MATRIX COMPLETION PROBLEMS IN MULTIDIMENSIONAL SYSTEMS
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Transcript of MATRIX COMPLETION PROBLEMS IN MULTIDIMENSIONAL SYSTEMS
MATRIX COMPLETION PROBLEMS IN MULTIDIMENSIONAL SYSTEMS
Wayne M. Lawton
Dept. of Mathematics, National University of Singapore
Block S14, Lower Kent Ridge Road, Singapore 119260
Zhiping Lin
School of Electrical and Electronic Engineering
Block S2, Nanyang Avenue
Nanyang Technological University, Singapore 639798
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 ?