Non-absolutely monotonic functions which preserve non-negative definiteness

Transcription

1 Non-absolutely monotonic functions which preserve non-negative definiteness Alexander Belton (Joint work with Dominique Guillot, Apoorva Khare and Mihai Putinar) Department of Mathematics and Statistics Lancaster University, United Kingdom QOP Meeting on Quantum Probability, Groups and Geometry IMPAN, Banach Centre, Warsaw 9th April 2015

2 Structure of the talk The Hadamard product Definiteness Schur s theorem Schoenberg s theorem Horn s theorem A new theorem Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 2 / 18

3 Two matrix products Notation The set of n n matrices with entries in a set K C is denoted M n (K). Products The vector space M n (C) is an associative algebra for at least two different products: if A = (a ij ) and B = (b ij ) then (AB) ij := n a ik b kj k=1 (standard) and (A B) ij := a ij b ij (Hadamard). Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 3 / 18

4 Definiteness Definition A matrix A M n (C) is non-negative definite if n x Ax = x i a ij x j 0 for all x C n. i,j=1 A matrix A M n (C) is positive definite if n x Ax = x i a ij x j > 0 for all x C n \{0}. i,j=1 Remark The subset of M n (K) consisting of non-negative-definite matrices is denoted M n (K) +.The set M n (C) + is a cone: closed under sums and under multiplication by elements of R +. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 4 / 18

5 Some consequences Symmetry If A M n (K) + then A t M n (K) + : note that Hermitianity 0 (x Ax) t = y A t y for all y = x C n. If A M n (C) + then A = A : note that x A x = (x Ax) = x Ax (x C n ) = A = A (polarisation). Non-negative eigenvalues If A M n (C) + = U diag(λ 1,...,λ n )U, where U M n (C) is unitary, then λ i = (U e i ) A(U e i ) 0. In particular, deta = λ 1 λ n 0 and tra = λ 1 + +λ n 0. Conversely, if A M n (C) is Hermitian and has non-negative eigenvalues then A = B B M n (C) +, where B = diag ( λ 1,..., λ n ) U. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 5 / 18

6 A characterisation of non-negative definiteness Definition Recall that a principal minor of a matrix A M n (K) is a k k matrix formed by deleting n k rows and the corresponding n k columns of A. Theorem 1 (Sylvester s criterion) A Hermitian matrix A M n (C) is non-negative definite if and only if each of its principal minors has non-negative determinant. Proof. Hint: det(λi A) = n ( 1) n k λ k k=0 i {1,...,n}, i =k deta i, where A i is the principal minor obtained from A by deleting the rows and columns labelled by elements of i. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 6 / 18

7 The Schur product theorem Theorem 2 (Schur, 1911) If A, B M n (C) + then A B M n (C) +. Proof. x (A B)x = tr(diag(x) B diag(x)a t) = tr ( (A t ) 1/2 diag(x) Bdiag(x)(A t ) 1/2). Corollary 3 If f(x) = n=0 a nx n is analytic on K and a n 0 for all n 0 then f[a] := ( f(a ij ) ) M n (C) + for all A = (a ij ) M n (K) + and all n 1. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 7 / 18

8 Schoenberg s theorem Theorem 4 (Schoenberg, 1942) If f : [ 1,1] R is (i) continuous and ( (ii) such that f[a] M n (R) + for all A M n [ 1,1] and all n 1 )+ then f is absolutely monotonic: f(x) = a n x n for all x [ 1,1], where a n 0 for all n 0. n=0 Proof (Christensen Ressel, 1978). ( {f : [ 1,1] R f(1) = 1, f[a] M n (R) + A M n [ 1,1], n 1} )+ is a Choquet simplex with closed set of extreme points {t n : n 0} {1 {1} 1 { 1},1 { 1,1} }. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 8 / 18

9 Horn s theorem Theorem 5 (Horn, 1969) If f : (0, ) R is continuous and such that f[a] M n (R) + for all A M n ( (0, ) )+ then (i) f C n 3( (0, ) ) and (ii) f (k) (x) 0 for all k = 0,...,n 3 and all x > 0. If, further, f C n 1( (0, ) ), then f (k) (x) 0 for all k = 0,..., n 1 and all x > 0. Remark (Guillot Khare Rajaratnam, 2014) The hypotheses may be weakened: it suffices to suppose R > 0 and f : (0,R) R is such that f[a] M n (R) + for all non-zero A of the form a1 n +xx t, where a [0,R) and x [0,(R a) 1/2 ). Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 9 / 18

10 A key inequality Notation For all s R and n 1 let the polynomial p(z;s,n) := s(1+z + +z n 1 ) z n (z C). Theorem 6 (B G K P, 2015) For any bounded set K C and any n 1 there exists a constant c(k,n) 1 such that p[a;s,n] = s ( 1 n +A+ +A n 1) A n M n (C) + for all s c(k,n) and all A M n (K) +. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 10 / 18

11 A lemma Lemma 7 If A, B M n (C) then det(a+b) = σ S n sgnσ = σ S n sgnσ = i {1,...,n} N (a iσ(i) +b iσ(i) ) i=1 i {1,...,n} i i detm i (A;B), a iσ(i) i i b iσ(i) where M i (A;B) is the matrix obtained by replacing the rows of A labelled by elements of i with the corresponding rows of B. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 11 / 18

12 A proof of Theorem 6 for rank-one matrices If x, y C n and A = xy t then 1 n +A+ +A n 1 = XY t, where X = (x i 1 j ) and Y = (y i 1 j ) are Vandermonde matrices. By the previous lemma, p(s) := detp[a;s,n] = det(sxy t A n ) n = ( 1) k s n k detm i (XY t ;A n ) k=0 i =k = a 0 s n a 1 s n 1 + +( 1) n a n, where a k := i =k detm i(xy t ;A n ). Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 12 / 18

13 Proof of Theorem 6 for rank-one matrices (ctd.) Since and a k = detm i (XY t ;A n ) = dk dz k det(xyt +za n ) i =k z=0 (XY t +za n ) ij = 1+x i y j + +x n 1 i y n 1 j +zxi n yn j, each coefficient a k is divisible by (x j x i )(y j y i ) = det(xy t ) = M (XY t ;A n ) = a 0. 1 i<j n Now suppose y = x and note that p(s) = det(sxx A n ) = det(sxx A n ) = p(s), so the coefficients a 0,..., a n are real. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 13 / 18

14 Proof of Theorem 6 for rank-one matrices (ctd.) The function f : C n M n (C) + C n2 ; x xx is proper, so c 0 (K,n) := sup { 1, (a k /a 0 )(x) : k = 0,...,n 1, x f 1 (K n2 ) } <. Thus if s 1 then n p(s) a 0 s n a 0 s n a k 1 a 0 s k nc 0(K,n) a 0 s n, s k=1 so p(s) 0 for all s nc 0 (K,n). Hence s(1 n +A+ +A n 1 ) A n M n (K) + for all s c 1 (K,n) := max{mc 0 (K,m) : m = 1,...,n}. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 14 / 18

15 Two more key lemmas Lemma 8 (FitzGerald Horn, 1977) Given A = (a ij ) M n (C) +, set z C n to equal (a in / a nn ) if a nn 0 and to be the zero vector otherwise.then A zz M n (C) + and the final row and column of this matrix are zero. Remark If A M n (K) +, where K K = B(0;R), then a ii a in = aii a a in a nn a in 2 0 = zz M n (K ). nn Lemma 9 If f : C C is entire then f(z) f(w) = 1 0 (z w)f(λz +(1 λ)w)dλ (z,w C). Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 15 / 18

16 Proof of Theorem 6 the general case We proceed by induction: let n 2 and suppose that, for any bounded set K C, there exists c(k,n 1) 1 such that p[a;s,n 1] M n 1 (C) + for all s c(k,n 1) and all A M n 1 (K) +. Let A M n (K) +. By the results on the previous slide, p[a;s,n] = p[zz ;s,n]+ 1 0 (A zz ) p [λa+(1 λ)zz ;s,n]dλ. Let M λ M n 1 (K ) + be the principal minor obtained by deleting the final row and column of λa+(1 λ)zz. Then n 2 p [M λ ;s,n] = np[m λ ;s/n,n 1]+s jm j λ M n 1(C) + j=1 wherever s nc(k,n 1). By the Schur product theorem, s c(k,n) := max{nc(k,n 1),c 1 (K,n)} = p[a;s,n] M n (C) +. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 16 / 18

17 The full result Corollary 10 Let K C be bounded and let n 1. If s c(k,n) then 2 k s k+1 (1 n +A+ A n 1 ) A n+k M n (C) + for all A M n (K) + and all k 0. Theorem 11 (B G K P, 2015) Given a bounded set K C and integers n 1 and m 0, there exists a constant h(k,n,m) > 0 such that, if f(x) = c 0 +c 1 x + +c n+m x n+m, then f[a] M n (C) + for all A M n (K) + whenever (i) c 0,..., c n 1 0 and (ii) min{c 0,...,c n 1 } h(k,n,m)max{ c l : c l < 0}. Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 17 / 18

18 References J. P. R. Christensen and P. Ressel, Functions operating on positive definite matrices and a theorem of Schoenberg, Trans. Amer. Math. Soc. 243 (1978) D. Guillot, A. Khare and B. Rajaratnam, Preserving positivity for rank-constrained matrices, technical report, Department of Mathematics, Stanford University. arxiv: , C. H. FitzGerald and R. A. Horn, On fractional Hadamard powers of positive definite matrices, J. Math. Anal. Appl. 61 (1977), R. A. Horn, The theory of infinitely divisible matrices and kernels, Trans. Amer. Math. Soc. 136 (1969) I. J. Schoenberg, Positive definite functions on spheres, Duke Math. J. 9 (1942) Alexander Belton (Lancaster University) Functions which preserve non-negativity QOP, 09iv15 18 / 18