1 Tournament Matrices with Extremal Spectral Properties 1 Stephen J. Kirkland Department of Mathematics and Statistics University of Regina Regina, Sa

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1 Tournament Matrices with Extremal Spectral Properties Stephen J. Kirkland Department of Mathematics and Statistics University of Regina Regina, Saskatchewan, Canada SS OA and Bryan L. Shader Department of Mathematics University ofwyoming Laramie, WY 80 ABSTRACT For a tournament matrix M of order n, we dene its walk space, W M,tobe SpanfM j : j =0;...;n g where is the all ones vector. We show that the dimension of W M equals the number of eigenvalues of M whose real parts are greater than =. We then focus on tournament matrices whose walk space has particularly simple structure, and characterize them in terms of their spectra. Specically, wecharacterize those tournament matrices such that M j is an eigenvector of M for some j 0. We also characterize the tournament matrices M such that J n M is a skew{ Hadamard matrix. Throughout, we illustrate our results with examples.. INTRODUCTION A tournament of order n is a (loopless) digraph T with vertices,,..., n such that exactly one of (i; j) and (j; i) is an arc of T ( i<jn). A tournament matrix of order n is a (0; ){matrix M =[m ij ] of order n such that M + M T = J n I n () where J n is the all ones matrix of order n and I n is the identity matrix of order n. Thus, a tournament matrix M is simply the adjacency matrix of a tournament T. There is an extensive literature on tournaments (see the This paper was written while both authors where postdoctoral members of the Institute for Mathematics and its Applications at the University of Minnesota.

2 KIRKLAND AND SHADER bibliographies in [BR8] and [M8]) and a growing literature on eigenvalues of tournament matrices (see for example, [DGKMP9], [F9], [GKS], [KS90], [K9], [MP90], [S9]). A square matrix A is reducible provided there exists a permutation matrix P such that PAP T has the form A O A where A and A are square (nonvacuous) matrices. The matrix A is irreducible provided it is not reducible. Equivalently, A is irreducible if and only if the digraph associated with A is strongly connected. Since many properties of reducible matrices can be studied in terms of irreducible matrices, we will focus our attention on irreducible tournament matrices. We denote the complex vector space of n by column vectors by C n and the all ones vector by. Let M be a tournament matrix of order n with corresponding tournament T. The vector s = M is called the score vector of M. Clearly T s = n, and the ith entry of s is the outdegree of vertex i in the tournament T. More generally, for any integer k the ith entry of M k equals the numberofwalks in T of length k which start at vertex i. We will call the subspace of C n spanned by the vectors fm j : j =0;;...;n g the walk space ofm, and denote it by W M. The walk polynomial of M is the unique monic polynomial p() of smallest degree such that p(m) = 0. Since W M is invariant under multiplication by M, it easy to see that p() is the minimum polynomial of the linear transformation obtained by restricting M to W M. In Section we investigate basic properties of the walk space and the walk polynomial of tournament matrices. It is known [BG8] that if is an eigenvalue of a tournament matrix M of order n, then Re() n : Our rst theorem asserts that the dimension of the walk space of M and the degree of the walk polynomial of M both equal the number of eigenvalues of M with real part not equal to =. In addition, we show that the orthogonal complementofw M in C n is the space spanned by the eigenvectors of M corresponding to the eigenvalues whose real parts equal =. In Section we consider matrices whose walk space has special properties. The tournament T is regular of degree t, provided the outdegree of each of its vertices is t. The tournament matrix M is a regular tournament

3 EXTREMAL TOURNAMENT MATRICES matrix provided T is regular. If T is a regular tournament of degree t then n =t+ and the score vector of M equals t. Clearly, W M has dimension if and only if M is a regular tournament matrix. A consequence of the results in Section is that the walk space of M has dimension if and only if there exist constants c and d such that M = cs + d, that is, if and only if the number of walks of length two starting at any given vertex can be determined by the outdegree of that vertex. Another consequence is that if W M has dimension, then M has exactly two eigenvalues whose real parts exceed =. Tournament matrices with exactly two eigenvalues with real part larger than = arise as the class of tournament matrices for which equality holds in a certain bound on the spectral radius [K9]. We use properties of the walk space to answer a question raised in [K9]. A special type of tournament matrix whose walk space has an interesting structure is one for which there exists an integer k such that the number of walks of length k starting at any given vertex is a constant multiple of the number of walks of length k starting at that vertex. Theorem shows that such tournament matrices can be completely characterized in terms of their spectra. The tournament T is doubly regular of degree t provided any twovertices of T jointly dominate precisely t vertices. It is easy to see that if T is doubly regular, then T is regular with degree t +. Thus, T is doubly regular of degree t if and only if M satises MM T = tj n +(t+)i n where n =t+. In [RB] the existence of a doubly regular tournament with degree t is shown to be equivalent to the existence of a skew{hadamard matrix of order t + (that is, a (; ){matrix Q of order t + such that Q I n isaskew{symmetric matrix and Q T Q =(t+)i t+ ). In Theorem we show that if M is an irreducible singular tournament matrix of order n with 0 as a simple eigenvalue, then M has at least distinct eigenvalues with equality if and only if J n M is a skew{hadamard matrix.. THE WALK SPACE AND THE WALK POLYNOMIAL We begin by proving a few basic properties of eigenvalues of tournament matrices. The assertions (i), (ii), (iii) and (iv) in Lemma below have been proven in [BG8], [BG8], [MP90] and [DGKMP9], respectively. However, because of their simplicity, we give complete proofs of these assertions. Lemma. Let M be a tournament matrix of order n and let z be an eigenvector of M with corresponding eigenvalue. Then (i) Re() =with equality if and only if z =0.

4 KIRKLAND AND SHADER (ii) Re() (n )= with equality only if M is a regular tournament matrix. (iii) If Re() = =, then has geometric multiplicity. (iv) If Re() = =, then the geometric and algebraic multiplicity of coincide. (v) If Re() = =, then z v =0for every vector v W M. Proof. see that By pre{ and post{multiplying () by z and z, respectively, we (Re()+)z z =jz j : () Assertion (i) is now an immediate consequence of (), and assertion (ii) follows from () and the Cauchy{Schwartz inequality. Consider eigenvectors x and y corresponding to the eigenvalue. Then applying () to the vector z =(x )y (y )x,we see that either z =0or Re() = =. Thus if Re() = =, then x and y are multiples of each other, and hence it follows that (iii) holds. To prove (iv) assume that Re() = =, and suppose to the contrary that the geometric multiplicityofis less than its algebraic multiplicity. Then there exist nonzero vectors v and w such that Mw = w; Mv = v + w and w v =0: Pre{ and post{multiplying () by w and v, respectively, and then simplifying we obtain (w v)+w w+w v=(w )( T v) w v: Since w v = 0 and since (i) implies that w =0,wehavew w=0,a contradiction. Therefore, (iv) holds. To prove (v) we assume that Re() = =. We show by induction on k that z M k = 0 for k =0;...;n. We have already seen that z =0. Suppose k and that z M k = 0. Then z M k = z (J n I n M T )M k = z T M k z M k z M k : Hence z M k = 0, and it follows that (v) holds. Theorem. Let M be a tournament matrix of order n and let l be the number of eigenvalues, counting multiplicity, of M with real part equal to =. Then (i) n = dim W M + l,

5 EXTREMAL TOURNAMENT MATRICES 5 (ii) the orthogonal complement of W M is the space spanned by the eigenvectors of M corresponding to eigenvalues with real part equal to =, and (iii) the walk polynomial of M is p() = Y ( j ) kj where the product is over all eigenvalues j of M with Re( j ) = = and where k j is the algebraic multiplicity of j. Proof. Let,,..., c be the distinct eigenvalues of M with real part equal to =, and let,,..., d be the distinct eigenvalues of M with with real part not equal to =. Also, let k, k,..., k d be the algebraic multiplicity of the eigenvalues,,..., d, respectively. It follows from (iii) and (iv) of Lemma that the minimum polynomial of M on C n is m() =( )( ) ( c )( ) k ( ) k ( d ) kd : For each integer j with j d, let r j () = m() ( j ) : Then r j (M ) = O, and the rows of r(m) are either zero vectors or left eigenvectors of M corresponding to the eigenvalue j. By (i) of Lemma (applied to M T )wehave r(m) = O, and hence r j (M) is a right eigenvector of M corresponding to the eigenvalue j. It now follows from (iii) of Lemma that f(m j I n ) m q j (M) :0mk j g is a basis for the generalized eigenspace E j of M corresponding to j, where q j () = m() : ( j ) kj Therefore, d j= E j W M : Since C n is the direct sum of the generalized eigenspaces of M, and since the algebraic and geometric multiplicities of the eigenvalues with real part

6 KIRKLAND AND SHADER equal to = coincide, the dimension of d j= E j equals n l. It follows from (iv) and (v) of Lemma, that dim W M n l. Therefore d j= E j = W M ; and (i) and (ii) hold. Clearly, the minimum polynomial of M restricted to W M is ( ) k ( ) k ( d ) kd ; and hence (iii) holds. Since the characteristic equation of any tournament matrix has integer coecients, it follows that = is not an eigenvalue of a tournament matrix. Further a tournament matrix has real entries, so its nonreal eigenvalues occur in complex conjugate pairs. As a result, the number of eigenvalues of a tournament matrix M having real part equal to = iseven. In particular, Theorem implies that dim W M and n have the same parity. We now describe an iterative method for nding the walk polynomial of a tournament matrix whose validity follows from Theorem.5 and the discussion in Section of [F9]. Let M be a tournament matrix of order n with score vector s, and suppose that the dimension of W M is k. Let A be the skew{symmetric matrix ( )(M M T )=M J n+ I n. It is not dicult to see that W M is spanned by the vectors, A, A,..., A k. We construct an orthonormal basis e, e,..., e k of W M by dening fe ;e ;...;e k g to be the orthonormal set of vectors obtained by applying the Gram{Schmidt process to the vectors, A, A,...,A k. Using the skew{symmetry of A, it is easy to verify that (i) e = p n, (ii) if k then and (iii) if k then e = Ae kae k = s ( n ) q ; s T n(n ) s e j = Aej ((e j ) T Ae j )e j kae j ((e j ) T Ae j )e j k (j =;...;k): () Let = kae k and j = kae j + j e j k q for j =;...;k. Simple s calculations show that = T s n(n ) n, and that j = (e j ) T Ae j+ for j =;...;k. Moreover, with respect to the basis e ;...;e k, the

7 EXTREMAL TOURNAMENT MATRICES linear transformation corresponding to M restricted to W M is given by tridiagonal matrix cm = n k k k The characteristic polynomial of M c + I n can be computed recursively n by setting q 0 () =, q ()=, and q j+ = q j + j q j () for j =;...k, and the walk polynomial of M equals q k ( +=). In applying the above method, it was not necessary to assume that the the dimension of W M was known beforehand. Using the fact that the vectors e ;...;e k constructed above are an orthonormal basis of W M,itis not dicult to show that 0=Ae k (e k Ae k )e k ; and hence k + is the rst positive integer j for which the numerator of the quantity on the right hand side of () is the zero vector. To illustrate this method for calculating the walk polynomial, consider M = Then s =(;;;;) T. According to the above algorithm, e == p 5(; ; ; ; ) T and e == p ( ; 0; 0; 0; ) T. Since A = 5 : we haveae = p (; ; ; ; )T, and hence (e ) T Ae = p 0. It follows that e = p 0 (; ; ; ; ) T.Further, Ae = p (9; 0; 0; 0; 9)T, 0 from which we nd that Ae (e Ae )e =0; 5 ; 5 :

8 8 KIRKLAND AND SHADER p and hence W M has dimension. Now = p, and =(e ) T Ae = 9 5 p0. Also, we see that q 0 () =,q ()= 5= and q = ( )+ 0 ( 5 )= : Finally, a few simplications yield that the walk polynomial of M is q ( +=) = : Note that the eigenvalues of M are (approximately) :8, :9 p ::90i, and i, and that as expected the rst three of these are the roots of. Theorem helps to answer a question posed in [K9]. In that paper, it is shown that if s T s<(n +n(n ) )=, then A has a real positive eigenvalue such that n +p n +(n ) s T s=n () and each of its remaining eigenvalues satisfy p Re() n n +(n ) s T s=n : (5) In addition, M has an eigenvalue and another eigenvalue for which equality holds in () and (5) if and only if M has at least n eigenvalues with real part equal to =, that is, if and only if the dim W M. The question asks whether or not there exist tournament matrices for which equality holds in exactly one of () and (5). The following argument shows that there do not exist such tournament matrices. Assume that M is a tournament matrix of order n, and that is an eigenvalue of M with corresponding eigenvector z, for which equality holds in either () or (5). Let s be the score vector of M, and assume that s T s>n(n ) = (otherwise, M is regular, and the result follows easily). Let v = z ( T z n ) From (), we nd that s T z ( n )T s T n(n ) s! s ( n ) s T z =(n ) T z: () :

9 EXTREMAL TOURNAMENT MATRICES 9 Using () and (), a number of algebraic manipulations yield (Re()) (n )Re() v v = z z( n Re) s T n(n ) s n(n ) + st s n z z(re() + )(Im()) s T n(n ) s : () Since equality holds in either () or (5), the rst term in () is zero, and it follows that must be a real eigenvalue. Thus v = 0, and since z is a linear combination of and s, the dimension of W M equals. Hence, by Theorem, M has n eigenvalues with real part equal to =. Consequently, from Theorem of [K9], M must have real eigenvalues, and (one of which is) such that yields equality in (), and yields equality in ().. EXTREMAL TOURNAMENTS In this section we study tournament matrices whose walk spaces have simple structure. We begin by discussing how tournament matrices with walk space of dimension arise in the study of the relationship between the score vector and the Perron{vector of tournament matrices. Let M be an irreducible tournament matrix of order n with spectral radius and score vector s. It is known [S9] that if M is singular, then s T s n (n )=, and (n )=. In addition, if s T s = n (n )=, then M is singular if and only if Ms =((n )=)s. Thus, the singular tournament matrices whose score vector has smallest possible length each have awalk space of dimension, and for such tournament matrices the score vector is an eigenvector. Motivated by this relationship between singular tournament matrices and tournament matrices whose score vector is a Perron{vector, we consider tournament matrices M for which the Perron{ vector is a vector of the form M k. The following theorem shows that such tournament matrices are completely characterized by their spectra. Theorem. Let M be a tournament matrix of order n, and suppose that k. Then the following are equivalent: (i) M k is an eigenvector of M, but M k is not. (ii) There isaconstant such that M k M k is a nonzero vector in the nullspace ofm. (iii) The eigenvalues of M are (n k)=, 0 with multiplicity k, and n k eigenvalues with real part equal to =.

10 0 KIRKLAND AND SHADER (iv) The walk polynomial of M is p() = k ( (n k )=). Proof. First assume (i). Then M k+ = M k for some number. This implies that M (M k M k ) = 0. Since M k is not an eigenvector of M, M k M k is nonzero, and thus (i) implies (ii). Now assume (ii). Then the walk polynomial p() ofmdivides k+ k = k ( ). Hence by Theorem, is an eigenvalue of M with Re() = = only if =0or=. Since M k M k = 0,we conclude that 0 is an eigenvalue of M with algebraic multiplicity k, and since M k = 0we conclude that is an eigenvalue of M with algebraic multiplicity. Because M is a real matrix, the assumptions imply that is real. That (iii) holds is now a consequence of the facts that the real part of each eigenvalue of M is at least = and that the trace of M, which is 0, is the sum of the eigenvalues of M. That (iii) implies (iv) is an immediate consequence of Theorem. Finally, assume (iv) holds. Then M(M k n k M k ) = 0, and M k is an eigenvector of M. Suppose that M k is an eigenvector of M. Then by what we have already shown M would have an eigenvalue of the form (n l)= for some l<k. Since such an eigenvalue is not a root of the walk polynomial, we have obtained a contradiction. Hence (iv) implies (i). It is not dicult to construct examples of tournament matrices of the type characterized in Theorem. Assume that n and k are positive integers of opposite parity, and let R be a regular tournament matrix of order n k. Let S be the tournament matrix of order k having 0 s on and above the main diagonal, and s below the main diagonal, and let J n k;k be the n k by k matrix of all s. Then M = S O J n k;k R is a tournament matrix of order n such that (n k )= is a simple eigenvalue, 0 is an eigenvalue of multiplicity k and the remaining n k eigenvalues each have real part equal to =. Thus M is of the type described in Theorem. However, M is a reducible matrix, and it would be more interesting to construct examples which were irreducible. While we are unable to construct such examples for general values of k, wenow discuss a class of irreducible examples for the case k =. Let p and q be nonnegative integers with q = 0, such that p + divides q(q + ). Let a =p+ and b =q+, and let m = ab +. Then m is a+ an odd integer. Let R be a regular tournament matrix of order m, R a regular tournament matrix of order am, and let B ;B ;...;B a, be (0,){ matrices of order m each of whose row and column sums equal (m b)=.

11 EXTREMAL TOURNAMENT MATRICES (Note that m>bsince q = 0, and that m are odd). The matrix M = R B... B a J m B T. R J m Ba T b is even since both m and b 5 is a tournament matrix of order (a +)m, and since both R and R are irreducible and since B is neither O nor J m, it follows that M is irreducible. ab(b ) It is easy to see that the rst m rows of M each contain ones, and each of the remaining rows contain exactly b(ab+) ones. Further, one veries that M =( ab )M. Thus is not an eigenvector of M, but M is. It follows from Theorem that ab and 0 are eigenvalues of M of algebraic multiplicitly, and that the remaining (a + )m eigenvalues each have real part equal to =. We note that the construction for the special case p = 0 (and hence a = ) appears in [MP90] and [S9]. We conclude this section by considering tournament matrices with few distinct eigenvalues. As mentioned in the introduction, it is shown in [DGKMP] that every tournament matrix of order n has at least distinct eigenvalues, and there is a correspondence between skew{hadamard matrices of order n and tournament matrices of order n + with exactly distinct eigenvalues. The following theorem shows that there is also a correspondence between skew{hadamard matrices of order n and tournament matrices of order n which have 0 as a simple eigenvalue and have exactly other distinct eigenvalues. Theorem. Let M be an irreducible tournament matrix of order n such that 0 is an eigenvalue with algebraic multiplicity. Then M has at least distinct eigenvalues, and M has exactly distinct eigenvalues if and only if J n M is a skew{hadamard matrix. Proof. Because M is a tournament matrix each of the main diagonal entries of M is 0. Since the trace of M is the sum of the squares of the eigenvalues of M and since the spectral radius > 0ofMis an eigenvalue of M, it follows that M has at least one nonreal eigenvalue. Because M is a real matrix, is also an eigenvalue of M. Hence M has at least four distinct eigenvalues. Now assume that M has exactly distinct eigenvalues, say,0,=a+bi,, and. Since is not real, we may assume that b>0. Then and

12 KIRKLAND AND SHADER both have algebraic multiplicity (n )=. In particular, n is even. The trace of M equals zero and is the sum of the eigenvalues of M, and thus 0=+(n )a: (8) Similarly, since the trace of M equals zero, we have 0= +(n )(a b ): These equations imply that a = n and b = p n n : (9) Since M is a(0;){matrix, the minimum polynomial m() of Mhas integer coecients and has the form m() =( )( ) k ( ) k for some integer k. Since an irreducible polynomial over the rational numbers has distinct roots, the minimum polynomial of over the rationals is either ( ) or( )( )( ). It follows that is rational (in the latter case we see that the coecient of is a = (n )=n ), and hence that is rational). Since M isa(0;){matrix, its eigenvalues are algebraic integers, and we conclude that is an integer. Since is an algebraic integer, so is a = +, and it follows from (9) that (n )= divides. Since a =, (8) now implies that =(n )=, a = =, and b = p n =. Let B = J n M. Then clearly, B isa(; ){matrix, and B I n isaskew{symmetric matrix. By Theorem, C n is a direct sum of the M {invariant subspaces W M and W where W is the space spanned by the eigenvectors of M corresponding to and. It follows from (i) of Lemma that W is invariant under multiplication by B, and that the eigenvalues of B on W are and. It follows from () and (i) of Lemma that if x is an eigenvector of M corresponding to, respectively, then x is an eigvenvector of M T corresponding to the eigenvalue, respectively. This implies that there is a basis of W which consists of common eigenvectors of B T and B. We conclude that the action of B T B of W is multiplication by kk =n. Clearly, W M is invariant under multiplication by B, and under multiplicationbyb T =M+I n J n. It follows from (i) of Theorem that the walk space of M has dimension and is spanned by

13 ni n = BB T (0) (M +M+nI) = nj n +MJ n J n M T () EXTREMAL TOURNAMENT MATRICES and s. Further, by Theorem, we havems = n s. Consequently, on the invariant subspace W M, B acts as follows (B) = n s and (B)s = n(n ) (n )s: Since B T =M J n +I n,wehave B T = (n ) +s and B T s = n(n ) +(n)s: A simple computation now shows that the action of B T B on W M is multiplication by n. Therefore, B T B = ni n, and J n M isaskew{hadamard matrix. Now suppose that B = J n M isaskew{hadamard matrix. Then = nj n MJ n J n M T +MM T : Since M is a tournament matrix MM T = MJ n M M. Substituting this identity into (0) and simplifying we obtain = n( T )+(s T ) (s T ): Post{multiplying both sides of () by we have Ms+s+n=(n )+(n)s (n(n )); and hence Ms =((n )=)s. In particular, this implies that M is singular. Post{multiplying () by M and Ms =((n )=)s, wehave M(M +M+nI n )=(n )s T ss T : Similarly, M (M +M+nI n )= n ((n )s T ss T ):

14 KIRKLAND AND SHADER It follows that M (M ( n )I n)(m +M+nI n )=O; and hence that M has at most distinct eigenvalues. Since a singular tournament matrix of order at n has at least distinct eigenvalues, M has exactly distinct eigenvalues. We can construct examples of tournament matrices of the type described in Theorem as follows. Suppose H is a Hadamard tournament matrix of order n and n 8, and let M = It can be veried directly that J n M is a skew{hadamard matrix, so that (M T M + I n )(M M T + I n )=ni n : Note that M is a reducible tournament matrix. To construct an irreducible tournament matrix with the desired properties, x i n, and let M i be the matrix obtained from M by \switching" its ith row and column, that is, M i is the same as M except the ith row ofm i equals the ith column of M, and the ith column of M i equals the ith row ofm. By considering score vectors it is easy to show that any principal submatrix of H of order n is irreducible. It now follows that M i is also irreducible. Further, M T i M i = D(M T M)D, where D is the diagonal matrix with a in its ith diagonal entry and s in each of its remaining diagonal entries. Thus we have H 5 : (J n M i )(J n M T i ) = (M T i M i + I n )(M i M T I + I n ) = D(M T M + I n )DD(M M T + I n )D = nd = ni n : Hence M i is an irreducible tournament matrix and such that J M i is a skew{hadamard matrix. Evidently, any further \switches" performed on M i would also maintain the desired skew{hadamard property, although they may not maintain irreducibility.

15 EXTREMAL TOURNAMENT MATRICES 5 For example, let H = and note that H is a Hadamard tournament matrix of order. Then M = and M = : Now M is irreducible, as is the matrix cm = : obtained from M by switching its third row and column. Thus both J 8 c M and J 8 M are skew{hadamard matrices, but since their score vectors are dierent, M c and M are not permutationally equivalent. However, a straightforward exercise reveals that M c is permutationally equivalent tom T. Thus, while the tournaments T b and T corresponding to M c and M, respectively, are not isomorphic, T b is isomorphic to the complement oft in the complete graph K 8.

16 KIRKLAND AND SHADER REFERENCES [BR8] L.W. Beineke and K.B. Reid. Tournaments, Selected Topics in Graph Theory I (L.W. Beineke and R.J. Wilson, eds.), Academic Press, New York, (98). [BG8] A. Brauer and I.C. Gentry. On the characteristic roots of tournament matrices. Bull. Amer. Math. Soc., (98) :{5. [BG] A. Brauer and I.C. Gentry. Some remarks on tournament matrices. Lin. Alg. and Appls., (9) 5:{8. [DGKMP9] D. decaen, D.A. Gregory, S.J. Kirkland, J.S. Maybee, and N.J. Pullman. Algebraic multiplicity of the eigenvalues of a tournament matrix. Lin. Alg. and Appls., (99) 9: 9{ 9. [F9] S. Friedland. Eigenvalues of almost skew{symmetric matrices and tournament matrices. IMA Preprint Series, (99) 9. [GKS] D.A. Gregory, S.J. Kirkland and B.L. Shader. Pick s inequality and tournaments. Lin. Alg. and Appls., to appear. [KS90] G.S. Katzenberger and B.L. Shader. Singular tournament matrices. Congr. Numer., (990) :{80. [K9] S.J. Kirkland. Hypertournament matrices, score vectors and eigenvalues. Lin. and Multilin. Alg., (99) 0:{. [MP90] J.S. Maybee and N.J. Pullman. Tournament matrices and their generalizations I.Lin. and Mulitilin. Alg., (990) 8:5{ 0. [M8] J.W. Moon. Topics on Tournaments, Holt, Rinehart and Winston, New York, (98). [RB] K. B. Reid, and E. Brown. Doubly regular tournaments are equivalent toskew{hadamard matrices. J. Combin. Theory Ser. A, (9) :{8. [R88] P. Rowlinson. On {cycles and 5{cycles in regular tournaments. Bull. London Math. Soc., (988) 8:5{9. [S9] B.L. Shader. On tournament matrices. Lin. Alg. and Appls., (99) {:5{8.