MATRIX COMPLETION PROBLEMS IN MULTIDIMENSIONAL SYSTEMS

Wayne M. Lawton

Dept. of Mathematics, National University of Singapore

Block S14, Lower Kent Ridge Road, Singapore 119260

wlawton@math.nus.edu.sg

Zhiping Lin

School of Electrical and Electronic Engineering

Block S2, Nanyang Avenue

Nanyang Technological University, Singapore 639798

ezplin@ntu.edu.sg

OUTLINE

1. Introduction

2. Continuous functions

3. Trigonometric polynomials

4. Stable rational functions

INTRODUCTION

is a Hermite ring if every unimodular row vector is the first row of a unimodular matrix (completion)

is unimodular if there exists

is a commutative ring with identityis a commutative ring with identityR 1mP RmQ R such that 1QP

mmM R is unimodular if 1 Mdet

R

INTRODUCTION

HERMITE RINGS INCLUDE

1. Polynomials over any field (Quillen-Suslin)

2. Laurent polynomials over any field (Swan)

3. Rings of formal power series over any field (Lindel and Lutkebohment)

4. Complex Banach algebras with contractible maximal ideal spaces (V. Ya Lin)

6. Principal ideal domains eg rational integers, stable rational functions of one variable (Smith)

DEGREE OF MAP OF SPHERE

THE DEGREE D(f) nn SS:f OF CONTINUOUS

is an integer that measures the direction and number of times the function winds the sphere onto itself.

)}sin,{(cosS1 )}sin,sincos,cos{(cosS2

k)k(D

EXAMPLES

0)D(constant 1)D(identity 1n1)(l)D(antipoda

D(f)D(h)h)D(f

HOMOTOPY

YX:f,f 21

21 f)F(1,,f)F(0,

are homotopic

if YX[0,1]:F

HOPF’S THEOREM Ifnn

21 SS:f,f then )D(f)D(fff 2121

)f(f 21

J. Dugundji, Topology, Allyn and Bacon, Boston, 1966.

DEFINITION

COROLLARY2121 ff(x)f)x(fx,

Proof. Consider ||tft)f(1||tft)f(1 2121

)nR(SRDefine

CONTINUOUS FUNCTIONS

Theorem 1.

is unimodular

For n even, a unimodular

)||||P(DD(P) PFor unimodular

1nP RThen

0P(x)x,

P define

1nP Radmits a matrix completion ,0D(P)

R is not Hermite since the identity function on nShas degree 1 and thus cannot admit a matrix completion.

hence

Proof

CONTINUOUS FUNCTIONS

and

Since of

D(Q).D(P)

be the second row of a matrix Let completion .1nP RP

nnnSc)g(1,P,)g(0,,SS[0,1]:g

linearly independent at each point, hence

1tt0L0

Hopf’s theorem implies there exists a homotopy

Q

1 Mdet MQ

Multiply the second and third rows of

M by 0.-D(Q)D(-Q)D(P) 1- to obtain

Choose x),g(tx),g(tx,1-kk

Construct 1L

MMM where )1SO(n)(k

xM x, satisfiesx),

1-kg(t(x)

kx)M,

kg(t and y(x)yM

k if

0.x),1-k

g(tx),k

yg(t y M is continuous and completes P.

)R(),R( nnrsrs TPLet

TRIGONOMETRIC POLYNOMIALSn

Z

continuous real-valued functions, trigonometric polynomials.

be the periodic symmetric

Isomorphic to rings of functions on the space obtained by

1X

nX

identifying andn

Tx x.-

2X

homeomorphic to interval [-1,1]under the map x2 cosx homeomorphic to sphere

2Sunder a map

22 XT:

that is 2-1 except at (.5,.5)}(.5,0),(0,.5),{(0,0),

)R(rsPLemma

RESULT

Proof This ring is isomorphic to the ring of real-valued functions on the interval

is a Hermite ring.

Choose a unimodularmR([-1,1])F

And approximate

by a continuously differentiable map

And use parallel transporting to extend to a map

1-mS]1,1[:H

|F|F/

SO(m)]1,1[:M

}{CC:p Define

}0{\

222 ])[()(p

zzz

WEIRSTRASS p-FUNCTION

Z}nm,|ni{m

by

where

Lemma

Proof.

wzorwzp(w)p(z)

.g-pg-p4p 3232

(z))pp(z),(z isomorphically onto the cubic curve in projective

maps the elliptic curve

C/

J. P. Serre, A Course in Arithmetic, Springer, New York, 1973, page 84.

space defined by the equation

2p2 S}{CCR 21

Define

WEIRSTRASS p-FUNCTION22

12 SR:p~

where2 is stereographic projection

]2w,v2,u2[w)( 12 z

ivuz1,vuw and

21211 ixx)x,(x ~ is

2Z periodic and defines 22 ST:

2k

12kk

2 z1)G(2kzp(z)

LAURENT EXPANSION

WEIERSTRASS p-FUNCTION

}0{\

2kkG

where

is the Eisentein series of index k for the lattice This provides an efficient computational algorithm.

RESULTSis isomorphic to the ring )R(S 2

And therefore is not a Hermite ring. Furthermore the ring

Theorem 2. )R( 2rsP

)R( 2rsT is not a Hermite ring.

Proof Define the map )(R)R(S: 2rs

2 Pby ).R(Sf,f)f( 2 Results for p imply that is a surjective isomorphism.

The second statement follows by perturbing a row having degree not equal to zero to obtain a unimodular row.

EXAMPLEEXAMPLE OF A UNIMODULAR ROW IN

32rs )R(T

)G-2.84(G)X,X(F12211

Proof Compute

THAT DOES NOT ADMIT A MATRIX COMPLETION

)G-2.51(G)X,X(F43212

)G(G2.56-)G(G55.310)X,X(F

4321213

)XX(cG),XX(cG214213

osos2211

XcG,XcG osos

|F|5.0F maps are never antipodal, hence

so these

1)(identity)D()D(D(F)

OPEN PROBLEMSPROBLEM 1

32rs )R(F T

is unimodular and has degree zero does it admit a matrix extension ?

If

PROBLEM 2

PROBLEM 3

Is the ring )R( 2sT

of symmetric trigonometric polynomials a Hermite ring ?

Is the ring )R( 2rT

of real-valued trigonometric polynomials a Hermite ring ?