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Lih Wang's and Dittert's conjectures on permanents

  • Divya K. Udayan and Kanagasabapathi Somasundaram EMAIL logo
Published/Copyright: May 22, 2024
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Special Matrices
From the journal Special Matrices Volume 12 Issue 1

Abstract

Let Ω n denote the set of all doubly stochastic matrices of order n . Lih and Wang conjectured that for n ≥ 3 , per ( t J n + ( 1 − t ) A ) ≤ t per J n + ( 1 − t ) per A , for all A ∈ Ω n and all t ∈ [ 0.5 , 1 ] , where J n is the n × n matrix with each entry equal to 1 n . This conjecture was proved partially for n ≤ 5 . Let K n denote the set of nonnegative n × n matrices whose elements have sum n . Let ϕ be a real valued function defined on K n by ϕ ( X ) = ∏ i = 1 n r i + ∏ j = 1 n c j - per X for X ∈ K n with row sum vector ( r 1 , r 2 , … r n ) and column sum vector ( c 1 , c 2 , … c n ) . A matrix A ∈ K n is called a ϕ -maximizing matrix if ϕ ( A ) ≥ ϕ ( X ) for all X ∈ K n . Dittert conjectured that J n is the unique ϕ -maximizing matrix on K n . Sinkhorn proved the conjecture for n = 2 and Hwang proved it for n = 3 . In this article, we prove the Lih and Wang partially for n = 6 . It is also proved that if A is a ϕ -maximizing matrix on K 4 , then A is fully indecomposable.

Keywords: permanents; doubly stochastic matrices; Lih-Wang conjecture; phi-maximizing matrix; Dittert’s conjecture
MSC 2010: 15A15

1 Introduction

The permanent of an n × n matrix A = ( a i j ) is defined by

per ( A ) = ∑ σ ∈ S n a 1 σ ( 1 ) a 2 σ ( 2 ) … a n σ ( n ) ,

where S n is the symmetric group of degree n .

Let Ω n denote the set of all doubly stochastic matrices of order n and let J n be an n × n matrix with each entry equal to 1 n .

For positive integers n and k with ( 1 ≤ k ≤ n ) , Q k , n denotes the set

{ ( i 1 , … , i k ) ∕ 1 ≤ i 1 ≤ ⋯ ≤ i k ≤ n , i 1 , i 2 , … i n are positive integers}. For α , β ∈ Q k , n , let A ( α ∕ β ) be the submatrix of A obtained by deleting the rows indexed by α and columns indexed by β and let A [ α ∕ β ] be the submatrix of A with rows and columns indexed by α and β , respectively.

For 1 ≤ k ≤ n , the k th order subpermanent of A is defined by

σ k ( A ) = ∑ α , β ∈ Q k , n per ( A [ α ∕ β ] ) .

In this article, we use the following results quoted by Minc [22]:

If A and B are two n × n matrices and 1 ≤ k ≤ n , then

per ( A ) = ∑ β ∈ Q k , n per ( A [ α ∕ β ] ) per ( A ( α ∕ β ) ) , for α ∈ Q k , n ,

and

per ( A + B ) = ∑ k = 0 n S k ( A , B ) ,

where S k ( A , B ) = ∑ α , β ∈ Q k , n per ( A [ α ∕ β ] ) per ( B ( α ∕ β ) ) , per ( A [ α ∕ β ] ) = 1 when k = 0 and per ( B ( α ∕ β ) ) = 1 when k = n .

Several authors have considered the problem of finding an upper bound for the permanent of a convex combination of J n and A , where A ∈ Ω n . In particular, Marcus and Minc [21] have proposed a conjecture, which states that if A ∈ Ω n , n ≥ 2 , then per n J n − A n − 1 ≤ per ( A ) , with equality holding if and only if A = J n , n ≥ 3 . They established that the conjecture is true for n = 2 , or if A is a positive semi-definite symmetric matrix, or if A is in a sufficiently small neighbourhood of J n . Wang [27] proved the Marcus-Minc conjecture for n = 3 with a revised statement of the case of equality, and he proposed a conjecture that per n J n + A n + 1 ≤ per ( A ) .

Rather than consider a particular convex combination of A with J n , Foregger [7] raised a question whether

(1) per ( t J n + ( 1 − t ) A ) ≤ per ( A )

holds for all A ∈ Ω n and 0 < t ≤ n n − 1 ? He proved that the inequality (1) holds for 0 < t ≤ 3 2 and A ∈ Ω 3 with equality if and only if A = J 3 or t = 3 2 and A = 1 2 ( I + P ) (up to permutations of rows and columns), where P is a full cycle permutation matrix. Later, Foregger himself proved [8] that the inequality (1) holds for n = 4 and t 0 < t ≤ 4 3 , where t 0 is approximately equal to 0.28095. Hutchinson [12] proved the Chollet conjecture, which states that per ( A ∘ B ) ≤ per ( A ) per ( B ) , for real 4 × 4 positive semi-definite matrices. Recently, Rodtes [23] proved the Chollet conjecture for full complex 4 × 4 matrices. Hutchinson [11] gave a counterexample to the Drury permanent conjecture concerning the permanent of positive semidefinite matrices.

Another conjecture in permanents is due to Holens [10] and Dokovic [5], namely, the Holens – Dokovic conjecture, which states that if A ∈ Ω n and k is an integer, 1 ≤ k ≤ n , then

(2) σ k ( A ) ≥ ( n − k + 1 ) 2 k n σ k − 1 ( A ) .

They proved this conjecture for k ≤ 3 . Kopotun [17] proved the inequality (2) for k = 4 and n ≥ 5 .

By Minc [22], inequality ( 2 ) is equivalent to the Monotonicity conjecture, which states that if A is any matrix in Ω n and 2 ≤ k ≤ n , then σ k ( X ) is monotonically increasing on the line segment from J n to A . For k = n , the monotonicity conjecture on permanents is equivalent to

per ( ( 1 − t ) J n + t A ) ≤ per ( A )

for all A ∈ Ω n and 0 ≤ t ≤ 1 [15,18]. Most interestingly, Wanless [28] disproved the Holens–Dokovic inequality (2).

Brualdi and Newman [1] considered the inequality

(3) per ( α J n + ( 1 − α ) A ) ≤ α per ( J n ) + ( 1 − α ) per ( A )

and showed that it did not always hold for all A ∈ Ω n and for all α ∈ [ 0 , 1 ] .

For n = 3 , Ko-wei and Wang [19] proved that (3) is true for all A ∈ Ω n and for all α ∈ [ 1 2 , 1 ] . This is known as the Lih Wang Conjecture, which is stated as follows:

Lih Wang conjecture: per ( α J 3 + ( 1 − α ) A ) ≤ α per ( J 3 ) + ( 1 − α ) per ( A ) for all α ∈ [ 1 2 , 1 ] . They also proposed the following conjecture: (quoted by Gi-Sang and Wanless [9]): The inequality (3) is true for all A ∈ Ω n and for all α ∈ [ 1 2 , 1 ] . For n = 4 , Foregger [8] proved that the inequality ( 3 ) holds for all A ∈ Ω 4 and for all α ∈ [ 0.63 , 1 ] . Kopotun [18] extended this inequality ( 3 ) to subpermanents. Subramanian and Somasundaram [25] proved that the inequality ( 3 ) holds for n = 4 and n = 5 with some conditions on A . A detailed survey of conjectures on permanent was given by Gi-Sang and Wanless [9] and Zhang [29]. In the next section, we prove that the Lih Wang conjecture is true for n = 4 and the inequality (3) is valid in [ β , 1 ] , where the value of β is about 0.486. Also, we prove the Lih Wang conjecture for n = 6 with some conditions on A in the interval [0.7836, 1]. The following theorem is due to Foregger [8].

Theorem 1.1

Let A ∈ Ω 4 and t 2 < t ≤ 1 , where t 2 is the unique real root of the polynomial 106 t 3 − 418 t 2 + 465 t − 153 . Then

per ( t J 4 + ( 1 − t ) A ) ≤ t per ( J 4 ) + ( 1 − t ) per A

with equality iff A = J 4 .

Foregger found that the value of t 2 is about 0.6216986477375 .

The following theorem is due to Subramanian and Somasundaram [25]

Theorem 1.2

Let A ∈ Ω n and 2 ≤ k ≤ n . If the polynomial ∑ r = 2 k r ( k − r ) ! n k − r n − r k − r 2 σ r ( A − J n ) t r − 2 has no root in ( 0 , 1 ) , then σ k ( A ) ≥ ( n − k + 1 ) 2 n k σ k − 1 ( A ) , k = 2 , 3 , … , n .

Another conjecture on permanents is Dittert’s conjecture [22, conjecture 28]. Let K n denote the set of all n × n real nonnegative matrices whose entries have sum n . Let ϕ denote a real valued function defined on K n by ϕ ( X ) = ∏ i = 1 n r i + ∏ j = 1 n c j − per ( X ) for X ∈ K n with row sum vector R X = ( r 1 , r 2 , … r n ) and column sum vector C X = ( c 1 , c 2 , … c n ) . For the same X , let ϕ i j ( X ) = ∏ k ≠ i r k + ∏ l ≠ j c l − per X ( i ∣ j ) .

A matrix A ∈ K n is called a ϕ -maximizing matrix on K n if ϕ ( A ) ≥ ϕ ( X ) for all X ∈ K n . An n × n matrix is called fully indecomposable if it does not contain an s × t zero submatrix with s + t = n .

Dittert conjectured that [22, conjecture 28] J n is the unique ϕ -maximizing matrix on K n . Sinkhorn [24] proved that every ϕ -maximizing matrix on K n has a positive permanent and also that the conjecture is true for n = 2 .

Hwang [13,14] investigated some properties of ϕ -maximizing matrices. The following Lemmas and theorems are from Hwang [14].

Lemma 1.1

[14] Let A be a ϕ -maximizing matrix on K n , and let 1 ≤ s < t ≤ n . If columns s and t of A have either the same sums or the same (0, 1) patterns, then the matrix obtained from A by replacing each of columns s and t by its average ( s i j and t i j become ( s i j + t i j ) ∕ 2 ), is also a ϕ -maximizing matrix on K n . A similar statement holds for rows.

Hwang also proved that Dittert’s conjecture is true for positive semidefinite symmetric matrices in K n and for matrices in a sufficiently small neighbourhood of J n in K n .

Lemma 1.2

[14] Let A = [ a i j ] be a ϕ -maximizing matrix on K n . Then

  • ϕ i j ( A ) = ϕ k l ( A ) if a i j > 0 and a k l > 0 .

  • ϕ i j ( A ) ≤ ϕ k l ( A ) if a i j = 0 and a k l > 0 .

Theorem 1.3

[14] If A = [ a i j ] is a ϕ -maximizing matrix on K n , then ϕ i j ( A ) = ϕ ( A ) if a i j > 0 and ϕ i j ( A ) ≤ ϕ ( A ) if a i j = 0 .

Theorem 1.4

[14] J 3 is the unique ϕ -maximizing matrix on K 3 .

Hwang [14] conjectured that one can at least prove that if A is a ϕ maximizing matrix on K n then A is fully indecomposable. Based on Hwang’s remark we prove that if A is a ϕ -maximizing matrix on K 4 then A is fully indecomposable.

Cheon and Wanless [3] proved that if A ∈ K n is partly decomposable, then ϕ ( A ) < ϕ ( J n ) and that if the zeroes in A ∈ K n form a block then A is not a ϕ -maximizing matrix. Cheon and Yoon [4] obtained some sufficient conditions for which the Dittert conjecture holds. More details Dittert function are in [2,16].

2 Lih Wang’s conjecture

In this section, first, we prove the Lih Wang conjecture completely for n = 4 . We improved the results of Theorems 1.1 and 1.2. Later in Theorem 2.2, we prove this conjecture partially for n = 6 .

Theorem 2.1

If A ∈ Ω 4 , then per ( α J 4 + ( 1 − α ) A ) ≤ α per ( J 4 ) + ( 1 − α ) per A for all α ∈ [ t , 1 ] , where the value of t is about 0.485.

Proof

Let f A ( α ) = α per ( J 4 ) + ( 1 − α ) per A − per ( α J 4 + ( 1 − α ) A ) .

As we know that per ( J 4 ) = 3 32 , substituting this value in f A ( α ) , we obtain

f A ( α ) = 3 32 α + ( 1 − α ) per A − per ( α J 4 + ( 1 − α ) A ) .

Now we have to prove that f A ( α ) ≥ 0 . Multiplying both sided by 32, it is enough to prove that

(4) f A ( α ) = 3 α + 32 ( 1 − α ) per ( A ) − 32 per ( α J 4 + ( 1 − α ) A ) ≥ 0 .

Let a i denote the i th column of A and e r ( x ) = e r ( x 1 , x 2 , … x n ) denote the r th elementary symmetric function of x = ( x 1 , x 2 … x n ) . That is, the sum of products of components taken r at a time, r = 1 , 2 , … n . Let T r ( A ) denote the set of n r sums of columns of A taken r at a time. That is, let T r ( A ) = { x = a i 1 + a i 2 + … + a i r ∣ ( i 1 , i 2 , … , i r ) is a r -subset of ( 1 , 2 , … n ) } , then

per A = ∑ T n ( A ) e n ( x ) − ∑ T n − 1 ( A ) e n ( x ) + … + ( − 1 ) n − 1 ∑ T 1 ( A ) e n ( x ) .

When A is doubly stochastic, we have

∑ T n ( A ) e n ( x ) = 1 .

Let B = α J 4 + ( 1 − α ) A . We use the formula of Eberlein and Govind [6, page 392] to calculate permanent of B .

per ( B ) = − 1 3 + 1 9 ∑ T 1 ( B ) ( − 4 e 2 + 9 e 3 − 18 e 4 ) (x) + 1 18 ∑ T 2 ( B ) ( 4 e 2 − 9 e 3 + 18 e 4 ) ( x ) .

The elements of T 1 ( B ) are the columns of B . Let b i denote the i th column of B . Then b i is of the form b i = α 1 4 1 4 1 4 1 4 + ( 1 − α ) a 1 i a 2 i a 3 i a 4 i , where a 1 i a 2 i a 3 i a 4 i is the i th column of A .

Now,

e 2 ( b i ) = e 2 α 4 + ( 1 − α ) a 1 i α 4 + ( 1 − α ) a 2 i α 4 + ( 1 − α ) a 3 i α 4 + ( 1 − α ) a 4 i = α 4 + ( 1 − α ) a 1 i α 4 + ( 1 − α ) a 2 i + α 4 + ( 1 − α ) a 1 i α 4 + ( 1 − α ) a 3 i + … = 3 8 α 2 + 3 4 α ( 1 − α ) + ( 1 − α ) 2 e 2 ( a i ) .

e 3 ( b i ) = α 4 + ( 1 − α ) a 1 i α 4 + ( 1 − α ) a 2 i α 4 + ( 1 − α ) a 3 i + α 4 + ( 1 − α ) a 1 i α 4 + ( 1 − α ) a 2 i α 4 + ( 1 − α ) a 4 i + … = α 3 16 + 3 α 2 ( 1 − α ) 16 + α ( 1 − α ) 2 e 2 ( a i ) 2 + ( 1 − α ) 3 e 3 ( a i ) .

Similarly,

e 4 ( b i ) = α 4 + ( 1 − α ) a 1 i α 4 + ( 1 − α ) a 2 i α 4 + ( 1 − α ) a 3 i α 4 + ( 1 − α ) a 4 i = α 4 256 + α 3 ( 1 − α ) 64 + α 2 ( 1 − α ) 2 e 2 ( a i ) 16 + α ( 1 − α ) 3 e 3 ( a i ) 4 + ( 1 − α ) 4 e 4 ( a i ) .

If x ∈ T 2 ( B ) , then x is of the form x = α 2 + ( 1 − α ) ( a 1 i + a 1 j ) α 2 + ( 1 − α ) ( a 2 i + a 2 j ) α 2 + ( 1 − α ) ( a 3 i + a 3 j ) α 2 + ( 1 − α ) ( a 4 i + a 4 j ) .

Then e 2 ( x ) = [ α 2 + ( 1 − α ) ( a 1 i + a 1 j ) ] [ α 2 + ( 1 − α ) ( a 2 i + a 2 j ) ] + [ α 2 + ( 1 − α ) ( a 1 i + a 1 j ) ] [ α 2 + ( 1 − α ) ( a 3 i + a 3 j ) ] + … = 3 2 α 2 + 3 α ( 1 − α ) + ( 1 − α ) 2 e 2 ( y ) , where y ∈ T 2 ( A ) .

e 3 ( x ) = [ α 2 + ( 1 − α ) ( a 1 i + a 1 j ) ] [ α 2 + ( 1 − α ) ( a 2 i + a 2 j ) ] [ α 2 + ( 1 − α ) ( a 3 i + a 3 j ) ] + [ α 2 + ( 1 − α ) ( a 1 i + a 1 j ) ] [ α 2 + ( 1 − α ) ( a 2 i + a 2 j ) ] [ α 2 + ( 1 − α ) ( a 4 i + a 4 j ) ] + … = α 3 2 + 3 2 α 2 ( 1 − α ) + α ( 1 − α ) 2 e 2 ( y ) + ( 1 − α ) 3 e 3 ( y ) , where y ∈ T 2 ( A ) .

e 4 ( x ) = [ α 2 + ( 1 − α ) ( a 1 i + a 1 j ) ] [ α 2 + ( 1 − α ) ( a 2 i + a 2 j ) ] [ α 2 + ( 1 − α ) ( a 3 i + a 3 j ) ] [ α 2 + ( 1 − α ) ( a 4 i + a 4 j ) ] = α 4 16 + ( α 3 ) ( 1 − α ) 4 + 1 4 α 2 ( 1 − α ) 2 e 2 ( y ) + 1 2 α ( 1 − α ) 3 e 3 ( y ) + ( 1 − α ) 4 e 4 ( y ) , where y ∈ T 2 ( A ) .

Therefore,

∑ i = 1 4 e 2 ( b i ) = 3 2 α 2 + 3 α ( 1 − α ) + ( 1 − α ) 2 ∑ i = 1 4 e 2 ( a i ) , ∑ i = 1 4 e 3 ( b i ) = α 3 4 + 3 4 α 2 ( 1 − α ) + 1 2 α ( 1 − α ) 2 ∑ i = 1 4 e 2 ( a i ) + ( 1 − α ) 3 ∑ i = 1 4 e 3 ( a i )

and

∑ i = 1 4 e 4 ( b i ) = 1 64 α 4 + 1 16 α 3 ( 1 − α ) + 1 16 α 2 ( 1 − α ) 2 ∑ i = 1 4 e 2 ( a i ) + 1 4 α ( 1 − α ) 3 ∑ i = 1 4 e 3 ( a i ) + ( 1 − α ) 4 ∑ i = 1 4 e 4 ( a i ) . ∑ T 2 ( B ) e 2 ( x ) = 18 α − 9 α 2 + ( 1 − α ) 2 ∑ T 2 ( A ) e 2 ( y ) , ∑ T 2 ( B ) e 3 ( x ) = − 6 α 3 + 9 α 2 + α ( 1 − α ) 2 ∑ T 2 ( A ) e 2 ( y ) + ( 1 − α ) 3 ∑ T 2 ( A ) e 3 ( y )

and

∑ T 2 ( B ) e 4 ( x ) = 3 α 3 2 − 9 α 4 8 + ( α ) 2 ( 1 − α ) 2 4 ∑ T 2 ( A ) e 2 ( y ) + α ( 1 − α ) 3 2 ∑ T 2 ( A ) e 3 ( y ) + ( 1 − α ) 4 ∑ T 2 ( A ) e 4 ( y ) .

Now, we use the formula of Eberlein and Govind ([6], equations (3.4) and (3.5)) for any A ∈ Ω 4 ,

∑ T 2 ( A ) e 2 ( x ) = 2 ∑ T 1 ( A ) e 2 ( x ) + 6 , ∑ T 2 ( A ) e 3 ( x ) = 2 ∑ T 1 ( A ) e 2 ( x ) .

By using the aforementioned two equations, we obtain

∑ T 2 ( B ) e 2 ( x ) = 18 α − 9 α 2 + ( 2 − 4 α + 2 α 2 ) ∑ i = 1 4 e 2 ( a i ) + 6 − 12 α + 6 α 2 , ∑ T 2 ( B ) e 3 ( x ) = − 6 α 3 + 9 α 2 + ( α − 2 α 2 + α 3 ) 2 ∑ i = 1 4 e 2 ( a i ) + 6 + ( 2 − 6 α + 6 α 2 − 2 α 3 ) ∑ i = 1 4 e 2 ( a i ) ,

and

∑ T 2 ( B ) e 4 ( x ) = 3 α 3 2 − 9 α 4 8 + α 2 − 2 α 3 + α 4 4 2 ∑ i = 1 4 e 2 ( a i ) + 6 + ( α − 3 α 2 + 3 α 3 − α 4 ) ∑ i = 1 4 e 2 ( a i ) + ( 1 − α ) 4 ∑ T 2 ( A ) e 4 ( y ) .

Therefore,

per ( B ) = − 1 3 + 1 9 − 4 3 α 2 2 + 3 α ( 1 − α ) + ( 1 − α ) 2 ∑ i = 1 4 e 2 ( a i ) + 9 α 3 4 + 3 α 2 ( 1 − α ) 4 + α ( 1 − α ) 2 2 ∑ i = 1 4 e 2 ( a i ) + ( 1 − α ) 3 ∑ i = 1 4 e 3 ( a i ) − 18 α 4 64 + α 3 ( 1 − α ) 16 + α 2 ( 1 − α ) 2 16 ∑ i = 1 4 e 2 ( a i ) + α ( 1 − α ) 3 4 ∑ i = 1 4 e 3 ( a i ) + ( 1 − α ) 4 ∑ i = 1 4 e 4 ( a i ) + 1 18 4 18 α − 9 α 2 + ( 2 − 4 α + 2 α 2 ) ∑ i = 1 4 e 2 ( a i ) + 6 − 12 α + 6 α 2 − 9 − 6 α 3 + 9 α 2 + ( 2 α − 4 α 2 + 2 α 3 ) ∑ i = 1 4 e 2 ( a i ) + 6 α − 12 α 2 + 6 α 3 + ( 2 − 6 α + 6 α 2 − 2 α 3 ) ∑ i = 1 4 e 2 ( a i ) + 18 3 α 3 2 − 9 α 4 8 + α 2 − 2 α 3 + α 4 2 ∑ i = 1 4 e 2 ( a i ) + 3 ( α 2 − 2 α 3 + α 4 ) 2 + ( α − 3 α 2 + 3 α 3 − α 4 ) ∑ i = 1 4 e 2 ( a i ) + ( 1 − α ) 4 ∑ T 2 ( A ) e 4 ( y ) .

Now,

f ( α ) = 3 α + 32 ( 1 − α ) − 1 3 + 1 9 ∑ T 1 ( A ) ( − 4 e 2 + 9 e 3 − 18 e 4 ) ( x ) + 1 18 ∑ T 2 ( A ) ( 4 e 2 − 9 e 3 + 18 e 4 ) ( x ) − 32 − 1 3 + 1 9 − 4 3 α 2 2 + 3 α ( 1 − α ) + ( 1 − α ) 2 ∑ i = 1 4 e 2 ( a i ) + 9 α 3 4 + 3 α 2 ( 1 − α ) 4 + α ( 1 − α ) 2 2 ∑ i = 1 4 e 2 ( a i ) + ( 1 − α ) 3 ∑ i = 1 4 e 3 ( a i ) − 18 α 4 64 + α 3 ( 1 − α ) 16 + α 2 ( 1 − α ) 2 16 ∑ i = 1 4 e 2 ( a i ) + α ( 1 − α ) 3 4 ∑ i = 1 4 e 3 ( a i ) + ( 1 − α ) 4 ∑ i = 1 4 e 4 ( a i ) + 1 18 4 18 α − 9 α 2 + ( 2 − 4 α + 2 α 2 ) ∑ i = 1 4 e 2 ( a i ) + 6 − 12 α + 6 α 2 − 9 − 6 α 3 + 9 α 2 + ( 2 α − 4 α 2 + 2 α 3 ) ∑ i = 1 4 e 2 ( a i ) + 6 α − 12 α 2 + 6 α 3 + ( 2 − 6 α + 6 α 2 − 2 α 3 ) ∑ i = 1 4 e 2 ( a i ) + 18 3 α 3 2 − 9 α 4 8 + α 2 − 2 α 3 + α 4 2 ∑ i = 1 4 e 2 ( a i ) + 3 ( α 2 − 2 α 3 + α 4 ) 2 + ( α − 3 α 2 + 3 α 3 − α 4 ) ∑ i = 1 4 e 2 ( a i ) + ( 1 − α ) 4 ∑ T 2 ( A ) e 4 ( y ) .

After simplifying, we obtain a fourth-degree polynomial in α ,

f ( α ) = 128 3 + α 67 − 80 ∑ i = 1 4 e 2 ( a i ) + 80 ∑ i = 1 4 e 3 ( a i ) − 192 ∑ i = 1 4 e 4 ( a i ) + 96 ∑ T 2 ( A ) e 4 (x) + α 2 − 120 + 148 ∑ i = 1 4 e 2 ( a i ) − 144 ∑ i = 1 4 e 3 ( a i ) + 384 ∑ i = 1 4 e 4 ( a i ) − 192 ∑ T 2 ( A ) e 4 (x) + α 3 68 − 88 ∑ i = 1 4 e 2 ( a i ) + 80 ∑ i = 1 4 e 3 ( a i ) − 256 ∑ i = 1 4 e 4 ( a i ) + 128 ∑ T 2 ( A ) e 4 (x) + α 4 − 15 + 24 ∑ i = 1 4 e 2 ( a i ) − 16 ∑ i = 1 4 e 3 ( a i ) + 64 ∑ i = 1 4 e 4 ( a i ) − 32 ∑ T 2 ( A ) e 4 (x) .

By using the MATLAB, we obtained the minimum values of the function f ( α ) for different values of α . We obtained minimum values of f ( α ) in different sub intervals of [0, 1], and the values are shown in Table 1.

Therefore, the function f ( α ) ≥ 0 in [ 0.5 , 1 ] . Hence, Lih-Wang conjecture is true for n = 4 , and in particular, the inequality (3) is true in [0.486, 1].□

Table 1

Minimum value

Interval Minimum at α Minimum value f ( α )
[ 0.5 , 1 ] 0.5 0.1250
[ 0.4 , 1 ] 0.4 ‒ 0.5472
[ 0.45 , 1 ] 0.45 ‒ 0.2654
[ 0.48 , 1 ] 0.48 ‒ 0.0452
[ 0.485 , 1 ] 0.45 ‒ 0.0044
[ 0.486 , 1 ] 0.486 0.0039
[ 0 , 1 ] 0.2545 ‒ 0.8675

Theorem 2.2

If A is a matrix in Ω 6 such that the polynomials

∑ r = 2 4 r ( 4 − r ) ! 6 4 − r 6 − r 4 − r 2 σ r ( A − J 6 ) t r − 2 , ∑ r = 2 5 r ( 5 − r ) ! 6 5 − r 6 − r 5 − r 2 σ r ( A − J 6 ) t r − 2 , and ∑ r = 2 6 r ( 6 − r ) ! 6 6 − r σ r ( A − J 6 ) t r − 2 have no roots in ( 0 , 1 ) , then per ( t J 6 + ( 1 − t ) A ) ≤ t per J 6 + ( 1 − t ) per A for all t ∈ [ 0.7836 , 1 ] .

Proof

t per J 6 + ( 1 − t ) per A − per ( t J 6 + ( 1 − t ) A ) = 5 324 t + ( 1 − t ) per A − ∑ r = 0 6 t r ( 1 − t ) 6 − r r ! 6 r σ 6 − r ( A ) = 5 324 t + ( 1 − t ) per A − ( 1 − t ) 6 per A + 1 6 t ( 1 − t ) 5 σ 5 ( A ) + 2 6 2 t 2 ( 1 − t ) 4 σ 4 ( A ) + 6 6 3 t 3 ( 1 − t ) 3 σ 3 ( A ) + 24 6 4 t 4 ( 1 − t ) 2 σ 2 ( A ) + 120 6 5 t 5 ( 1 − t ) σ 1 ( A ) + 720 6 6 t 6 = 5 324 t + per A ( 1 − t ) [ 1 − ( 1 − t ) 5 ] − 1 6 t ( 1 − t ) 5 σ 5 ( A ) − 1 18 t 2 ( 1 − t ) 4 σ 4 ( A ) − 1 36 t 3 ( 1 − t ) 3 σ 3 ( A ) − 1 54 t 4 ( 1 − t ) 2 σ 2 ( A ) − 5 54 t 5 ( 1 − t ) − 5 324 t 6 = 5 324 t ( 1 − t ) ( 1 + t + t 2 + t 3 + t 4 ) − 5 54 t 5 ( 1 − t ) + per A ( 1 − t ) ( 5 t − 10 t 2 + 10 t 3 − 5 t 4 + t 5 ) − 1 6 t ( 1 − t ) 5 σ 5 ( A ) − 1 18 t 2 ( 1 − t ) 4 σ 4 ( A ) − 1 36 t 3 ( 1 − t ) 3 σ 3 ( A ) − 1 54 t 4 ( 1 − t ) 2 σ 2 ( A ) = 5 324 t ( 1 − t ) ( 1 + t + t 2 + t 3 − 5 t 4 ) + per A t ( 1 − t ) ( 5 − 10 t + 10 t 2 − 5 t 3 + t 4 ) − 1 6 t ( 1 − t ) 5 σ 5 ( A ) − 1 18 t 2 ( 1 − t ) 4 σ 4 ( A ) − 1 36 t 3 ( 1 − t ) 3 σ 3 ( A ) − 1 54 t 4 ( 1 − t ) 2 σ 2 ( A ) = t ( 1 − t ) F A ( t ) ,

where

F A ( t ) = 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) + per A ( 5 − 10 t + 10 t 2 − 5 t 3 + t 4 ) − 1 6 ( 1 − t ) 4 σ 5 ( A ) − 1 18 t ( 1 − t ) 3 σ 4 ( A ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) .

It is enough to prove that F A ( t ) ≥ 0 for all t ∈ [ 0.7836 , 1 ] . Since the theorem holds good for t = 1 , it is sufficient to consider t < 1 .

F A ( t ) = ( 5 − 10 t + 10 t 2 − 5 t 3 + t 4 ) per A − 1 36 σ 5 ( A ) + 1 36 ( 5 − 10 t + 10 t 2 − 5 t 3 + t 4 ) σ 5 ( A ) − 1 6 ( 1 − t ) 4 σ 5 ( A ) − 1 18 t ( 1 − t ) 3 σ 4 ( A ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) .

By Theorem 1.2,

F A ( t ) ≥ 1 36 ( − 1 + 14 t − 26 t 2 + 19 t 3 − 5 t 4 ) σ 5 ( A ) − 1 18 t ( 1 − t ) 3 σ 4 ( A ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) = 1 36 ( − 1 + 14 t − 26 t 2 + 19 t 3 − 5 t 4 ) σ 5 ( A ) − 2 15 σ 4 ( A ) + 1 270 σ 4 ( A ) ( − 1 + 14 t − 26 t 2 + 19 t 3 − 5 t 4 ) − 1 18 t ( 1 − t ) 3 σ 4 ( A ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) .

By Theorem 1.2, F A ( t ) ≥ 1 270 σ 4 ( A ) ( − 1 + 14 t − 26 t 2 + 19 t 3 − 5 t 4 ) − 1 18 t ( 1 − t ) 3 σ 4 ( A ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) .

This implies that

F A ( t ) ≥ 1 270 σ 4 ( A ) { − 1 + 14 t − 26 t 2 + 19 t 3 − 5 t 4 − 15 t ( 1 − 3 t + 3 t 2 − t 3 ) } − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) = 1 270 σ 4 ( A ) ( − 1 − t + 19 t 2 − 26 t 3 + 10 t 4 ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) = 1 270 ( − 1 − t + 19 t 2 − 26 t 3 + 10 t 4 ) σ 4 ( A ) − 3 8 σ 3 ( A ) + 3 2,160 σ 3 ( A ) ( − 1 − t + 19 t 2 − 26 t 3 + 10 t 4 ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 )

Again by Theorem 1.2,

F A ( t ) ≥ 3 2,160 σ 3 ( A ) ( − 1 − t + 19 t 2 − 26 t 3 + 10 t 4 ) − 1 36 t 2 ( 1 − t ) 2 σ 3 ( A ) − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) = σ 3 ( A ) − 3 2,160 − 3 2,160 t − 3 2,160 t 2 + 42 2,160 t 3 − 30 2,160 t 4 − 1 54 t 3 ( 1 − t ) σ 2 ( A ) + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 )

From Dokovic [5] and Tverberg [26], we know that for A ∈ Ω n ,

σ 2 ( A ) = 1 2 ∑ i , j = 1 n a i j 2 + 1 2 n ( n − 2 )

and

σ 3 ( A ) = 2 3 ∑ i , j = 1 n a i j 3 + 1 2 ( n − 4 ) ∑ i , j = 1 n a i j 2 + 1 6 n ( n 2 − 6 n + 10 ) .

Therefore,

F A ( t ) ≥ 1 2,160 ( − 3 − 3 t − 3 t 2 + 42 t 3 − 30 t 4 ) 2 3 ∑ i , j = 1 6 a i j 3 + ∑ i , j = 1 6 a i j 2 + 10 − 1 54 t 3 ( 1 − t ) 1 2 ∑ i , j = 1 6 a i j 2 + 12 + 5 324 ( 1 + t + t 2 + t 3 − 5 t 4 ) = 1 3,240 ( − 3 − 3 t − 3 t 2 + 42 t 3 − 30 t 4 ) ∑ i , j = 1 6 a i j 3 + 1 2,160 ∑ i , = 1 6 a i j 2 ( − 3 − 3 t − 3 t 2 + 22 t 3 − 10 t 4 ) + 1 648 ( 1 + t + t 2 − 8 t 3 + 4 t 4 ) = ∑ i = 1 6 { 1 3,240 ( − 3 − 3 t − 3 t 2 + 42 t 3 − 30 t 4 ) ∑ j = 1 6 a i j 3 + 1 2,160 ∑ j = 1 6 a i j 2 ( − 3 − 3 t − 3 t 2 + 22 t 3 − 10 t 4 ) } + 1 648 ( 1 + t + t 2 − 8 t 3 + 4 t 4 ) .

It is easy to see that the coefficient of ∑ j = 1 6 a i j 3 is nonnegative for all t ∈ [ 0.6530 , 1 ] and ( ∑ j = 1 6 a i j 2 ) 2 ≤ ∑ j = 1 6 a i j 3 for i = 1 , 2 , … 6 (Kopotun [17]).

Therefore, for all t ∈ [ 0.6530 , 1 ] ,

F A ( t ) ≥ ∑ i = 1 6 1 3,240 ( − 3 − 3 t − 3 t 2 + 42 t 3 − 30 t 4 ) ∑ j = 1 6 a i j 2 2 + 1 2,160 ( − 3 − 3 t − 3 t 2 + 22 t 3 − 10 t 4 ) ∑ j = 1 6 a i j 2 + 1 648 ( 1 + t + t 2 − 8 t 3 + 4 t 4 ) .

Put x = ∑ j = 1 6 a i j 2 and f ( x ) = 1 3,240 ( − 3 − 3 t − 3 t 2 + 42 t 3 − 30 t 4 ) x 2 + 1 2,160 ( − 3 − 3 t − 3 t 2 + 22 t 3 − 10 t 4 ) x .

Now, f ′ ( x ) = 2 3,240 x ( − 3 − 3 t − 3 t 2 + 42 t 3 − 30 t 4 ) + 1 2,160 ( − 3 − 3 t − 3 t 2 + 22 t 3 − 10 t 4 ) .

It is easy to see that for t ∈ [ 0.7836 , 1 ] , f ′ ( x ) ≥ 0 .

London and Minc (Lemma 1, [20]) showed that x = ∑ j = 1 6 a i j 2 ≥ 1 6 .

Hence, f ( x ) is an increasing function on [ 1 6 , ∞ ) .

That is, f ( x ) ≥ f ( 1 6 ) = − 1 3,888 − 1 3,888 t − 1 3,888 t 2 + 1 486 t 3 − 1 972 t 4

Therefore, t ∈ [ 0.7836 , 1 ]

F A ( t ) ≥ 6 − 1 3,888 − 1 3,888 t − 1 3,888 t 2 + 1 486 t 3 − 1 972 t 4 + 1 648 ( 1 + t + t 2 − 8 t 3 + 4 t 4 ) = 0 .

Hence, the theorem.□

The following two corollaries are consequence of the aforementioned theorem and theorems in [25].

Corollary 2.1

If A is a normal matrix in Ω 6 whose eigenvalues lie in the sector [ − π 10 , π 10 ] and the polynomial ∑ r = 2 6 r ( 6 − r ) ! 6 6 − r σ r ( A − J 6 ) t r − 2 has no root in ( 0 , 1 ) , then per ( t J 6 + ( 1 − t ) A ) ≤ t per J 6 + ( 1 − t ) per A for all t ∈ [ 0.7836 , 1 ] .

Proof

The proof follows since if A is a normal matrix in Ω 6 whose eigenvalues lie in the sector [ − π 10 , π 10 ] , then the polynomial ∑ r = 2 5 r ( 5 − r ) ! 6 5 − r 6 − r 5 − r 2 σ r ( A − J 6 ) t r − 2 has no roots in (0, 1).□

Corollary 2.2

If A is a normal matrix in Ω 6 whose eigenvalues lie in the sector [ − π 12 , π 12 ] then per ( t J 6 + ( 1 − t ) A ) ≤ t per J 6 + ( 1 − t ) per A for all t ∈ [ 0.7836 , 1 ] .

Proof

The proof follows since if A is a normal matrix in Ω 6 whose eigenvalues lie in the sector [ − π 12 , π 12 ] , then the polynomials ∑ r = 2 5 r ( 5 − r ) ! 6 5 − r 6 − r 5 − r 2 σ r ( A − J 6 ) t r − 2 and ∑ r = 2 6 r ( 6 − r ) ! 6 6 − r σ r ( A − J 6 ) t r − 2 have no roots in (0, 1).□

3 Dittert’s conjecture for n = 4

Hwang [14] conjectured that if A is a ϕ maximizing matrix on K n , then A is fully indecomposable. Based on this we prove that if A is a ϕ -maximizing matrix on K 4 , then A is fully indecomposable. We denote R A = ( r 1 , … , r n ) and C A = ( c 1 , … , c n ) are, respectively, the row and column sum vectors of A .

Theorem 3.1

If A is a ϕ -maximizing matrix on K 4 , then A is fully indecomposable.

Proof

Let A = [ a i j ] be a ϕ -maximizing matrix on K 4 . First, we assume that A is not fully indecomposable. Without loss of generality, we may assume that a 12 = a 13 = a 14 = 0 . Now we intend to show that ( a 21 , a 31 , a 41 ) ≠ ( 0 , 0 , 0 ) . Suppose a 21 = a 31 = a 41 = 0 . Let B = A ( 1 ∣ 1 ) and a 11 = a . Then A = a I 1 ⊕ B , ϕ ( A ) = a ϕ ( B ) and ϕ 11 ( A ) = ϕ ( B ) . By Theorem 1.3, a = 1 , and hence, B ∈ K 3 and by Theorem 1.4, B = J 3 , and hence, ϕ ( A ) = ϕ ( B ) = ϕ ( J 3 ) < ϕ ( J 4 ) , which is a contradiction. Thus, ( a 21 , a 31 , a 41 ) ≠ ( 0 , 0 , 0 ) .

If a 21 > 0 , a 31 > 0 and a 41 > 0 , by Lemma 1.2, ϕ 21 ( A ) = ϕ 31 ( A ) = ϕ 41 ( A ) , ϕ 21 ( A ) = r 1 r 3 r 4 + c 2 c 3 c 4 -per A ( 2 ∣ 1 ) , and ϕ 31 ( A ) = r 1 r 2 r 4 + c 2 c 3 c 4 -per A ( 3 ∣ 1 ) . Therefore, ϕ 21 ( A ) = ϕ 31 ( A ) = 0 implies a r 4 ( r 3 − r 2 ) = 0 , and hence, r 2 = r 3 . Then by Lemma 1.1 (the matrix A 1 obtained from A by replacing the second and third rows by its average and A 1 is also a ϕ -maximizing matrix on K 4 ), we assume that A 1 is of the form

A 1 = a 0 0 0 x y z t x y z t p q r s ,

where a x y z t p q r s > 0 . Since the second, third and fourth columns have the same (0, 1) patterns, by Lemma 1.1, we assume that A 1 is of the form

A 2 = a 0 0 0 x y y y x y y y p q q q ,

where a x y p q > 0 . Now since the second, third and fourth rows have the same (0, 1) patterns, by Lemma 1.1 we can assume that A 2 is of the form

A 3 = a 0 0 0 x y y y x y y y x y y y ,

where a x y > 0 . Suppose x ≥ y . We construct A 4 from A 3 such that R A 4 = R A 3 , C A 4 = C A 3 , and per A 4 = per A 3 ,

A 4 = a 0 0 0 x y y y x + y 0 y y x − y 2 y y y .

The matrix A 4 is also a ϕ -maximizing matrix on K 4 since R A 4 = R A 3 , C A 4 = C A 3 , and per A 4 = per A 3 . Now a 33 > 0 and a 42 > 0 , but ϕ 42 ( A 4 ) − ϕ 33 ( A 4 ) = a y 2 > 0 , which is a contradiction to Lemma 1.2. Hence x < y . We construct A 5 from A 3 such that R A 5 = R A 3 , C A 5 = C A 3 , and per A 5 = per A 3 ,

A 5 = a 0 0 0 x y y y 2 x y y y − x 0 y y y + x .

The matrix A 5 is a ϕ -maximizing matrix on K 4 since R A 5 = R A 3 , C A 5 = C A 3 , and per A 5 = per A 3 . So, by Lemma 1.1, we construct the matrix A 6 from A 5 by taking average of third and fourth columns of A 5 .

A 6 = a 0 0 0 x y y y 2 x y y − x 2 y − x 2 0 y y + x 2 y + x 2

is also a ϕ -maximizing matrix on K 4 . R A 6 = R A 5 , C A 6 = C A 5 , but per A 6 = a ( 6 y 3 − y x 2 2 ) < 6 a y 3 = per A 5 . This implies that ϕ ( A 6 ) > ϕ ( A 5 ) = ϕ ( A 3 ) , which is a contradiction. Hence, at least one of a 21 , a 31 , a 41 is zero.

Suppose a 21 > 0 and a 31 = a 41 = 0 . Hence A has the following form:

A = a 0 0 0 b B 0 0

with a b > 0 . We define a matrix to be a positive matrix if all its entries are positive. We claim that B is not a positive matrix. Because if B > 0 , then we may say

A = a 0 0 0 b x x x 0 y y y 0 z z z .

Since the third and fourth rows have the same (0, 1) patterns, we construct the matrix A ˜ from A by taking the average of third and fourth rows of A .

A ˜ = a 0 0 0 b x x x 0 y y y 0 y y y ,

with a b x y > 0 . ϕ 11 ( A ˜ ) − ϕ 21 ( A ˜ ) = 3 y 2 ( 3 b − 3 a + 7 x ) . From Lemma 1.2, ϕ 11 ( A ˜ ) = ϕ 21 ( A ˜ ) , this implies 3 a = 3 b + 7 x . For a sufficiently small ε > 0 , consider the perturbation matrix A ε ˜ ,

A ε ˜ = a − ε ε 0 0 b + ε x − ε x x 0 y y y 0 y y y .

Then R A ε ˜ = R A ˜ , C A ε ˜ = C A ˜ but per A ε ˜ = 6 a x y 2 − 32 3 ε x y 2 + O ( ε 2 ) < 6 a x y 2 = per ( A ˜ ) which gives us ϕ ( A ε ˜ ) > ϕ ( A ˜ ) , which is a contradiction. Therefore, B is not a positive matrix.

Suppose a 21 > 0 and a 31 > 0 and a 41 = 0 . Hence, A has the form

A = a 0 0 0 b B c 0

with a c b > 0 . Suppose B > 0 , then

A = a 0 0 0 b x x x c y y y 0 z z z .

Let x < y and for a sufficiently small ε > 0 , consider the perturbation matrix A ε ˜ ,

A ε ˜ = a 0 0 0 b + ε x − ε x x c − ε y + ε y y 0 z z z .

per A ε ˜ = 6 a x y z + 2 a z ε ( x − y ) and per ( A ˜ ) = 6 a x y z . As x < y gives per A ε ˜ < per ( A ˜ ) , which gives us ϕ ( A ε ˜ ) > ϕ ( A ˜ ) , which is a contradiction. Therefore, B is not a positive matrix.

Since a per B = per A > 0 , we have per B > 0 , and hence, without loss of generality, we assume that

B = x l m 0 y v 0 0 z ,

where x y z l m v > 0 . Consider another perturbation matrix of A

A ˆ = a 0 0 0 b − ε x l m + ε 0 0 y v ε 0 0 z − ε .

A ˆ is a matrix in K 4 whose permanent value is a x y ( z − ε ) , which is strictly less than per A = a x y z . So ϕ ( A ˆ ) is strictly greater than ϕ ( A ) , which is a contradiction to our assumption that A is ϕ -maximizing matrix. So A is fully indecomposable.□

Acknowledgements

The authors would like to thank the anonymous reviewers for their valuable suggestions, which improved the presentation of the article.

  1. Authors contributions: All the authors contributed equally.

  2. Conflict of interest: The authors declare no conflicts of interest.

  3. Data availability statement: No data availability for this article since the authors not used any data.

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Received: 2023-09-15
Revised: 2024-04-15
Accepted: 2024-04-15
Published Online: 2024-05-22

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/spma-2024-0006
Keywords for this article
permanents; doubly stochastic matrices; Lih-Wang conjecture; phi-maximizing matrix; Dittert’s conjecture
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Articles in the same Issue

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  2. The diameter of the Birkhoff polytope
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