# On Some Properties for the Sequence of Brualdi-Li Matrices

Authors: Xiaogen Chen
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

LetB2n denote the Brualdi-Li matrix, and let ρ2n=ρ(B2n) denote the Perron value of the Brualdi-Li matrix of order 2n. We prove that 2n(n-1/2-ρ2n) is monotonically decreasing for all n and ρ2n<n-1/2-(e2-1)/4(e2+1)n, where e=2.718281828459045….


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## Body

## 1. Introduction

A tournament matrix of ordern is a (0,1) matrix T satisfying the equation T+Tt=J-I, where J is the all ones matrix, I is the identity matrix, and Tt is the transpose of T. If n is odd, then the maximum possible Perron value of a tournament matrix T of order n is (n-1)/2, attained when the row sums of T are all equal [1]. However, if n is even, it is difficult to attain. Brualdi and Li [2] conjectured that the matrix
(1)ℬ2n=(UnUntI+UntUn),
where Un is strictly upper triangular tournament matrix (all of its entries above the main diagonal are equal to one), that is,(2)Un=(011⋯1001⋯1⋮⋱⋱⋱⋮00⋯0100⋯00)n×n,
is the tournament matrix of order 2n which maximizes the Perron value. This conjecture has recently been confirmed in [3].The matrixℬ2n has been dubbed by the Brualdi-Li matrix [4]. Let ρ2n=ρ(B2n) denote the Perron value of the Brualdi-Li matrix of order 2n.It is well known thatn-1<ρ2n<n.Friedland [4] has paid particular attention to the behavior of the Perron value of ℬ2n. Friedland and Katz [5] posed the problem of determining the behavior of the sequence 2n(n-1/2-ρ2n). Kirkland [6] conjectured that 2n(n-1/2-ρ2n) is monotonically decreasing for all n. We investigate this conjecture and obtain the following results.Theorem 1.
Letn≥2 be an integer, and let ρ2n=ρ(ℬ2n) be the Perron value of the Brualdi-Li matrix ℬ2n. Then 2n(n-1/2-ρ2n) is monotonically decreasing for all n.By Theorem1 and Lemma 4, one can easily prove the following corollaries.Corollary 2.
Letn≥2 be an integer, and let ρ2n=ρ(ℬ2n) be the Perron value of the Brualdi-Li matrix ℬ2n. Then
(3)ρ2n<n-12-e2-14(e2+1)n,
where e=2.718281828459045….In [6], Kirkland proved that
(4)limn→+∞n3(n-12-γ4n-ρ2n)=γ216+e2(3γ+1)12(e2+1)2,
where γ=(e2-1)/(e2+1).We make the following conjecture and remark that it is supported by computations forn≤100.Conjecture 3.
n
3
(
n
-
1
/
2
-
γ
/
4
n
-
ρ
2
n
) is monotonically decreasing for all n>1.We give some lemmas in Section2 and prove Theorem 1 in Section 3.


## 2. Some Lemmas

### 2.1. Intermediate Value Theorem

Iff is a real-valued continuous function on the closed interval [a,b], and μ is a number between f(a) and f(b), then there is ξ∈(a,b) such that f(ξ)=μ. where (a,b) is open interval.For two variables, the Implicit Function Theorem is as follows.


### 2.2. Implicit Function Theorem

If a real-valued functionF is defined on an open disk containing point (x0,y0), satisfying(i)
F(x0,y0)=0,(ii)
∂F/∂y|(x0,y0)≠0,(iii)
∂F/∂x and ∂F/∂y are continuous on the open disk,then the equation F(x,y)=0 defines y as a function of x near the point (x0,y0) and the derivative of this function is given by
(5)dydx=-∂F/∂x∂F/∂y,
where ∂F/∂x and ∂F/∂y indicate the partial derivatives of F with respect to x and y.Lemma 1 (see [6]).
Letn≥2 be an integer, and let ρ2n=ρ(ℬ2n) be the Perron value of the Brualdi-Li matrix ℬ2n. Then
(6)(2ρ2n2+2(1-n)ρ2n+1-n)((1+ρ2n-1)2n+1)-1=0.LetF=F(m,ρ)=(2ρ2+2(1-m)ρ+1-m)((1+ρ-1)2m+1)-1 and D={(m,ρ)∣m>2, m-1<ρ<m, m and ρ are real numbers}, then the partial derivatives of F are
(7)∂F∂m=-(2ρ+1)((1+ρ-1)2m+1)+2(2ρ2+2(1-m)ρ+1-m)×(1+ρ-1)2mln(1+ρ-1),∂F∂ρ=2(2ρ-m+1)((1+ρ-1)2m+1)-2mρ-2(2ρ2+2(1-m)ρ+1-m)×(1+ρ-1)2m-1.
Evidently, ∂F/∂m and ∂F/∂ρ are derivable on the open disk D, and ∂F/∂ρ=2(2ρ-m+1)+2((2ρ-m+1)(1+ρ-1) -mρ-2(2ρ2+2(1-m)ρ+1-m))·(1+ρ-1)2m-1>0, (m,ρ)∈D. By the Implicit Function Theorem, the equation F(m,ρ)=0 defines ρ as a function in terms of m and the derivative of this function is given by
(8)dρdm=-∂F/∂m∂F/∂ρ.Lemma 2.
Letm≥3 be a real number, and let ρ be a function in terms of m satisfying the equation F(m,ρ)=0. Then
(9)m-12-15m<ρ<m-12-1053m,
where F(m,ρ)=(2ρ2+2(1-m)ρ+1-m)((1+ρ-1)2m+1)-1, (m,ρ)∈D.Proof.
For fixed real numberm≥3, F(m,ρ) is continuous real function in terms of ρ.
Letρ_=m-1/2-1/5m and ρ¯=m-1/2-10/53m. It is easy to show that the real functions y1(x)=(1+(1/(x-1/2-1/5x)))2x and y2(x)=(1+(1/(x-1/2-10/53x)))2x are monotonically decreasing for real number x>2. Thus
(10)F(m,ρ_)=(2ρ_2+2(1-m)ρ_+1-m)×((1+ρ_-1)2m+1)-1=(2(m-12-15m)2+2(1-m)(m-12-15m)+1-m(m-12-15m)2)·((1+1m-1/2-1/5m)2m+1)-1=(12-25+252m2)×((1+1m-1/2-1/5m)2m+1)-1≤(12-25+252·32)×((1+13-1/2-1/15)6+1)-1=-0.0319⋯<0.
Notice that(11)limx→+∞(1+1x-1/2-10/53x)2x=e2=7.38905609893065…,
where e=2.718281828459045…. We have
(12)F(m,ρ¯)=(2(m-12-1053m)2+2(1-m)×(m-12-1053m)+1-m(m-12-1053m)2)·((1+1m-1/2-10/53m)2m+1)-1=(12-2053+200(53m)2)×((1+1m-1/2-10/53m)2m+1)-1>(12-2053)(e2+1)-1>0.
HenceF(m,ρ_)F(m,ρ¯)<0. By the Intermediate Value Theorem, there is ρ∈(ρ_,ρ¯) such that F(m,ρ)=0; that is,
(13)m-12-15m<ρ<m-12-1053m.
We are done.Lemma 3.
Letm≥3 be a real number, and let ρ be a function in terms of m satisfying the equation F(m,ρ)=0, and let G=G(m,ρ)=2ρ2+2(1-m)ρ+1-m, (m,ρ)∈D. Then
(14)(110+225m2)(93106-2002089m2)<G(m,ρ)(1-G(m,ρ))<(910-225m2)(13106+2002089m2).Proof.
For fixed real numberm≥3, G(m,ρ) is continuous real function in terms of ρ. Let ρ_=m-1/2-1/5m and ρ¯=m-1/2-10/53m; then
(15)G(m,ρ_)=110+225m2,G(m,ρ¯)=13106+2002089m2.
By Lemma 2, we have
(16)G(m,ρ_)<G(m,ρ)=2ρ2+2(1-m)ρ+1-m<G(m,ρ¯).
Hence
(17)110+225m2<G(m,ρ)<10106+2002089m2;
that is,
(18)0<93106-2002089m2<1-G(m,ρ)<910-225m2.
We have the desired result.Lemma 4 (see [6]).
There is anM such that the sequence 2n(n-1/2-ρ2n) decreases monotonically to its limit (e2-1)/2(e2+1) for n>M.


## 2.1. Intermediate Value Theorem

Iff is a real-valued continuous function on the closed interval [a,b], and μ is a number between f(a) and f(b), then there is ξ∈(a,b) such that f(ξ)=μ. where (a,b) is open interval.For two variables, the Implicit Function Theorem is as follows.


## 2.2. Implicit Function Theorem

If a real-valued functionF is defined on an open disk containing point (x0,y0), satisfying(i)
F(x0,y0)=0,(ii)
∂F/∂y|(x0,y0)≠0,(iii)
∂F/∂x and ∂F/∂y are continuous on the open disk,then the equation F(x,y)=0 defines y as a function of x near the point (x0,y0) and the derivative of this function is given by
(5)dydx=-∂F/∂x∂F/∂y,
where ∂F/∂x and ∂F/∂y indicate the partial derivatives of F with respect to x and y.Lemma 1 (see [6]).
Letn≥2 be an integer, and let ρ2n=ρ(ℬ2n) be the Perron value of the Brualdi-Li matrix ℬ2n. Then
(6)(2ρ2n2+2(1-n)ρ2n+1-n)((1+ρ2n-1)2n+1)-1=0.LetF=F(m,ρ)=(2ρ2+2(1-m)ρ+1-m)((1+ρ-1)2m+1)-1 and D={(m,ρ)∣m>2, m-1<ρ<m, m and ρ are real numbers}, then the partial derivatives of F are
(7)∂F∂m=-(2ρ+1)((1+ρ-1)2m+1)+2(2ρ2+2(1-m)ρ+1-m)×(1+ρ-1)2mln(1+ρ-1),∂F∂ρ=2(2ρ-m+1)((1+ρ-1)2m+1)-2mρ-2(2ρ2+2(1-m)ρ+1-m)×(1+ρ-1)2m-1.
Evidently, ∂F/∂m and ∂F/∂ρ are derivable on the open disk D, and ∂F/∂ρ=2(2ρ-m+1)+2((2ρ-m+1)(1+ρ-1) -mρ-2(2ρ2+2(1-m)ρ+1-m))·(1+ρ-1)2m-1>0, (m,ρ)∈D. By the Implicit Function Theorem, the equation F(m,ρ)=0 defines ρ as a function in terms of m and the derivative of this function is given by
(8)dρdm=-∂F/∂m∂F/∂ρ.Lemma 2.
Letm≥3 be a real number, and let ρ be a function in terms of m satisfying the equation F(m,ρ)=0. Then
(9)m-12-15m<ρ<m-12-1053m,
where F(m,ρ)=(2ρ2+2(1-m)ρ+1-m)((1+ρ-1)2m+1)-1, (m,ρ)∈D.Proof.
For fixed real numberm≥3, F(m,ρ) is continuous real function in terms of ρ.
Letρ_=m-1/2-1/5m and ρ¯=m-1/2-10/53m. It is easy to show that the real functions y1(x)=(1+(1/(x-1/2-1/5x)))2x and y2(x)=(1+(1/(x-1/2-10/53x)))2x are monotonically decreasing for real number x>2. Thus
(10)F(m,ρ_)=(2ρ_2+2(1-m)ρ_+1-m)×((1+ρ_-1)2m+1)-1=(2(m-12-15m)2+2(1-m)(m-12-15m)+1-m(m-12-15m)2)·((1+1m-1/2-1/5m)2m+1)-1=(12-25+252m2)×((1+1m-1/2-1/5m)2m+1)-1≤(12-25+252·32)×((1+13-1/2-1/15)6+1)-1=-0.0319⋯<0.
Notice that(11)limx→+∞(1+1x-1/2-10/53x)2x=e2=7.38905609893065…,
where e=2.718281828459045…. We have
(12)F(m,ρ¯)=(2(m-12-1053m)2+2(1-m)×(m-12-1053m)+1-m(m-12-1053m)2)·((1+1m-1/2-10/53m)2m+1)-1=(12-2053+200(53m)2)×((1+1m-1/2-10/53m)2m+1)-1>(12-2053)(e2+1)-1>0.
HenceF(m,ρ_)F(m,ρ¯)<0. By the Intermediate Value Theorem, there is ρ∈(ρ_,ρ¯) such that F(m,ρ)=0; that is,
(13)m-12-15m<ρ<m-12-1053m.
We are done.Lemma 3.
Letm≥3 be a real number, and let ρ be a function in terms of m satisfying the equation F(m,ρ)=0, and let G=G(m,ρ)=2ρ2+2(1-m)ρ+1-m, (m,ρ)∈D. Then
(14)(110+225m2)(93106-2002089m2)<G(m,ρ)(1-G(m,ρ))<(910-225m2)(13106+2002089m2).Proof.
For fixed real numberm≥3, G(m,ρ) is continuous real function in terms of ρ. Let ρ_=m-1/2-1/5m and ρ¯=m-1/2-10/53m; then
(15)G(m,ρ_)=110+225m2,G(m,ρ¯)=13106+2002089m2.
By Lemma 2, we have
(16)G(m,ρ_)<G(m,ρ)=2ρ2+2(1-m)ρ+1-m<G(m,ρ¯).
Hence
(17)110+225m2<G(m,ρ)<10106+2002089m2;
that is,
(18)0<93106-2002089m2<1-G(m,ρ)<910-225m2.
We have the desired result.Lemma 4 (see [6]).
There is anM such that the sequence 2n(n-1/2-ρ2n) decreases monotonically to its limit (e2-1)/2(e2+1) for n>M.


## 3. Proof of Theorem1

In Section2, F=F(m,ρ)=(2ρ2+2(1-m)ρ+1-m)((1+ρ-1)2m+1)-1 and G=G(m,ρ)=2ρ2 +2(1-m)ρ+1-m, (m,ρ)∈D, and ρ is the function in terms of m satisfying the equation F(m,ρ)=0. Clearly, F and G are the functions in terms of m.Letm≥3 be real number. Since F(m,ρ)=0, we have
(19)(1+ρ-1)2m=1-G(m,ρ)G(m,ρ),dFdm=0.
Further
(20)dFdm=ddm(G((1+ρ-1)2m+1))=Gddm((1+ρ-1)2m+1)+((1+ρ-1)2m+1)dGdm=2G(1+ρ-1)2m(ln(1+ρ-1)-mρ(1+ρ)dρdm)+((1+ρ-1)2m+1)((4ρ+2-2m)dρdm-2ρ-1)=2(1-G)(ln(1+ρ-1)-mρ(1+ρ)dρdm)+1G((4ρ+2-2m)dρdm-2ρ-1)=1G(mρ(1+ρ)2(1-G)Gln(1+ρ-1)-2(1-G)Gmρ(1+ρ)dρdm+(4ρ+2-2m)dρdm-2ρ-1mρ(1+ρ))=1ρ(1+ρ)G×(dρdm2ρ(1+ρ)(1-G)Gln(1+ρ-1)-2m(1-G)Gdρdm+ρ(1+ρ)((4ρ+2-2m)dρdm-2ρ-1))=1ρ(1+ρ)G×(dρdm2ρ(1+ρ)(1-G)Gln(1+ρ-1)+2m(G-1)Gdρdm+ρ(1+ρ)(4ρ+2-2m)dρdm-ρ(1+ρ)(2ρ+1)dρdm)=0.
Thus
(21)(2m(G-1)G+ρ(1+ρ)(4ρ+2-2m))dρdm=2ρ(1+ρ)(G-1)Gln(1+ρ-1)+ρ(1+ρ)(2ρ+1).
We have
(22)dρdm=(2ρ(1+ρ)(G-1)Gln(1+ρ-1)+ρ(1+ρ)(2ρ+1))×(2m(G-1)G+ρ(1+ρ)(4ρ+2-2m))-1.
Hence
(23)dρdm-2+2ρ+12m=2ρ(1+ρ)(G-1)Gln(1+ρ-1)+ρ(1+ρ)(2ρ+1)2m(G-1)G+ρ(1+ρ)(4ρ+2-2m)+2ρ+1-4m2m=(2m(2ρ(1+ρ)(G-1)Gln(1+ρ-1)+ρ(1+ρ)(2ρ+1)))×(2m(ρ(1+ρ)(4ρ+2-2m)-2m(1-G)G))-1+((2m(G-1)G+ρ(1+ρ)(4ρ+2-2m))×(2ρ+1-4m))×(2m(ρ(1+ρ)(4ρ+2-2m)-2m(1-G)G))-1.
Let A=ρ(1+ρ)(4ρ+2-2m)-2m(1-G)G and
(24)B=2m(2ρ(1+ρ)(G-1)Gln(1+ρ-1)+ρ(1+ρ)(2ρ+1){(1+ρ-1)})+(2m(G-1)G+ρ(1+ρ)(4ρ+2-2m))×(2ρ+1-4m);
then
(25)dρdm-2+2ρ+12m=B2mA,B=4mρ(1+ρ)(G-1)Gln(1+ρ-1)+2mρ(1+ρ)(2ρ+1)+2m(2ρ+1-4m)(G-1)G+ρ(1+ρ)(4ρ+2-2m)(2ρ+1-4m)=2mρ(1+ρ)(2ρ+1)+ρ(1+ρ)(4ρ+2-2m)(2ρ+1-4m)+4mρ(1+ρ)(G-1)Gln(1+ρ-1)+2m(2ρ+1-4m)(G-1)G=2ρ(1+ρ)(m(2ρ+1)+(2ρ+1-m)(2ρ+1-4m))+2mG(1-G)×(4m-2ρ-1-2ρ(1+ρ)ln(1+ρ-1)).
Let C=m(2ρ+1)+(2ρ+1-m)(2ρ+1-4m) and D=4m-2ρ-1-2ρ(1+ρ)ln(1+ρ-1), then
(26)B=2ρ(1+ρ)C+2mG(1-G)D.
Obviously,
(27)C=m(2ρ+1)+(2ρ+1-m)(2ρ+1-4m)=(2ρ+1-2m)2>0.
By Lemma 2 and Taylor’s Theorem, we have
(28)D=4m-2ρ-1-2ρ(1+ρ)ln(1+ρ-1)>4m-2ρ-1-2ρ(1+ρ)(1ρ-12ρ2+13ρ3)=4m-4ρ-2+13ρ-23ρ2>4m-4(m-12-1053m)-2+13ρ-23ρ2=13ρ(1-2ρ)+4053m.
Due to ρ>m-1≥2, then
(29)D>0.
By Lemmas 2 and 3,
(30)A=ρ(1+ρ)(4ρ+2-2m)-2m(1-G)G>(m-1)m(4(m-1)+2-2m)-2m(910-225m2)(13106+2002089m2)=2m((m-1)2-(910-225m2)(13106+2002089m2))>0.
By Lemma 3 and (25)–(30), we have B/2mA>0 for m≥3. Hence (d/dm)(2m(m-1/2-ρ))<0, where m≥3 is a real number.To sum up, we obtain the following result.Letm≥3 be a real number, and let ρ be a function of m satisfying the equation F(m,ρ)=0, (m,ρ)∈D; then 2m(m-1/2-ρ) is monotonically decreasing function for all m.Note thatρ2n=ρ(ℬ2n) is the Perron value of the Brualdi-Li matrix ℬ2n, by Lemma 1, ρ2n satisfy the equation
(31)F(n,ρ2n)=(2ρ2n2+2(1-n)ρ2n+1-n)×((1+ρ2n-1)2n+1)-1=0,
where n≥3 is an integer. It is easy to see that 2n(n-1/2-ρ2n) is monotonically decreasing for all n≥3.Since(32)2×2(2-12-ρ2×2)=4(1.5-1.3953)>2×3(3-12-ρ2×3)=6(2.5-2.4340),
evidently, it follows that 2n(n-1/2-ρ2n) is monotonically decreasing for all integers n≥2. We complete the proof of  Kirkland’s conjecture.



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