Solution to Problem 73-17: A Hadamard-type bound on the coefficients of a determinant of polynomials

Solution to Problem 73-17: A Hadamard-type bound on the

coefficients of a determinant of polynomials

Citation for published version (APA):

Lossers, O. P. (1974). Solution to Problem 73-17: A Hadamard-type bound on the coefficients of a determinant of polynomials. SIAM Review, 16(3), 394-395. https://doi.org/10.1137/1016065, https://doi.org/10.1137/1016064

DOI:

10.1137/1016065 10.1137/1016064

Document status and date: Published: 01/01/1974 Document Version:

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394 PROBLEMS AND SOLUTIONS

where M is r r and nonsingular.

In

what follows, it will be assumed without

loss ofgenerality that A is already partitioned as above, for ifRAC BG is a

nonnegative rank factorization ofRAC, then A

(RrB)(GCr)

is a nonnegative

rankfactorizationofA.

If P is a real matrix with r columns, then define c(p)=

{Pxlx

>=

0}

and

(P)

{x

>=

OlPx

>=

0}.

Note that (#(P) and c(p) are polyhedral cones.

A

cone ,f_willbe saidtobesolid ifthereexists anonsingular matrixNofrankrsuch that

C(N)

_.

,.cT,

and simplicial ifthere exists such anN for which equality holds.

THEORF,M.

_If

A is an m x n nonnegative matrix

_of

rank r where r

_satisfies

0 < r < min

{m,

n}

and A is partitioned as stated previously, then A has a

non-negative rank

_{factorization if}

andonly

_if

there existsasimplicialcone

,

satisfying

N([M,

MQ])

_

,cT

_

c(p).

Proof.

If such an ,9 exists, then

,

oN(X),

where X is r x r and

non-singular. Thus X

>=

0 and PX

>=

O. Since N([M, MQ])_

Y,

there exist

non-negative matrices Yand Z such that M X Yand

MQ

XZ. Then A has the nonnegative rank factorization

A .(Y,Z). PX

Conversely, suppose

is a nonnegative rank factorization ofA, where X and Yare r x r. Since M is

nonsingular, then so are the nonnegative matrices X and Y. Let 5T

c(X).

Then ,cT is simplicial and also 5T

_

off(p) since PX Wand Wis nonnegative.

Since M XYand

MQ

XZ,

COROLLARY.

Every

rank and rank 2 nonnegative matrix has a nonnegative

rank

_{factorization.}

Proof.

Every

solid polyhedral cone in R and R2 _{is simplicial.}

In light of the corollary, the minimum dimensions and rank possible for a

nonnegativematrix A having no nonnegative rank factorization are 4 x 4and 3, respectively. The following is such a matrix"

0 0

A= 0 0

0 0 1

0 0

Proof.

P

[-1,

1,

1].

(P)isnot simplicial,while (#([M, MQ]) (P). A. BFN-ISRAF,L (The Technion, Haifa, Israel) also gave a similar

counter-example.

Problem 73-17, A Hadamard-Type Bound on the

_Coefficients

_of

a Determinant

of

Polynomials, by

A.

J. GOLDSTFIN and R. L. GRAHAM (Bell Telephone

Laboratories).

PROBLEMS AND SOLUTIONS 395

assertsthat

Idet

A[

<=

laijI

2

H(A).

i=1 j=l

Inrecentstudiesoncoefficientgrowthingreatest commondivisoralgorithms

for polynomials,

W.

S. Brown

[1]

was led to inquireabout possible analoguesof this inequality for the caseinwhich theentriesof the matrixarepolynomials.

Let

A(x)=

(Aij(x))

be a matrix whose elements are polynomials and let

ao,

al,"" be the coefficients of the polynomial representation of det

A(x).

If

W

(wij),

where wij denotes the sum of the absolute values of the coefficients

of

Aj(x),then

show that

(

lakl2)

1/2

=<

H(W).

REFERENCES

[1] W.S. BROWN,OnEuclid’salgorithm and thecomputationofpolynomialgreatestcommondivisors, J.Assoc. Comput. Mach., 18(1971),pp.478-504.

[2] F. R. GANTMACHER, TheTheoryofMatrices,vol. 1,Chelsea,NewYork,1960.

Solution by O. P. LOSSERS (Technological University, Eindhoven, the

Netherlands).

Since

_IA,l(e91

W,l, it follows from Hadamard’s inequality that

Idet

A(eit)[

2

<=

]Akl(eit)l

2 Wkl

(H(W))

2.

k=l/=1 k=l /=1

However,

f:

f

eikt)

ilt))

2

2z

Idet (e")l

dt

((

a

(

Ztl e-

dt-

lal

Hence

Idet A(e")l

dt <_

-n

(H(W))

dt=

(H(W))

2

Also solved byA. A. JAGERS(TechnischeH ogeschool Twente, Enschede, the