A quantitative form of Schoenberg's theorem in fixed dimensionbelton/www/notes/25vii17.pdf ·...

A quantitative form of Schoenberg’s theorem

in fixed dimension

Alexander Belton

(Joint work with Dominique Guillot,Apoorva Khare and Mihai Putinar)

Department of Mathematics and StatisticsLancaster University, United Kingdom

[email protected]

http://www.maths.lancs.ac.uk/~belton/

Iowa State University

25th July 2017

http://www.maths.lancs.ac.uk/~belton/

Structure of the talk

The Hadamard product

Positive definiteness

Schur’s theorem

Schoenberg’s theorem

Horn’s theorem

A new theorem and its proof

Alexander Belton (Lancaster University) A quantitative Schoenberg’s theorem ILAS 2017, 25vii17 2 / 22

Two matrix products

Notation

The set of n × n matrices with entries in a set K ⊆ C is denoted Mn(K ).

Products

The vector space Mn(C) is an associative algebra for at least two differentproducts:if A = (aij) and B = (bij ) then

(AB)ij :=

n∑

k=1

aikbkj (standard)

and (A ◦ B)ij := aijbij (Hadamard).


Positive definiteness. I

Definition

A matrix A ∈ Mn(C) is positive semidefinite if

x∗Ax =

n∑

i ,j=1

xi aij xj > 0 for all x ∈ Cn.

A matrix A ∈ Mn(C) is positive definite if

x∗Ax =

n∑

i ,j=1

xi aij xj > 0 for all x ∈ Cn \ {0}.


Positive definiteness. II

Remark

The subset of Mn(K ) consisting of positive semidefinite matrices isdenoted Mn(K )+.The set Mn(C)+ is a cone: closed under sums and undermultiplication by elements of R+.

Symmetry

If A ∈ Mn(K )+ then AT ∈ Mn(K )+: note that

0 6 (x∗Ax)T = y∗ATy for all y = x ∈ Cn.

Hermitianity

If A ∈ Mn(C)+ then A = A∗: note that

x∗A∗x = (x∗Ax)∗ = x∗Ax (x ∈ Cn).


Positive definiteness. III

Non-negative eigenvalues

IfA ∈ Mn(C)+ = U∗ diag(λ1, . . . , λn)U,

where U ∈ Mn(C) is unitary, then λi = (U∗ei )∗A(U∗ei ) > 0.

Conversely, if A ∈ Mn(C) is Hermitian and has non-negative eigenvaluesthen A = B∗B ∈ Mn(C)+, where

B = A1/2 = U∗ diag(

√

λ1, . . . ,√

λn

)

U.

Corollary

If A ∈ Mn(C)+ thendetA = λ1 · · ·λn > 0

andtrA = λ1 + · · ·+ λn > 0.


A characterisation of positive semidefiniteness

Definition

A principal minor of an n× n matrix is the determinant of the k × k matrixformed by deleting n − k rows and the corresponding n − k columns.

Theorem 1 (Sylvester’s criterion)

A Hermitian matrix A ∈ Mn(C) is positive semidefinite if and only if each

of its principal minors is non-negative.

Proof.

For the converse, proceed by induction. If the n × n matrix A is notpositive semidefinite but all of its principal minors are non-negative then ithas two negative eigenvalues with unit eigenvectors x and y. The unitvector z = λx+ µy can be taken orthogonal to (0, . . . , 0, 1)⊥ and is suchthat

z∗Az = |λ|2x∗Ax+ |µ|2y∗Ay < 0.


The Schur product theorem

Theorem 2 (Schur, 1911)

If A, B ∈ Mn(C)+ then A ◦ B ∈ Mn(C)+.

Proof.

x∗(A ◦ B)x = tr(diag(x)∗B diag(x)AT)

= tr(

(AT )1/2 diag(x)∗B diag(x)(AT )1/2)

.


Schoenberg’s theorem

Corollary 3

If f (z) =∑

∞

n=0 anzn is analytic on K and an > 0 for all n > 0 then

f [A] :=(

f (aij))

∈ Mn(C)+ for all A = (aij) ∈ Mn(K )+ and all n > 1.

Theorem 4 (Schoenberg, 1942)

If f : [−1, 1] → R is

(i) continuous and

(ii) such that f [A] ∈ Mn(R)+ for all A ∈ Mn

(

[−1, 1])

+and all n > 1

then f is absolutely monotonic:

f (x) =

∞∑

n=0

anxn for all x ∈ [−1, 1], where an > 0 for all n > 0.


Horn’s theorem

Theorem 5 (Horn, 1969)

If f : (0,∞) → R is continuous and such that f [A] ∈ Mn(R)+whenever A ∈ Mn

(

(0,∞))

+then

(i) f ∈ Cn−3(

(0,∞))

and

(ii) f (k)(x) > 0 for all k = 0, . . . , n − 3 and all x > 0.

If, further, f ∈ Cn−1(

(0,∞))

, then

f (k)(x) > 0 for all k = 0, . . . , n − 1 and x > 0.

Remark (Guillot–Khare–Rajaratnam, 2014)

For Theorem 5, it suffices to suppose ρ > 0 and f : (0, ρ) → R is such that

f [a1n + xxT ] ∈ Mn(R)+ for all a ∈ [0, ρ) and x ∈ [0, (ρ− a)1/2).


A new theorem

Theorem 6 (B–G–K–P, 2015)

Let ρ > 0, n > 1, c = (c0, . . . , cn−1) ∈ Rn and

f (z) =

n−1∑

j=0

cjzj + c ′zm,

where c ′ ∈ R and m > 0. The following are equivalent.

(i) f [A] ∈ Mn(C)+ for all A ∈ Mn

(

D(0, ρ))

+.

(ii) f [A] ∈ Mn(R)+ for all A ∈ Mn

(

(0, ρ))

+with rank at most one.

(iii) Either c ∈ Rn+ and c ′ ∈ R+, or c ∈ (0,∞)n and c ′ > −C(c;m, ρ)−1,

where

C(c;m, ρ) :=

n−1∑

j=0

(

m

j

)2(m − j − 1

n − j − 1

)2ρm−j

cj.


Proof of Theorem 6 with m > n

Lemma 7 (B–G–K–P, 2015)

If f : D(0, ρ) → R is analytic, where ρ > 0, and such that f [A] ∈ Mn(R)+whenever A ∈ Mn

(

(0, ρ))

+has rank at most one, then the first n non-zero

Taylor coefficients of f are strictly positive.

Proof of Theorem 6

Ir is immediate that (i) implies (ii).

To see (ii) implies (iii), note that Lemma 7 gives that c0, . . . , cn−1 > 0, sowe assume c ′ < 0 < c0, . . . , cn−1 and show that c ′ > −C(c;m, ρ)−1. If

p(z ; t, c,m) := t(c0 + · · · + cn−1zn−1)− zm (z ∈ C, t ∈ R)

then (ii) implies that

det p[uuT ; |c ′|−1, c,m] > 0 for all u ∈(

0,√ρ)n.


A determinant expressed with Schur polynomials

Given a non-increasing n-tuple k = (kn > · · · > k1) ∈ Zn+, let

sk(x1, . . . , xn) :=det(x

kj+j−1i )

det(x j−1i )

and Vn(x1, . . . , xn) := det(x j−1i ).

Theorem 8

Let m > n > 1. If µ(m, n, j) := (m − n+ 1, 1, . . . , 1, 0, . . . , 0), where there

are n − j − 1 ones and j zeros, and c ∈ (F×)n then, for all u, v ∈ Fn,

det p[uvT ; t, c,m]

= tn−1 Vn(u)Vn(v)(

t −n−1∑

j=0

sµ(m,n,j)(u)sµ(m,n,j)(v)

cj

)

n−1∏

j=0

cj .


Proof of Theorem 6, m > n, (ii) =⇒ (iii) (ctd.)

It follows from the previous working that

0 6 det p[uuT ; |c ′|−1, c,m]

= |c ′|1−n Vn(u)2(

|c ′|−1 −n−1∑

j=0

sµ(m,n,j)(u)2

cj

)

n−1∏

j=0

cj .

Taking uj =√ρ δj , where the δj are distinct and tend to 1 from below, it

follows that

|c ′|−1>

n−1∑

j=0

sµ(m,n,j)(1, . . . , 1)2 ρm−j

cj= C(c;m, ρ).


Proof of Theorem 6, m > n, (iii) =⇒ (i)

It remains to prove that (iii) implies (i). By Schur’s Theorem 2, we mayassume that −C(c;m, ρ)−1 6 c ′ < 0 < c0, . . . , cn−1.

Key step

The key step is to show that f [A] ∈ Mn(C)+ if A ∈ Mn

(

D(0, ρ))

+has

rank at most one.

Given this, the proof concludes by induction. The n = 1 case is immediate,since every 1× 1 matrix has rank at most one. Assume the n− 1 caseholds. Since

f (z) = |c ′| p(z ; |c ′|−1, c,m),

it suffices to show p[A] := p[A; |c ′|−1, c,m] ∈ Mn(C)+ if A ∈ Mn(C)+.


Two more lemmas

Lemma 9 (FitzGerald–Horn, 1977)

Given A = (aij) ∈ Mn(C)+, set z ∈ Cn to equal (ain/

√ann) if ann 6= 0 and

to be the zero vector otherwise.Then A− zz∗ ∈ Mn(C)+ and the final row

and column of this matrix are zero.

Furthermore, if A ∈ Mn

(

D(0, ρ))

+then zz∗ ∈ Mn

(

D(0, ρ))

+, since

∣

∣

∣

aii ainain ann

∣

∣

∣= aiiann − |ain|2 > 0.

Lemma 10

If F : C → C is differentiable then

F (z) = F (w) +

∫ 1

0(z − w)F ′(λz + (1− λ)w)dλ (z ,w ∈ C).


Proof of Theorem 6: m > n, (iii) =⇒ (i)

By Lemmas 9 and 10,

p[A] = p[zz∗] +

∫ 1

0(A− zz∗) ◦ p′[λA+ (1− λ)zz∗]dλ.

Now,

p′(z ; |c ′|−1, c,m) = mp(z ; |c ′|−1/m, (c1, . . . , (n − 1)cn−1),m − 1),

which is positive semidefinite, by the inductive assumption, since

m C(

(c1, . . . , (n − 1)cn−1);m − 1, ρ) 6 C(c;m, ρ).

Hence p[A] is positive semidefinite, as required.


Proof of the key step

Fix u ∈ Cn such that A = uu∗ ∈ Mn

(

D(0, ρ))

+. For k = 1, . . . , n, let

Ck :=

k−1∑

j=0

sµ(m−n+k,k,j)(1, . . . , 1)2 ρm−n+k−j

cn−k+j

and note that

0 < C1 = ρm−n+1/cn−1 < · · · < Cn = C(c;m, ρ).

We claim that, for k = 1, . . . , n, every principal k × k submatrix of

B = Ck(cn−k1n + · · ·+ cn−1A)◦k−1)− A◦m−n+k

is positive semidefinite. When k = n, the matrix B is equal to Cn f [A]with c ′ = −C(c;m, ρ)−1, and we are done.


Principal submatrices of the matrix B

Again we proceed by induction. If k = 1 then

B = C1cn−11n − A◦m−n+1 = ρm−n+11N − A◦m−n+1.

The principal 1× 1 submatrices equal ρm−n+1 − |ui |2(m−n+1), and theseare non-negative since |ui |2 6 ρ.

Suppose the k − 1 case holds. For any non-empty k ⊆ {1, . . . , n}, let Bk

denote the principal submatrix of B with rows and columns labelled byelements of k, and similarly for uk.

If k has cardinality j < k then

Bk > Ck(cn−jA◦k−jk + · · ·+ cn−1A

◦k−1k )− A◦m−n+k

k

> A◦k−jk ◦

(

Cj(cn−j1j + · · ·+ cn−1A◦j−1k )− A

◦m−n+jk

)

> 0,

by induction.


Principal submatrices of the matrix B (continued)

Finally, if k has cardinality k then

Bk = p[uku∗

k;Ck , (cn−1, . . . , cn−k ),m − n + k]

and Theorem 8 gives that

detBk = Cn−1k |Vk(uk)|2

(

Ck −k−1∑

j=0

|sµ(m−n+k,k,j)(u)|2cn−k+j

)

k∏

j=1

cn−j .

It now follows from the triangle inequality and the fact that Schurpolynomials have non-negative coefficients that detBk > 0.

The end!

Thank you for your attention


References

A. Belton, D. Guillot, A. Khare and M. Putinar, Matrix positivity

preservers in fixed dimension. I, Adv. Math. 298 (2016) 325–368.

A. Belton, D. Guillot, A. Khare and M. Putinar, Matrix positivity preservers

in fixed dimension, C. R. Math. Acad. Sci. Paris 354 (2016) 143–148.

C. H. FitzGerald and R. A. Horn, On fractional Hadamard powers of

positive definite matrices, J. Math. Anal. Appl. 61 (1977) 633–642.

D. Guillot, A. Khare and B. Rajaratnam, Preserving positivity for

rank-constrained matrices, Trans. Amer. Math. Soc. 369 (2017)6105–6145.


References

R. A. Horn, The theory of infinitely divisible matrices and kernels, Trans.Amer. Math. Soc. 136 (1969) 269–286.

I. J. Schoenberg, Positive definite functions on spheres, Duke Math. J. 9(1942) 96–108.

J. Schur, Bemerkungen zur Theorie der beschrankten Bilinearformen mit

unendlich vielen Veranderlichen, J. Reine Angew. Math. 140 (1911) 1–28.


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