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Extremal digraphs with given clique number
Linear Algebra and its Applications 439 (2013) 328–345
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Extremal digraphs with given clique number S.W. Drury a,∗ , Huiqiu Lin b a
Department of Mathematics and Statistics, McGill University, Burnside Hall, 805 Ouest, rue Sherbrooke Street West, Montréal, Québec, Canada H3A 2K6
b
Department of Mathematics, East China Normal University, Shanghai 200062, China
ARTICLE INFO
ABSTRACT
Article history: Received 21 August 2012 Accepted 19 March 2013 Available online 12 April 2013
We classify those digraphs with a given number of vertices and a given clique number that maximize the Perron root of the adjacency matrix. © 2013 Elsevier Inc. All rights reserved.
Submitted by R.A. Brualdi AMS classification: 05C20 05C50 15A15 Keywords: Digraph Clique number Tournament Characteristic polynomial Perron root
1. Introduction In this paper, we deal with simple graphs and simple digraphs without loops. The clique number of a graph H is the largest integer d such that the complete graph on d vertices is a subgraph of H. The clique number of a digraph G is the largest integer d such that the complete digraph on d vertices is a subgraph of G. − → A tournament is a digraph in which for any two distinct vertices u and v, exactly one of uv and − → vu is an arc. The adjacency matrix of a tournament is called a tournament matrix. A tournament on n vertices that maximizes the Perron root of its adjacency matrix among all such tournaments will be ∗ Corresponding author. E-mail address: [email protected] (S.W. Drury) 0024-3795/$ - see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.03.024
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
329
called extremal. It has long been known that if n is odd, the extremal tournaments are precisely the ones that are regular, i.e. have indegree and outdegree (n − 1)/2 at each vertex. For n even, the extremal tournaments are those which are isomorphic to the Brualdi–Li tournament. This was conjectured by Brualdi and Li [4] and proved by Drury [5, Theorem 1]. In the case n odd, it is part of the folklore of the subject that for n = 1, 3, 5, there is only one extremal tournament up to isomorphism, but for n 7, there may be many. For an excellent survey of spectral estimates for digraphs, see Brualdi [3]. For n and d fixed, let = nd and r = n − d. Let Vj , j = 1, . . . , d be disjoint vertex sets where |Vj | = + 1 for j = 1, . . . , r and |Vj | = for j = r + 1, . . . , d. Let G0 be a digraph with vertex d set Vj with all possible arcs between Vj and Vk for j = k and the induced subdigraph G0 [Vj ] an j=1
extremal tournament for j = 1, . . . , d. If n > 6d, then two such G0 need not be isomorphic, but the Perron roots of their adjacency matrices are known to be the same. Such a digraph G0 has n vertices and clique number d. The primary objective of this article is to solve Lin et al. [8, Problem 2] by establishing the following theorem. Theorem 1. Let 1 d n. If G is a digraph with n vertices and clique number d, then ρ(G) Unless n = 2 and d = 1 equality occurs only if G is isomorphic to some G0 . 1
ρ(G0 ).
Note that if d = 0, G has no vertices and that the clique number of G cannot exceed the number of vertices, so the theorem does not make sense unless 1 d n. We use I to denote the identity matrix and J to denote the matrix in which every entry is 1. The shape of these matrices should be deduced from the context. We denote by 1 the vector of ones. For any graph or digraph, we will use the same notation for its adjacency matrix in the hope that no confusion will arise. We are interested in maximizing the Perron root of the adjacency matrix of a digraph G with n vertices and clique number d. In view of [7, Corollary 8.1.19], we may always assume that for any two − → − → distinct vertices u and v in G, either uv or vu is an arc in G. This is because if there are no arcs between − → − → u and v, then adding just one of the arcs uv or vu does not decrease the Perron root (and in many circumstances increases it) without increasing the clique number. Let us assume that this has been done. Then we define the associated graph H of G to be an undirected − → − → graph in which uv is an edge if and only if both uv and vu are arcs in G. Clearly, H has the same clique number as G. A theorem of Nikiforov [10, Theorem 1], states that if H is a graph with n vertices and clique number d then
ρ(H ) ρ(H0 ) where H0 is the d-partite Turán graph on n vertices. Furthermore, equality obtains only when H is isomorphic to H0 . Restating this more aggressively, if H is not isomorphic to H0 then we have ρ(H ) ν(n, d) < ρ(H0 ). Here ν(n, d) is the best possible upper bound for max{ρ(H ); H has n vertices, clique number d and H obtainable by Nikiforov’s method. 2 Since G + G II] now gives
ρ(G)
1 2
(ρ(H ) + n − 1)
1 2
∼ = H0 }
= H + J − I, a theorem of Bendixson [1, Théorème
(ν(n, d) + n − 1)
(1)
1 At issue here is that there are two digraphs with n = 2 and d = 1, one with no arcs and one with a single arc. Both have zero Perron root. d 2 In fact ν(n, d) is the largest real root of the equation det(λI − H0 ) + (j − 1)λn−j = 0, cf. Corollary 13 below, but using
the full strength of Zykov’s result.
j =2
330
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
and if it should be that 1 2
(ν(n, d) + n − 1) < ρ(G0 )
(2)
then G cannot have a maximal Perron root. Turned around this says that if G does have a maximal Perron root, then its associated H is isomorphic to H0 , a huge step in the right direction. In most cases (i.e. unless r = 0 and odd), ρ(G0 ) < 12 (ρ(H0 ) + n − 1), so establishing (2) or something similar requires delicate estimates. Unfortunately the inequality (2) fails for general n and d (for example in the case n = 15 and d = 8, 12 (ν(15, 8)+15−1) ≈ 13·517876 and ρ(G0 ) ≈ 13·517834). Our strategy is to build a theory parallel to that of Nikiforov that estimates ρ(H + J ) directly. In the second section, we establish formulæ relating to Schur complements. In the third section we establish the results that are needed to complete the proof of Theorem 1 once it is known that the associated graph is H0 . In the fourth section we estimate ρ(H + J ). In the remainder of the paper we make the necessary delicate estimates to complete the proof of Theorem 1. 2. Schur complement identities Proposition 2. Let A be a p × p real matrix, B a q × q real matrix and in this section, let J denote the p × q matrix in which every entry is one. Then ⎛
−J B
⎛
1 Adj ⎝
−J
A
det ⎝
⎠ = det(A)det(B) − (1 Adj(A)1)(1 Adj(B)1)
(3)
⎞
−J
A
⎞
⎠ 1 = det(A)(1 Adj(B)1) + (1 Adj(A)1)det(B) + 2(1 Adj(A)1)(1 Adj(B)1) (4)
−J B
and in all cases where it makes sense ⎛ 1
−J
⎝ A
−J B
⎞−1 ⎠
1
=
1 A−1 1 + 1 B−1 1 + 2(1 A−1 1)(1 B−1 1)
1 − (1 A−1 1)(1 B−1 1)
.
(5)
Proof. We see that (5) follows from (3) and (4). Next, we observe that (3) and (4) are polynomial identities. It therefore suffices to prove them when A−1 and B−1 exist and have sufficiently small norm. Now ⎞⎛ ⎛ ⎞ ⎛ ⎞ Ip A −J A − JB−1 J −J 0 ⎠⎝ ⎝ ⎠=⎝ ⎠ 0 B −J B −B−1 J Iq so that
⎛
det ⎝
A
−J
−J B
⎞ ⎠
since A−1 JB−1 J has rank one. Thus (3) follows. We also have ⎛ ⎝
A
−J
−J B
⎞−1 ⎠
= det(A)det(B)det(I − A−1 JB−1 J ) = det(A)det(B) 1 − (1 A−1 1)(1 B−1 1)
⎛
=⎝
(A − JB−1 J )−1
A−1 J (B − J A−1 J )−1
B−1 J (A − JB−1 J )−1
(B − J A−1 J )−1
⎞ ⎠.
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331
Now observe that ⎛
1 (A − JB
−1 −1
J
)
−1
1=1A
(I − JB−1 J A−1 )−1 1 = 1 A−1 ⎝
= (1 A−1 1)
∞ k=0
∞ k=0
⎞
(JB−1 J A−1 )k ⎠ 1
−1
(1 A−1 1)k (1 B−1 1)k = (1 A−1 1) 1 − (1 A−1 1)(1 B−1 1)
Together with similar calculations for the other pieces of the left hand side of (5), this yields (4) using (3). 3. Estimates on tournaments The objective of this section is the following theorem. Theorem 3. For T an n × n tournament matrix, T0 an extremal n m × m nonnegative matrix, let ⎞
⎛ M0
=⎝
Then ρ(M0 )
T0 J J H
⎞
⎛
⎠ and M
=⎝
× n tournament matrix and H be an
T J J H
⎠.
ρ(M ). In case of equality, T is extremal.
The n × n Pick matrix W is the skew symmetric matrix with ones above the main diagonal and minus ones below the main diagonal. The reader may refer to Gregory et al. [6] for information about this matrix and its significance to tournaments. It is well known that det(W ) = 0 if n is odd and det(W ) = 1 if n is even. The following theorem [5, Theorem 6] which is implicitly proved in [5] will be needed. Theorem 4. Let ν > 1 be fixed and let Q denote the set of all skew-symmetric real matrices Q with entries bounded above in absolute value by 1. Let Z = Q + ν J a positive matrix. Then the minimum possible value of ρ(Z ) as Q runs over Q is attained when Q = W and any Q which gives rise to a minimum value of ρ(Z ) is necessarily permutationally similar to W . Lemma 5. Let n
γ =μ
4 be even, 0 < μ < 1 and let
(1 + μ)n + (1 − μ)n > μ. (1 + μ)n − (1 − μ)n
Then
1 (I
− μW )−1 1 = γ −1
and ρ(μW
(6)
+ γ J ) = 1.
Proof. The identity (6) is obtained by writing
1 (I
− μW )−1 1 =
1 Adj(I
− μW )1 μ|Y |−1 det(W Y) = Y |X | det(I − μW ) μ det ( W X) X
.
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
where in the denominator the sum is over all subsets X of {1, 2, . . . , n} and in the numerator, the determinants are of bordered matrices, ⎛ W Y
=
⎝
1
0
−1 WY
⎞ ⎠
the sum extending again over all subsets Y of {1, 2, . . . , n}. We note that the matrices WX and W Y are themselves Pick matrices so that their determinants depend on whether X and Y have odd or even cardinality. We leave the details to the reader. Let v be the positive vector with entries vk
=
1−μ
k
1+μ
.
Calculations show that (μW μ < 1.
+ γ J )v = v and the result follows from [7, Corollary 8.1.30] since 0 <
The adjusted tournament matrix A of a tournament has the form A = 2T + I where T is the adjacency matrix. It was introduced in [5] in order to access the corresponding skew tournament matrix S = A − J = T − T and also to simplify calculations involving characteristic polynomials. We denote by A0 an extremal adjusted tournament matrix of the same size, S0 = A0 − J and μ0 = ρ(A0 )−1 . It is well known that μ0 (n − 1)−1 , so that for n 4
μ0 cot
π 2n
2n
(n − 1)π
8 3π
< 1.
(7)
Proposition 6. Let n 4 be even. Let Q be an n × n skew tournament matrix. Let u be a vector with all its entries greater than or equal to 1. Then
1 (I
− μW )−1 1 u (I − μQ )−1 u
for all μ with 0
(8)
< μ < μ0 and if equality obtains for any such μ then Q is permutationally similar to W .
Proof. By Lemma 5, we need to show that
γ −1 u (I − μQ )−1 u which is equivalent to det(I
− μQ − γ u ⊗ u ) 0.
(9)
Now by [7, Corollary 8.1.19], Theorem 4 and Lemma 5
ρ(μQ + γ u ⊗ u ) ρ(μQ + γ J ) ρ(μW + γ J ) = 1
(10)
with equality only if Q is permutationally similar to W . At this point the proof splits into two cases.
• μQ + γ u ⊗ u has two or more real characteristic roots counted according to multiplicity that are greater than or equal to 1. A theorem of Bendixson [1, Théorème II] asserts that all the characteristic roots of μQ + γ u ⊗ u have positive real part. So if there were two or more such roots, then we
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
would necessarily have tr(μQ + γ u ⊗ u ) 2. But this leads to γ u 2 estimate [11, Inequality(17)] or [6, Theorem 1.1] and (7) u (I
− μQ )−1 u
u 2
1 + μ2 cot
π
2
2n
2γ −1 2 π 1 + μ2 cot 2n
333
2. But then, using Pick’s
> γ −1 .
• μQ + γ u ⊗ u has one or fewer real characteristic roots counted according to multiplicity that are greater than or equal to 1. Then by (10) there is exactly one such root. It follows that (9) holds by considering the sign of the characteristic polynomial. Finally, if γ −1 = u (I − μQ )−1 u, then ρ(μQ to W by Theorem 4.
+ γ J ) = 1 and Q is necessarily permutationally similar
For n even, we use the methods in [5]. Then s = S 1 is a vector with odd integer entries summing to zero. Let D be the diagonal matrix with entries sgn(sj ), for j = 1, . . . , n and let Q = DSD. Then Q is again a skew-symmetric tournament matrix. Let uj = |sj | 1 define a vector u with entries greater than or equal to unity. If this is done for S0 instead of S we find that the corresponding Q0 is permutationally similar to the Pick matrix W and the corresponding u vector is 1. Proposition 7. For S an n × n skew tournament matrix and S0 an extremal n × n skew tournament matrix, the inequality
1 (I
− μS0 )−1 1 1 (I − μS)−1 1
holds for all μ with 0
(11)
< μ < μ0 . Further if for any such μ equality obtains in (11) then S is extremal.
Proof. The proof depends on whether n is odd or even.
• Case n is odd. Since S0 1 = 0, we note that 1 (I − μS0 )−1 1 = n. On the other hand, we have
1 (I
•
− μS)−1 1 =
∞ k=0
μk 1 Sk 1 =
∞ k=0
μ2k 1 S2k 1 = n − μ2 s (I − μ2 S2 )−1 s
(12)
where s = S 1. But the operator (I − μ2 S 2 )−1 is strictly positive definite so that (11) holds with equality only if s = 0 which is the regular case. Case n is even. Using (12), we see that the desired inequality is equivalent to
(S0 1) (I − μS0 )−1 (S0 1) s (I − μS)−1 s. But this is identical to
1 (I
− μQ0 )−1 1 u (I − μQ )−1 u
(13)
since S0 1 has ±1 entries. Finally, Q0 is permutationally similar to W so (13) follows by Proposition 6. Clearly if equality obtains in (11) it also obtains in (8). This forces Q to be permutationally similar to W . So, after applying a permutational similarity to S, we can assume that Q = W . But then Ws = Qs = DSDs = DS 1 = Ds = 1. But W is invertible so s (and hence D) is uniquely determined. We can then compute S = DQD = DWD exactly and show that it is permutationally equivalent to S0 . We leave this chore as an exercise for the reader.
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Lemma 8. Let n 3 and let A be an n × n adjusted tournament matrix with related skew tournament matrix S = A − J. Then
1 (I
for 0
− μA)−1 1 = μ−1 ϕ μ1 (I − μS)−1 1
(14)
< μ < μ0 where ϕ(x) = x/(1 − x).
Proof. We have
1 (I
− μA)−1 1 = 1 (I − μS − μJ )−1 1
−1
= 1 (I − μS)(I − μ(I − μS)−1 J ) ⎛
=1 =
⎝
∞ k=0
∞ k=0
−1
μ (I − μS) k
k
J)
k+1
μk 1 (I − μS)−1 1
= μ−1 ϕ μ1 (I − μS)−1 1 for |μ| sufficiently small. We note that for 0 0
< μ1 (I − μS)−1 1 = 1 −
det(I det(I
1
⎞
−1 ⎠
(I − μS)
1
(15)
< μ < μ0 , (I − μA)−1 exists and also
− μA) < 1. − μS)
(16)
See [5, Section 2, first display] for a proof of the equality in (16). Thus (14) extends to the range claimed since both sides are rational functions of μ. Theorem 9. For T an n × n tournament matrix and T0 an extremal n × n tournament matrix, the inequality 1
(λI − T0 )−1 1 1 (λI − T )−1 1
holds for all λ
(17)
> ρ(T0 ). Further if for any λ > ρ(T0 ) equality obtains in (17) then T is extremal.
Proof. If n = 1 or n = 2, then there is nothing to prove, so we may assume that n 3. Tournament and adjusted tournament matrices are related by A = I + 2T, therefore, (17) becomes 1
for λ
((2λ + 1)I − A0 )−1 1 1 ((2λ + 1)I − A)−1 1
> ρ(T0 ) and this is clearly equivalent to
1 (I
− μA0 )−1 1 > 1 (I − μA)−1 1
for all μ with 0 < μ < μ0 = ρ(A0 )−1 . But by Lemma 8 and the fact that ϕ is strictly increasing on 0 < x < 1, this is equivalent to (11). It is also clear that if equality obtains in (17) then it also obtains in (11). Proof of Theorem 3. By (3) we have det(λI
− M ) = det(λI − T ) det(λI − H ) 1 − (1 (λI − T )−1 1)(1 (λI − H )−1 1) .
(18)
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
This identity holds pointwise for λ functions. For λ > ρ(T ) we can expand 1
(λI − T )−1 1 =
∞ k=0
335
> max(ρ(T ), ρ(H )) and also in the sense of equality of rational
λ−k−1 1 T k 1
and it follows that 1 (λI − T )−1 1 is positive and decreasing on this interval. Similarly we have, 1 (λI − H )−1 1 is positive and decreasing on λ > ρ(H ) and it follows that
(1 (λI − T )−1 1)(1 (λI − H )−1 1) is positive and decreasing on λ 1
= (1 (λI − T )−1 1)(1 (λI − H )−1 1)
has exactly one solution in λ Similarly 1
> max(ρ(T ), ρ(H )). Thus, the equation
> max(ρ(T ), ρ(H )) and this solution is ρ(M ).
= (1 (λI − T0 )−1 1)(1 (λI − H )−1 1)
has exactly one solution in λ
(19)
(20)
> max(ρ(T0 ), ρ(H )) and this solution is ρ(M0 ). If ρ(M ) > ρ(M0 ), then
(1 (ρ(M0 )I − T )−1 1)(1 (ρ(M0 )I − H )−1 1) > (1 (ρ(M )I − T )−1 1)(1 (ρ(M )I − H )−1 1) =1 = (1 (ρ(M0 )I − T0 )−1 1)(1 (ρ(M0 )I − H )−1 1) leading to 1 (ρ(M0 )I − T )−1 1) > (1 (ρ(M0 )I − T0 )−1 1) contradicting Theorem 9. Hence ρ(M0 ) and in case of equality, equality also holds in (17) so that T is extremal. 4. Excursions and estimates on H
+ J for an undirected graph H
Unless indicated to the contrary in the balance of this article n and d are integers with 2
= nd and r = n − d.
ρ(M )
d n,
The results in this section are inspired by the work of Nikiforov [9,10]. We introduce the concept of an excursion. A p-excursion in a graph H is a sequence of vertices v1 , v2 , . . . , vp in H (not necessarily distinct) together with a sequence m1 , m2 , . . . , mp−1 of modes of transportation. The mode of transportation from vp to vp+1 is mp . The modes are chosen from the set {step, jump}. If mp is a step, then vp vp+1 must be an edge in the graph H. In particular, since our graphs have no loops, vp = vp+1 in this case. If mp is a jump, then there is no restriction whatever and the case vp = vp+1 is allowed. We denote by ep (v) the number of p-excursions starting at v and by Ep the total number of p-excursions. We denote by kp (v) the number of p-cliques containing the vertex v and by Kp the total number of p-cliques in the graph. Of course excursions are actually walks in the multigraph with adjacency matrix H + J. Thus, all the machinery of walks [2, Section 1.8] applies. In particular we have that the entries of the vector (H + J )p−1 1 are (ep (v))v∈V and 1 (H + J )p−1 1 = Ep . Furthermore, we have lim
q→∞
Ep+q Eq
= ρ(H + J )p .
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Lemma 10. Let H be an undirected graph with clique number d (kp (v)ek+1 (v) − kp+1 (v)ek (v)) (2p − 1)Kp Ek .
p 1. For any k 1, (21)
v∈V
Proof. Fix a p-clique S in V . For each v ∈ S, there are three types of (k + 1)-excursions starting at v. There are those that start with a jump, those that start with a step to another element of S and those that start with a step to an element outside S. Let Outv be the set of all k-excursions that begin with a neighbour of v outside S. Then we have v∈S
ek+1 (v) = pEk
+ (p − 1)
v∈S
ek (v) +
v∈S
|Outv |
ek (v) + (p − 1) Outv + Outv . pEk + (p − 1) v∈S v∈S v∈S
Now
Outv is contained in the set of all k-excursions starting outside S, so
v∈S
ek+1 (v) pEk + (p − 1) ek (v) + (p − 1) ek (v) + Outv v∈S v∈S v∈S v∈ /S = (2p − 1)Ek + Outv . v∈S
But
Outv is the set of all k-excursions starting at a vertex u such that S
v∈S
(22)
∪ {u} is a (p + 1)-clique.
Therefore summing (22) over all p-cliques S gives v∈V
kp (v)ek+1 (v)
(2p − 1)Kp Ek +
v∈V
kp+1 (v)ek (v)
as required. Theorem 11. Let H be an undirected graph with n vertices and clique number d. Then
ρ(H + J )n
d j=1
(2j − 1)Kj (H )ρ(H + J )n−j
(23)
where Kj (H ) is the number of j-cliques in H. We note that K1 (H ) = n is the number of vertices in H and that K2 (H ) is the number of edges. By convention K0 (H ) = 1 and Kj (H ) = 0 if j < 0. Proof. Let m be a large integer. In (21), replace k by m side is a telescoping sum, giving v∈V
k1 (v)em (v) − kd+1 (v)em−d (v)
d
− p and sum from p = 1 to d. The left-hand
(2p − 1)Kp Em−p .
p=1
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
But k1 (v)
= 1 and kd+1 (v) = 0 for all v, so we get
Em
337
d p=1
(2p − 1)Kp Em−p
and therefore d Em−p (2p − 1)Kp . 1 Em p=1 Passing to the limit as m 1
d p=1
→ ∞ gives
(2p − 1)Kp ρ(H + J )−p
which is equivalent to the desired conclusion. Theorem 12. Let H be a complete d-partite undirected graph with n vertices. Then det
λI − (H + J ) = λn − =
d j=1
(2j − 1)Kj (H )λn−j
(24)
d j=0
(1 − 2j)Kj (H )λn−j
(25)
and
1 Adj
λI − (H + J ) 1 =
d j=0
jKj (H )λn−j .
(26)
In particular, this theorem applies to the d-partite Turán graph with n vertices. Proof. We prove this by induction on d. The induction hypothesis is both (25) and (26). It is easy to check that if d = 1, then these formulæ are correct. Let H be a d-partite graph with part sizes 1 , 2 , . . . , d , and let H be the same graph with the last part missing. We denote Kj the number of j-cliques in H and Kj the number of j-cliques in H . Then we have Kj
= Kj + d Kj−1 , j = 1, . . . , d − 1
and
Kd
= d Kd −1
since each j-clique in H is either a j-clique in H or is determined by a point in the last part of H together with a j − 1 clique in H . The induction step is now completed using det
λI − (H + J ) = λd −1 (λ − d )det λI − (H + J ) − 4d λd −1 1 Adj λI − (H + J ) 1
and
1 Adj
λI − (H + J ) 1 = d λd −1 det λI − (H + J ) + (λ + 3d )λd −1 1 Adj λI − (H + J ) 1
which formulæ can be derived from (3) and (4). We leave the details to the interested reader. A result of Zykov [12,13] asserts that if H has n vertices, clique number d and is not isomorphic to H0 , the corresponding Turán graph, then Kj (H )
< Kj (H0 )
338
for j for j
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
= 2, . . . , d. We have the following corollary using only K2 (H ) K2 (H0 ) − 1 and Kj (H ) Kj (H0 ) = 3, . . . , d, (24) and (23).
Corollary 13. Let d 2. Let H be an undirected graph with n vertices, clique number d and not isomorphic to H0 . Then det ρ(H + J )I − (H0 + J ) + 3ρ(H + J )n−2 0. Proof. Let λ
= ρ(H + J ). Then from (23), − K1 (H )λn−1 − 3K2 (H )λn−2 −
0 λn
j=3
(2j − 1)Kj (H )λn−j
− K1 (H0 )λn−1 − 3 K2 (H0 ) − 1 λn−2
λn =
d
−
d j=3
(2j − 1)Kj (H0 )λn−j
d j=0
(1 − 2j)Kj (H0 )λn−j + 3λn−2
= det λI − (H0 + J ) + 3λn−2 by (25). Theorem 14. Let H0 be a d-partite Turán graph with n vertices. Then det
λI − (H0 + J ) is equal to
λn−d (λ + )d−1 λ − (2d − 1) = 0 and
in case r
λn−d (λ + )d−r −1 (λ + + 1)r −1 λ2 − (2n − 2 − 1)λ − ( + 1)(2d − 1) in case 0
det
1
(27)
< r < d.
Proof. First, consider H a p-partite Turán graph with pq vertices. Then H Thus, the eigenvalues of H are easy to figure and we obtain
In case r
+ J = (2Jp,p − Ip ) ⊗ Jq,q .
λI − (H + J ) = λp(q−1) (λ + q)p−1 (λ − (2p − 1)q)
= 0, we set p = d and q = and we are done. Also we have
λI − (2Jp,p − Ip ) ⊗ Jq,q
−1
1
=
pq
λ − (2p − 1)q
because 1 is an eigenvector of (2Jp,p − Ip ) ⊗ Jq,q . We first set p = r and q = and q = . The results are combined to get (27) essentially using (18).
+ 1 and second p = d − r
Corollary 15. Let H0 be a d-partite Turán graph with n vertices. Then ρ(H0 + J ) is equal to (2d − 1) 2n − in case r = 0 and equal to the positive root of the quadratic equation
=
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
q(λ) in case 0
≡ λ2 − (2n − 2 − 1)λ − ( + 1)(2d − 1) = 0
339
(28)
< r < d.
If r > 0 we find that q(2n − ) all cases 2n − − 1
= 2r ( + 1) > 0 and q(2n − − 1) = −2(d − r ) < 0 so that in
< ρ(H0 + J ) 2n − .
(29)
Combining Corollary 13 and Theorem 14 we obtain Corollary 16. Let n d 2. Let H be an undirected graph with n vertices, clique number d and not isomorphic to H0 . Then, denoting λ = ρ(H + J ) we have in case r = 0
λn−d (λ + )d−1 λ − (2d − 1) + 3λn−2 0 and in case 0
(30)
λn−d (λ + )d−r −1 (λ + + 1)r −1 q(λ) + 3λn−2 0
(31)
where q(λ) is defined in (28). Unfortunately, Corollary 16 is not in a suitable form. Therefore, we next prove Proposition 17. Let n d 2. Let H be an undirected graph with n vertices, clique number d and not isomorphic to H0 . Then in case r = 0
ρ(H + J ) + ρ(H + J ) − (2d − 1) +
and in case 0
q(ρ(H
+ J )) +
9
9 5
0
5
0
(32)
(33)
where q(λ) is defined in (28). Proof. Let λ = ρ(H + J ). We deal first with the case r we see that it suffices to show that 3λn−2
λn−d (λ
+
)d−2
or equivalently
λ+ λ
d−2
5 3
.
9 5
= 0. Comparing Corollary 16 and Proposition 17,
340
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
But
λ+ λ
d−2
= 1+ λ
provided that 2n − 4
d−2
1 (d − 2) 5 n − 2 exp = exp e2 λ λ 3
λ. In the contrary case, 2n − 4 > λ,
(λ + ) λ − (2d − 1) +
9 5
(n + )(−3) +
9 5
0
since 1 and λ = ρ(H + J ) ρ(J ) = n and (32) is verified directly. Turning to the case r > 0, we proceed analogously. It will suffice to show that
λ+ λ
d−r −1
λ++1 λ
r −1
5 3
.
But
λ+ λ
d−r −1
λ++1 λ
r −1
exp
n − 2 − 1
λ
provided that 2n − 4 − 2 λ. In the contrary case 2n increasing on [n, ∞) and that λ n. Thus q(λ)
1
e2
5 3
− 4 − 2 > λ, we first observe that q is
q(2n − 4 − 2) = (9 − 6d)2 + (9 − 4d − 4r ) − 2(r − 1)
after a calculation. Now d
2, r 1 and 1 so that q(λ) −6 which implies (33).
5. Estimates on G0 In this section we find a lower bound for ρ(G0 ). It is actually more convenient to work with ρ(2G0 + I ) = 2ρ(G0 ) + 1. It is perhaps worth noting that in case r = 0, odd, 1 is a Perron vector for 2G0 + I and we find ρ(2G0 + I ) = ρ(H0 + J ) in this case. In general, we have ρ(2G0 + I ) ρ(H0 + J ) from Bendixson’s Theorem [1, Théorème II], but we need to know that ρ(2G0 + I ) is not too much smaller than ρ(H0 + J ). The main theorem of this section is the following. Theorem 18. Let β
= ρ(2G0 + I ). Then
β 3 − ρ(H0 + J )β 2 + ρ(H0 + J ) 0. Lemma 19. Let 0
ϕ(y) ≡
y < 1 then 2y
(1 − y2 )
Proof. We have ϕ(0) Theorem.
+ ln(1 − y) − ln(1 + y) 0.
(34)
= 0 and ϕ (y) = 4y2 (1 − y2 )−2 0. The result follows from the Mean Value
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Lemma 20. Let 0
< x 1, 0 ν 1, then
(1 + ν x)
1 x
− (1 − ν x)
1 x
1 x
ν.
(1 + ν x)
341
+ (1 − ν x)
1 x
(35)
Consequently
(1 + μ) − (1 − μ) μ (1 + μ) + (1 − μ) for 0
(36)
μ −1 and 1 an integer.
Proof. The result is clear if x = 1, so we assume 0 Then a calculation yields that
< x < 1. Let h(x) denote the left hand side of (35).
1
1
∂h (1 + ν x) x (1 − ν x) x = 2x−2 2 ϕ(ν x) 0 1 1 ∂x (1 + ν x) x + (1 − ν x) x since 0 ν x < 1 and by Lemma 19. Therefore h(x) is increasing in x, yielding h(x) second assertion follows by a substitution. Lemma 21. If A0 is an extremal × adjusted tournament matrix, then for 0
1 (I
− μA0 )−1 1 =
h(1) = ν . The
μ < ρ(A0 )−1 ,
m
(37)
1 − μm
where
m
⎧ ⎨
=⎩
if
−(1−μ) − μ ((11+μ) if +μ) +(1−μ)
is odd,
Proof. If is odd, then 1 (I − μS0 )−1 1 then, as in [5],
1 (I
(38)
is even.
= since S0 1 = 0 and the result follows from (15). If is even,
− μS0 )−1 1 = − μ2 1 (I − μW )−1 1 = m
by (6) and again the result follows from (15). Lemma 22. Let Y1 and Y2 be invertible matrices, and let yj ⎛ Y
=
⎝
Then det(Y )
Y1
−J
−J Y2
⎞ ⎠.
= 0 if and only if y1 y2 = 1, equivalently
y1 1 + y1
+
y2 1 + y2
= 1,
= 1 Yj−1 1 for j = 1, 2. Let also
342
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
and in case det(Y ) −1
1Y
= 0,
=y
1
where y
y1
=
1+y
1 + y1
y2
+
1 + y2
.
Proof. This follows immediately from (3) and (5). Lemma 23. Let d ⎛
⎞
Y1
Y
=
2 and
⎜ ⎜ ⎜ −J ⎜ ⎜ . ⎜ . ⎜ . ⎝ −J
−J · · · −J ⎟ ⎟ Y2 · · · −J ⎟ ⎟ .. . . .. ⎟ ⎟ . . . ⎟ ⎠
−J . . . Yd .
Suppose that all proper principal block submatrices of Y are nonsingular and that yj 1, . . . , d. Then det(Y ) = 0 if and only if
1
=
d
yj
(39)
1 + yj j=1
and in case det(Y ) −1
1Y
1
= 1 Yj−1 1 for j =
= 0,
=y
where y 1+y
=
d
yj
1 + yj j=1
.
Proof. This follows from Lemma 22 by an induction argument. We say that a matrix T is a block tournament matrix if it has the form ⎛
⎞
T1 J
T
where d
=
⎜ ⎜ ⎜ J T2 ⎜ ⎜ . . ⎜ . . ⎜ . . ⎝ J J
··· J ⎟ ⎟ ··· J ⎟ ⎟ . . .. ⎟ . . ⎟ ⎟ ⎠ . . . Td
2 and T1 , T2 , . . . , Td be tournament matrices of sizes 1 , 2 , . . . , d .
(40)
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
343
Corollary 24. Let d 2 and T be a block tournament matrix as in (40) with T1 , T2 , . . . , Td extremal tournament matrices. Let μ = (2ρ(T ) + 1)−1 and let mj be related to j as in (38) then
=
1
d
2mj μ
j=1
1 + mj μ
.
(41)
Proof. Decoding (37) with A0 1
(ρ(T )I − Tj )−1 1 =
= 2Tj + I yields 2mj μ
1 − mj μ
μ < ρ(Aj )−1 . Substituting in (39) gives the result.
since 0
Proof of Theorem 18. Since d 2, for each j, the matrix I + 2T contains a (j + 1)×(j + 1) submatrix with row sums greater than or equal to j + 1. It follows that 2ρ(T ) + 1 j + 1 and that j < μ−1 so that (36) applies. The function x → x(1 + x)−1 is increasing on x 0. By (36) and (38), we have mj j (1 − μ2 ). Therefore, from (41) 1
Let ν
d
2j μ(1 − μ2 )
j=1
1 + j μ(1 − μ2 )
.
= μ(1 − μ2 ). In case r = 0, this gives 1 (2d − 1)ν and in case r > 0 it gives 1 − (2n − 2 − 1)ν
− ( + 1)(2d − 1)ν 2 0
after a calculation. Alternatively, this can be written q(ν −1 ) cases, this implies that ν −1 ρ(H0 + J ). But β = μ−1 , so 1
β
1−
1
β2
=ν
0 where q(λ) is defined in (28). In both
1
ρ(H0 + J )
which yields the desired inequality. 6. Final resolution We denote ρ = ρ(H0 + J ) and β = ρ(2G0 + I ). Let obtainable in view of (32) or (33), whichever applies. Proposition 25. Let n
5 and define γ = ρ −
11 −1 ρ . 10
α be the best upper bound for ρ(H + J )
Then α
< γ < β.
Proof. Let ψ(t ) = t 3 − ρ t 2 + ρ . Then it can be checked that ψ(γ ) < 0 unless ρ < 4·9 which implies n 4 by (29). Therefore ψ(γ ) < ψ(β) by Theorem 18. Now ψ (t ) = 3t 2 − 2ρ t and so ψ is increasing on t > 23 ρ . We need to know that both β and γ lie in this range. For γ this follows since
ρ>
√
3·3 again by (29). For β , we use Theorem 18 in the form
β+
1
β2 − 1
to deduce that β
+
1
β +1
=
β3 ρ β2 − 1
> 23 ρ since clearly β 2. We may conclude that γ < β .
344
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Next we show that α
< γ . In case r = 0, ρ = 2n − and a calculation shows that
(γ + )(γ − ρ) +
9
=
5
280n2
− 500n + 121 + 1802 >0 100(2n − )2
since 2 n. One easily checks that γ > n − , the value at which t → (t + )(t − ρ) + 95 takes its minimum, so α < γ in this case by (32). For 0 < r < d we write q(λ) = (λ − ρ)(λ − ρ ) where ρ is the negative root of q(λ) = 0. Then q(γ ) +
9 5
=−
11 10ρ
ρ − ρ −
11 −1 ρ 10
9
+ . 5
This is positive if
−ρ + ρ +
11 −1 ρ
10
+
18 11
ρ>0
and therefore if
(ρ + ρ ) − But ρ
4 11
ρ > 0.
+ ρ can be read off Eq. (28). Hence, we verify 11(2n − 2 − 1) − 4(2n − )
= 14n − 18 − 11 5n − 11 > 0
using ρ 2n − from (29), n 2 and assuming n 3. Again, one easily checks that γ > n − − 12 , the value at which ψ takes its minimum. Hence α < γ by (33). So α < γ in both cases and result is proved. Corollary 26. For 2
d n, α < β .
Proof. The cases n = 2, 3, 4 follow by making explicit numerical estimates. The case n from Proposition 25.
5 follows
We state some lemmas that will be needed for the uniqueness assertion in Theorem 1. We leave the proofs to the reader. Lemma 27. Let n
3 and T an n × n extremal tournament matrix. Then T is irreducible.
Lemma 28. Let d
2 and T a block tournament matrix as in (40). Then T is irreducible.
Lemma 29. Let A be an n × n irreducible matrix with nonnegative entries. Let B have entries such that 0 bjk ajk for 1 j, k n. Suppose that B = A. Then ρ(B) < ρ(A).
− →
Proof of Theorem 1. A digraph G is said to be full if for every pair of distinct vertices u and v either uv − → or vu is a directed edge in G. Suppose first that G is a full digraph with n vertices and clique number d and suppose that ρ(G) is maximal among all digraphs with n vertices and clique number d. If d = 1, then G is a tournament and therefore an extremal tournament and the result is proved. Hence we may assume that 2 d n. Now let H be the associated undirected graph. Then either H is isomorphic to the d-partite Turán graph on n vertices or it is not. If it is not, then ρ(H + J ) satisfies (32) or (33), whichever applies. But
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
G +G But
345
= H +J −I and it follows from Bendixson’s Theorem [1, Théorème II] that ρ(G) ρ( 12 (H +J −I )). ρ
1 2
(H + J − I ) =
1 2
1
(α − 1) < (β − 1) = ρ(G0 ) 2
and ρ(G) cannot be maximal. This contradicts the supposition that H is not isomorphic to H0 . But in case H is isomorphic to H0 , the adjacency matrix of G is of the form (40) and with the blocks as in H0 . It now follows by successively applying Theorem 3 that each of the tournaments Tj (j = 1, . . . , d) figuring is extremal. Now we consider the case that ρ(G) is maximal, but G not full. Then it is easy to construct G1 G with G1 full, G1 = G and G and G1 having the same associated graph. Then G1 has n vertices and clique number d and ρ(G1 ) ρ(G). Hence ρ(G1 ) is also maximal and it follows that G1 is a G0 . But unless n = 2 and d = 1 it follows from Lemmas 27 and 28 that G0 and hence G1 is strongly connected. But G1 G and G1 = G. It follows now from Lemma 29 that ρ(G) < ρ(G1 ) contradicting the maximality of ρ(G). Acknowledgements We thank the referee for many helpful comments and for simplifying the proof of Proposition 25. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
I. Bendixson, Sur les racines d’une équation fondamentale, Acta Math. 25 (1902) 359–365. D. Cvetkovic, ´ M. Doob, H. Sachs, Spectra of Graphs, VEB Deutscher Verlag der Wissenschaften, Berlin, 1980. R.A. Brualdi, Spectra of digraphs, Linear Algebra Appl. 432 (2010) 2181–2213. R.A. Brualdi, Q. Li, Problem 31, Discrete Math. 43 (1983) 329–330. S.W. Drury, Solution of the conjecture of Brualdi and Li, Linear Algebra Appl. 436 (2012) 3392–3399. D.A. Gregory, S.J. Kirkland, B.L. Shader, Pick’s inequality and tournaments, Linear Algebra Appl. 186 (1993) 15–36. R.A. Horn, C.R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1985. H. Lin, J. Shu, Y. Wu, G. Yu, Spectral radius of strongly connected digraphs, Discrete Math. 312 (24) (2012) 3663–3669. V. Nikiforov, Some inequalities for the largest eigenvalue of a graph, Combin. Probab. Comput. 11 (2002) 179–189. V. Nikiforov, Bounds on graph eigenvalues II, Linear Algebra Appl. 427 (2007) 183–189. G. Pick, Über die Wurzeln der characteristischen Gleichung von Schwingungsproblemen, Z. Angew. Math. Mech. 2 (1922) 353–357. [12] A.A. Zykov, On some properties of linear complexes, Mat. Sb. N.S. 24 (66) (1949) 163–188, in Russian. [13] A.A. Zykov, On some properties of linear complexes, Amer. Math. Soc. Transl. Ser. 1 (7) (1952) 417–449.
Contents lists available at SciVerse ScienceDirect
Linear Algebra and its Applications journal homepage: w w w . e l s e v i e r . c o m / l o c a t e / l a a
Extremal digraphs with given clique number S.W. Drury a,∗ , Huiqiu Lin b a
Department of Mathematics and Statistics, McGill University, Burnside Hall, 805 Ouest, rue Sherbrooke Street West, Montréal, Québec, Canada H3A 2K6
b
Department of Mathematics, East China Normal University, Shanghai 200062, China
ARTICLE INFO
ABSTRACT
Article history: Received 21 August 2012 Accepted 19 March 2013 Available online 12 April 2013
We classify those digraphs with a given number of vertices and a given clique number that maximize the Perron root of the adjacency matrix. © 2013 Elsevier Inc. All rights reserved.
Submitted by R.A. Brualdi AMS classification: 05C20 05C50 15A15 Keywords: Digraph Clique number Tournament Characteristic polynomial Perron root
1. Introduction In this paper, we deal with simple graphs and simple digraphs without loops. The clique number of a graph H is the largest integer d such that the complete graph on d vertices is a subgraph of H. The clique number of a digraph G is the largest integer d such that the complete digraph on d vertices is a subgraph of G. − → A tournament is a digraph in which for any two distinct vertices u and v, exactly one of uv and − → vu is an arc. The adjacency matrix of a tournament is called a tournament matrix. A tournament on n vertices that maximizes the Perron root of its adjacency matrix among all such tournaments will be ∗ Corresponding author. E-mail address: [email protected] (S.W. Drury) 0024-3795/$ - see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.03.024
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
329
called extremal. It has long been known that if n is odd, the extremal tournaments are precisely the ones that are regular, i.e. have indegree and outdegree (n − 1)/2 at each vertex. For n even, the extremal tournaments are those which are isomorphic to the Brualdi–Li tournament. This was conjectured by Brualdi and Li [4] and proved by Drury [5, Theorem 1]. In the case n odd, it is part of the folklore of the subject that for n = 1, 3, 5, there is only one extremal tournament up to isomorphism, but for n 7, there may be many. For an excellent survey of spectral estimates for digraphs, see Brualdi [3]. For n and d fixed, let = nd and r = n − d. Let Vj , j = 1, . . . , d be disjoint vertex sets where |Vj | = + 1 for j = 1, . . . , r and |Vj | = for j = r + 1, . . . , d. Let G0 be a digraph with vertex d set Vj with all possible arcs between Vj and Vk for j = k and the induced subdigraph G0 [Vj ] an j=1
extremal tournament for j = 1, . . . , d. If n > 6d, then two such G0 need not be isomorphic, but the Perron roots of their adjacency matrices are known to be the same. Such a digraph G0 has n vertices and clique number d. The primary objective of this article is to solve Lin et al. [8, Problem 2] by establishing the following theorem. Theorem 1. Let 1 d n. If G is a digraph with n vertices and clique number d, then ρ(G) Unless n = 2 and d = 1 equality occurs only if G is isomorphic to some G0 . 1
ρ(G0 ).
Note that if d = 0, G has no vertices and that the clique number of G cannot exceed the number of vertices, so the theorem does not make sense unless 1 d n. We use I to denote the identity matrix and J to denote the matrix in which every entry is 1. The shape of these matrices should be deduced from the context. We denote by 1 the vector of ones. For any graph or digraph, we will use the same notation for its adjacency matrix in the hope that no confusion will arise. We are interested in maximizing the Perron root of the adjacency matrix of a digraph G with n vertices and clique number d. In view of [7, Corollary 8.1.19], we may always assume that for any two − → − → distinct vertices u and v in G, either uv or vu is an arc in G. This is because if there are no arcs between − → − → u and v, then adding just one of the arcs uv or vu does not decrease the Perron root (and in many circumstances increases it) without increasing the clique number. Let us assume that this has been done. Then we define the associated graph H of G to be an undirected − → − → graph in which uv is an edge if and only if both uv and vu are arcs in G. Clearly, H has the same clique number as G. A theorem of Nikiforov [10, Theorem 1], states that if H is a graph with n vertices and clique number d then
ρ(H ) ρ(H0 ) where H0 is the d-partite Turán graph on n vertices. Furthermore, equality obtains only when H is isomorphic to H0 . Restating this more aggressively, if H is not isomorphic to H0 then we have ρ(H ) ν(n, d) < ρ(H0 ). Here ν(n, d) is the best possible upper bound for max{ρ(H ); H has n vertices, clique number d and H obtainable by Nikiforov’s method. 2 Since G + G II] now gives
ρ(G)
1 2
(ρ(H ) + n − 1)
1 2
∼ = H0 }
= H + J − I, a theorem of Bendixson [1, Théorème
(ν(n, d) + n − 1)
(1)
1 At issue here is that there are two digraphs with n = 2 and d = 1, one with no arcs and one with a single arc. Both have zero Perron root. d 2 In fact ν(n, d) is the largest real root of the equation det(λI − H0 ) + (j − 1)λn−j = 0, cf. Corollary 13 below, but using
the full strength of Zykov’s result.
j =2
330
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
and if it should be that 1 2
(ν(n, d) + n − 1) < ρ(G0 )
(2)
then G cannot have a maximal Perron root. Turned around this says that if G does have a maximal Perron root, then its associated H is isomorphic to H0 , a huge step in the right direction. In most cases (i.e. unless r = 0 and odd), ρ(G0 ) < 12 (ρ(H0 ) + n − 1), so establishing (2) or something similar requires delicate estimates. Unfortunately the inequality (2) fails for general n and d (for example in the case n = 15 and d = 8, 12 (ν(15, 8)+15−1) ≈ 13·517876 and ρ(G0 ) ≈ 13·517834). Our strategy is to build a theory parallel to that of Nikiforov that estimates ρ(H + J ) directly. In the second section, we establish formulæ relating to Schur complements. In the third section we establish the results that are needed to complete the proof of Theorem 1 once it is known that the associated graph is H0 . In the fourth section we estimate ρ(H + J ). In the remainder of the paper we make the necessary delicate estimates to complete the proof of Theorem 1. 2. Schur complement identities Proposition 2. Let A be a p × p real matrix, B a q × q real matrix and in this section, let J denote the p × q matrix in which every entry is one. Then ⎛
−J B
⎛
1 Adj ⎝
−J
A
det ⎝
⎠ = det(A)det(B) − (1 Adj(A)1)(1 Adj(B)1)
(3)
⎞
−J
A
⎞
⎠ 1 = det(A)(1 Adj(B)1) + (1 Adj(A)1)det(B) + 2(1 Adj(A)1)(1 Adj(B)1) (4)
−J B
and in all cases where it makes sense ⎛ 1
−J
⎝ A
−J B
⎞−1 ⎠
1
=
1 A−1 1 + 1 B−1 1 + 2(1 A−1 1)(1 B−1 1)
1 − (1 A−1 1)(1 B−1 1)
.
(5)
Proof. We see that (5) follows from (3) and (4). Next, we observe that (3) and (4) are polynomial identities. It therefore suffices to prove them when A−1 and B−1 exist and have sufficiently small norm. Now ⎞⎛ ⎛ ⎞ ⎛ ⎞ Ip A −J A − JB−1 J −J 0 ⎠⎝ ⎝ ⎠=⎝ ⎠ 0 B −J B −B−1 J Iq so that
⎛
det ⎝
A
−J
−J B
⎞ ⎠
since A−1 JB−1 J has rank one. Thus (3) follows. We also have ⎛ ⎝
A
−J
−J B
⎞−1 ⎠
= det(A)det(B)det(I − A−1 JB−1 J ) = det(A)det(B) 1 − (1 A−1 1)(1 B−1 1)
⎛
=⎝
(A − JB−1 J )−1
A−1 J (B − J A−1 J )−1
B−1 J (A − JB−1 J )−1
(B − J A−1 J )−1
⎞ ⎠.
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
331
Now observe that ⎛
1 (A − JB
−1 −1
J
)
−1
1=1A
(I − JB−1 J A−1 )−1 1 = 1 A−1 ⎝
= (1 A−1 1)
∞ k=0
∞ k=0
⎞
(JB−1 J A−1 )k ⎠ 1
−1
(1 A−1 1)k (1 B−1 1)k = (1 A−1 1) 1 − (1 A−1 1)(1 B−1 1)
Together with similar calculations for the other pieces of the left hand side of (5), this yields (4) using (3). 3. Estimates on tournaments The objective of this section is the following theorem. Theorem 3. For T an n × n tournament matrix, T0 an extremal n m × m nonnegative matrix, let ⎞
⎛ M0
=⎝
Then ρ(M0 )
T0 J J H
⎞
⎛
⎠ and M
=⎝
× n tournament matrix and H be an
T J J H
⎠.
ρ(M ). In case of equality, T is extremal.
The n × n Pick matrix W is the skew symmetric matrix with ones above the main diagonal and minus ones below the main diagonal. The reader may refer to Gregory et al. [6] for information about this matrix and its significance to tournaments. It is well known that det(W ) = 0 if n is odd and det(W ) = 1 if n is even. The following theorem [5, Theorem 6] which is implicitly proved in [5] will be needed. Theorem 4. Let ν > 1 be fixed and let Q denote the set of all skew-symmetric real matrices Q with entries bounded above in absolute value by 1. Let Z = Q + ν J a positive matrix. Then the minimum possible value of ρ(Z ) as Q runs over Q is attained when Q = W and any Q which gives rise to a minimum value of ρ(Z ) is necessarily permutationally similar to W . Lemma 5. Let n
γ =μ
4 be even, 0 < μ < 1 and let
(1 + μ)n + (1 − μ)n > μ. (1 + μ)n − (1 − μ)n
Then
1 (I
− μW )−1 1 = γ −1
and ρ(μW
(6)
+ γ J ) = 1.
Proof. The identity (6) is obtained by writing
1 (I
− μW )−1 1 =
1 Adj(I
− μW )1 μ|Y |−1 det(W Y) = Y |X | det(I − μW ) μ det ( W X) X
.
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
where in the denominator the sum is over all subsets X of {1, 2, . . . , n} and in the numerator, the determinants are of bordered matrices, ⎛ W Y
=
⎝
1
0
−1 WY
⎞ ⎠
the sum extending again over all subsets Y of {1, 2, . . . , n}. We note that the matrices WX and W Y are themselves Pick matrices so that their determinants depend on whether X and Y have odd or even cardinality. We leave the details to the reader. Let v be the positive vector with entries vk
=
1−μ
k
1+μ
.
Calculations show that (μW μ < 1.
+ γ J )v = v and the result follows from [7, Corollary 8.1.30] since 0 <
The adjusted tournament matrix A of a tournament has the form A = 2T + I where T is the adjacency matrix. It was introduced in [5] in order to access the corresponding skew tournament matrix S = A − J = T − T and also to simplify calculations involving characteristic polynomials. We denote by A0 an extremal adjusted tournament matrix of the same size, S0 = A0 − J and μ0 = ρ(A0 )−1 . It is well known that μ0 (n − 1)−1 , so that for n 4
μ0 cot
π 2n
2n
(n − 1)π
8 3π
< 1.
(7)
Proposition 6. Let n 4 be even. Let Q be an n × n skew tournament matrix. Let u be a vector with all its entries greater than or equal to 1. Then
1 (I
− μW )−1 1 u (I − μQ )−1 u
for all μ with 0
(8)
< μ < μ0 and if equality obtains for any such μ then Q is permutationally similar to W .
Proof. By Lemma 5, we need to show that
γ −1 u (I − μQ )−1 u which is equivalent to det(I
− μQ − γ u ⊗ u ) 0.
(9)
Now by [7, Corollary 8.1.19], Theorem 4 and Lemma 5
ρ(μQ + γ u ⊗ u ) ρ(μQ + γ J ) ρ(μW + γ J ) = 1
(10)
with equality only if Q is permutationally similar to W . At this point the proof splits into two cases.
• μQ + γ u ⊗ u has two or more real characteristic roots counted according to multiplicity that are greater than or equal to 1. A theorem of Bendixson [1, Théorème II] asserts that all the characteristic roots of μQ + γ u ⊗ u have positive real part. So if there were two or more such roots, then we
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
would necessarily have tr(μQ + γ u ⊗ u ) 2. But this leads to γ u 2 estimate [11, Inequality(17)] or [6, Theorem 1.1] and (7) u (I
− μQ )−1 u
u 2
1 + μ2 cot
π
2
2n
2γ −1 2 π 1 + μ2 cot 2n
333
2. But then, using Pick’s
> γ −1 .
• μQ + γ u ⊗ u has one or fewer real characteristic roots counted according to multiplicity that are greater than or equal to 1. Then by (10) there is exactly one such root. It follows that (9) holds by considering the sign of the characteristic polynomial. Finally, if γ −1 = u (I − μQ )−1 u, then ρ(μQ to W by Theorem 4.
+ γ J ) = 1 and Q is necessarily permutationally similar
For n even, we use the methods in [5]. Then s = S 1 is a vector with odd integer entries summing to zero. Let D be the diagonal matrix with entries sgn(sj ), for j = 1, . . . , n and let Q = DSD. Then Q is again a skew-symmetric tournament matrix. Let uj = |sj | 1 define a vector u with entries greater than or equal to unity. If this is done for S0 instead of S we find that the corresponding Q0 is permutationally similar to the Pick matrix W and the corresponding u vector is 1. Proposition 7. For S an n × n skew tournament matrix and S0 an extremal n × n skew tournament matrix, the inequality
1 (I
− μS0 )−1 1 1 (I − μS)−1 1
holds for all μ with 0
(11)
< μ < μ0 . Further if for any such μ equality obtains in (11) then S is extremal.
Proof. The proof depends on whether n is odd or even.
• Case n is odd. Since S0 1 = 0, we note that 1 (I − μS0 )−1 1 = n. On the other hand, we have
1 (I
•
− μS)−1 1 =
∞ k=0
μk 1 Sk 1 =
∞ k=0
μ2k 1 S2k 1 = n − μ2 s (I − μ2 S2 )−1 s
(12)
where s = S 1. But the operator (I − μ2 S 2 )−1 is strictly positive definite so that (11) holds with equality only if s = 0 which is the regular case. Case n is even. Using (12), we see that the desired inequality is equivalent to
(S0 1) (I − μS0 )−1 (S0 1) s (I − μS)−1 s. But this is identical to
1 (I
− μQ0 )−1 1 u (I − μQ )−1 u
(13)
since S0 1 has ±1 entries. Finally, Q0 is permutationally similar to W so (13) follows by Proposition 6. Clearly if equality obtains in (11) it also obtains in (8). This forces Q to be permutationally similar to W . So, after applying a permutational similarity to S, we can assume that Q = W . But then Ws = Qs = DSDs = DS 1 = Ds = 1. But W is invertible so s (and hence D) is uniquely determined. We can then compute S = DQD = DWD exactly and show that it is permutationally equivalent to S0 . We leave this chore as an exercise for the reader.
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Lemma 8. Let n 3 and let A be an n × n adjusted tournament matrix with related skew tournament matrix S = A − J. Then
1 (I
for 0
− μA)−1 1 = μ−1 ϕ μ1 (I − μS)−1 1
(14)
< μ < μ0 where ϕ(x) = x/(1 − x).
Proof. We have
1 (I
− μA)−1 1 = 1 (I − μS − μJ )−1 1
−1
= 1 (I − μS)(I − μ(I − μS)−1 J ) ⎛
=1 =
⎝
∞ k=0
∞ k=0
−1
μ (I − μS) k
k
J)
k+1
μk 1 (I − μS)−1 1
= μ−1 ϕ μ1 (I − μS)−1 1 for |μ| sufficiently small. We note that for 0 0
< μ1 (I − μS)−1 1 = 1 −
det(I det(I
1
⎞
−1 ⎠
(I − μS)
1
(15)
< μ < μ0 , (I − μA)−1 exists and also
− μA) < 1. − μS)
(16)
See [5, Section 2, first display] for a proof of the equality in (16). Thus (14) extends to the range claimed since both sides are rational functions of μ. Theorem 9. For T an n × n tournament matrix and T0 an extremal n × n tournament matrix, the inequality 1
(λI − T0 )−1 1 1 (λI − T )−1 1
holds for all λ
(17)
> ρ(T0 ). Further if for any λ > ρ(T0 ) equality obtains in (17) then T is extremal.
Proof. If n = 1 or n = 2, then there is nothing to prove, so we may assume that n 3. Tournament and adjusted tournament matrices are related by A = I + 2T, therefore, (17) becomes 1
for λ
((2λ + 1)I − A0 )−1 1 1 ((2λ + 1)I − A)−1 1
> ρ(T0 ) and this is clearly equivalent to
1 (I
− μA0 )−1 1 > 1 (I − μA)−1 1
for all μ with 0 < μ < μ0 = ρ(A0 )−1 . But by Lemma 8 and the fact that ϕ is strictly increasing on 0 < x < 1, this is equivalent to (11). It is also clear that if equality obtains in (17) then it also obtains in (11). Proof of Theorem 3. By (3) we have det(λI
− M ) = det(λI − T ) det(λI − H ) 1 − (1 (λI − T )−1 1)(1 (λI − H )−1 1) .
(18)
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
This identity holds pointwise for λ functions. For λ > ρ(T ) we can expand 1
(λI − T )−1 1 =
∞ k=0
335
> max(ρ(T ), ρ(H )) and also in the sense of equality of rational
λ−k−1 1 T k 1
and it follows that 1 (λI − T )−1 1 is positive and decreasing on this interval. Similarly we have, 1 (λI − H )−1 1 is positive and decreasing on λ > ρ(H ) and it follows that
(1 (λI − T )−1 1)(1 (λI − H )−1 1) is positive and decreasing on λ 1
= (1 (λI − T )−1 1)(1 (λI − H )−1 1)
has exactly one solution in λ Similarly 1
> max(ρ(T ), ρ(H )). Thus, the equation
> max(ρ(T ), ρ(H )) and this solution is ρ(M ).
= (1 (λI − T0 )−1 1)(1 (λI − H )−1 1)
has exactly one solution in λ
(19)
(20)
> max(ρ(T0 ), ρ(H )) and this solution is ρ(M0 ). If ρ(M ) > ρ(M0 ), then
(1 (ρ(M0 )I − T )−1 1)(1 (ρ(M0 )I − H )−1 1) > (1 (ρ(M )I − T )−1 1)(1 (ρ(M )I − H )−1 1) =1 = (1 (ρ(M0 )I − T0 )−1 1)(1 (ρ(M0 )I − H )−1 1) leading to 1 (ρ(M0 )I − T )−1 1) > (1 (ρ(M0 )I − T0 )−1 1) contradicting Theorem 9. Hence ρ(M0 ) and in case of equality, equality also holds in (17) so that T is extremal. 4. Excursions and estimates on H
+ J for an undirected graph H
Unless indicated to the contrary in the balance of this article n and d are integers with 2
= nd and r = n − d.
ρ(M )
d n,
The results in this section are inspired by the work of Nikiforov [9,10]. We introduce the concept of an excursion. A p-excursion in a graph H is a sequence of vertices v1 , v2 , . . . , vp in H (not necessarily distinct) together with a sequence m1 , m2 , . . . , mp−1 of modes of transportation. The mode of transportation from vp to vp+1 is mp . The modes are chosen from the set {step, jump}. If mp is a step, then vp vp+1 must be an edge in the graph H. In particular, since our graphs have no loops, vp = vp+1 in this case. If mp is a jump, then there is no restriction whatever and the case vp = vp+1 is allowed. We denote by ep (v) the number of p-excursions starting at v and by Ep the total number of p-excursions. We denote by kp (v) the number of p-cliques containing the vertex v and by Kp the total number of p-cliques in the graph. Of course excursions are actually walks in the multigraph with adjacency matrix H + J. Thus, all the machinery of walks [2, Section 1.8] applies. In particular we have that the entries of the vector (H + J )p−1 1 are (ep (v))v∈V and 1 (H + J )p−1 1 = Ep . Furthermore, we have lim
q→∞
Ep+q Eq
= ρ(H + J )p .
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Lemma 10. Let H be an undirected graph with clique number d (kp (v)ek+1 (v) − kp+1 (v)ek (v)) (2p − 1)Kp Ek .
p 1. For any k 1, (21)
v∈V
Proof. Fix a p-clique S in V . For each v ∈ S, there are three types of (k + 1)-excursions starting at v. There are those that start with a jump, those that start with a step to another element of S and those that start with a step to an element outside S. Let Outv be the set of all k-excursions that begin with a neighbour of v outside S. Then we have v∈S
ek+1 (v) = pEk
+ (p − 1)
v∈S
ek (v) +
v∈S
|Outv |
ek (v) + (p − 1) Outv + Outv . pEk + (p − 1) v∈S v∈S v∈S
Now
Outv is contained in the set of all k-excursions starting outside S, so
v∈S
ek+1 (v) pEk + (p − 1) ek (v) + (p − 1) ek (v) + Outv v∈S v∈S v∈S v∈ /S = (2p − 1)Ek + Outv . v∈S
But
Outv is the set of all k-excursions starting at a vertex u such that S
v∈S
(22)
∪ {u} is a (p + 1)-clique.
Therefore summing (22) over all p-cliques S gives v∈V
kp (v)ek+1 (v)
(2p − 1)Kp Ek +
v∈V
kp+1 (v)ek (v)
as required. Theorem 11. Let H be an undirected graph with n vertices and clique number d. Then
ρ(H + J )n
d j=1
(2j − 1)Kj (H )ρ(H + J )n−j
(23)
where Kj (H ) is the number of j-cliques in H. We note that K1 (H ) = n is the number of vertices in H and that K2 (H ) is the number of edges. By convention K0 (H ) = 1 and Kj (H ) = 0 if j < 0. Proof. Let m be a large integer. In (21), replace k by m side is a telescoping sum, giving v∈V
k1 (v)em (v) − kd+1 (v)em−d (v)
d
− p and sum from p = 1 to d. The left-hand
(2p − 1)Kp Em−p .
p=1
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
But k1 (v)
= 1 and kd+1 (v) = 0 for all v, so we get
Em
337
d p=1
(2p − 1)Kp Em−p
and therefore d Em−p (2p − 1)Kp . 1 Em p=1 Passing to the limit as m 1
d p=1
→ ∞ gives
(2p − 1)Kp ρ(H + J )−p
which is equivalent to the desired conclusion. Theorem 12. Let H be a complete d-partite undirected graph with n vertices. Then det
λI − (H + J ) = λn − =
d j=1
(2j − 1)Kj (H )λn−j
(24)
d j=0
(1 − 2j)Kj (H )λn−j
(25)
and
1 Adj
λI − (H + J ) 1 =
d j=0
jKj (H )λn−j .
(26)
In particular, this theorem applies to the d-partite Turán graph with n vertices. Proof. We prove this by induction on d. The induction hypothesis is both (25) and (26). It is easy to check that if d = 1, then these formulæ are correct. Let H be a d-partite graph with part sizes 1 , 2 , . . . , d , and let H be the same graph with the last part missing. We denote Kj the number of j-cliques in H and Kj the number of j-cliques in H . Then we have Kj
= Kj + d Kj−1 , j = 1, . . . , d − 1
and
Kd
= d Kd −1
since each j-clique in H is either a j-clique in H or is determined by a point in the last part of H together with a j − 1 clique in H . The induction step is now completed using det
λI − (H + J ) = λd −1 (λ − d )det λI − (H + J ) − 4d λd −1 1 Adj λI − (H + J ) 1
and
1 Adj
λI − (H + J ) 1 = d λd −1 det λI − (H + J ) + (λ + 3d )λd −1 1 Adj λI − (H + J ) 1
which formulæ can be derived from (3) and (4). We leave the details to the interested reader. A result of Zykov [12,13] asserts that if H has n vertices, clique number d and is not isomorphic to H0 , the corresponding Turán graph, then Kj (H )
< Kj (H0 )
338
for j for j
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
= 2, . . . , d. We have the following corollary using only K2 (H ) K2 (H0 ) − 1 and Kj (H ) Kj (H0 ) = 3, . . . , d, (24) and (23).
Corollary 13. Let d 2. Let H be an undirected graph with n vertices, clique number d and not isomorphic to H0 . Then det ρ(H + J )I − (H0 + J ) + 3ρ(H + J )n−2 0. Proof. Let λ
= ρ(H + J ). Then from (23), − K1 (H )λn−1 − 3K2 (H )λn−2 −
0 λn
j=3
(2j − 1)Kj (H )λn−j
− K1 (H0 )λn−1 − 3 K2 (H0 ) − 1 λn−2
λn =
d
−
d j=3
(2j − 1)Kj (H0 )λn−j
d j=0
(1 − 2j)Kj (H0 )λn−j + 3λn−2
= det λI − (H0 + J ) + 3λn−2 by (25). Theorem 14. Let H0 be a d-partite Turán graph with n vertices. Then det
λI − (H0 + J ) is equal to
λn−d (λ + )d−1 λ − (2d − 1) = 0 and
in case r
λn−d (λ + )d−r −1 (λ + + 1)r −1 λ2 − (2n − 2 − 1)λ − ( + 1)(2d − 1) in case 0
det
1
(27)
< r < d.
Proof. First, consider H a p-partite Turán graph with pq vertices. Then H Thus, the eigenvalues of H are easy to figure and we obtain
In case r
+ J = (2Jp,p − Ip ) ⊗ Jq,q .
λI − (H + J ) = λp(q−1) (λ + q)p−1 (λ − (2p − 1)q)
= 0, we set p = d and q = and we are done. Also we have
λI − (2Jp,p − Ip ) ⊗ Jq,q
−1
1
=
pq
λ − (2p − 1)q
because 1 is an eigenvector of (2Jp,p − Ip ) ⊗ Jq,q . We first set p = r and q = and q = . The results are combined to get (27) essentially using (18).
+ 1 and second p = d − r
Corollary 15. Let H0 be a d-partite Turán graph with n vertices. Then ρ(H0 + J ) is equal to (2d − 1) 2n − in case r = 0 and equal to the positive root of the quadratic equation
=
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
q(λ) in case 0
≡ λ2 − (2n − 2 − 1)λ − ( + 1)(2d − 1) = 0
339
(28)
< r < d.
If r > 0 we find that q(2n − ) all cases 2n − − 1
= 2r ( + 1) > 0 and q(2n − − 1) = −2(d − r ) < 0 so that in
< ρ(H0 + J ) 2n − .
(29)
Combining Corollary 13 and Theorem 14 we obtain Corollary 16. Let n d 2. Let H be an undirected graph with n vertices, clique number d and not isomorphic to H0 . Then, denoting λ = ρ(H + J ) we have in case r = 0
λn−d (λ + )d−1 λ − (2d − 1) + 3λn−2 0 and in case 0
(30)
λn−d (λ + )d−r −1 (λ + + 1)r −1 q(λ) + 3λn−2 0
(31)
where q(λ) is defined in (28). Unfortunately, Corollary 16 is not in a suitable form. Therefore, we next prove Proposition 17. Let n d 2. Let H be an undirected graph with n vertices, clique number d and not isomorphic to H0 . Then in case r = 0
ρ(H + J ) + ρ(H + J ) − (2d − 1) +
and in case 0
q(ρ(H
+ J )) +
9
9 5
0
5
0
(32)
(33)
where q(λ) is defined in (28). Proof. Let λ = ρ(H + J ). We deal first with the case r we see that it suffices to show that 3λn−2
λn−d (λ
+
)d−2
or equivalently
λ+ λ
d−2
5 3
.
9 5
= 0. Comparing Corollary 16 and Proposition 17,
340
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
But
λ+ λ
d−2
= 1+ λ
provided that 2n − 4
d−2
1 (d − 2) 5 n − 2 exp = exp e2 λ λ 3
λ. In the contrary case, 2n − 4 > λ,
(λ + ) λ − (2d − 1) +
9 5
(n + )(−3) +
9 5
0
since 1 and λ = ρ(H + J ) ρ(J ) = n and (32) is verified directly. Turning to the case r > 0, we proceed analogously. It will suffice to show that
λ+ λ
d−r −1
λ++1 λ
r −1
5 3
.
But
λ+ λ
d−r −1
λ++1 λ
r −1
exp
n − 2 − 1
λ
provided that 2n − 4 − 2 λ. In the contrary case 2n increasing on [n, ∞) and that λ n. Thus q(λ)
1
e2
5 3
− 4 − 2 > λ, we first observe that q is
q(2n − 4 − 2) = (9 − 6d)2 + (9 − 4d − 4r ) − 2(r − 1)
after a calculation. Now d
2, r 1 and 1 so that q(λ) −6 which implies (33).
5. Estimates on G0 In this section we find a lower bound for ρ(G0 ). It is actually more convenient to work with ρ(2G0 + I ) = 2ρ(G0 ) + 1. It is perhaps worth noting that in case r = 0, odd, 1 is a Perron vector for 2G0 + I and we find ρ(2G0 + I ) = ρ(H0 + J ) in this case. In general, we have ρ(2G0 + I ) ρ(H0 + J ) from Bendixson’s Theorem [1, Théorème II], but we need to know that ρ(2G0 + I ) is not too much smaller than ρ(H0 + J ). The main theorem of this section is the following. Theorem 18. Let β
= ρ(2G0 + I ). Then
β 3 − ρ(H0 + J )β 2 + ρ(H0 + J ) 0. Lemma 19. Let 0
ϕ(y) ≡
y < 1 then 2y
(1 − y2 )
Proof. We have ϕ(0) Theorem.
+ ln(1 − y) − ln(1 + y) 0.
(34)
= 0 and ϕ (y) = 4y2 (1 − y2 )−2 0. The result follows from the Mean Value
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Lemma 20. Let 0
< x 1, 0 ν 1, then
(1 + ν x)
1 x
− (1 − ν x)
1 x
1 x
ν.
(1 + ν x)
341
+ (1 − ν x)
1 x
(35)
Consequently
(1 + μ) − (1 − μ) μ (1 + μ) + (1 − μ) for 0
(36)
μ −1 and 1 an integer.
Proof. The result is clear if x = 1, so we assume 0 Then a calculation yields that
< x < 1. Let h(x) denote the left hand side of (35).
1
1
∂h (1 + ν x) x (1 − ν x) x = 2x−2 2 ϕ(ν x) 0 1 1 ∂x (1 + ν x) x + (1 − ν x) x since 0 ν x < 1 and by Lemma 19. Therefore h(x) is increasing in x, yielding h(x) second assertion follows by a substitution. Lemma 21. If A0 is an extremal × adjusted tournament matrix, then for 0
1 (I
− μA0 )−1 1 =
h(1) = ν . The
μ < ρ(A0 )−1 ,
m
(37)
1 − μm
where
m
⎧ ⎨
=⎩
if
−(1−μ) − μ ((11+μ) if +μ) +(1−μ)
is odd,
Proof. If is odd, then 1 (I − μS0 )−1 1 then, as in [5],
1 (I
(38)
is even.
= since S0 1 = 0 and the result follows from (15). If is even,
− μS0 )−1 1 = − μ2 1 (I − μW )−1 1 = m
by (6) and again the result follows from (15). Lemma 22. Let Y1 and Y2 be invertible matrices, and let yj ⎛ Y
=
⎝
Then det(Y )
Y1
−J
−J Y2
⎞ ⎠.
= 0 if and only if y1 y2 = 1, equivalently
y1 1 + y1
+
y2 1 + y2
= 1,
= 1 Yj−1 1 for j = 1, 2. Let also
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
and in case det(Y ) −1
1Y
= 0,
=y
1
where y
y1
=
1+y
1 + y1
y2
+
1 + y2
.
Proof. This follows immediately from (3) and (5). Lemma 23. Let d ⎛
⎞
Y1
Y
=
2 and
⎜ ⎜ ⎜ −J ⎜ ⎜ . ⎜ . ⎜ . ⎝ −J
−J · · · −J ⎟ ⎟ Y2 · · · −J ⎟ ⎟ .. . . .. ⎟ ⎟ . . . ⎟ ⎠
−J . . . Yd .
Suppose that all proper principal block submatrices of Y are nonsingular and that yj 1, . . . , d. Then det(Y ) = 0 if and only if
1
=
d
yj
(39)
1 + yj j=1
and in case det(Y ) −1
1Y
1
= 1 Yj−1 1 for j =
= 0,
=y
where y 1+y
=
d
yj
1 + yj j=1
.
Proof. This follows from Lemma 22 by an induction argument. We say that a matrix T is a block tournament matrix if it has the form ⎛
⎞
T1 J
T
where d
=
⎜ ⎜ ⎜ J T2 ⎜ ⎜ . . ⎜ . . ⎜ . . ⎝ J J
··· J ⎟ ⎟ ··· J ⎟ ⎟ . . .. ⎟ . . ⎟ ⎟ ⎠ . . . Td
2 and T1 , T2 , . . . , Td be tournament matrices of sizes 1 , 2 , . . . , d .
(40)
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
343
Corollary 24. Let d 2 and T be a block tournament matrix as in (40) with T1 , T2 , . . . , Td extremal tournament matrices. Let μ = (2ρ(T ) + 1)−1 and let mj be related to j as in (38) then
=
1
d
2mj μ
j=1
1 + mj μ
.
(41)
Proof. Decoding (37) with A0 1
(ρ(T )I − Tj )−1 1 =
= 2Tj + I yields 2mj μ
1 − mj μ
μ < ρ(Aj )−1 . Substituting in (39) gives the result.
since 0
Proof of Theorem 18. Since d 2, for each j, the matrix I + 2T contains a (j + 1)×(j + 1) submatrix with row sums greater than or equal to j + 1. It follows that 2ρ(T ) + 1 j + 1 and that j < μ−1 so that (36) applies. The function x → x(1 + x)−1 is increasing on x 0. By (36) and (38), we have mj j (1 − μ2 ). Therefore, from (41) 1
Let ν
d
2j μ(1 − μ2 )
j=1
1 + j μ(1 − μ2 )
.
= μ(1 − μ2 ). In case r = 0, this gives 1 (2d − 1)ν and in case r > 0 it gives 1 − (2n − 2 − 1)ν
− ( + 1)(2d − 1)ν 2 0
after a calculation. Alternatively, this can be written q(ν −1 ) cases, this implies that ν −1 ρ(H0 + J ). But β = μ−1 , so 1
β
1−
1
β2
=ν
0 where q(λ) is defined in (28). In both
1
ρ(H0 + J )
which yields the desired inequality. 6. Final resolution We denote ρ = ρ(H0 + J ) and β = ρ(2G0 + I ). Let obtainable in view of (32) or (33), whichever applies. Proposition 25. Let n
5 and define γ = ρ −
11 −1 ρ . 10
α be the best upper bound for ρ(H + J )
Then α
< γ < β.
Proof. Let ψ(t ) = t 3 − ρ t 2 + ρ . Then it can be checked that ψ(γ ) < 0 unless ρ < 4·9 which implies n 4 by (29). Therefore ψ(γ ) < ψ(β) by Theorem 18. Now ψ (t ) = 3t 2 − 2ρ t and so ψ is increasing on t > 23 ρ . We need to know that both β and γ lie in this range. For γ this follows since
ρ>
√
3·3 again by (29). For β , we use Theorem 18 in the form
β+
1
β2 − 1
to deduce that β
+
1
β +1
=
β3 ρ β2 − 1
> 23 ρ since clearly β 2. We may conclude that γ < β .
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S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
Next we show that α
< γ . In case r = 0, ρ = 2n − and a calculation shows that
(γ + )(γ − ρ) +
9
=
5
280n2
− 500n + 121 + 1802 >0 100(2n − )2
since 2 n. One easily checks that γ > n − , the value at which t → (t + )(t − ρ) + 95 takes its minimum, so α < γ in this case by (32). For 0 < r < d we write q(λ) = (λ − ρ)(λ − ρ ) where ρ is the negative root of q(λ) = 0. Then q(γ ) +
9 5
=−
11 10ρ
ρ − ρ −
11 −1 ρ 10
9
+ . 5
This is positive if
−ρ + ρ +
11 −1 ρ
10
+
18 11
ρ>0
and therefore if
(ρ + ρ ) − But ρ
4 11
ρ > 0.
+ ρ can be read off Eq. (28). Hence, we verify 11(2n − 2 − 1) − 4(2n − )
= 14n − 18 − 11 5n − 11 > 0
using ρ 2n − from (29), n 2 and assuming n 3. Again, one easily checks that γ > n − − 12 , the value at which ψ takes its minimum. Hence α < γ by (33). So α < γ in both cases and result is proved. Corollary 26. For 2
d n, α < β .
Proof. The cases n = 2, 3, 4 follow by making explicit numerical estimates. The case n from Proposition 25.
5 follows
We state some lemmas that will be needed for the uniqueness assertion in Theorem 1. We leave the proofs to the reader. Lemma 27. Let n
3 and T an n × n extremal tournament matrix. Then T is irreducible.
Lemma 28. Let d
2 and T a block tournament matrix as in (40). Then T is irreducible.
Lemma 29. Let A be an n × n irreducible matrix with nonnegative entries. Let B have entries such that 0 bjk ajk for 1 j, k n. Suppose that B = A. Then ρ(B) < ρ(A).
− →
Proof of Theorem 1. A digraph G is said to be full if for every pair of distinct vertices u and v either uv − → or vu is a directed edge in G. Suppose first that G is a full digraph with n vertices and clique number d and suppose that ρ(G) is maximal among all digraphs with n vertices and clique number d. If d = 1, then G is a tournament and therefore an extremal tournament and the result is proved. Hence we may assume that 2 d n. Now let H be the associated undirected graph. Then either H is isomorphic to the d-partite Turán graph on n vertices or it is not. If it is not, then ρ(H + J ) satisfies (32) or (33), whichever applies. But
S.W. Drury, H. Lin / Linear Algebra and its Applications 439 (2013) 328–345
G +G But
345
= H +J −I and it follows from Bendixson’s Theorem [1, Théorème II] that ρ(G) ρ( 12 (H +J −I )). ρ
1 2
(H + J − I ) =
1 2
1
(α − 1) < (β − 1) = ρ(G0 ) 2
and ρ(G) cannot be maximal. This contradicts the supposition that H is not isomorphic to H0 . But in case H is isomorphic to H0 , the adjacency matrix of G is of the form (40) and with the blocks as in H0 . It now follows by successively applying Theorem 3 that each of the tournaments Tj (j = 1, . . . , d) figuring is extremal. Now we consider the case that ρ(G) is maximal, but G not full. Then it is easy to construct G1 G with G1 full, G1 = G and G and G1 having the same associated graph. Then G1 has n vertices and clique number d and ρ(G1 ) ρ(G). Hence ρ(G1 ) is also maximal and it follows that G1 is a G0 . But unless n = 2 and d = 1 it follows from Lemmas 27 and 28 that G0 and hence G1 is strongly connected. But G1 G and G1 = G. It follows now from Lemma 29 that ρ(G) < ρ(G1 ) contradicting the maximality of ρ(G). Acknowledgements We thank the referee for many helpful comments and for simplifying the proof of Proposition 25. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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