Examensarbete Total positivity and oscillatory kernels: an...

Examensarbete

Total positivity and oscillatory kernels: an overview, andapplications to the spectral theory of the cubic string

Marcus Kardell

LiTH-MAT-EX–2010/18–SE

Total positivity and oscillatory kernels: an overview, andapplications to the spectral theory of the cubic string

Department of Mathematics, Linkopings Universitet

Marcus Kardell

LiTH-MAT-EX–2010/18–SE

Examensarbete: 30 hp

Level: Advanced

Supervisor: Hans Lundmark,Department of Mathematics, Linkopings Universitet

Examiner: Hans Lundmark,Department of Mathematics, Linkopings Universitet

Linkoping: Aug 10 2010

Abstract

In the study of the Degasperis-Procesi differential equation, an eigenvalue prob-lem called the cubic string occurs. This is a third order generalization of thesecond order problem describing the eigenmodes of a vibrating string. In thisthesis we study the eigenfunctions of the cubic string for discrete and continu-ous mass distributions, using the theory of total positivity, via a combinatorialapproach with planar networks.

Keywords: Cubic string, total positivity, oscillatory kernels, planar networks.

URL: http://urn.kb.se/resolve?urn=urn:nbn:se:liu:diva-58005

Kardell, 2010. v

Acknowledgements

First I would like to thank my supervisor, Hans Lundmark, for providing mewith reading material, pedagogical explanations of difficult subjects, and forcatching errors in my proofs and the early drafts. Thanks also to my opponentAnton Hoghall, for reading a daunting report. Finally, thanks to family andfriends, who supported me by attending my final presentation or otherwiseencouraged me to do a good job.

Thank you.

Kardell, 2010. vii

Contents

1 Introduction 1

2 Oscillatory matrices and their eigenvalue properties 52.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 The Cauchy–Binet Theorem . . . . . . . . . . . . . . . . . . . . . 62.3 Oscillatory matrices . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Associated matrices . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Perron’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.6 Eigenvalues of oscillatory matrices . . . . . . . . . . . . . . . . . 14

3 Integral eigenvalue problems with oscillatory kernels 153.1 Oscillatory kernels . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Associated kernels . . . . . . . . . . . . . . . . . . . . . . . . . . 163.3 An integral analogue of Perron’s Theorem . . . . . . . . . . . . . 183.4 Eigenvalues and eigenfunctions of oscillatory kernels . . . . . . . 19

4 Planar networks 234.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Lindstrom’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . 244.3 Constructing a planar network with a given weight matrix . . . . 25

5 Summary and conclusion 37

Kardell, 2010. ix

x Contents

Chapter 1

Introduction

In the theory of small oscillations, the vibrations u(y, t) of a string with massdensity g(y) can be described by the second order partial differential equationuyy = g(y)utt. Separation of variables u(y, t) = Φ(y)τ(t), combined with thecondition that the ends are fixed at y = ±1, gives the following second ordereigenvalue problem

Φyy(y) = zg(y)Φ(y), y ∈ (−1, 1), Φ(±1) = 0.

The object of interest in this thesis is a third-order generalization of thisproblem, called the cubic string, which in its basic form is the differential equa-tion

−Φyyy(y) = zg(y)Φ(y), y ∈ (−1, 1).

When g(y) is a discrete measure – a linear combination of Dirac distributions –the ordinary string is piece-wise linear, since the second derivative is zero exceptat the mass points. The cubic string is instead piece-wise a quadratic polyno-mial. When the mass distribution is continuous, things get more complicated.

The main difference between the problems is that the cubic string uses threeboundary conditions. In [1], the Dirichlet-like boundary conditions

Φ(−1) = Φy(−1) = Φ(1) = 0

were studied, and this is the case we will study. Note that this is not a self-adjoint problem. For Neumann-like boundary conditions, see [2].

There are no known physical applications of the cubic string so far. Thehistorical motivation for the cubic string comes from the Degasperis–Procesi(DP) and Camassa–Holm (CH) shallow water equations. These are the PDEs

ut − uxxt + (b+ 1)uux = buxuxx + uuxxx, (x, t) ∈ R,

where b = 2 is CH and b = 3 is DP. The DP equation was studied in [3], wherethe wave equation

ψx(x)− ψxxx(x) = zmψ(x), (1.1)

indirectly arose. (See [3] for more details.) Here we just note that this waveequation is equivalent to the cubic string problem via the transformation

y = tanhx

2, ψ(x) =

2Φ(y)

1− y2, m(x) =

�1− y2

2

�3

g(y).

Kardell, 2010. 1

2 Chapter 1. Introduction

Note that the variable transformation is invertible, and that the mass distri-bution m(x) is still positive if g(y) is. We can use the wave equation (1.1)to derive a system of linear equations, which will tell us a lot about how theeigenfunctions of the cubic string behave.

We start by taking the Fourier transform of each side. We get

iω(1 + ω2)Õψ(ω) = Ôzmψ =: bf =⇒ bψ(ω) =1

iω(1 + ω2)bf =⇒

bψ(ω) =

Éπ

2

ÛZ x

−∞e−|x| dx · bf

The integral can be calculated asZ x

−∞e−|x|dx =

¨ex x ≤ 0

2− e−x x ≥ 0=: g(x).

Transforming back, we get the integral equationbψ(ω) =1

2

√2π bf · bg =⇒ ψ(x) =

1

2(f ∗ g)(x) =

z

2

Z ∞−∞

g(x− s)m(s)ψ(s) ds.

When m(s) is a discrete measure,

m(s) = 2nXj=1

mjδ(x− xj), x1 < x2 < · · · < xn,

this integral equation breaks down to a system of n equations:

ψ(x) = znXj=1

g(x− xj) ·mjψ(xj) =⇒�ψ(x1)ψ(x2)

...ψ(xn)

�= G ·

�m1ψ(x1)m2ψ(x2)

...mnψ(xn)

�,

where

G =

�g(x1 − x1) g(x1 − x2) . . . g(x1 − xn)g(x2 − x1) g(x2 − x2) . . . g(x2 − xn). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .g(xn − x1) g(xn − x2) . . . g(xn − xn)

�.

We know that g(xi − xi) = g(0) = 1. When i < j, g(xi − xj) = exi−xj , andotherwise g(xi − xj) = 2 − exj−xi . Defining Eij = exi−xj we see that G takesthe form

Gn =

�1 E12 E13 . . . E1n

2− E12 1 E23 . . . E2n

2− E13 2− E23 1 . . . E3n

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2− E1n 2− E2n 2− E3n . . . 1

�

3

for each n. To study the eigenvalues of this matrix, we are going to need thetheory of oscillatory matrices and kernels, which is presented in Chapter 2 and3 respectively, relying heavily on the original work [4]. An important steppingstone is Perron’s Theorem, which is proven in detail. In Chapter 4 we show thatG is oscillatory for each n, by constructing a planar network which has G as itsweight matrix.

Note that G was already known to be oscillatory, by the more general theoryin [5], but here we have chosen to take a more combinatorial approach to thisproblem. The matrix G appeared in the original article [3] about the DP equa-tion, but no one seems to have studied this particular planar network before.

4 Chapter 1. Introduction

Chapter 2

Oscillatory matrices andtheir eigenvalue properties

In this chapter we study the simpler case of oscillatory matrices and their eigenvalueproperties, then in the next chapter we move on the more general case of oscillatorykernels.

2.1 Preliminaries

We first remind the reader of some facts from linear algebra. Let A be a squarematrix

A =

�a11 a12 . . . a1na21 a22 . . . a2n. . . . . . . . . . . . . . . . . . . .an1 an2 . . . ann

�.

We are going to represent such a matrix as ||aik||n1 . The transpose of A is writtenas At = ||aki||n1 . When At = A, the matrix is called symmetric. The notationfor the determinant of A will be detA, |A|, or |aik|n1 . The rank of a matrix (themaximal number of linearly independent rows or columns) is written rank A.

The eigenvalues of A are calculated as roots to the characteristic equation|A − λI| = 0, where I is the identity matrix. We can write I = ||δik||n1 usingthe Kronecker delta δik (1 if i = k, otherwise 0). Since transposing a matrixdoesn’t change its determinant, we see that |A− λI| = |(A− λI)t| = |At − λI|,which implies that A and At have the same eigenvalues.

Minors (subdeterminants) of A will be notated

A

�s1 s2 . . . spt1 t2 . . . tp

�,

where (s1, s2, . . . , sp), and (t1, t2, . . . , tp), are the indices determining what rowsand columns, respectively, are kept from A. In connection with this we letMnp = {(i1, i2, . . . , ip) : 1 ≤ i1 < i2 < · · · < ip ≤ n} be the set of all tuples of

p distinct ordered indices, chosen from the integers 1 to n. A minor where allsj = tj is called principal.

Kardell, 2010. 5

6 Chapter 2. Oscillatory matrices and their eigenvalue properties

Closely related to minors are cofactors, written

Aik = (−1)i+kA

�1 . . . i− 1 i+ 1 . . . k k + 1 . . . n1 . . . i− 1 i . . . k − 1 k + 1 . . . n

�,

which are used when expanding a determinant with respect to a row or column;for example

|A| =nXk=1

aikAik

where we have expanded along row i.Differentiating a determinant is possible with the use of the product rule of

differentiation. The determinant is a sum of products of elements. Differentiat-ing with respect to each element gives us

d

dx|aik|n1 =

nXi=1

nXk=1

Aikd

dxaik.

Finally, we are also going to need a few facts about permutations, i.e. bi-jections from {1, 2, . . . , n} to itself. There are exactly n! such functions on nelements. One interpretation of a permutation σ is that it changes the orderof the elements (1, 2, . . . , n) to (σ(1), σ(2), . . . , σ(n)). Multiplication of two per-mutations σ and τ is defined as the composition of the two functions, σ(τ(x)).Permutations that only swap two elements are called transpositions. All permu-tations can be expressed as a product of transpositions in many different ways,but one can show that the number of factors for a given permutation is alwaysodd or always even. One can then define the sign of a permutation, sgn(σ) =(−1)k, where k is the number of factors of σ.

We are going to use permutations to write determinants in a very succinctway.

Definition 1. Let A be a square matrix ||aik||n1 . Then

|A| =Xσ

sgn(σ)

nYi=1

aiσ(i)

!,

where the sum is over all permutations σ on {1, 2, . . . , n}.

2.2 The Cauchy–Binet Theorem

In linear algebra the basic relation |AB| = |A||B| for square matrices is stud-ied. The Cauchy–Binet Theorem is a generalization, handling both square andrectangular matrices.

Theorem 1. Let A be a p × n matrix, and B an n × p matrix. Let C = AB.Then

|C| =Xk∈Mn

p

A

�1 2 . . . pk1 k2 . . . kp

�B

�k1 k2 . . . kp1 2 . . . p

�.

2.2. The Cauchy–Binet Theorem 7

Proof. The sum is over the�np

�ways of choosing indices k1 to kp. Note that if

p > n, then the sum is empty. This is consistent with the ranks of the matrices,since rank C is at most equal to the smallest rank of its factors, and both A andB has rank less than or equal to n. Consequently, C, which is a p × p matrix,is rank deficient, which means that |C| = 0.

If p = n, the sum has only one term. The relation is then the well-known|C| = |A||B|.

Let us now look at the case where p < n (note that the following argumentis valid also for p = n). Looking closer at C, we have

|C| =

�� c11 c12 . . . c1pc21 c22 . . . c2p. . . . . . . . . . . . . . . . . .cp1 cp2 . . . cpp

�� ,which according to the rules of matrix multiplication is equal to��

Pnk1=1 a1k1bk11

Pnk1=1 a1k1bk12 . . .

Pnk1=1 a1k1bk1pPn

k2=1 a2k2bk21Pnk2=1 a2k2bk22 . . .

Pnk2=1 a2k2bk2p

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Pnkp=1 apkpbkp1

Pnkp=1 apkpbkp2 . . .

Pnkp=1 apkpbkpp

�� .Thanks to the multilinearity of determinants, we can write this as

nXk1=1

nXk2=1

· · ·nX

kp=1

a1k1a2k2 . . . apkp

�� bk11 bk12 . . . bk1pbk21 bk22 . . . bk2p. . . . . . . . . . . . . . . . . . . . .bkp1 bkp2 . . . bkpp

�� .Each term that has ki = kj for some i and j will be zero, because the corre-sponding rows in the determinant are equal, and thus linearly dependent. In theremaining terms, all ki’s are different, which means we can introduce (for eachterm in the sum) a function σ, which is from {1, 2, . . . , p} onto {k1, k2, . . . , kp},via the relation σ(i) = ki. We get

|C| =Xσ

pYi=1

aiσ(i)

�� bσ(1)1 bσ(1)2 . . . bσ(1)pbσ(2)1 bσ(2)2 . . . bσ(2)p. . . . . . . . . . . . . . . . . . . . . . . . . .bσ(p)1 bσ(p)2 . . . bσ(p)p

�� . (2.1)

Note that σ is not really a permutation in a strict sense, since it is not ontoitself, but it still behaves just like a normal permutation. (It is isomorphic toone such.)

Let us transpose the rows in each minor, in a way such that the indices σ(i)appear ordered from lowest to highest, i.e. kmin = min{ki} ends up on the firstrow, and so on. Each transposition changes the sign of the determinant, so foreach term this process yields a sign factor, equal to the sign of σ.

|C| =Xσ

sgn(σ)

pYi=1

aiσ(i)

�� bkmin1 . . . bkminp

. . . . . . . . . . . . . . . . . . .bkmax1 . . . bkmaxp

�� .


The extra sign factor is exactly what we need to convert this sum of productsto minors of A. We can group the terms according to the range of σ, or thevalues {kmin, . . . , kmax}. Then, using Definition 1 on each such group, we get

|C| =Xk∈Mn

p

A

�1 . . . p

kmin . . . kmax

�B

�kmin . . . kmax

1 . . . p

�,

which is nothing but the desired relation.

The Cauchy–Binet Theorem also gives us a way to deal with minors of aproduct.

Corollary 1. An arbitrary minor of C = AB can be computed with

C

�s1 . . . skt1 . . . tk

�=Xr

A

�s1 . . . skr1 . . . rk

�B

�r1 . . . rkt1 . . . tk

�.

Proof. The matrix behind the minor

C

�s1 . . . skt1 . . . tk

�is the product of�

as11 . . . as1n. . . . . . . . . . . . . . . .ask1 . . . askn

�and

�b1t1 . . . b1tk. . . . . . . . . . . . . . .bnt1 . . . bntk

�.

Applying the Cauchy–Binet Theorem to this product gives us the corollary.

2.3 Oscillatory matrices

Definition 2. A matrix A = ||aik||n1 is called totally non-negative (TNN) if allminors of A are non-negative. A is called totally positive (TP) if all minorsare positive. If A is totally non-negative, and some power of it, Aκ, is totallypositive, A is called oscillatory. The smallest such κ is then called the exponentof A.

Theorem 2. {TP matrices} ⊂ {Oscillatory matrices} ⊂ {TNN matrices}.Proof. The inclusions follow immediately from the definition. That the inclu-sions are strict follows from simple examples. Let

A =

�0 10 1

�, B =

�1 1 01 1 10 1 1

�.

A is TNN, but not TP. A is also not oscillatory since A2 = A, which means thatno power of A can be TP.

B is also TNN, since all minors are found to be 0 or 1. We can calculate

B2 =

�2 2 12 3 21 2 2

�and verify that all minors are positive, which means that B is oscillatory (butnot TP).

2.4. Associated matrices 9

Using the Cauchy–Binet Theorem, we can find more properties of these threeclasses of matrices.

Theorem 3. The product of two TNN (TP) matrices is TNN (TP).

Proof. From Cauchy–Binet, each minor of the product is a sum of products ofminors from the factors. When the factors are TNN, the sum must be greaterthan or equal to zero. In the TP case, the sum is positive.

Theorem 4. The product of a TNN matrix A and a TP matrix is TP if andonly if A is non-singular.

Proof. The necessity is trivial. The sufficiency follows from expanding the prod-uct using the Cauchy–Binet theorem. In fact, if A is non-singular, at least oneminor is positive, making the sum on the right-hand side positive.

Theorem 5. The product of an oscillatory matrix A and a TP matrix is TP.Thus, if an oscillatory matrix A has exponent κ, An is TP for all n ≥ κ.

Proof. This follows from the previous theorem, and that the relation 0 <det(Aκ) = (detA)κ implies that an oscillatory matrix must be non-singular.

2.4 Associated matrices

Before we can say anything about the eigenvalues of an oscillatory matrix weneed to lay some groundwork. The definitions and theorems below will thenhave their analogues when we move on to oscillatory kernels.

An important tool to study the eigenvalues of oscillatory matrices will beassociated matrices. Let A be an (arbitrary) n×n matrix ||aik||n1 . For 1 ≤ p ≤ n,let Mn

p = {(i1, i2, . . . , ip) : 1 ≤ i1 < i2 < · · · < ip ≤ n} be the set of all tuplesof p distinct, ordered indices as before. Note that the elements in Mn

p can beordered lexicographically1, and numbered from 1 to N =

�np

�. Let s, t represent

two such elements (s1, s2, . . . , sp) and (t1, t2, . . . , tp) respectively. We can thendefine as,t as the minor

A

�s1 s2 . . . spt1 t2 . . . tp

�.

Definition 3. Ap = ||as,t||N1 is the p-th associated matrix of A.

Let

A =

�2 1 34 1 21 2 3

�.

A1 is of course equal to A itself. A3 is the matrix consisting of only one element,

namely det A. To calculate A2 we need to compute the�32

�2= 9 minors of size

2, with row and column indices chosen from {(1, 2), (1, 3), (2, 3)}, correspondingto s, t ∈ {1, 2, 3}. For example,

a1,2 = A

�1 21 3

�=

�� 2 34 2

�� = −8.

1Given two tuples s = (s1, s2, . . . , sp) and t = (t1, t2, . . . , tp), s precedes t lexicographicallyif the first non-zero difference (s1 − t1), (s2 − t2), . . . is negative.


Note that the associated matrix of a totally positive matrix has positiveelements. We can also note the following two properties.

Theorem 6. Let A, B and C be matrices with their respective associated ma-trices Ap, Bp and Cp.

i) If C = AB then Cp = ApBp.

ii) If A = B−1 then Ap = B−1p .

Proof. Property i) follows immediately from the Cauchy–Binet Theorem. Infact, each element cs,t of Cp is a minor of C, from which follows that

cs,t =NXr=1

as,rbr,t.

If we apply property i) to the relation AB = I we get ApBp = Ip. But Ip isnothing but another unit matrix Ip = ||δst||N1 . (If s 6= t, then at least one rowof the minor es,t consists only of zeros.) Consequently, ApBp = I, and propertyii) follows.

The main theorem regarding associated matrices is the Kronecker Theorem.

Theorem 7 (Kronecker). Let A be an n× n matrix, with the complete systemof eigenvalues λ1, λ2, . . . , λn. Then the eigenvalues of Ap consist of all possibleproducts of p distinct λk.

Proof. Each matrix A is similar to a triangular matrix T with the same eigen-values as A, i.e., for some matrix P ,

A = PTP−1 = P

�λ1 ∗ ∗ ∗0 λ2 ∗ ∗

0 0. . . ∗

0 0 0 λn

�P−1.

Applying property i) and ii) from Theorem 6, we get Ap = PpTpP−1p . Let us

study the elements of Tp,

ts,t = T

�s1 s2 . . . spt1 t2 . . . tp

�.

For s = t, it is easy to see that ts,s = λs1λs2 . . . λsn := Λs. For s > t, thereexists a number r such that sr > tr, which means that

T

�s1 s2 . . . spt1 t2 . . . tp

�=

��λs1 ∗ ∗ ∗ ∗ ∗

0. . . ∗ ∗ ∗ ∗

0 0 λsr−1∗ ∗ ∗

0 0 0 0 ∗ ∗

0 0 0 0. . . ∗

0 0 0 0 0 λsn

�� = 0.

2.5. Perron’s Theorem 11

The conclusion is that Tp is also a triangular matrix

Tp =

�Λ1 ∗ ∗ ∗0 Λ2 ∗ ∗

0 0. . . ∗

0 0 0 ΛN

�,

with the same eigenvalues Λs = λs1λs2 . . . λsn as the matrix Ap. This completesthe proof.

2.5 Perron’s Theorem

An important theorem of this chapter is Perron’s Theorem. This concerns anymatrix with positive elements, and has been subject to many different proofs.This proof follows [4]. For an interesting survey of alternative proofs, see [6].

Theorem 8 (Perron). If all the elements of a matrix A = ||aik||n1 are positive,then A has a positive, simple eigenvalue ρ, whose absolute value is larger thanthat of all other eigenvalues of A. To this eigenvalue corresponds an eigenvectorwith positive coordinates.

Proof. When A is a 1 × 1 matrix, the theorem is true. We are going to do aproof by induction, so let us assume that the theorem is true for matrices of size(n− 1)× (n− 1) and smaller. Let

Dm(λ) = |λδik − aik|m1 (m = 1, . . . , n)

be determinants of different size. Expanding the largest one, Dn(λ), along thelast column we get

Dn(λ) = (λ− ann)Dn−1(λ) +n−1Xi=1

(−ain)Ain(λ),

where

Ain(λ) = (−1)i+n[λI −A]

�1 . . . i− 1 i+ 1 . . . n1 . . . i− 1 i . . . n− 1

�(2.2)

is the cofactor of λδin − ain. Since i < n, all of the minors in (2.2) have thesame last row, which means that we can expand each of them with respect tothis row. Note that this is now the (n− 1)st row. We get

Ain(λ) = (−1)i+nn−1Xk=1

(−1)n−1+k(−ank)

[λI −A]

�1 . . . i− 1 i+ 1 . . . n− 11 . . . k − 1 k + 1 . . . n− 1

�.

From this, by moving some sign factors around, we obtain

Ain(λ) =n−1Xk=1

ankA(n−1)ik (λ),


where A(n−1)ik is the cofactor of λδik − aik in Dn−1(λ). Putting this together we

have

Dn(λ) = (λ− ann)Dn−1(λ)−n−1Xi,k=1

ainankA(n−1)ik (λ).

According to the induction hypothesis, the truncated matrix ||aik||n−11 has amaximal, positive eigenvalue which we can denote by ρn−1. Using this value,which is a root to Dn−1(λ) = 0, we get

Dn(ρn−1) = −n−1Xi,k=1

ainankA(n−1)ik (ρn−1).

Here we are going to need a second induction hypothesis, that at step n − 1

we have A(n−1)ik (λ) > 0 when λ ≥ ρn−1. It is not meaningful to talk about an

eigenvalue ρ0, but we consider the second hypothesis true for n = 1.Using this, Dn(ρn−1) turns out to be negative, since we know all elements

of the matrix A to be positive, in particular the elements ain and ank. On theother hand,

limλ→∞

Dn(λ) = +∞.

Dn(λ) is a continuous function, so the equation Dn(λ) = 0 must have at leastone positive root larger than ρn−1. We denote the largest of these roots by ρn.

Let us study the cofactors and principal minors Aii(λ). Note that we couldhave begun by expanding Dn(λ) along any row i and column i, leaving us withthe first term (λ − aii)Aii(λ) instead. Aii(λ) would play the role of Dn−1(λ),and we would find that ρn is larger than the largest eigenvalue of each of theseprincipal minors. So for λ ≥ ρn, the cofactors Aii(λ) cannot change sign.Expanding them in terms of λ, the first term turns out to be λn−1, so

limλ→∞

Aii(λ) = +∞.

The conclusion is that all cofactors Aii(λ) > 0 when λ ≥ ρn. This gives usanother part of the theorem. Indeed, by the product rule of differentiation,

D′n(ρn) =nX

i,k=1

δikAik(ρn) =nXi=1

Aii(ρn) > 0,

thus ρn is a simple root of Dn(λ) = 0.Now we need to deal with the other cofactors Aik(λ), where i 6= k. Expand-

ing them, with respect to row k and column i, similarly to how we did beforeand keeping close track of all the sign factors, we get

Aik(λ) = akiC(λ) +Xp,q

Cpq(λ)apiakq (p, q 6= i, k)

where C(λ) is the principal minor where rows and columns i, k are deleted,and Cpq(λ) is the cofactor of λδpq − apq in C(λ). Note especially that the firstterm has no minus sign, which is because the element (−aki) in this cofactor issituated at row k−1 or column i−1 depending on which of the indices is largest.This compensates the minus sign from (−aki) when expanding the cofactor.

2.5. Perron’s Theorem 13

Note also that there seems to be a minor error in [4], where they havereversed the order of the indices k and q in akq.

By the induction hypotheses, C(λ) and Cpq(λ) are positive when λ ≥ ρn, sowe conclude that Aik(λ) > 0 for λ ≥ ρn. This implies that Aik(λ) is positivealso when i 6= k, which completes the induction step for hypothesis two.

It is now possible to construct an eigenvector with positive coordinates. Letu be a vector with coordinates uk equal to the cofactors A1k(ρn). Expandingthe first row of the determinant in the characteristic equation, we get

|δikρn − aik|n1 = 0 =⇒nXk=1

(δ1kρn − a1k)uk = 0.

Replacing the first row of that determinant with a copy of another row i ∈{2, . . . , n}, another determinant equal to zero appears. For each i, we can againexpand with respect to the first row. Since all other rows are as before, thisresults in the same cofactors, but with other coefficients multiplying them. Thus

nXk=1

(δikρn − aik)uk = 0

holds for each i ∈ {1, . . . , n}. This is nothing but the vector equation

||δikρn − aik||n1 · u = 0 =⇒ ρnu = Au.

We also need to compare ρn to the absolute value of the other eigenvaluesof A. Let (v, λ) be an eigenpair of the transposed matrix At, with ρn 6= λ. Notethat λ then also is an eigenvalue of A. We have the equations

Au = ρnu =⇒nXk=1

aikuk = ρnui (i = 1, 2, . . . , n), (2.3)

Atv = λv =⇒nXi=1

aikvi = λvk (k = 1, 2, . . . , n). (2.4)

Applying the triangle inequality to (2.4),

|λ||vk| =

�� nXi=1

aikvi

�� ≤ nXi=1

aik|vi| (k = 1, 2, . . . , n), (2.5)

with equality if and only if all the complex numbers vi have the same argument.Combining (2.3) and (2.5) into

|λ|nXk=1

uk|vk| ≤nXk=1

uk

nXi=1

aik|vi| =nXi=1

|vi|nXk=1

aikuk = ρn

nXi=1

|vi|ui

we see that ρn = |λ| if and only if we have equality in (2.5). But equality therewould make λ > 0 by (2.4), implying ρn = λ which contradicts the assumption.Thus, ρn must be strictly larger than all other |λ|. This completes the inductionstep for the first hypothesis, which means we have proven the theorem.


2.6 Eigenvalues of oscillatory matrices

We are ready to prove the main theorem of this section.

Theorem 9. The eigenvalues of an oscillatory matrix A = ||aik||n1 are simpleand positive:

λ1 > λ2 > · · · > λn > 0.

Proof. First, consider the special case that A is totally positive. Then, for eachq ≤ n, the associated matrix Aq has positive elements. As the reader mighthave guessed, we are going to apply Perron’s theorem to this matrix.

Numbering the eigenvalues of A according to absolute value, |λ1| ≥ |λ2| ≥· · · ≥ |λn|, Kronecker’s theorem combined with Perron’s theorem gives us thatAq has an eigenvalue λ1λ2 . . . λq > 0, which is larger than the absolute value ofall other products of p λk’s. Specifically, λ1λ2 . . . λq > |λ1λ2 . . . λq−1λq+1|. Thefirst inequality implies that all λq > 0, and from the second we get λq > λq+1.These combine to λ1 > λ2 > · · · > λn > 0.

If, instead, A is oscillatory, there is some exponent κ, such that Aκ is totallypositive. Since A is oscillatory, Aκ+1 = A ·Aκ is also totally positive, accordingto Theorem 5 of Section 1. From this, we get two chains of inequalities, λκ1 >λκ2 > · · · > λκn > 0 and λκ+1

1 > λκ+12 > · · · > λκ+1

n > 0.From λκi > 0 and λκ+1

i > 0 we get that each λi > 0, which we can use onthe inequalities λκi > λκi+1 to get that, for each i ∈ {1, . . . , n − 1}, λi > λi+1,giving us the desired properties of the eigenvalues.

Chapter 3

Integral eigenvalueproblems with oscillatorykernels

In this chapter we deal with the integral generalization of oscillatory matrices,namely oscillatory kernels. We show that their eigenvalues behave similarly to thematrix case, and that their eigenfunctions have certain oscillatory properties.

3.1 Oscillatory kernels

We are now going to study an integral eigenvalue problem of the form

ϕ(x) = λ

Z b

aK(x, s)ϕ(s) dσ(s).

This is sometimes called a Fredholm equation of the second kind [8]. Here σ(s)is a strictly increasing function on (a, b), i.e. dσ(s) > 0. This corresponds tothe mass distribution m(x) from the introduction being positive. K(x, s) is acontinuous, oscillatory kernel, which is defined as follows.

Definition 4. K(x, s) is oscillatory if each sampling of it is oscillatory, i.e. ifthe matrix ||K(xi, xk)||n1 is oscillatory for each choice of a < x1 < · · · < xn < b.

We also introduce the following notation for determinants of samplings ofK(x, s):

K

�x1 x2 . . . xns1 s2 . . . sn

�=

�� K(x1, s1) K(x1, s2) . . . K(x1, sn)K(x2, s1) K(x2, s2) . . . K(x2, sn). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .K(xn, s1) K(xn, s2) . . . K(xn, sn)

�� .The oscillatory property of K implies three important determinantal propertieswhich we will often refer to.

Theorem 10. If K(x, s) is oscillatory, the following holds: 1

1These three properties are actually necessary and sufficient, but we will not use this fact.

Kardell, 2010. 15

16 Chapter 3. Integral eigenvalue problems with oscillatory kernels

i) K(x, s) > 0, a < x, s < b

ii) K

�x1 x2 . . . xns1 s2 . . . sn

�≥ 0, a <

x1 < x2 < · · · < xns1 < s2 < · · · < sn

< b

iii) K

�x1 x2 . . . xnx1 x2 . . . xn

�> 0, a < x1 < x2 < · · · < xn < b

Proof. The determinant in property ii) is a minor of an oscillatory matrix. Bydefinition, oscillatory matrices must be totally non-negative, so the minor isgreater than or equal to zero.

Property iii) contains the determinant of an oscillatory matrix. This mustalways be greater than zero, otherwise no power of the matrix could be totallypositive.

Property i) follows from looking at a 2× 2 matrix�K(x1, x1) K(x1, x2)K(x2, x1) K(x2, x2)

�.

For this matrix to be oscillatory, K(x1, x2) and K(x2, x1) must be non-zero(positive). If they were not, the matrix would be triangular, and so would allpowers of it. Thus no power would be totally positive. Since x1 < x2 werechosen arbitrarily, property i) must be valid for x 6= s. The case x = s iscovered by property iii).

3.2 Associated kernels

In analogy with the matrix case, we can call the determinants

Kn(X,S) = K

�x1 x2 . . . xns1 s2 . . . sn

�the n-th associated kernel of K(x, s). Here, X = (x1, x2, . . . , xn) and S =(s1, s2, . . . , sn) fulfill the conditions

a <x1 < x2 < · · · < xns1 < s2 < · · · < sn

< b.

Let us denote the set of such n-tuples as Mn. The associated kernels have thefollowing important property:

Theorem 11. Let K(x, s), L(x, s) and N(x, s) be kernels related by

K(x, s) =

Z b

aL(x, t)N(t, s) dσ(t).

Then for their respective associated kernels Kn(X,S), Ln(X,S) and Nn(X,S),

Kn(X,S) =

ZMn

Ln(X,T )Nn(T, S) dσ(T )

holds, where X,S, T ∈Mn and dσ(T ) = dσ(t1) . . . dσ(tn).

3.2. Associated kernels 17

Proof. This can be shown by the following calculations. The associated kernelKn(X,S) is equal to

Kn(X,S) = |K(xi, sk)|n1 =

��Z b

aL(xi, t)N(t, sk) dσ(t)

��n1

.

On each row i, to be able to break out the integral sign, we index the integrationvariable t with any permutation π(i), i.e.,

Kn(X,S) =

��Z b

aL(xi, tπ(i))N(tπ(i), sk) dσ(tπ(i))

��n1

=

=

Z b

a. . .

Z b

a

��L(xi, tπ(i))N(tπ(i), sk)��n1dσ(T ).

From each row in the determinant, we can also take out a factor L(xi, tπ(i));Z b

a. . .

Z b

aL(x1, tπ(1)) . . . L(xn, tπ(n))

��N(tπ(i), sk)��n1dσ(T ).

Rearranging the rows of what is left in the determinant, we can get the indicesin increasing order. This gives us a determinant of the form Nn(T, S), but todo this, we must compensate with a sign factor depending on the permutationπ. The expression then becomesZ b

a. . .

Z b

asgn(π)

nYi=1

L(xi, tπ(i))Nn(T, S) dσ(T ).

Note that the value of this integral does not depend on π. Therefore, we canform the average over all permutations π and still get the same value. Thismeans that we have

Kn(X,S) =1

n!

Z b

a. . .

Z b

a

Xπ

sgn(π)nYi=1

L(xi, tπ(i))Nn(T, S) dσ(T ) =

=1

n!

Z b

a. . .

Z b

aLn(X,T )Nn(T, S) dσ(T ).

The n-dimensional cube [a, b]n that we are integrating over can be seen as theunion of n! wedges

{t1 ≤ t2 ≤ · · · ≤ tn} ∪ {t2 ≤ t1 ≤ · · · ≤ tn} ∪ · · · ∪ {tn ≤ · · · ≤ t2 ≤ t1}.

Each of these give the same contribution to the integral over the whole cube,since each sign change in Ln(X,Π(T )) is cancelled be the same sign changein Nn(Π(T ), S). This means that we can choose just one of these wedges tointegrate over n! times, thus cancelling the factor 1

n! . We choose the wedge Mn

of course, and the result is the desired relation

Kn(X,S) =

ZMn

Ln(X,T )Nn(T,K) dσ(T ).


With help from this, we can show the following theorem, which is the integralanalogue of Kronecker’s theorem.

Theorem 12. If the integral equation

ϕ(x) = λ

Z b

aK(x, s)ϕ(s) dσ(s)

has a complete system of orthonormal eigenfunctions ϕ0(x), ϕ1(x), . . . with cor-responding eigenvalues λ0, λ1, . . . then the integral equation for the associatedkernel Kn(X,S),

Φ(X) = Λ

ZMn

Kn(X,S)Φ(S) dσ(S),

has a complete system of orthonormal eigenfunctions consisting of determinants

Φp(X) = |ϕpi(xk)|ni,k=1, X ∈Mn

with corresponding eigenvalues Λp = λp1λp2 . . . λpn where 0 ≤ p1 < p2 < · · · <pn are running over all possible products of λ’s.

Proof. We have

|ϕpi(xk)|n1 =

��λpi Z b

aK(xk, s)ϕpi(s) dσ(s)

��n1

which by the calculations of the previous theorem is equal to

λp1 . . . λpn

ZMn

K(X,S)|ϕpi(sk)|n1 dσ(S).

Thus, we know the eigenfunctions and eigenvalues to be correct. What aboutthe orthonormality? We have

(Φp(S),Φq(S)) =

ZMn

|ϕpi(sk)|n1 |ϕqi(sk)|n1 dσ(S) =

=

��Z b

aϕpi(s)ϕqi(s) dσ(s)

��n1

=Xτ

sgn(τ)nYi=1

δpiτ(qi).

The products in the expression above are 1 if the sequences pi and τ(qi) areequal for each entry. Since pi and qi are strictly increasing, all permutationsexcept the identity permutation will give a zero contribution to the sum. Theorthonormality follows.

3.3 An integral analogue of Perron’s Theorem

To prove the eigenvalue properties of our integral eigenvalue problem, we need atheorem which corresponds to Perron’s Theorem in the matrix case. Note thatthe following theorem can also be applied to associated kernels Kn(X,S). Thearguments then vary in the n-dimensional space Mn instead of an interval.

3.4. Eigenvalues and eigenfunctions of oscillatory kernels 19

Theorem 13. If the integral equation

ϕ(x) = λ

Z b


has a continuous kernel K(x, s) satisfying K(x, s) ≥ 0 and K(x, x) > 0, then theabsolutely smallest eigenvalue λ0 is positive, simple, and has a strictly smallerabsolute value than the other eigenvalues. The corresponding eigenfunctionϕ0(x) has no zeros in (a, b).

We will not present a full proof of this theorem here. In [4], a proof is givenfor the case when K(x, s) is symmetric, i.e., K(x, s) = K(s, x), though that isnot the case for our kernel. Indeed, in the introduction we showed a samplingGn, which clearly is non-symmetric.

The general theory of these integral equations dates back to Fredholm [7],where a generalization of determinants, called the Fredholm determinant, isdefined and studied. It turns out that the Fredholm determinant, which is afunction of the parameter λ, is zero if and only if λ is an eigenvalue of thekernel. This is used to prove several analogues to theorems from linear algebra,for example that K(x, s) and K(s, x) have the same eigenvalues. For a niceexposition of the theory, see [8]. A complete proof of Theorem 13 in the non-symmetric case is given in [9].

3.4 Eigenvalues and eigenfunctions of oscillatorykernels

Now we just have to combine the previous theorems to receive the desired prop-erties for eigenvalues of an oscillatory kernel.

Theorem 14. If K(x, s) is a continuous, oscillatory kernel, its eigenvalues arepositive and distinct, i.e. 0 < λ1 < λ2 < · · · <∞.

Proof. Let the eigenvalues of K be numbered2 according to their absolute values|λ0| ≤ |λ1| ≤ . . . with corresponding orthonormal eigenfunctions

ϕ0(x), ϕ1(x), . . .

According to Theorem 12, the eigenvalues of Kn(X,S) are all possible prod-ucts λp1 . . . λpn . The smallest eigenvalue must then be λ0λ1 . . . λn−1, with thecorresponding eigenfunction |ϕi−1(xk)|n1 .

Theorem 13 gives that the n-th associated kernel’s eigenvalue λ0λ1 . . . λn−1is positive, and less than |λ0λ1 . . . λn−2λn| for each n. From this, the eigenvaluecondition 0 < λ0 < λ1 < . . . follows.

We can also say something about the eigenfunctions of K(x, s). First, adefinition.

Definition 5. The functions ϕ0(x), ϕ1(x), . . . , ϕn−1(x) are called a Chebyshevsystem if each non-trivial linear combination of them has at most n− 1 zeros.

2In [8] it is proven that the set of eigenvalues of K is at most enumerable.


There is a useful criterion for determining when a number of functions forma Chebyshev system.

Theorem 15 (Chebyshev criterion). The functions ϕ0(x), ϕ1(x), . . . , ϕn−1(x)form a Chebyshev system if and only if the determinant D = |ϕi(xk)|n−1i,k=0 isdifferent from zero (of constant sign) for all samplings a ≤ x1 < · · · < xn ≤ b.

Proof. Assume that there is a sampling (x1, . . . , xn) such that D = 0. Then, forsome solution ci, not all equal to zero,

D = |ϕi(xk)|n−1i,k=0 = 0 ⇐⇒ Dt = |ϕk(xi)|n−1i,k=0 = 0 ⇐⇒�ϕ0(x0) ϕ1(x0) . . . ϕn−1(x0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .ϕ0(xn−1) ϕ1(xn−1) . . . ϕn−1(xn−1)

�� c0c1...

cn−1

�=

�0...0

�⇐⇒

⇐⇒n−1Xi=0

ciϕi(xk) = 0, k = 0, 1, . . . , n− 1.

Clearly, when D is zero, there exists a non-trivial linear combination with nzeros x0 through xn−1, which is equivalent to ϕ0(x), . . . , ϕn−1(x) not being aChebyshev system. When no such linear combination exists, D is non-zero.Since it is a continuous function on a connected set, D can not change sign.

The following theorem shows one of the main properties of Chebyshev sys-tems. There are also other oscillatory properties of these systems, but we willnot go into these here. Again, see [4].

Theorem 16. If a sequence of functions ϕ0(x), ϕ1(x), ϕ2(x), . . . is orthonormal,and the first n functions form a Chebyshev sequence for each n, then the functionϕj(x) has exactly j zeros in (a, b) for any j.

Proof. By the definition of a Chebyshev system, ϕj(x) can have at most j zeros.Assume it has p ≤ j zeros, α1 < · · · < αp. Let αp+1 be an arbitrary point in theinterval (a, b) and study the determinant D(αp+1) = |ϕi(αk+1)|p0. D(αp+1) thusdefines a function D(x) which is a linear combination of ϕ0(x), ..., ϕp(x). FromTheorem 15 we know that D is different from zero when αp+1 6= α1, . . . , αp.(When αp+1 is equal to any other αi, D is of course zero, since two columns areequal.) Letting αp+1 slide from b to a, we can swap two columns each time itpasses through a zero, to keep the ordering of the points α. This means thatwe can still apply Theorem 15 in each interval, though the column switchinggives us a sign change between each interval. It follows that D(αp+1) has thesame p zeros as ϕj(x). Both functions change sign at each zero, which meansthat (D,ϕj) 6= 0. Thus, D has a component in the ϕj-direction. Consequently,p ≥ j, which combined with the Chebyshev criterion gives us that ϕj has exactlyj zeros.

Next, we show that eigenfunctions of an oscillatory kernel form a Chebyshevsystem.

Theorem 17. Given an oscillatory kernel K(x, s), the first n functions in thesequence of its eigenfunctions ϕ0(x), ϕ1(x), . . . form a Chebyshev system for eachn.

3.4. Eigenvalues and eigenfunctions of oscillatory kernels 21

Proof. Apply Theorem 12 and 13 (the analogues of Kronecker and Perron) tothe associated kernel Kn(X,S). The eigenfunction corresponding to its small-est eigenvalue is |ϕi−1(xk)|n1 , and has no zeros. The Chebyshev criterion thenimplies that ϕ0(x), ϕ1(x), . . . , ϕn−1(x) form a Chebyshev system for each n.

Now we have all the parts needed for the theorem we set out to prove.

Theorem 18. If K(x, s) is a continuous oscillatory kernel, and σ(s) is a strictlyincreasing function, then the integral equation

ϕ(x) = λ

Z b


has the following three properties:

i) All eigenvalues are positive and simple; 0 < λ0 < λ1 < λ2 < . . .

ii) The eigenfunction ϕ0(x) corresponding to the smallest eigenvalue λ0 hasno zeros in (a, b).

iii) The eigenfunction ϕj(x) corresponding to the eigenvalue λj has exactly jzeros in (a, b) and changes sign at each.

Proof. Property i) is Theorem 14. Theorem 17 shows that the eigenfunctionsform a Chebyshev system, and thus have the properties ii) and iii).

Let us just note that the theorem could be generalized to an arbitrary non-decreasing σ(x). The completely discrete case (where dσ is a sum of Diracdeltas) breaks down to the case with oscillatory matrices, handled in Chapter2. When dσ is a mix of these cases, the proof gets a bit more technicallycomplicated, but can still be followed. The interested reader is referred toSection 4.4 in [4].

Chapter 4

Planar networks

In this chapter we show the connection from TNN/Oscillatory/TP matrices to planarnetworks, and show that G is totally nonnegative.

4.1 Preliminaries

A planar network is a graph consisting of vertices and edges. Edges are directedconnections from one vertex to another, and may not cross other edges; also,no loops are allowed. This means that there must be vertices without anyincoming edges, which we can call sources. Vertices without outgoing edges arecalled sinks.

In this chapter, we assume that each network has n sources and sinks each,which we place to the left and right respectively, numbered bottom to top. Othervertices will be placed between them in such a way that all edges are directedleft to right. A series of edges connecting a source and a sink is called a path.

To each edge is assigned a real number, called its weight. The weight ofa path is defined as the product of the weights of its edges. The sum of theweights of all paths connecting source i and sink k is denoted wik. The weightmatrix of a network is the matrix W = ||wik||n1 .

Example: Let us study the following simple network.

1 1

2 2

a1 1

1 1

b c

d

Here n = 2. The entries w11, w12 and w21 correspond to only one possible path,and will consist of only one term. For w22 we can choose two paths, abc or d.The weight matrix is �

a acab abc+ d

�.

A collection of paths, connecting a set of sources to a set of sinks, can also beassigned a weight. This is defined as the product of the weights of the paths. Wewill want to look at vertex-disjoint paths, i.e., paths that are non-intersectingand non-touching. In the example above, the only collection of vertex-disjointpaths connecting sources {1, 2} and sinks {1, 2} has weight ad.

Kardell, 2010. 23

24 Chapter 4. Planar networks

We will mostly deal with totally connected networks. That is, for each set ofp sources and p sinks, there exists at least one collection of vertex-disjoint pathsconnecting those sources and sinks. The previous example showed a totallyconnected 2× 2 network. By inspection, the following 4× 4 network also turnsout to be totally connected. From now on we assume that all edges go from leftto right.

1 1

2 2

3 3

4 4

The same construction will be used throughout this chapter. It is not hard toimagine what the network looks like for larger n. In this construction, an edgewill be called essential if it is diagonal, or if it is one of the horizontal edges inthe middle of the network. Note that there are exactly n2 such edges. Theseare the weights that we want to manipulate later on. All other weights will beset to 1, and will not be displayed in the graphs.

4.2 Lindstrom’s Lemma

We want to show that the matrix G from the introduction is oscillatory. Thefirst part will be to show that it is TNN. Calculating all minors of arbitrary sizeis unfeasible, since there are

�2nn

�− 1 minors of an n × n matrix. Luckily, this

number can be reduced quite a lot. Lindstrom’s Lemma links the minors of aweight matrix to path collections of a planar network.

Theorem 19 (Lindstrom’s Lemma). A minor

W

�i1 i2 . . . ipk1 k2 . . . kp

�of the weight matrix of a planar network is equal to the sum of weights of allcollections of vertex-disjoint paths that connect the sources {i1, i2, . . . , ip} to thesinks {k1, k2, . . . , kp}.

Proof. It is enough to prove the lemma for the determinant of the entire weightmatrix, which corresponds to choosing all sources and sinks. Minors can thenbe handled in exactly the same way.

The determinant is equal to

|W |n1 =Xσ

sgn(σ)

nYi=1

wiσ(i)

!,

where the sum is over all permutations on {1, 2, . . . , n}. The productQni=1 wiσ(i)

breaks down to a sum over all collections of paths π = (π1, . . . , πn) such that πi

4.3. Constructing a planar network with a given weight matrix 25

connects source i to sink σ(i). Contributions from collections of vertex-disjointpaths comes from the identity permutation, which has sign +1. Our task is toshow that all other terms cancel.

To do this we deform the network a bit, if necessary, to guarantee that novertices lie on the same vertical line. This means that we can number the verticesfrom left to right in a well-defined way. We can then define an involution onthe non-vertex-disjoint collections of paths, that takes the leftmost vertex wheretwo paths coincide and switch the parts of those paths that lie to the left ofthat vertex. This preserves the weight of the collection, but changes the sign ofthe permutation σ associated to this term. Pairing terms in this way gives usthe desired cancellation of all non-vertex-disjoint collections.

The following corollaries are straightforward applications of Lindstrom’sLemma.

Corollary 2. If a planar network has non-negative weights, then its weightmatrix is TNN.

Corollary 3. If a totally connected network has positive weights, then its weightmatrix is TP.

This means that if we can show that our matrix G is the weight matrix ofa planar network with non-negative weights, it is TNN. This is the goal of thenext section.

Note that in [10] it is shown that our construction of totally connectednetworks generates all TP matrices, and each TP matrix is the weight matrixof such a network with positive weights on each essential edge. Since we donot expect our matrix to be TP, we will probably get a network with zeros insome weights, but it turns out that we can still construct a network such thatits weight matrix equals G.

4.3 Constructing a planar network with a givenweight matrix

From the introduction, we have for each n the matrix

Gn =

�1 E12 E13 . . . E1n

2− E12 1 E23 . . . E2n

2− E13 2− E23 1 . . . E3n

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2− E1n 2− E2n 2− E3n . . . 1

�.

where Eij = exi−xj . We are going to show that this matrix is the weight matrixof a planar network with non-negative weights. With a change of variables,where Eij = 1−Qij , we can write our matrix as

Gn =

�1 1−Q12 1−Q13 . . . 1−Q1n

1 +Q12 1 1−Q23 . . . 1−Q2n

1 +Q13 1 +Q23 1 . . . 1−Q3n

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 +Q1n 1 +Q2n 1 +Q3n . . . 1

�.


Remember that x1 < · · · < xn, which means that both Eij and Qij are between0 and 1. We are going to switch between these forms depending on which onegives the cleanest calculations. We start with the case n = 2.

Theorem 20. G2 is totally positive.

Proof. We need to match the network

1 1

2 2

ab c

d

with the matrix

G2 =

�1 1−Q12

1 +Q12 1

�.

Moving from source 1 to sink 1, we have only one choice of paths, namely a.This path must be equal to the element g11, so a = 1. The same reasoning givesus ab = 1 +Q12 and ac = 1−Q12, from which we get b and c immediately. Tofind d, we note that the weight of the collection of vertex-disjoint paths from{1, 2} to {1, 2} is equal to ad = 1 · d = d. We calculate the determinant of G2

to be

detG2 = 1− (1 +Q12)(1−Q12) = Q212.

Lindstrom’s Lemma tells us that d must be equal to this value. We draw theconclusion that G2 is the weight matrix of the network

1 1

2 2

11 +Q12 1−Q12

Q212

which is a totally connected planar network with positive weights. G2 is thus atotally positive matrix.

We could of course have found out this result just by looking at the matrixelements and the determinant, but the network will be needed later on, as wesee that things are not this simple for G3 and higher. Let us denote the weightsof each essential edge up to n = 5 with a letter:

1 1

2 2

3 3

4 4

5 5

a = 1b = 1 +Q12 c = 1−Q12

d = Q212

e f g h

ij k l m n o

pq r s t u v x y

z


We will keep referring back to this picture several times, sometimes referringto the edges themselves by these same letters. With the help from Lindstrom’sLemma, we will try to find the weights of all these edges, and hope to see apattern which extends to arbitrary size. We start with the rightmost diagonal(consisting of c, h, o, and y), corresponding to the first row of G.

Theorem 21. The weights of the rightmost diagonal (c, h, o, etc.) are positive.

Proof. We get the following system of equations8>>>><>>>>:a = 1

ac = 1−Q12

ach = 1−Q13

acho = 1−Q14

. . .

,

which has the solution

a = 1, c = 1−Q12, h =1−Q13

1−Q12, o =

1−Q14

1−Q13, . . .

Note that we can simplify h, o, and the following weights in their diagonal, since

h =1−Q13

1−Q12=E13

E12=ex1−x3

ex1−x2= ex2−x3 = E23 = 1−Q23.

In the same way, we get o = 1−Q34 and so on. We see that all weights on thisdiagonal are positive.

In the same way, we can study the leftmost diagonal.

Theorem 22. The weights of the leftmost diagonal (b, e, j, etc.) are positive.

Proof. Here we get the system8>>>><>>>>:a = 1

ab = 1 +Q12

abe = 1 +Q13

abej = 1 +Q14

. . .

.

The solution is

a = 1, b = 1 +Q12, e =1 +Q13

1 +Q12, j =

1 +Q14

1 +Q13, . . .

which unfortunately cannot be simplified further, but we do see that theseweights are positive.

On the right half of the network, things remain relatively simple thanks tothe fact that we could simplify the weights on the outer diagonal. Indeed, thefollowing theorem holds.


Theorem 23. The remaining weights in the right half of the network are zero.

Proof. We start with the edge g. To calculate its weight we use that Eii =exi−xi = 1, and see that the minor

G

�1 22 3

�=

�� E12 E13

E22 E23

�� =

�� E12 E12E23

E22 E22E23

�� = E23

�� E12 E12

E22 E22

��is equal to zero. The only collection of vertex-disjoint paths from {1, 2} to {2, 3}uses the following links;

1 1

2 2

3 3

ac

dg

where g is the only unknown weight. We draw the conclusion that g must bezero. In the same way, all 2× 2 minors from {1, 2} to {i, i+ 1} will turn out tobe zero, but that will not help us find out more weights along the diagonal of g.(We get equations like 0 = 0 ·n, etc.) Instead, we need a more clever argument.

Let us look at all possible paths from 3 to 4. We know that moving fromsource 3 to sink 3, or more generally, source i to sink i, will add up to a totalweight of 1. Thus, moving first from 3 to 3, and then up to 4 along edge o

1 1

2 2

3 3

4 4

m n o

1

gives the total weight 1 ·o = o which we calculated to be 1−Q34. This is exactlythe weight we want in total from source 3 to sink 4, so the contributions frompaths using edges m and n should add up to zero. The only way to do this,while using non-negative weights, is to let the weights m and n both be zero.

With the same argument we can show, by induction, that if we put zeroson all diagonal edges in the right half of the network (except for the rightmostdiagonal, which we already calculated); the entries above the main diagonal inG match the possible paths in the network. Thus, u, v, and x should be put tozero, and so on. An arbitrary source i will then only have one possible path tosink k, provided that i < k. The weight on this path will be

(1−Qi,i+1)(1−Qi+1,i+2) . . . (1−Qk−1,k) = Ei,i+1 . . . Ek−1,k = Eik = 1−Qik

as desired.

Now we are going to deal with the horizontal edges in the middle of thenetwork. We already calculated a = 1 and d = Q2

12, and might guess that thepattern continues with i = Q2

23, p = Q234, and so on. This is indeed the case.


Theorem 24. The n-th essential horizontal edge has positive weight Q2n−1,n,

for all n ≥ 2.

Proof. We show that the guess for p is correct, using an argument easily ex-tended to an arbitrary such edge.

Thanks to the zeros we only get two contributions when moving from source4 to sink 4. The first is p, and the second is o times the total weight movingfrom 4 to 3,

1 1

2 2

3 3

4 4

0 0 o

p

which is 1 +Q34. This gives us p by the following simple calculation:

p+ o · (1 +Q34) = 1 =⇒ p = 1− (1−Q34)(1 +Q34) = Q234.

The reader can probably see how this calculation works for arbitrary n.

So far, it’s been relatively simple to calculate the weights, and we have foundsimple patterns, which show that certain weights are positive. On the left halfof the network, things get a bit more complicated. We will need to do f first.

Theorem 25. The weight f is positive.

Proof. To get an equation determining f , we look at path collections from {2, 3}to {1, 2}. The only vertex-disjoint one is adbf .

1 1

2 2

3 3

ab

df

By Lindstrom’s Lemma, this value is equal to the minor

G

�2 31 2

�=

�� 1 +Q12 11 +Q13 1 +Q23

�� =

(1 +Q12)(1 +Q23)− (1 +Q13) = 1 +Q12 +Q23 +Q12Q23 − 1−Q13 =

1− E12 + 1− E23 + 1− E12 − E23 + E12E23 − 1− 1 + E13 =

2(1− E12 − E23 + E13) = 2(1− E12)(1− E23) = 2Q12Q23.

Thus, we get

f =2Q12Q23

abd=

2Q12Q23

(1 +Q12)Q212

=2Q23

(1 +Q12)Q12,

which is positive.


Above f in the second diagonal, we can calculate a sequence of positiveweights in the following way.

Theorem 26. The weights of the second leftmost diagonal (k, r, etc.) arepositive.

Proof. Moving up along the second diagonal, we need to look at path collectionsfrom {i, i+ 1} to {1, 2}, for i ≥ 3. These minors are equal to

G

�i i+ 11 2

�=

�� 1 +Q1,i 1 +Q2,i

1 +Q1,i+1 1 +Q2,i+1

�� = 2Q12Qi,i+1(1−Q2,i),

by a calculation similar to that in the previous proof. We then calculate k andr;

k =2Q12Q34(1−Q23)

abdef= · · · = Q34(1 +Q12)(1−Q23)

Q23(1 +Q13)(1−Q22),

r =2Q12Q45(1−Q24)

abdefjk= · · · = Q45(1 +Q13)(1−Q24)

Q34(1 +Q14)(1−Q23).

Now the pattern for the second diagonal should be apparent. Since the weightsform a telescoping product, it follows by an induction argument that this patternholds higher up along the diagonal.

An even trickier task is to prove the pattern for edges l, t and upwards.(Those diagonal edges in the left half of the network that are closest to themiddle, to be precise). To find a path collection that includes edge l we choosesources {2, 3, 4} and sinks {1, 2, 3}. The only collection of vertex-disjoint pathshas weight adibfl:

1 1

2 2

3 3

4 4

ab

df

il

Then, to get an equation for t, we choose sources {2, 3, 4, 5} and sinks {1, 2, 3, 4},and so on. Apparently, we would need to be able to calculate the minors

Gn

�2 3 . . . n1 2 . . . n− 1

�.

Calculating these minors with Mathematica up to n = 6 showed a promisingpattern, namely 2Q12Q

223Q

234 . . . Q

2n−2,n−1Qn−1,n. We are going to prove this

now.

Theorem 27. The diagonal edges in the left half of the network closest to themiddle are positive.


Proof. It is actually possible to calculate the desired minor directly. Let exi = zi.With this definition,

1−Qij = Eij =zizj,

and

1 +Qij = 2− Eij = 2− zizj

=2zj − zizj

.

We calculate the minor for n = 6, but exactly the same row and column opera-tions could be used to calculate this minor for arbitrarily large n. We have

G

�2 . . . 61 . . . 5

�=

��1 +Q12 1 1−Q23 1−Q24 1−Q25

1 +Q13 1 +Q23 1 1−Q34 1−Q35

1 +Q14 1 +Q24 1 +Q34 1 1−Q45

1 +Q15 1 +Q25 1 +Q35 1 +Q45 11 +Q16 1 +Q26 1 +Q36 1 +Q46 1 +Q56

�� =

=

��

2z2 − z1z2

1z2z3

z2z4

z2z5

2z3 − z1z3

2z3 − z2z3

1z3z4

z3z5

2z4 − z1z4

2z4 − z2z4

2z4 − z3z4

1z4z5

2z5 − z1z5

2z5 − z2z5

2z5 − z3z5

2z5 − z4z5

1

2z6 − z1z6

2z6 − z2z6

2z6 − z3z6

2z6 − z4z6

2z6 − z5z6

��.

Factoring out the denominators z2 to zn from each row, and z3 to zn−1 fromcolumn 3 through 5, gives us the following determinant;

1

z2z23z24z

25z6

��2z2 − z1 z2 z22 z22 z22

2z3 − z1 2z3 − z2 z23 z23 z23

2z4 − z1 2z4 − z2 (2z4 − z3)z3 z24 z24

2z5 − z1 2z5 − z2 (2z5 − z3)z3 (2z5 − z4)z4 z25

2z6 − z1 2z6 − z2 (2z6 − z3)z3 (2z6 − z4)z4 (2z6 − z5)z5

�� .We now subtract column 2 from column 1, and see that we get z2 − z1 in eachposition in the first column. Factoring this out, we get

z2 − z1z2z23z

24z

25z6

��1 z2 z22 z22 z22

1 2z3 − z2 z23 z23 z23

1 2z4 − z2 (2z4 − z3)z3 z24 z24

1 2z5 − z2 (2z5 − z3)z3 (2z5 − z4)z4 z25

1 2z6 − z2 (2z6 − z3)z3 (2z6 − z4)z4 (2z6 − z5)z5

�� .We leave the first row as it is, and subtract the first row from the second, the


second from the third, and so on. This yields

z2 − z1z2z23z

24z

25z6

��1 z2 z22 z22 z22

0 2(z3 − z2) z23 − z22 z23 − z22 z23 − z220 2(z4 − z3) 2(z4 − z3)z3 z24 − z23 z24 − z230 2(z5 − z4) 2(z5 − z4)z3 2(z5 − z4)z4 z25 − z240 2(z6 − z5) 2(z6 − z5)z3 2(z6 − z5)z4 2(z6 − z5)z5

�� .Expanding along the first column gives only one term. From the remainingcofactor we can take out a factor 2 from the first column, and a factor of theform (zi+1 − zi) from each row. We have now reduced our determinant to thefollowing expression;

2Q5i=1(zi+1 − zi)z2z23z

24z

25z6

�� 1 z3 + z2 z3 + z2 z3 + z21 2z3 z4 + z3 z4 + z31 2z3 2z4 z5 + z41 2z3 2z4 2z5

�� .Once again, we subtract the first row from the second row, and so on, to getthe final step

G

�2 . . . 61 . . . 5

�=

2Q5i=1(zi+1 − zi)z2z23z

24z

25z6

�� 1 z3 + z2 z3 + z2 z3 + z20 z3 − z2 z4 − z2 z4 − z20 0 z4 − z3 z5 − z30 0 0 z5 − z4

��=

2(z2 − z1)(z3 − z2)2(z4 − z3)2(z5 − z4)2(z6 − z5)

z2z23z24z

25z6

= 2Q12Q223Q

234Q

245Q56.

We have proven our claim about the minors from {2, . . . , n} to {1, . . . , n − 1}.Using this result, by an induction argument similar to the one regarding thesecond left diagonal,

l =Q34

Q23, t =

Q45

Q34, . . .

Fortunately, the remaining edges will turn out to be zero, but proving thisrequires a little finesse.

Theorem 28. The remaining weights in the left half of the network are zero.

Proof. We start with the edge s. This edge is included in the only vertex-disjointpath collection from {3, 4, 5} to {1, 2, 3}.


1 1

2 2

3 3

4 4

5 5

ab

df

il

e

k

s

This corresponds to the minor

G

�3 4 51 2 3

�=

�� 2− E13 2− E23 2− E33

2− E14 2− E24 2− E34

2− E15 2− E25 2− E35

�� ,which is the determinant of a sum of two matrices, each with rank 1:�� 2 2 2

2 2 22 2 2

�+

�−E13 −E23 −E33

−E14 −E24 −E34

−E15 −E25 −E35

�� .Indeed, the first matrix obviously has all columns linearly dependent. In thesecond matrix, all columns are multiples of the first one, because of how Eij isdefined. The rank of a sum of two matrices can never exceed the sum of theranks of the terms, so our minor has rank less than or equal to two. Thus, itcannot have full rank, and must be zero. Because of this, one of the links in thepath collection must be zero, and the only unknown link at this point is s. Weconclude that s is zero.

Let us take a close look at the left half of the network. We now have a ”gap”in the structure.

1

2

3

4

5

6x1 x2

Note that there is then no way to find a path collection that distinguishes x1from x2. Thus, we can let x1 be zero, and try to calculate x2. (By the samereasoning, all edges in the diagonal above x1 can bet set to zero.) To include


x2 in a unique path collection, we need to use sources {2, 4, 5, 6} and sinks{1, 2, 3, 4}.

1

2

3

4

5

6x2

Provided that the corresponding minor is zero, x2 must be zero. In that casewe need to look at the minor from {2, 3, 5, 6, 7} to {1, 2, 3, 4, 5}, then from{2, 3, 4, 6, 7, 8} to {1, 2, 3, 4, 5, 6}, and so on. We draw the network for n = 8 toconvince ourselves that the pattern continues.

1

2

3

4

5

6

7

8

We see that below the three highest sources, exactly one source is excluded fromthe pattern, then all sources down to number 2 are included. So what we needto show is that the minors

G

�2 . . . n− 4 n− 2 n− 1 n1 . . . . . . . . . . . . . . . . . . . . . . . . . n− 2

�are zero. There is actually a very simple argument for this. Expanding theminor along the three last rows, we see from the structure of the matrix thatall 3× 3 minors from these rows are zero, with the same argument about ranks


as before. Summing over these minors times their cofactors of course gives us azero sum, and the proof is complete.

Concluding this section, we see that all links of the network we have con-structed are non-negative. G is the weight matrix of this network, so we havepresented a complete proof of the main result of this chapter.

Theorem 29. The matrix G is totally non-negative.

Below is a table of all the calculated non-zero weights, for easy reference.

1 1

2 2

3 3

4 4

5 5

ab c

de f h

ij k l o

pq r t y

a = 1 c = 1−Q12 b = 1 +Q12 f =2Q23

(1 +Q12)Q12l =

Q34

Q23

d = Q212 h = 1−Q23 e =

1 +Q13

1 +Q12k =

Q34(1 +Q12)(1−Q23)

Q23(1 +Q13)(1−Q22)t =

Q45

Q34

i = Q223 o = 1−Q34 j =

1 +Q14

1 +Q13r =

Q45(1 +Q13)(1−Q24)

Q34(1 +Q14)(1−Q23). . .

p = Q234 y = 1−Q45 q =

1 +Q15

1 +Q14. . . . . .

Chapter 5

Summary and conclusion

To show that our totally non-negative matrix is oscillatory we need only ashort additional argument. We note that multiplying weight matrices of twonetworks is the same as first concatenating the networks and then calculatingthe resulting weight matrix. This is nothing but the Cauchy–Binet Theorem indisguised form. Then we realize that by concatenating a large enough numberof networks of the kind we just calculated, there will be enough room in thenetwork to find a collection of vertex-disjoint paths with non-zero weight forany sets of sources and sinks. This network then has a totally positive weightmatrix, which is a power of G. We have proven the following.

Theorem 30. G is an oscillatory matrix.

There is a more formal criterion proven in [4] which tells us that a totallynon-negative matrix A is oscillatory if (and only if) it is invertible and haspositive elements above and below the main diagonal. This makes sense whenwe think about A as the weight matrix of a planar network, as those elementsguarantees that one can move at least one step up or down from each node.This is necessary if the concatenated network is to be totally positive. For moreoscillatory criteria, see [10].

Since G is oscillatory for arbitrary samplings x1 < x2 < · · · < xn, thekernel g(s − x) that gave rise to it must be oscillatory. This means that theeigenfunctions ψ(x) of the integral equation

ψ(x) =z

2

Z ∞−∞

g(s− x)m(s)ψ(s) ds

have the properties proven in Chapter 3.

Kardell, 2010. 37

38 Chapter 5. Summary and conclusion

Bibliography

[1] H. Lundmark, J. Szmigielski, Degasperis–Procesi peakons and the discretecubic string, International Mathematics Research Papers, (2005), no. 2,53–116.

[2] J. Kohlenberg, H. Lundmark, J. Szmigielski, The inverse spectral problemfor the discrete cubic string, Inverse Problems 23 (Feb 2007), 99–121.

[3] A. Degasperis, D. D. Holm, A. N. W. Hone, A new integrable equation withpeakon solutions, Theoret. and Math. Phys. 133 (2002), no. 2, 1463–1474.

[4] F. R. Gantmacher, M. G. Krein, Oscillation Matrices and Kernels andSmall Vibrations of Mechanical Systems, revised ed., American Mathemat-ical Society, Providence, Rhode Island, (2002).

[5] M. G. Krein, Sur les fonctions de Green non-symetriques oscillatoires desoperateurs differentiels ordinaires C. R. (Doklady) Acad. Sci. URSS (N.S.)(1939), no. 25, 643–646

[6] C. R. MacCluer, The many proofs and applications of Perron’s Theorem,SIAM Rev. 42 (2000), no. 3, 487–498.

[7] I. Fredholm, Sur une classe d’equations fonctionnelles, Acta Math. 23(1903), 365–390.

[8] F. Smithies, Integral Equations, Cambridge University Press, New York,(1958).

[9] R. Jentzsch, Uber Integralgleichungen mit positivem Kern, J. Reine Angew.Math. 141 (1912), 235–244

[10] S. Fomin and A. Zelevinsky, Total positivity: tests and parametrizations,Math. Intelligencer 22 (2000), no. 1, 23–33.

Kardell, 2010. 39

40 Bibliography

Copyright

The publishers will keep this document online on the Internet - or its possi-ble replacement - for a period of 25 years from the date of publication barringexceptional circumstances. The online availability of the document implies apermanent permission for anyone to read, to download, to print out single copiesfor your own use and to use it unchanged for any non-commercial research andeducational purpose. Subsequent transfers of copyright cannot revoke this per-mission. All other uses of the document are conditional on the consent of thecopyright owner. The publisher has taken technical and administrative mea-sures to assure authenticity, security and accessibility. According to intellectualproperty law the author has the right to be mentioned when his/her work isaccessed as described above and to be protected against infringement. For ad-ditional information about the Linkoping University Electronic Press and itsprocedures for publication and for assurance of document integrity, please referto its WWW home page: http://www.ep.liu.se/

Upphovsratt

Detta dokument halls tillgangligt pa Internet - eller dess framtida ersattare- under 25 ar fran publiceringsdatum under forutsattning att inga extraordi-nara omstandigheter uppstar. Tillgang till dokumentet innebar tillstand forvar och en att lasa, ladda ner, skriva ut enstaka kopior for enskilt bruk ochatt anvanda det oforandrat for ickekommersiell forskning och for undervisning.Overforing av upphovsratten vid en senare tidpunkt kan inte upphava dettatillstand. All annan anvandning av dokumentet kraver upphovsmannens med-givande. For att garantera aktheten, sakerheten och tillgangligheten finns detlosningar av teknisk och administrativ art. Upphovsmannens ideella ratt in-nefattar ratt att bli namnd som upphovsman i den omfattning som god sedkraver vid anvandning av dokumentet pa ovan beskrivna satt samt skydd motatt dokumentet andras eller presenteras i sadan form eller i sadant sammanhangsom ar krankande for upphovsmannens litterara eller konstnarliga anseende elleregenart. For ytterligare information om Linkoping University Electronic Pressse forlagets hemsida http://www.ep.liu.se/

c© 2010, Marcus Kardell

Kardell, 2010. 41

Examensarbete Total positivity and oscillatory kernels: an...

Documents

Transcript of Examensarbete Total positivity and oscillatory kernels: an...