutationally similar to the Brualdi–Li matrix. Suppose that T is singular but not transitive with eigenvalues ρ (the spectral radius), 0 of multiplicity k, and other nonzero eigenvalues xj + iyj , 1 6 i, j 6 n − k − 1. Then  det (T + I ) =(ρ + 1)   =(ρ + 1)   >ρ  n−k−1 Y  j =1 xj + 1 n−k−1 Y  xj2 j =1 +  1/2 + yj2  1/2  xj2 + 2xj + 1 + yj2  n−k−1 Y  j =1
2 yj2 1/2   . But this last term has the same absolute value as that of the coefficient of λk in the characteristic polynomial of T. In particular,  n−k−1 Y  ρ j =1 1/2  xj2 + yj2  is a positive integer, so it is at least 1. Hence det (T + I ) > 1 if T is not transitive.  Theorem 5.11. Suppose that T is a singular n × n tournament matrix. Then |det MT | 6 (n − 1) nn/2 , 2n−1 118 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 with √ equality holding if and only if the eigenvalues of T are (n − 2)/2, 0 and −1/2 ± (i n − 1)/2, the first two having multiplicity 1, and the last two each having multiplicity (n − 2)/2. Proof. From Corollary 5.2, | det MT | = (n − 1) det (T + I ). Let the eigenvalues of T be 0, ρ (the spectral radius), and xj + iyj , 1 6 j 6 n − 2. Since the trace of T and T 2 are both 0,  2 n−2 n−2 n−2 n−2 n−2 X X X X X xj2 . (5.4) xj  + xj2 =  yj2 = ρ 2 + xj and ρ=− j =1 j =1 j =1 j =1 j =1 Using the fact that each xj > −1/2, we have that ρ 6 (n − 2)/2. This and the arithmetic–geometric mean inequality imply that det(T + I )=(ρ + 1)  n−2 Yq j =1 1 + 2xj + xj2 + yj2 (n−2)/2 n−2 n 1 X (1 + 2xj + xj2 + yj2 ) . 6 2 n−2 (5.5) j =1 If ρ = 0, we’re done. Otherwise, ρ > 1. Observe that n−2 X (1 + 2xj + xj2 + yj2 ) j =1 =n−2+2 n−2 X =n−2+2 n−2 X j =1 j =1 xj + n(n − 2) = + 4 n(n − 2) + 4 n−2 X j =1 j =1 xj2 j =1 xj + 2 n−2 X = n−2 X n−2 X j =1 + n−2 X yj2 j =1  2 n−2 X xj  xj2 +  j =1 (xj + 1/2) 2xj + n−2 X k=1 (n − 4) xk − 2   (n − 4) . (xj + 1/2) 2xj − ρ − 2 Now 0 = trace(T ) = ρ + xj + X k =j / xk > 2xj − n−3 2 ! (5.6) C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 119 and ρ > 1. Hence, 2xj − ρ − (n − 3)/2 < −1 for each j , and by (5.6) we conclude that n−2 X n(n − 2) (1 + 2xj + xj2 + yj2 ) 6 . 4 j =1 The desired upper bound on | det MT | now readily follows from (5.5). Suppose that (n − 1) nn/2 . 2n−1 From the argument above |det MT | = nn/2 , 2n−1 ρ = (n − 2)/2, each xj is −1/2, and (from the case of equality for the arithmetic and geometric means) all the yj2 ’s are the same. The expressions for the eigenvalues √ of T now follow. Conversely, if T has only the eigenvalues (n − 2)/2, −1/2 ± (i n − 1)/2 and 0 with the specified multiplicities, then the equality holds.  det(T + I ) = Remark 3. From [12, Theorem 4], T has eigenvalues (n − 2)/2, 0, and −1/2 ± √ (i n − 1)/2 (the last pair with multiplicity (n − 2)/2) if and only if J − 2T is a skew–Hadamard matrix. Conjecture 5.12. For any matrix MT of order 2n in the class Mn , |det MT | 6 (n − 1) nn/2 . 2n−1 Remark 4. Observe that both the conjectured bound and the upper bound of Corollary 5.4 are asymptotic to nn/2+1 /2n−1 for large n. Lemma 5.13. If T is an n × n tournament matrix, then det(T + I ) 6 (n + 1)(n+1)/2 . 2n Proof. This follows from [19, Theorem 5.2].  Our last result confirms Conjecture 5.12 for the case that the associated T is reducible. Theorem 5.14. Suppose that T is an n × n reducible tournament matrix. Then |det MT | 6 (n − 1) nn/2 . 2n−1 120 C. Eschenbach et al. / Linear Algebra and its Applications 306 (2000) 103–121 Proof. Without loss of generality, write   0 T , T = 1 J T2 where T1 is k × k and where T2 is (n − k) × (n − k). If 1 6 k 6 2 or (n − 2) 6 k 6 (n − 1), then T is singular, and we are done by Theorem 5.11, so henceforth we take 3 6 k 6 n − 3. By an exhaustive search for the cases n = 3, 4, 5, 6, 7, and 8 (using the technique of [17] for generating these tournaments) the upper bound holds. Thus we can assume that n > 9. From the proof of Corollary 5.2, it follows that |det MT | 6 2 (n − 1) det (T + I ) , so we need only show that det (T + I ) 6 nn/2 /2n . 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