Solution to Problem 75-15: An eigenvalue problem

Solution to Problem 75-15: An eigenvalue problem

Citation for published version (APA):

Lossers, O. P. (1976). Solution to Problem 75-15: An eigenvalue problem. SIAM Review, 18(3), 502-505. https://doi.org/10.1137/1018092

DOI:

10.1137/1018092

Document status and date: Published: 01/01/1976

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Now we assume that A and

B

are distinct. The convergence proof is simplifiedbysending

A

andB to+ 1 bythelinear transformation

(4)

x,

=1/2(A-B)Z+1/2(A

+B).

Thesequenceofapproximationssatisfies

(5)

Zn+l

--1/2(Zn -’-z-l),

n=1,2,...

and thus the limit points satisfy a quadratic equationwith roots + 1. [Editorial

note: Subsequently, the author shows that Re

_Z1

>

0

limit

+

1, Re

_Z1

<

0=),

limit=-1 and Re

_Z1

0=),no limit.

A

simpler proof follows easily from the explicitsolution of

(5),

i.e.,

Z.+l-

(Z,

+

1)2n

-F(Z,

1)

2"

-(z,

+

Thus the imaginaryaxisrepresents theset ofZthatdoesnotconvergeand

(4)

mapsthis ontotherightbisector of

A

and

B

in theoriginalcoordinates.

Theoriginal trial roots did notnecessarily satisfy

(2).

Theset ofsuch roots, that give

x]

and

x

ontherightbisectorof

A

and

B,

isfoundby taking

x’,

iC(A-B)+1/2(A

+B)

forCreal.Thenusing the definitionofx andeliminatingAandB bymeans of

weobtain

P(x,)

(x,

A

)(xi

B),

1, 2, 1

4P(x,)P(x2)

4C ---7

[P(xl)

+

P(x2)-(xl

x2)2]

2-1,

contingent upon

_x

# x2. Thus trial roots making the expression on the right positive real forma setof measure zerothatdoesnotgiveconvergence.Of course,

x

x2mustjointhis set as

(1)

is notdefined.

Also solved partially F. CARTY (North

Canton,

Ohio),

T.

O. ESPELID (Universitet I

Bergen, Bergen, Norway)

and theproposers.

Problem75-15,

An

EigenvalueProblem, by E. WASSERSTROM (IsraelInstituteof Technology,Haifa, Israel).

Let d 0

D=

0

_d2

0 0

T=

-1 2 -1

d3

0 -1 2

whered,

d2

and

_d3

arepositiveand

_d3

-<

_dl.

_{Show that}if

_d.

< d/3,

thenthereare two other positive diagonal matrices

_D1

and

_D2

such that

D,

_D

and

_D2

are distinct but

DT,

_DT

and

_D2T

have the same eigenvalues. Show also that if

d3

>

d/3

and

_D1

is apositivediagonalmatrixdistinct from

D,

then

DT

and

_D

T musthavedifferent setsof eigenvalues.

Illustration. Forthe threematricesD=diag

(5.5596,

1.4147,

1.5257),

_D1

diag

(5.1030,

2.4288,

0.9682)

and

_D2

=diag

(2.9782,

4.6565,

0.8653),

correct to

thegivenfigures, the eigenvaluesof

DT, D T

and

_D2

T

arethe same, i.e.,

At

1,

A2

4,

_A3

12. Onthe otherhand,with

D

=diag

(3.4530,

1.4584,

1.5887),

the eigenvalues of

DT

are

₁₁

1,

₁₂

4 and

₁₃

8, and there is no otherpositive diagonal matrix

_D

such that theeigenvaluesof

_D

T

arethesame.

Remark. This problem arises from the discretizationof the inverse eigen-valueproblem

d2y/dx

2 Ap(x)y, y(0)

y(1)

=0. Foragiven spectrum,A,one is

then required to find the densityfunction p(x).

(See B.

M. Levitan and M. G.

Gasymov,

Determination

o[

a

_differential

equationbymeans

_of

twospectra,Uspehi Mat. Nauk., 19

(1964),

pp.

3-63.)

Editorialnote,The

proposer’s

solutionisessentiallyanumerical one.Itwould be desirabletogiveananalyticsolution.

Solutionby

O.

P. LOSSERS (Technical University, Eindhoven, the

Nether-lands).

A.

The eigenvaluesof

DT

aretheroots ofthe equationin

(1)

3-2aA2+b-4c

=0, where

a

_{d + de}

+

d3, b

3d2(d

+

d3)

+

4dd3, c

dd2d3.

Eliminating

_dl

and

_d3

betweentheequalities (1),wefindthat

_d2

is arootofthe equation

(2)

3x3-3ax2+bx-4c

=0

in x.

Apart

from d2,thisequationhastwo rootssatisfying

(3)

f(x)

3x2-

3(d

+d3)x

+4dd3

0.

Thefunction

_f(x)

has m -(3dl-

d3)(d-3d3)

as its absolute minimum.

We

assume that throughout

d3<=da.

The roots of

(3)

are imaginary if d3>-jd. It followsthat theredoes not exist inthiscase a3

x

3 diagonalmatrixMwith real

elements such that

MT

has the same eigenvalues as DT.

We

suppose, further

d3<-jdl. Then m<0. Since

f(d)=f(d3)=dld3>O,

we infer that

(3)

has two differentrealrootsbetween

_d3

and

_d.

Let

62

beone ofthese roots; thenwehave

d3<62<d.

B.

We

proposetoprovethat thereexistreallytwopositive numbers

6

and

₆₃

with

₆₃

<

1/2

61

such that

(4)

₊₂₊₃₌

a,

32(

+

3)

₊₄₃

b,

We

observe in thefirstplacethat theconditions

(5)

61

-[-

₆₂

--

₆₃

a

^

616263

c

togetherwiththefact that

₆₂

is a solution of

(2)

imply the secondofthe equalities

(4).

Forweobtain, under the said conditions:

Thereexist differentpositive numbers 6, and

63

satisfying

(5)

iff

(61-63)2=(a-62)2-46’c

>0, that is, iff

Introducing thefunctionq(x)

x(a

-x)2-4c,

wetherefore havetoprove 0. Tothisend,weobserve:

q(0)

o(a)

-4c

<

0;

q(d3)=d3(d,-d3)2>O;

q(d) d,(d2- d3) O;

()>0.

It follows that the equation q(x)=0 has three different real roots _Xl

<

X2

<

X3

satisfying:

0<Xl

<

d3

<

d,

<=x2<a

<x3.

Therefore

_d3

<

x

<

d, implies q(x)

>

0. Since we knowfrom

A

that

d3

<

₆₂

<

_dl

weobtainq9(62)>0. Theexistence of

61

and

₆₃

thus beingascertained wechose

ournotationsuch that 0

<

63

<

6,.

C. Itremains toprovethat

63

<

1/2

6, and that thematrixdiag(61,62,

63)

is not equaltoD. Itis notpossibletoderivethis

[rom

the dataprovided bythe proposer theproblem. Itwillbenecessarytomaketheadditionalassumptionthat

(2)

hasno

multipleroot.Inthiscase,

_d2

is not a solutionof

(3)

andtherefore

_d2

62.

Hence diag

(6,,

62,

63)

D.

Moreover,

the equation

3x2-3(6,

+

63)x

₊₄₆₁₆₃

=0 hastwo

differentrealroots. Thismeansthat

O<

9(6,

+

63)2-486,63

3(36,-63)(6,-

363),

whence in view of

₆₃

<

6, the missing inequality"

63

<

1/261.

Now

we have com-pletely provedthat thereexist twodifferent positivediagonalmatrices

D,

and

_D2

other thanD andofpreciselythesame nature as

D,

such that

DT, D,

Tand

_D2

T have thesameeigenvalues.

We

havetoshow thatouradditionalassumptionisin fact anecessaryone.Ifwedropit,

’/2

mayhappentobeamultipleroot of

(2).

Then

(6)

3d-3d2(d,

+d3)+4d,d3=O

and

(5)

with

₆₂

_d2

leadsto6, dl,

63

d3.

Therefore

_D1

diag

(6,,

62,

63)

D.

Apart

fromd2,the equation

(3)

hasd,

+

d3- d2

as aroot. Sinceitiseasytoverify that d,

+ d3-d2

-d2, we may expect that this root providesus with adiagonal matrix

_D2

diag(61, d,

+

d3-d2,

63)

D

and ofthesame nature asD. Tofind

₆₁

and 63, we have the equations: 6,

₊₆₃

2d2,

(dl

+

d3-d2)6,63 c. This means

that

₆₁

and

₆₃

aretheroots of thequadraticequationin z: (d,

+d3-d2)zZ-2d2(d,

+d3-d2)z +d,d2d3=O. Thisleads,in viewof

(6),

to

2dd3

2d,d3

(7)

6,

Here

wehave, insteadoftherequired inequalityt3

<

gtl, theequality

63

61/3.

The theoremisthereforeintheredactiongivenbytheproposernot uncondition-allytrue.

Numericalexample inwhich

(2)

hasa multipleroot.

D=diag(dl, d2, d3)=diag

(9,

3,

2)

:z

a 14, b 171, c 54. Equation

(2)

becomes:

3(x3-4xZ+57x-72)=Oz(x-3)Z(x-8)=O.

Starting

9

from

_D2

(tl,8,

6,)

we find

61

+63=6,

6163

27/4.

Hence

61

=,

63=;

63

1/2tl.

Also solved by

A. A.

JAGERS (Technische Hogeschool

Twente,

Enschede, theNetherlands),who also pointedoutthat theproblemis notquite correct, and theproposer.