aa r X i v : . [ m a t h . R A ] J un SOLUTIONS TO TWO PROBLEMS ON PERMANENTS
PETTER BR ¨AND´EN
Abstract.
In this note we settle two open problems in the theory of perma-nents by using recent results from other areas of mathematics. Both problemswere recently discussed in Bapat’s survey [2]. Bapat conjectured that certainquotients of permanents, which generalize symmetric function means, are con-cave. We prove this conjecture by using concavity properties of hyperbolicpolynomials. Motivated by problems on random point processes, Shirai andTakahashi raised the problem: Determine all real numbers α for which the α -permanent (or α -determinant) is nonnegative for all positive semidefinitematrices. We give a complete solution to this problem by using recent resultsof Scott and Sokal on completely monotone functions. It turns out that theconjectured answer to the problem is false. Bapat’s conjecture on quotients of permanents
Recently Gurvits [6] successfully used hyperbolic polynomials to prove inequal-ities for permanents and determinants. In this section we show how a conjecture(Conjecture 1.1 below) of Bapat on the concavity of certain quotients of perma-nents follows from concavity properties of hyperbolic polynomials. Recall that if A = ( a ij ) ni,j =1 is a matrix, then the permanent of A is defined byper( A ) = X σ ∈ S n n Y i =1 a iσ ( i ) , where S n is the symmetric group on { , . . . , n } . Conjecture 1.1 (Bapat, [1]) . Let b , b , . . . , b k be fixed vectors in R n ++ := (0 , ∞ ) n ,where ≤ k < n . The function x per( b , . . . , b k , x, . . . , x )per( b , b , . . . , b k , x, . . . , x ) (1) is concave on R n ++ . A motivation for Conjecture 1.1 is the case when b , b , . . . , b k are all equal tothe vector of all ones. Then (1) is equal to a constant multiple of x e n − k ( x ) e n − k − ( x ) , (2) Mathematics Subject Classification.
Primary: 15A15, 15A45; Secondary: 26B25.
Key words and phrases.
Permanent, α -permanent, α -determinant, positivity, hyperbolic poly-nomial, complete monotonicity, symmetric function mean.PB is a Royal Swedish Academy of Sciences Research Fellow supported by a grant from theKnut and Alice Wallenberg Foundation. where e k ( x ) is the k ’th elementary symmetric function in the variables x = ( x , . . . , x n ).The function (2) is a symmetric function mean and such are known to be concave[7].A homogeneous polynomial h ( x ) ∈ R [ x , . . . , x n ] is hyperbolic with respect toa vector e ∈ R n if h ( e ) = 0, and if for all x ∈ R n the univariate polynomial t h ( x + et ) has only real zeros, see [3, 5, 8, 9]. Here are some examples ofhyperbolic polynomials:(1) Let h ( x ) = x · · · x n . Then h ( x ) is hyperbolic with respect to any vector e ∈ R n that has no coordinate equal to zero: h ( x + et ) = n Y j =1 ( x j + e j t ) . (2) Let x = ( x ij ) ni,j =1 be a matrix of variables where we impose x ij = x ji .Then det( x ) is hyperbolic with respect to I = diag(1 , . . . , t det( x + tI ) is the characteristic polynomial of the symmetric matrix x , soit has only real zeros.(3) Let h ( x ) = x − x − · · · − x n . Then h is hyperbolic with respect to(1 , , . . . , h is hyperbolic with respect to e , and of degree d . We may write h ( x + et ) = h ( e ) d Y j =1 ( t + λ j ( x )) , where λ ( x ) ≤ · · · ≤ λ d ( x ). The hyperbolicity cone is the setΛ ++ = Λ ++ ( e ) = { x ∈ R n : λ ( x ) > } . G˚arding [5] proved that hyperbolicity cones are convex. The hyperbolicity conesfor the examples above are:(1) Λ ++ ( e ) = { x ∈ R n : x i e i > i } .(2) Λ ++ ( I ) is the cone of symmetric positive definite matrices.(3) Λ ++ (1 , , . . . ,
0) is the
Lorentz cone (cid:26) x ∈ R n : x > q x + · · · + x n (cid:27) . Let h ( x , . . . , x n ) be a homogeneous polynomial of degree d . Let v j = ( v j , . . . , v nj ) T for 1 ≤ j ≤ d . The complete polarized form of h may be defined as the form H : ( R n ) d → R defined by H ( v , . . . , v d ) = 1 d ! d Y j =1 n X i =1 v ij ∂∂x i ! h ( x ) , see G˚arding [5]. Note that H is multilinear, symmetric in v , . . . , v d , and H ( v, . . . , v ) = h ( v ). Now if h = x · · · x n , then H ( v , . . . , v n ) = 1 n ! per( v , . . . , v n ) . (3) WO PROBLEMS ON PERMANENTS 3
For u = ( u , . . . , u n ) T ∈ R n let D u = P ni =1 u i ∂/∂x i . It follows that H ( v , . . . , v k , u, . . . , u ) = ( d − k )! d ! 1( d − k )! d − k Y j =1 n X i =1 u i ∂∂x i ! D v D v · · · D v k h ( x )= ( d − k )! d ! D v D v · · · D v k h ( u ) . Lemma 1.2 (G˚arding, [5]) . Suppose that h is hyperbolic, and that v ∈ Λ ++ . Then D v h is hyperbolic and its hyperbolicity cone contains Λ ++ . The following lemma was proved in [3, Corollary 4.6], see also [9] where Lemma 1.3is strengthened.
Lemma 1.3 (Bauschke et al. , [3]) . Suppose that h is hyperbolic and that v ∈ Λ ++ .The function x h ( x ) D v h ( x ) is concave on Λ ++ . In view of (3) we see that Conjecture 1.1 is the special case of Corollary 1.4 when h = x · · · x n . Corollary 1.4.
Let h be a hyperbolic polynomial in R [ x , . . . , x n ] and let b , b , . . . , b k be fixed vectors in Λ ++ . The function x H ( b , . . . , b k , x, . . . , x ) H ( b , b , . . . , b k , x, . . . , x ) (4) is concave on Λ ++ .Proof. Suppose that the degree of h is d . By Lemma 1.2 the polynomial g ( x ) := H ( b , . . . , b k , x, . . . , x ) = ( d − k )! d ! D b · · · D b k h ( x )is hyperbolic with hyperbolicity cone containing Λ ++ . The function (4) is equal to g ( x ) /D b g ( x ), and so the proof follows from Lemma 1.3. (cid:3) Remark . If h ( x ) = det( x ), acting on symmetric matrices of size n × n , then thecomplete homogenized form is the mixed discriminant H ( A , . . . , A n ) = 1 n ! ∂ n ∂x · · · ∂x n det n X i =1 x i A i ! . Hence if A , . . . , A k are fixed positive definite matrices, then the function A H ( A , . . . , A k , A, . . . , A ) H ( A , A , . . . , A k , A, . . . , A )is concave on the cone of positive definite matrices. This also holds for complexhermitian matrices since the determinant on complex hermitian matrices is againhyperbolic. P. BR¨AND´EN α -permanents and complete monotonicity The α -permanent, introduced by Vere–Jones [13], interpolates between the de-terminant and the permanent. Let α ∈ R and A = ( a ij ) be an n × n matrix. The α - permanent and α - determinant of A are defined byper α ( A ) = X σ ∈ S n α c ( σ ) n Y i =1 a iσ ( i ) and det α ( A ) = α n per /α ( A ) , where c ( σ ) denotes the number of disjoint cycles of σ . Motivated by problems onrandom point processes, Shirai and Takahashi [11, 12] posed the following problem: Problem 2.1.
For which α ∈ R is (1) det α ( A ) ≥ for all real symmetric positive semidefinite matrices A ? Let D R be the set of such α ’s. (2) det α ( A ) ≥ for all complex hermitian positive semidefinite matrices A ?Let D C be the set of such α ’s. Shirai [12] proved that {± / ( m + 1) : m ∈ N } ⊆ D C ⊆ {− / ( m + 1) : m ∈ N } ∪ [0 ,
1] and {− / ( m + 1) : m ∈ N } ∪{ / ( m + 1) : m ∈ N } ⊆ D R ⊆ {− / ( m + 1) : m ∈ N } ∪ [0 , . Moreover, Shirai and Takahashi conjectured:
Conjecture 2.2 (Shirai and Takahashi, [11, 12]) . D R = {− / ( m + 1) : m ∈ N } ∪ [0 , and D C = {− / ( m + 1) : m ∈ N } ∪ [0 , . We shall see that Conjecture 2.2 is false, infact
Theorem 2.3. D R = {− / ( m + 1) : m ∈ N } ∪ { / ( m + 1) : m ∈ N } ∪ { } and D C = {± / ( m + 1) : m ∈ N } ∪ { } . The proof of Theorem 2.3 relies on recent results on complete monotonicity dueto Scott and Sokal [10].For n = ( n , . . . , n m ) ∈ N m and A = ( a ij ) mi,j =1 , let A [ n ] be the | n | × | n | matrix,where | n | = P mi =1 n i , obtained by replacing the ( i, j )’th entry of A by an n i × n j matrix whose entries are all equal to a ij . The next theorem is a generalization ofthe MacMahon Master Theorem. Theorem 2.4 (Foata and Zeilberger, [4]; Vere-Jones, [13]) . Let A = ( a ij ) mi,j =1 , X = diag( x , . . . , x m ) and α ∈ R . Then det( I − XA ) − α = X n ∈ N m per α ( A [ n ]) x n n ! and det( I − αXA ) − /α = X n ∈ N m det α ( A [ n ]) x n n ! , where x n = x n · · · x n m m and n ! = n ! · · · n m ! . WO PROBLEMS ON PERMANENTS 5
Remark . If A is a hermitian positive semidefinite m × m matrix and n ∈ N m ,then A [ n ] is positive semidefinite. Indeed, if y is the column vector with entries y ij for 1 ≤ i ≤ m and 1 ≤ j ≤ n i , then y T A [ n ] y = x T Ax ≥ x = ( x , . . . , x m ) T has entries x i = P n i j =1 y ij .Recall that a C ∞ -function f : R m ++ → R is completely monotone if( − | n | ∂ n ∂x n · · · ∂ n m ∂x n m m f ( x ) ≥ , for all n ∈ N m and x ∈ R m ++ . For m ≥ C ( m ) = N ∪ { x ∈ R : x ≥ m − } and R ( m ) = { x/ x ∈ C ( m ) } . Theorem 2.6 (Scott and Sokal, [10]) . Let A , . . . , A n be m × m real or complexhermitian matrices, and form the polynomial P ( x ) = det n X i =1 x i A i ! . Assume that P and β ≥ . (1) If A , . . . , A n are real symmetric and positive semidefinite, then P − β iscompletely monotone for all β ∈ R ( m ) . If A , . . . , A n span the space of m × m symmetric matrices, then P − β fails to be completely monotone foreach β R ( m ) . (2) If A , . . . , A n are complex hermitian positive semidefinite, then P − β is com-pletely monotone for all β ∈ C ( m ) . If A , . . . , A n span the space of m × m complex hermitian matrices, then P − β fails to be completely monotone foreach β C ( m ) . We will use the following elementary but useful lemma.
Lemma 2.7.
Let A be an m × n matrix and B be an n × m matrix, then det( I − AB ) = det( I − BA ) . Proof of Theorem 2.3.
We prove Theorem 2.3 for real symmetric matrices, theproof for complex hermitian matrices is almost identical.The cases left to consider is for α ∈ [0 , \ { / ( k + 1) : k ∈ N } . Then β = 1 /α / ∈ R ( m ) for some m ∈ N . Choose positive semidefinite rank one matrices A , . . . , A n that span the space of real symmetric m × m matrices. Then P ( x ) = det n X i =1 x i A i ! is not completely monotone by Theorem 2.6. Hence there is a vector y ∈ R n ++ such that x P ( y − x ) − β fails to have only nonnegative Taylor coefficients. Since A := P ni =1 y i A i is positive definite we may write A = B − / B − / for some positivedefinite matrix B . Let B i = B / A i B / and write B i = u i u Ti for some vector u i ∈ R m . Collect the u i ’s as columns in an m × n matrix U . Then, by Lemma 2.7, P ( y − x ) = det A − n X i =1 x i A i ! = det( A ) det I − n X i =1 x i B i ! = det( A ) det( I − U XU T ) = det( A ) det( I − XU T U ) . P. BR¨AND´EN
By Theorem 2.4 and the fact that P ( y − x ) − β has at least one negative Tay-lor coefficient, there is a vector n ∈ N n such that per β ( U T U [ n ]) <
0, so thatdet α ( U T U [ n ]) <
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Department of Mathematics, Royal Institute of Technology, SE-100 44 Stockholm,Sweden
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