Three different ways There are three different ways to show that ρ(A) is a simple eigenvalue of an...

Three different ways

There are three different ways to show that

ρ(A) is a simple eigenvalue of an irreducible

nonnegative matrix A:

.1

.

).(

,

.

..

.,

..)(

.

.1)(

:1

isAofmultiplegeometrictheHence

AoflityirreducibithescontradictThis

Atoingcorrespond

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positivenotbut

vesemipositiisyxtsRthen

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tsAtoingcorrespondRyreigenvecto

anotherhasAThennotSuppose

isAofmultiplegeometricthe

thatshowtoFirst

Method

n

).(max)(

,)()(

))((

)(

,0))((

.0

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.

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AA

smallsufficientisifyAAy

yAAy

xyIAASince

ythatassumemayweHence

xythenelsufficientIf

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xyIAAtsRyThen

notSuppose

orderofAtoingcorrespondreigenvecto

generatingnohasAthatshowtoremainsIt

n

.)(

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)](1)(

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11

1

AofeigenvaluesimpleaisAHence

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AIofeigenvaluesimpleaisAI

AIofeigenvaluesimpleaisAI

TheoremsPerronBy

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AIA

A

Athen

AthenAILet

AAIClaim

Method

nn

n

1000

1010

0001

)(

0))((:3

,11,13,12,11,1

2232221

1131211

321

3223231

1131211

321

3223231

2212221

21

22221

11211

nnnnnnn

n

n

nnnnn

n

n

nnnnn

n

n

nnnn

n

n

A

A

aataaa

aaata

aaaat

ataaa

aataa

aaaat

ataaa

aataa

aaata

ataa

aata

aaat

dt

dtc

ActhatshowToMethod

0))(())(())((

0))((

1))((dim))((

2))((

,,10))((

,,10))((

,,1)()(3.3.2

,0

))(())((

\)()(

01

0

1

1

AcAcActhen

ksomeforAcHence

nAIANnAIArankBut

nAIArankthen

nkforActhatSuppose

nkforActhen

nkforAACorollaryby

ASince

AcAc

knAAwheretctcthen

kk

k

k

k

k

k

A

n

kAA

A

A

A

k

n

kAA

k

n

kAA

1000

1010

0001

)(

0))((:4

,11,13,12,11,1

2232221

1131211

321

3223231

1131211

321

3223231

2212221

21

22221

11211

nnnnnnn

n

n

nnnnn

n

n

nnnnn

n

n

nnnn

n

n

A

A

aataaa

aaata

aaaat

ataaa

aataa

aaaat

ataaa

aataa

aaata

ataa

aata

aaat

dt

dtc

ActhatshowToMethod

RandzzAzAxxAAx

wherexzAIAadj

AcAIAadjAIA

AIAAIAadj

tcAtIAtIadjAtIAtIAtIadj

nkforAc

thatshowtosufficientisIt

AcAc

knAAwheretctcthen

T

T

A

A

A

n

kAA

k

n

kAA

k

k

k

;,0,)(;0,)(

,))((

0))(())(())((

))()()((

)()det()()()()(

,,10))((

))(())((

\)()(

1

1

0))(())((

,,10))((0

1))((dim))((

2))((

)(1min

0))((,0

.0

,,10))((

,,1)()(,3.3.2

,,1))(())((

1

n

kAA

A

A

k

Akk

AcActhen

nkforAcsoandHence

nAIANnAIArankBut

nAIArankthen

zeroareAIAofnsizeoforsallsoand

AIAadjthenSuppose

then

nkforAc

nkforAACorollaryBy

nkforAcAIAadj

k

k

k

k

Exercise

0BASuppose that

If A is irreducible and A ≠B, then

)()( BA

)()(

,,0

0)(

)(,9.4.2Re

,0.)(

)()(),()(

)()(3.3.2,0

)(

00)(',0

BAHence

ioncontradictaBAthenuBut

uBAthen

uAAumarkby

AandeirreduciblisASinceuAAuthen

uAuBBuAuthenBASuppose

BACorollarybyBASince

uBButhatsuch

uandBThmsPerronbyBSince

Remark

R

nn wwwwww 2121

such that

with equality iff the nonzero wis have the

argument i.e.

njeww ijj ,,1

Lemma 2.4.24 (Wielandt’s Lemma) p.1

nMCA ,

CA

Let

Suppose that A is irreducible, nonnegative,

)()( CA

and Then

(a)

Lemma 2.4.24 (Wielandt’s Lemma) p.2

1DADei

)(Aei (b) Suppose that C has an eigenvalue

then C is of the form

ID

where D is a diagonal matrix that satisfies

CA

A

A

zzCzzCzA

CASince

zCztsCzthenCLeta n

)(

)(

)(

,

..0),()(

IDandeediagDwhere

zD

z

z

z

e

e

e

ez

ez

ez

z

z

z

zLet

zandCAceCA

zCA

markbyzAzCzA

zAzCzzCzAzCz

tsCzthenCAeLetb

n

nn

ii

ni

i

i

in

i

i

n

ni

),,(

00sin,

0

9.4.2Re,)(

)(

..0),()()(

1

2

1

2

1

2

1

2

1

2

1

1

1

22112211

1

1

1(*),

..

(*)

,,1

,

)(

)(

DADeC

ACDDe

AFF

eBy

jeff

jezfzftsRthen

zfzfzfzfzfzf

nrFor

zFzFthen

ACFthenCDDeFLet

zAzAzCDDe

zDAezCDzCz

i

i

i

irjrj

ijrjjrj

nrnrrnrnrr

i

i

i

Exercise 2.4.25 p.1

kkn AAI 1

,nMA and let k be a positive integer.

(i)

.AA

Prove that

where the equality holds if and only if

AA

AAIsomefor

A

AAI

AAI

A

Asomefor

thenAIIf

i

kn

kkk

kkn

kkn

kkk

k

kn

,

111

1)2(

1

111

1

,

)1()(

Exercise 2.4.25 p.2

A

kA1

the algebraic multiplicity of

(i) Prove also that when the equality holds,

algebraic multiplicity of

as an eigenvalue of A is equal to the

as an eigenvalue of A. kn AI

mmHence

impossibleiswhich

AeAeA

eA

ksomeforeA

A

tsAAthenmmthatSuppose

mmthatobviousisIt

AIofeigenvalueanasAof

tymultipliciebraicathebemandAofeigenvalue

anasAoftymultipliciebraicathebemLet

i

ki

ki

ki

ki

kk

kn

k

1111

11

1,,111

11

..,,

.

1

lg

lg

)3()(

22

2

2

Exercise 2.4.25 p.3

and Perron’s theorem to deduce that if A

(ii) Use the result of part (i), Theorem 2.4.4

A is a simple eigenvalue of A.

is an irreducible nonnegative matrix , then

.),(

.

'0

,)(11

AofeigenvaluesimpleaalsoisAipartBy

eigenvaluesimpleais

AITheoremsPerronbyandAI

thenmatrixenonnegativeirreduciblisAIfiinn

Exercise 2.4.26 p.1

BABA ,0

nMBA , Prove that if

(i) Let

BA

and if A is irreducible

Hint: Use Remark 2.4.9.

then

.,

00

00

9.4.2Re'

,

..00,

..0,

BATherefore

ioncontradictaBAHence

uandBABut

uBAuanduAAu

markandThmsPerronofpartfirsttheBy

enonnegativandeirreduciblisASince

uAuBBuAu

tsuTheoremFronbeniusPerronBy

BAthatSuppose

BAHence

AA

zBzzBzA

zBztsCzBanyFor n

Exercise 2.4.26 p.2

if A is an irreducible nonnegative matrix,

(ii) Use the result of part (i) to deduce that

)()( AB

different from A.

for any principal submatrix B of A,

then

)(00

0)()()(

000

0

**

*

,,

**

*

,

BB

APPAiby

BBAPP

andenonnegativeirreduciblisAPPthen

BAPP

thatsuchPmatrixpemutationaexiststhere

thenAofsubmatrixprincipalabeBLet

T

T

T

T

Exercise 2.4.27 p.1

nA ItcAtItBtBAtI

Tcyzt

nMA

relation

where y,z are Perron vectors of A and

Let

TA

be an irreducible nonnegative

matrix. (i) Let B(t) =adj(tI-A). By using the

multiplicity of ρ(A) as an eigenvalue of A is 1,

at t=ρ(A) and the fact that the geometric

show that B(ρ(A)) is of the form

respectively, and c is a nonzero real constant.

.

..,

,

00

0

..0,

,

22

TTT

TT

TTTTTTT

TTTn

TT

T

TT

A

TT

cyzczyyuABHence

y

yvcwhereczv

y

yvzv

yzvyuyzvuyzvyu

zvABandyuABtsRvuThen

AtoingcorrespondAofreigenvecto

orvectorzeroeitherisABofcolumnevery

andAtoingcorrespondAofreigenvecto

orvectorzeroeitherisABofcolumnevery

ABAIAandABAIA

AcAIAABABAIA

zAzAandyAAytszy

matrixenonnegativeirreduciblisASince

Exercise 2.4.27 p.2

iA

0))(( AcA

)()()(1

ttrBtctcn

iAA i

is the (n-1)-square principal

the result of part (i) to deduce that

(ii)Show that

)(A

and column. (These relations are true for a

where

and hence

submatrix obtained from A by deleting its row

general square complex matrix A.) Then use

is a simple eigenvalue of A.

n

iAA

n

iA

n

n

nnnn

n

nnnn

n

nnnn

n

n

A

n

iAA

tctcthen

tcaata

aaat

ataa

aaat

ataa

aata

ataa

aata

aaat

dt

dtc

tctcthatshowTo

i

i

i

1

1

22221

11211

21

11211

21

22221

21

22221

11211

1

)()(

)(

100

010

001

)(

)()(

n

iA

n

iii

Aiii

n

iA

tctBttrB

tcAtItB

AtIadjtBSince

tcttrBthatshowTo

i

i

i

11

1

)()()(

)()det()(

),()(

)()(

.

)(0))((

1)(dim)(

2)(,1

)(min

0))((,,1

0))(()(

0))(())((

,0))((

.0))((

Aofeigenvalue

simpleaisAsoandAcHence

nAIANnAIArankBut

nAIArankthenzeroarenorderof

AIAofsubmatrixoforallthen

ABnjianyforthen

csoandcyzABiBy

AcAtrB

thenActhatSuppose

ActhatshowTo

A

ij

T

A

A

A

Exercise 2.4.27 p.3

)(A

)(tcA

))(( AcA the largest real root of

(iii) Using the fact that

0)( AB cannot be negative. Deduce that

is equal to

explain why

0))(())((

0))((0

))(())((0,)(lim

)(arg)(

)()(3.3.2

0))((

))(())((0)(

1

0

1

00

00

0

0

n

iAA

A

TiiAA

t

Ai

i

A

n

iAA

AcAcTherefore

iallforAcandcHence

yzcABAchavewetc

andtcofrootrealestltheisASince

AACorollarybyand

nisomeforActhen

AcAciiBy

i

i

ii

i

i

i

Exercise 2.4.28

nMA

ttrBtcA )(

matrix. Let B(t) have the same meaning as

Let

0))(( AcA

Exercise 2.4.26 (ii) and the relation

be an irreducible nonnegative

given in Exercise 2.4.27. Use the result of

(which was established)

to show that

0))(())(())((

,,10))((

,,1)()(),(26.4.2

1

n

iAA

A

i

AcAtrBAc

niActhen

niAAiiExerciseBy

i

i

Exercise 2.4.29 p.1

Let A be an irreducible nonnegative matrix

with index of imprimitivity h

(i) Prove that trace(A)=0 whenever h>1

0)(

0

0

0

:1)(

1

1,

,1

12

1

PAPtrAtrthen

A

A

A

PAP

thatsuchPnpermutatioaistherethen

cyclickisA

Methodi

k

kk

0)(,0

)()(

:2)(

2

21

2

12

2

AtrsoandethenhIf

AtreADDetrAtrthen

ADDeA

thatsuchDmatrixdiagonalthen

similardiagonallyareAeandA

Methodi

hi

hi

hi

hi

hi

0)(

0

,11

0..,

0

,

:3)(

000

AtrTherefore

iallforaHence

ioncontradictahsoandlengthof

itselftoivertexfromwalkclosedaistherethen

nisomeforaeinotSuppose

iallfora

thatshowtosufficientisitenonnegativisASince

Methodi

ii

ii

ii

Exercise 2.4.29 p.2



(ii) Prove that h is a divisor of the number

of nonzero eigenvalues of A (counting

algebraic multiplicities)

Aofseigenvalue

nonzeroofnumbertheofdivisoraishTherefore

heformtheof

isAofseigenvaluenonzeroofsetthehence

Ahe

AanyforThen

AeA

AetosimilardiagonallyisA

cyclichisASinceii

hi

hi

hi

hi

1,,1,0

0\)(1,,1,0

)(0

)()(

,)(

2

2

2

2

Exercise 2.4.29 p.3



(iii) Prove that if A is nonsingular, nxn and

n is a prime, then h=1 or n.

.1

.)(

)(0,sin)(

norkHence

primeaisnButnkiibyand

AthengularnonisAIfiii

Theorem2.4.22(2nd part of the Frobenius Thm)

0A

1,,1,0);(2

mAe mi

Given irreducible matrix

with m distinct eigenvalues with moduloρ(A)

Then

(i) the peripheral spectrum of A is

Theorem2.4.22(2nd part of the Frobenius Thm)

)()(2

AAe mi

2m

(ii)

000

0

000

00

000

~

1

,1

23

12

m

mm

P

A

A

A

A

A

(iii) If

Apply Wielandt’s Lemma to prove

keA

ee

kr

AGofcircuitsallforr

unityofrootrthprimitivea

sayunityofrootaise

AGofcircuitsallfor

AeAeAe

AGofcircuitsallforAeA

andAeGAGAeAthen

RsomeforeAWrite

CAtakingbyLemmasWielandtApply

AAanyConsider

kAofindexcyclicthe

thenAGofityimprimitivofindexthekLet

ki

kiri

i

iii

i

iiD

i

,,2,1)(

11

)(

,,

)(

)()()(

)()()(

),()(~

)(

'

)(),(

),(

2

Another proof in next page

1,,2,1)(

1sgn

)(

.sgn,

1sgn

sgnsgnsgn

)(

sgnsgnsgn0

,,2,1

0)(

,

)(

..0

).()(

).(

2

22112211

keA

henceandkr

AGinlengths

circuitofdivisorcommonaisrsoandr

unityofrootrthprimitiveaissay

zzThen

ivertexcontainingAGincircuitabeLet

zza

zzazazazazaza

niFor

zandzAAzzA

enonnegativandeirreduciblisASince

zAzAzzA

zAztsCzThen

AwithALet

AGofityimprimitivofindexthebekLet

ki

k

ii

ijij

ininiininii

n

1sgn,0

.)(sgn

)sgn(

)sgn(

)sgn(

)sgn()sgn()sgn(

2

1

2211

21

zdefinethenzIf

IDandDzzzthen

z

z

z

DLet

zzzzzz

zzzzLet

n

Tnn

Tn

Fully Cyclic

kA )(

1,0,)( nCzAzzA

The peripheral spectrum of A is called

fully cyclic if

then kzz sgn is an eigenvector of A

corresponding to for all integers k.

Fully Cyclic p.2

Note that for k=0, the latter condition

becomes zAzA )(

Corollary 2.4.30


then the index of imprimitivity of A is equal

to the spectral index of A.

.1,,1,0(

,,

)(),()(

,.

,,,

,18.4.2

)()(,'

.,,,

,)(

2

2

22

21

2

21

212121

meAofspectralperipheralthe

imposibleiswhich

unityofrootmthaisethenAof

eigenvalueperipheralaisAeAeASince

kmthatSupposekmthen

kmhenceandnnnnnn

ofdivisorcommonaismExerciseBy

AeATheoremsFrobeniusBy

Aofindexspectraltheisk

andAofityimprimitivofindexthebemLet

zerofromdifferentarecccand

nnnwheretctctcttcLet

mi

ki

ki

ki

SS

Sk

mi

S

k

kn

knnn

A

S

SS

k

Corollary 2.4.31

The peripheral spectrum of an irreducible

nonnegative matrix is fully cyclic

.

)(

sgn

)()(

,

sgn,,sgn

,,,)(

,'

)'(:1

1

1222111

1

1

cyclicfullyisAofspectrumperipheraltheHence

ZjAtoingcorrespond

AofreigenvectoaniszzDthen

ZjzDAzDeA

zADezADthen

ZjDADeA

relationtheestablishweyInductivel

DADeDDADeDeDADeAthen

zzdiagDwhere

zzDDADeAzAzA

haveweLemmasWielandtBy

RsomeforeWrite

LemmasWielandtByMethod

j

jj

jjjij

jijj

jjij

iiii

n

i

i

Exercise 2.4.32

zPzP )(

that the peripheral spectrum of P is fully

If

Let P be a square nonnegative matrix. Prove

cyclic if and only if P satisfies the following:

z is a corresponding eigenvector, then

is a peripheral eigenvalue of P and

.)(,0

)sgn,,(sgn

)(

,

.)(..0

)(

,

""

1

zPzPhavewekTake

zzdiagDwhere

ZkzDePzPD

cyclicfullyisspectrumperipheraltheSince

zePPztsCzand

RsomeforePwritecanwethen

PofeigenvalueperipheralabeLet

n

kikk

in

i

Theorem 2.4.33

nMA

kk

T

A

A

A

APP

*

0

22

11

If

then there is a permutation matrix P such that

where iiA is 1x1 or is irreducible.

Frobenius normal form

22

11

2211

22

11

*

**

0*

~

,,

.

,

*

0~,

.,

A

A

thenreducibleisAthatassumemayOtherwise

doneisproofthisthen

eirreduciblareAandAIf

A

AAthenreducibleisAIf

doneisproofthistheneirreduciblisAIf

P

P

Example

eirreduciblisAwhere

A

A

A

V

V

V

VVV

ii

33

22

11

3

2

1

321

*

0

V1 V2

V3

Three components

all are maximal Strongly

connected component

Remark p.1

kk

T

A

A

A

APP

*

0

22

11

If A is reducible , nonnegative, and

there is a permutation matrix P such that

where iiA is 1x1 or is irreducible.

Remark p.2

)()()()( 2211 kkAAAA

)(max)(1

iiki

AA

then

form of )(A

The peripheral eigenvalue of A is of the

times a root of unity and

Theorem 2.4.34

0.. mAtsZm

0AGiven

A is primitive if and only if

.

,.1

,,

.

)(

matrix

positivenotisAofpowerpositiveeverythenBut

matrixcyclichatosimilarnallypermutatioisA

ThmFrobeniusBythengreater

ishAofityimprimitivofindextheThen

primitivenotisAthatSuppose

eirreduciblbemustA

.0..

arg,0

.).(

).(

lim

mod

,

m

kT

T

Tk

k

AtsZm

ellysufficientkforA

AsoxyBut

positiveishenceandArespAofvector

Perronaisyrespxwhere

xyA

AThen

Aofseigenvalueotherthangreaterulo

withAofeigenvaluesimpleaisAAlso

eirreduciblisAthenprimitiveisAIf

Exercise 2.4.35

The spectral radius of a nonnegative matrix

A is positive if and only if G(A) contains at

least a circuit.

Exercise 2.4.36

)(A

00004

00080

02000

00400

30021

A

(i) Find the Frobenius normal form of the

matrix

(ii) Compute

Exercise 2.4.37

0

pn )( 1

nixxaxa ipnini ,,1,,,11

matrix and let

vector x such that

Let A be an nxn irreducible nonnegative

p1

and a positive

Then there exists

where is defined to be

pn

i

pi

1

1

Introduce A Semiring 0; aRaR

abba

baba ,max

are associative,commutative

R+ form a semiring under

On

and

introduce

by:

distributes over

0 is zero element

and

and

A B ⊕ p.1

nmnppm MBAMBMA ,

410

023

010

,

112

100

013

BA

kjikk

ij baBA max

kjkk

j xaxA max

A B ⊕ p.2

423

410

033

BA

412

123

013

BA

Circuit Geometric mean

13221)( iiiiii k

aaaA

121: iiii k circuit

kiiiiiik

kaaaA

11

13221)(

is called circuit geometric mean

Lemma 2.4.38

0xxA


x is a semipositive vector and

such that

)(0 Aandx then

where μ(A) is the maximum circuit

geometric mean.

γ is called a max eigenvalue and x is the corresponding max eige

nvector

.0

0

,0

00

,0

0)1(

22

1211

2211

2221

1211

1

1

xHence

eirreduciblisAfactthescontradictwhich

reducibleisAsoandA

AAA

haveweandxxASince

MAandMAwhere

AA

AAAformtheofisA

andxxxassumemay

nisomeforxthatSuppose

xxxthatshowTo

knk

Tk

i

Tn

).(

)()(

)(

max

max

max

max

),,(,,),(),,(

)(

max

)()2(

1

13221

11

111

22332

11221

AHence

AGincircuitanyforA

Athen

xxaxa

xxaxa

xxaxa

xxaxa

haveweiiiiii

arcswithAGincircuitanyFor

nkxxaxxA

AthatshowTo

kkk

kkkkk

iiiii

iiiii

iiiii

iiiii

k

kk

.

.

.1deg

,.

),(

)(

)()3(

toequal

meangeometriccircuithascircuitThatcircuita

containsitHenceleastatreeouthas

vertexeverydigraphthisInxxa

thatsuchjiarcsthoseofconsistswhich

AGofsubdigraphtheConsider

AthatshowTo

ijij

Plus max algebra R {-∞}∪

yxyx ,max

yx

yx

eeyx

eeyx

)exp(

)exp(

yxyx

RR:exp

They are isometric

Brouwer’s Fixed Point Theorem

ppfthatsuchCp )(

nRLet C be a compact convex subset of

and f is a continuous map from C to C, then

Theorem 2.4.39 Max version of the Perron-Frobenius Theorem

xAxA )(

If A be an irreducible nonnegative matrix

then there is a positive vector x such that

.)(

038.4.2

,

.)(

int'

,.

;0,

)(:

10:

1

1

1

1

1

n

ii

n

ii

n

ii

n

ii

n

ii

n

xAA

andxthatLemmafromfollowsIt

xxAxATherefore

xxfthatsuchxexiststhere

TheoremPoFixedsBrouwerby

henceandcontinuousisfAlsodefinedwell

isfsoyAeirreduciblisASince

yA

yAyfbyfmapthe

defineandyandyRyLet

Fact

ADD 1diagonal entries such that

has the constant row sums.

there is a diagonal matrix D with positive

Given an irreducible nonnegative matrix A

Remark 2.4.40

ADD 1diagonal entries such that in

Theorem 2.4.39 is equivalent to the assertion

there is a diagonal matrix D with positive

that given an irreducible nonnegative matrix A

the maximum entry in each row is the same.

).()(max

maxmax

,

111

11

1

AxAxxAxxax

xaxADD

iFix

x

x

DTake

iiiijijj

i

jijij

ijj

n

nRyx ,

nn yxyxyx ,,, 2211

nk 1If

Let k be a fixed positive integer,

:yx kT the sum of the k largest

, then let us define

components in

then

iii

T yxyx max1

n

iiin

T yxyx1

nn RxRMA ),(

niyaayA kiniik ,,1

nk 1If

Let k be a fixed positive integer,

yA k by

, then let us define

then

xAxA 1 AxxA n

Theorem 2.4.41

xxA k

nk 1and let

constant


Then there exists a positive vector x and a

such that

.0,

)(

int'

,.

0,

)(

:

1

1

1

xandxyAxATherefore

xxfthatsuchxexiststhere

TheoremPoFixedsBrouwerby

henceandcontinuouisfAlsodefinedwellisf

soandyAeirreduciblisASince

yyA

yAyf

byfDefine

n

iikk

n

iik

n

iik

k

Remark p.2

,0

XxxTxT )()(

FXT :space and

Then for any

Let X be a Topology space, F is a Banach

FXT :

is continuous

map such that T(X) is precompact in F.

there is a continuous

map of finite rank s.t.

Combinatorial Spectral Theory of Nonnegative Matrices

Three different ways There are three different ways to show that ρ(A) is a simple eigenvalue of an...

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