Solution to Problem 72-10*: Conjectured monotonicity of a matrix

Solution to Problem 72-10*: Conjectured monotonicity of a

matrix

Citation for published version (APA):

Lossers, O. P. (1973). Solution to Problem 72-10*: Conjectured monotonicity of a matrix. SIAM Review, 15(2), 387-390. https://doi.org/10.1137/1015045

DOI:

10.1137/1015045

Document status and date: Published: 01/01/1973

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Thus for

a< t thereis a unique minimum,y(x) 0;

a > r thereis nominimum,i.e. infJ(y)

-while for

a 7 the functions y(x)= b sin

(rex/a)

satisfy J(y)= 0 (for all b,- < b

< oe)and0 infJ(y).

Alsosolvedinessentiallythe samewaybyK.PARI,:_(NASA,LangleyResearch

Center), O. RtJEHR(Michigan Technological University) and the proposer. How-ever, only the proposer showed that J has no maxima, by observing that if

(I),(x)

sin

(nnx/a),

then

J((I),)

(a/2)((nn/a)

2 _1)-,

o as n

,

for alla > 0.

II. SolutionbyR. T. SHIELD(University ofIllinois).

The substitution y u sin

(nx/a) (cf.

Jacobi’s method of multiplicative vari-ation

[1])

is permissiblein viewofthe properties of y(x). The integral becomes

and theextrema fora < n, a nanda > ncanbedeterminedby inspection.

REFERENCE

[1] R. COtJRANa"ANDD.HILBERT,Methods_ofMathematicalPhysics,vol.I,Interscience,NewYork,

1953,p.458.

Problem72-10", ConjecturedMonotonicity

of

aMatrix,by R.

A.

USMANI (Univer-sity of Manitoba, Manitoba, Winnipeg, Canada).

Consider A

_(aij),

afive-band matrix of order N suchthat

5 +

i, i=j=l,N,

6 +/;i, =j 2,3,

-4,

_li-Jl-

1, 1, li-jl-- 2, 0, otherwise,

where_i 0. Itis conjectured that the matrix Ais monotone,i.e., if

A-1

(aj),

then

_a/>

0for all and j.

Remark. Itisknown thatthe matrix

_A0

obtainedfrom A by setting_:i 0 is monotone. For

_A0

j

z,

where J

_(Jmn)

such that

m n,

Jrnn

1,

_Im

_nl

1,

and if

J-x

(j,), then n(N

+

m)

(u+ 1)

m(N

+

n) (N+ 1) > O, m

_>-

n, >0,

m<n.

ThusJandconsequently

_A0

are monotonematrices.ThematrixA arisesin

com-putation of thenorm oftheinverseofafourth orderdifferentialoperator.

Solutionby O. P.LOSSERS (Technological University, Eindhoven, the Nether-lands).

The conjecture is false. Let el 2 N 2,2 > 0, and let A A 2I; then

A-X

2-(I

+

2-A)

-

2-(I

2-XA

+

R), where Rij

---O(,-2),

for all and j

(2- o).

Then

a.*.,,,+2

2-1(6i,i+2

,,,-

_it_

O(2-2))

/-2

₊

O(/-3),

which is less than 0 for sufficiently large 2.

Solution by A. A.JAGERS(TwenteUniversity ofTechnology, Enschede, the Netherlands).

A

straightforward calculation of

A-

for N

__<

4 shows (i) that A is always

monotoneifN

__<

2,(ii) that,for N 3,Ais monotone ifand onlyif;2

-

10 and

(iii) that, for N 4, A is monotone if and only _{if -2}

--

;3 5. For larger N the conditions becomemore complicated. Anyhow weshall see later that, ifN

>=

3, then

_a3

< 0,as soon as ei 0 forall 4= 2 and ;2 islarge enough. So, ifN

>=

3,

the conjecture does not even hold if we restrict ourselves to the casewhere only

one ofthe _:i ispositive. Itis worthwhile to consider thisphenomenon inamore

generalsetting"

Let B

(bij)

be a matrix of order N with det(B)4= 0. Choose a fixed

k, 1

__<

k

__<

N. Let

B(e.k)=

_(bi(Zk))

be the matrix given by

bkk(k)=

bkk +

k and

bij(e,k)

bij

if 4= kor j 4: k. Denotethe ijth entry ofthe inverse matrix of by

_b(e

k)

and

_b(0)

by

_b.

Supposethat

_bk*

> 0. Then we have the followingtwo

propositions.

An

unconventional but easy way of provingthe second part ofthe first propositionis by successively solving the N2 _{differential equations}

b(,

_)b(:).

*" The second propositionis aconsequence of thefirst.

PROPOSITION 1.

(i) det

(BO:k))

(1

_{+ bk*"}

ok)

det

(B)

4= O.

(ii)

+ bk*.c

k

* * h,b,

with

_Aikj

_bkkbij

_{ik kj"}

PROPOSITION 2.

_If

Bismonotone,then

B(ek)

is

monotone]br

all[;k 0

if

andonly

Notethat

_Aikk

_Aj

0.

Now adirectcalculation usingthegivenJm,"* shows thatin theoriginal prob-lemthe quantity correspondingto

_A3

is equal to

1 N

’j---’3

(N

+

1 k)2

(N

₊

1)2

which isless than0. Thisproves theannouncedstatementconcerning

_a’3.

On the other hand,it follows at once fromthe corollary to the proposition below,that atleast all diagonalentries

_a/are

positive if_ei >0 for all i.

PROPOSITION 3.

_If

B is symmetricandpositivedefinite, then

b*,i(ek)

> 0

_for

all andr.

>=

O.

Proof

Since B is symmetric and positive definite, a symmetric matrix M

exists such that B

m

2.

Then the condition

Aik/

_>_

0 is equivalent to Schwarz’s inequality appliedtothe ithand kth columnvectorof

M-1.

Now

B(ek)

is again symmetric and positive definite. (Note that det

(B(e.k)

4:0forall :k

0.)

SowemayapplyProposition3another time,this timechanging another diagonal entry,etc. Wefinallyobtainthe followingcorollary.

COROLLARY.

_If

B is symmetric and positivedefinite, then the diagonalentries

of

the inverse C-1

of

the matrix C obtained

_from

B by adding l,i(O) tO the ith

diagonal entry(i 1,2,...,N) are all positive.

Solutionby A. N. WILLSON, JR.(University of California, Los Angeles). The stated conjectureis false. This is easily verified by considering the case in which N 3, with r. 11, i= 1,2,3. In this case the elements

(A-1)13

(A-X)3

< 0. Also,it isclearthat the conjecture cannotbe trueforany matrix

of the type specified having order N > 2, as it would contradict the following theorem (which is proved in

[1]).

THEOREM. ThematrixA issuch that

(A

+

D)-

>__

0

_for

everydiagonalmatrix

D having positive elements on the main diagonal only

_if

the off-diagonal elements

of

A are all nonpositive.

REFERENCE

Ill A. N.WILLSOY, JR.,Pairs_ofmatricesandmatrices_ofmonotonekind,SIAMJ.Numer.Anal., 10

(1973).

Also solvedby L.ELSNER (InstitutftirAngewandteMathematik

I,

Erlangen-Nurnberg), I.FARICAS (University of

Toronto,

Canada), M. J. MARSDEN (Univer-sity of Pittsburgh), C. C. ROUSSEAU (Memphis StateUniversity), O. G. RUEHR (Michigan Technological University),J.N. SHOOSIITH (NASA, LangleyResearch Center), R.

A.

SWEET (National CenterforAtmosphericResearch),J. J.

VANDER-CRAFT(UniversityofMaryland), L. RUBENFELD(RensselaerPolytechnic Institute) submittedasolutionco-authoredbysixNSF undergraduateresearch participants: M.

FRAME,

D.

HALE,

C.

KELLER,

D. RUTLEDGE, S. SENTOFFandD. WAGNER.

By [1],

it is sufficient to construct a 2m+x 2m+

1(__

1,1) matrix W for which det W- 2(’/1)2". Let A denote the 2 x 2 matrix and define

W= (R)"

+IA,

the (m+ 1)th Kronecker power of A. W satisfies detW

(det

A)(’/x)2- _2(./

2-which provides a desired matrix.

Problem 72-11,Limit

_of

anIntegral,by N. MULLINEUXandJ. R. REED(University of

Aston,

Birmingham,England).

Determine

! lim log

Isin

t-

13/21

dt

-o sin

_13/2

sin

The problem arose in the numerical inversion of singular integral equations

such asthose which occur inaerodynamics and problems associated with plane

transmission lines.

Editorial note. Most solvers

(see

laterdiscussion) noted that the integral as

given above does not exist since for fixed e, 0, the integrand behaves like

-

log

{(13/2)sin

13/2}

as 0. The error is a typographical one since in the original proposal the term

_Isin

t/2l

appears as

_Isin

(t

13/2)1.

In the following solution ofO. P. LOSSERS (Technological University of Eindhoven, Eindhoven,

the

Netherlands)

the latter expression is used. Replacing 13 by 213 we get

log Isin(t 13)1 dt

,2

sin13 sin ₀

dt log

Icos

cot13.sin

tl

sin

logcost---

+

10gll cot13.tan

tl

sn sn

Since (logcost)/sin

t/2

+

O(t

3)

(t

0), we have that the first term is

O(132),

(13

0).

By

substitution ofv cot13.tan we get

fcote.tan2e

log

_l1

_vl

I lim O(132

--e--,0 _d0

(1

+

/)2tan2

_13)-1/2

21ogll_/)ldv=

v 4

In the solutions by N. M. BLACHMAN (Sylvania Electronics Systems), L. KHATCHATOORIANTZ (California State College, Los Angeles), S. L. PAVERI-FONTANA(University of California,Berkeley), Z. C. MOTTELERandO. G.RUEHR (Michigan Technological University),J. W. RIESE(Kimberly-ClarkCorporation), J.D.TALMAN(University of

Western

Ontario,London,Canada)andP. TH. L. M.

VANWOERKOM (National

Aerospace

Laboratories,Amsterdam,theNetherlands),

it was remarked that the integral as originally stated does not exist. Various

alternative statementsof the problem weremade. We donot listthem all.

Most

solverschoseto replacesin

e/2

by

e/2

(onecouldaswellreplace

e/2

by sin

e/2)

in