vol3_p13-22. 160KB Jun 04 2011 12:06:13 AM

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The Electronic Journal of Linear Algebra.

A publication of the International Linear Algebra Society. Volume 3, pp. 13-22, February 1998.

ISSN 1081-3810. http://math.technion.ac.il/iic/ela

ELA

ON NONNEGATIVE OPERATORSANDFULLYCYCLIC

PERIPHERALSPECTRUM

K.-H. FORSTERy

AND B. NAGY z

DedicatedtoHans Schneideron the occasionof his seventiethbirthday Abstract. In this note the propertiesof the peripheralspectrumof a nonnegativelinear operator A(for which the spectral radius is a pole of its resolvent) in a complex Banach lattice are studied. It is shown, e.g., that the peripheral spectrum of a natural quotient operator is always fully cyclic. We describe when the nonnegative eigenvectors corresponding to the spectral radiusrspan the kernel N(r,A). Finally, we apply our results to the case of a nonnegative matrix, and show that they sharpen earlier results by B.-S. Tam [Tamkang J. Math.21:65{70, 1990] on such matrices and full cyclicity of the peripheral spectrum.

AMSsubject classications. 47B65, 47A10, 15A48

Key words. Banach lattice, lattice ideal, nonnegative operator, peripheral spectrum, fully cyclic, nonnegative matrix

1. Intro duction. It follows from results of H.H. Schaefer [6, I.2.6 and V.4.6],

that in the nite dimensional caseCj n

and in certain Banach function lattices a non-negative operatorAhas a fully cyclic peripheral point spectrum i for all2

j

C with

jj=r(A)

x2N(,A) impliesjxj2N(r(A),A);

herer(A) denotes the spectral radius ofA; see also B.-S. Tam [8, Lemma 2.1].

In this note we consider nonnegative operators A in a Banach lattice for which

the spectral radius r(A) is a pole of its resolvent. We give necessary and sucient

conditions that for a given2 j

C withjj=r(A) the inclusionfjxj:x2N(,A)g N(r(A),A) holds; in particular, we give necessary and sucient conditions that N(r(A),A) has a basis of nonnegative eigenvectors, and is a sublattice, respectively.

As examples show (see [8, Example 2.7]), the inclusion above is, even in the matrix case, not only a property of the spectrum and the associated directed graph of

A. This will be very clear from Theorem 3.5. On the other hand, Theorem 4.2 in the

matrix case and under the assumption that the nonnegative vectors inN(r(A),A)

span this kernel will show that the property ofN(r(A),A) being a sublattice can

be characterized by properties of the reduced and of the singular graphs ofA.

The main method of the investigation is the systematic application of an idea going back to Lotz and Schaefer (cf. [6]), and successfully developed by Greiner [2], [4]. The closed lattice idealsI

0and I

1, dened below in terms of the Laurent expansion

Received by the editors on 27 November 1997. Accepted for publication on 5 February 1998. Handling Editor: Daniel Hershkowitz

yDepartment of Mathematics, Technical University Berlin, Sekr. MA 6-4, Strasse des 17. Juni 135, D-10623 Berlin, Germany (foerster@math.tu-berlin.de).

zDepartment of Analysis, Institute of Mathematics, Technical University Budapest, H-1521 Bu-dapest, Hungary (bnagy@math.bme.hu).

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14 K.-H.ForsterandB.Nagy

of the resolvent ofAaround the spectral radius, their quotientI

hand the restrictions

of (or the induced operator by) A are the most important technical means of the

study, and a number of the results presented here nd their natural formulation in this terminology.

In Section 4 we compare the results of Section 3 on nonnegative operators in Banach lattices with those of B.-S. Tam [8] on nonnegative matrices. One of the main results here is Theorem 4.2, which shows that for a nonnegative square matrixAthe

kernelN(r,A

1) for the restriction A

1 of

Ato the lattice idealI

1 is a sublattice i

there does not exist a (strongly connected equivalence) class of A having access to

two dierent distinguished basic classes (as expressed in the already standard graph theoretic terminology). The results of Section 4 sharpen the main results of Tam [8, Theorems 2.4 and 2.5] on matrices with fully cyclic peripheral spectrum.

2. Denitions and Preliminaries.

In the followingA denotes a nonnegative

operator in a complex Banach latticeE. We assume that its spectral radiusr=r(A)

is a pole of order p (1) of its resolvent R(:;A), i.e., we have for some >0 and

operatorsQ

k (

k=,p;,p+ 1;:::) the Laurent expansion

R(;A) = 1 X

k =,p

(,r) k

Q

k if 0

<j,r j<:

Note that Q

,p is nonnegative, since

A is nonnegative. Following G. Greiner [2], [4,

Chapters B III and C III], we dene

I 1=

fx2E: (Q ,p

jxj=:::=Q ,3

jxj=)Q ,2

jxj= 0g

(henceI 1=

E ifp= 1), and

I 0=

fx2E:Q ,1

jxj= 0g:

Note thatI

0 can be trivial. Then we have (see [4, p. 174 and p. 303]) that I

1and I

0

are closed ideals of E, and they are invariant under A andQ k (

k=,p;,p+ 1;:::).

IfA 1 and

A

0 denote the restrictions of Ato I

1 and I

0, respectively, then

r(A 1) =

r(A) is a pole of R(;A

1) of order 1; r(A

0)

< r(A); here we set r(A 0) =

,1 if I 0=

f0g:

SinceI 0

6=I

1, the quotient space I

h= I

1 =I

0 is well-dened and is a Banach lattice.

Further,A

1induces uniquely a nonnegative operator A

hin I

hsuch that q

h A

1= A

h q

h,

where q

h denotes the quotient map I

1 ! I

1 =I

0. Then (see [4, p. 174 and p. 303]) r(A

h) = r(A

1) =

r(A) = r, r is a pole of order 1 ofR(;A

h), and the residuum of R(;A

h) at

r(which is induced byQ ,1 in

I

h) is strictly positive in the sense that the

zero element is the only nonnegative element inI

hwhich it maps to the zero element.

Sincer is a pole of order 1 ofR(;A

h), the associated residuum is a projection with

kernelR(r,A

h) = range of r,A

h. Therefore

R(r,A

h) cannot contain any nonzero

elementy ofI

h for which either

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CyclicPeripheralSp ectra 15

Let J be the closed ideal of Ih generated by N(r ,A

h). J is Ah-invariant, since

N(r,A

h) is Ah-invariant. Let Ai= Ah

jJ (i.e. the restriction of A

hto J). Further,

let Ij = Ih=J and Aj = Ah=J (i.e. the operator induced by Ah in Ij). We set

r(Aj) =

,1if I j=

f0g.

The important basic connections between dierent kernels and spectral radii of the operators dened above will be collected in the proposition below. The following useful lemma, which is not new, will be applied in the proof of the proposition and several times at other places of this note.

Lemma 2.1. LetT be a linear map in a vector spaceV and letM be aT-invariant linear submanifold of V. LetTjM denote the restriction of T toM, letT=M denote

the linear map induced by T in the quotient space V=M, and let qM : V

! V=M

denote the quotient map. Then the following hold.

(I)If TjM is bijective, thenq

M maps the kernel N(T)bijectively onto N(T=M).

(II)If T=M is injective, thenN(T) = N(T jM).

Proof. (I) Since qMT = (T=M)qM, the quotient map qM maps N(T) into

N(T=M). Let z 2 N(T=M). Take v 2 V such that q

M(v) = z, then qM(Tv) =

T=M z = 0, i.e., Tv 2M. Since TjM is surjective, we have Tv = (TjM)w for some

w2M V . Thus u = v,w2N(T) and q

M(u) = qM(v) = z. If u

2N(T) satises

qM(u) = 0, then u

2M and (TjM)u = 0. Since TjM is injective, we obtain u = 0.

(II) Clearly we have N(TjM)N(T). Let u2N(T). Then T=M q

M(u) = qM(Tu) =

0, and therefore qM(u) = 0, since T=M is injective. Thus u

2M\N(T) = N(TjM).

Proposition 2.2. Let the assumptions and notations preceeding Lemma 2.1 hold. Then the following hold.

(I) r(A) = r(A1) = r(Ah) = r(Ai) > max fr(A

j);r(A0)

g, and r(A) is a pole of order

1 of R(;A 1);R(

;A

h)and R( ;A

i), respectively.

(II)For all2 j

Cwithjj= r(A) = r we haveN(,A)N(,A

1); N(

,A h) =

N(,A

i); qhmapsN( ,A

1)bijectively ontoN( ,A

h),dimN(r ,A

1)

dimN(,

A1) = dimN(

,A

h) = dimN(

,A i), and

fjzj: z 2N(,A h)

gN(r,A h). It

follows thatN(r,A

h)is a sublattice ofIh.

Proof. (I) As seen above, the residuum Qh;,1of R( ;A

h) at r is strictly positive.

Then its restriction Qi;,1to J is also strictly positive (note that

f0g6= N(r,A h)

J). This implies the three equalities. Since R(Qh;,1) = N(r ,A

h)

J, it follows

that R(;A

j) is holomorphic at r. Therefore r(Aj) < r(A), since Aj is a nonnegative

operator in the Banach lattice Ij [3, Proposition 4.1.1.i]. Note that we have dened

r(Aj) =

,1if I j=

f0g.

(II) The rst inclusion is evident. For the second statement we apply Lemma 2.1

(II) with V = Ih, M = J and T =

,A

h. Note that T=M =

,A

j is injective,

since r(Aj) < r(A) =

jj. The third statement follows from Lemma 2.1 (I) if we

set V = I1, M = I0 and T =

,A

1. Note that T

jM = ,A

0 is bijective, since

r(A0) < r(A) =

jj. The equality dimN(,A

1) = dimN(

,A

h) = dimN(

,A i)

is now evident. We shall prove dimN(r,A 1)

dimN(r,A

i), which needs a proof

only if m = dimN(r,A

1) <

1. By [4, C-III, Lemma 3.13], the ideal J is the

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Aik= Ai jJ

k are irreducible, and r(Aik) = r(A) is a pole of R( ;A

ik) (k = 1;:::;m).

The eigenspaces N(r,A

ik) are one-dimensional; see [6, V.

x 5]. By [6, Corollary

to Theorem V.5.4], dimN(,A

ik)

1. A

i is the direct sum of the restrictions

Aik. Then N(

,A

i) is the direct sum of the N( ,A

ik) (k = 1;:::;m). Therefore

dimN(,A

i)

m = dimN(r,A

1). For the proof of the next statement take

z2N(,A

h). Let y = (r ,A

h)

jzj. Then yrjzj,jA hz

jj(,A h)z

j= 0. By a

remark preceding Lemma 2.1, then y = 0.

3. ResultsandPro ofs. We shall always assume that A is a nonnegative linear

operator in a Banach lattice E, and its spectral radius is a pole of its resolvent. We shall freely use the concepts and notations of Section 2. The next lemma is crucial for the main results of this note.

Lemma 3.1. Let jj= r(A) = r. For each z2N(,A

h) there exists a unique

nonnegative w in N(r,A

1) such that qh(w) =

jzj; for the unique u2 N(,A 1)

satisfying qh(u) = z it follows thatqh(w) = qh(

juj)and wjuj.

Proof. Let z 2 N(,A

h). By Proposition 2.2(II), there exists a unique u 2

N(,A

1) with qh(u) = z. Proposition 2.2(II) implies

jzj2N(r,A

h). Since qhis a

lattice homomorphism, we get qh((r ,A

1)

juj) = (r,A h)qh(

juj) = (r,A h)

jzj= 0, i.e.

(r,A 1)

juj2I

0. Since (r ,A

1)

jujj(,A 1)u

j= 0, there exists a unique nonnegative

u0 in I0 such that (r ,A

0)u0 =

,(r,A 1)

juj (notice that r(A

0) < r(A) = r, thus

r,A

0 has a nonnegative inverse). Then w =

juj+ u

0 is a vector we are looking

for. If v2 I

1 satises qh(v) = qh(

juj), then juj,v 2 I

0. If further v

2N(r,A 1),

then juj+ u 0

,v2I 0

\N(r,A

1) = N(r ,A

0). But r ,A

0 is injective, therefore

v =juj+ u

0= w, i.e. w is unique as stated.

Proposition 3.2. Under our general assumptions the following hold.

(I) N(r,A)\E +

N(r,A 1).

(II) span(N(r,A)\E

+) = N(r ,A

1).

(III)the nonnegative eigenvectors ofAcorresponding torspan the eigenspaceN(r,A)

i N(r,A)I

1 i N(r

,A) = N(r,A

1).

(IV) if N(r,A

1) is nite dimensional, then N(r ,A

1) has a basis of nonnegative

eigenvectors of A1 corresponding tor.

Proof. (I) u2N(r,A) is equivalent to R(;A)u = (,r)

,1u if 0 <

j,rj<

for some > 0. Therefore u2N(r,A)\E

+ implies u

2fx2E : Q ,2

jxj= 0g= I 1.

(II) Let u2N(r,A

1). Then its real and its imaginary parts belong to N(r ,A

1).

Therefore we assume w.l.o.g. that u is real. Choose w as in Lemma 3.1. Then u = 1

2(w + u) ,

1 2(w

,u), wu2N(r,A

1) and w

u2E +.

(III) The rst part follows from N(r,A

1)= N(r

,A)\I

1 and (II), the second part

ist clear.

(IV) follows simply from (II).

Remark 3.3. In connection with (III) and (IV) recall that U.G. Rothblum [5]

proved that in the case dim E <1 the generalized eigenspace N((r,A)

p) (where

p is the order of the pole r) always has a basis of nonnegative vectors. However, [1, Example 18] shows that in the general case (dim E =1) the corresponding statement

can be false even if p = 2 and dim N((r,A) 2) = 2.

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equivalent.

2 are disjoint vectors in

N(r,A

h), then the unique nonnegative w

2be disjoint vectors in

N(r, A

2 are disjoint [3, Theorem

1.1.1.vi], By Lemma 3.1, we get w

modulus 1, since w

1+ w

2 is nonnegative and j

1) is a sublattice. This is equivalent to the

disjointness ofw 1 and

sublattice of I

h (see Proposition 2.2(II)), z

+ and z

, (the positive and the negative

parts of z, respectively) belong to N(r,A

h). Applying Lemma 3.1, we choose

nonnegative w

2belong to

N(r,A

1), we obtain u=w

1 ,w

2, by Proposition 2.2(II). Now z

+and z

,are disjoint [3, Theorem 1.1.1.iv],

thereforew 1and

w

2are disjoint, by assumption. But then

juj=w

For the rest of this section we assume thatI

0is a projection band in

E. Note that

this holds in any Banach lattice with order continuous norm [3, Corollary 2.4.4.xii, Theorem 2.4.2.ii), and Theorem 1.2.9.], and a fortiori inCj

`. Then

h with the kernel of the band projection of I

means disjoint sum in the lattice sense. Since I 0

isA

1-invariant, we can represent A

1 with respect to this direct sum as a triangular

operator matrix as follows,

A

Theorem 3.5. In addition to the general assumptions we assume that I 0 is a

0. The following assertions are equivalent.

(I)juj2N(r,A

j = 0, by Proposition 2.2(II). Therefore juj 2 N(r,A

From this inequality it follows immediately that (I) and (II) are equivalent. The second equality in (II) means thatA

0h: I

h !I

0is a lattice homomorphism

onN(,A

h). However, even if (I) holds, A

0h need not be a lattice homomorphism

ofallofI

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Example 3.6. Let

A=A

j. We cannot expect that we have here equality, even

if statement (I) in Theorem 3.5 holds, as is shown by the following example.

Example 3.7. Let

A=A

C. Therefore statement (I) of Theorem 3.5 holds for

= 2. Let u = [u

In the next section we will use the following result.

Proposition 3.8. In addition to the general assumptions we assume that I 0 is

jj=r(A) =rthe following assertions are equivalent.

(I)N(,A)I

is now (identied with) the band projection ofI 1 onto

I h.

(II))(I) follows from Proposition 3.2(I); notice that we use for this implication only

the general assumptions.

4. The matrix case.

In the last section of this note we want to compare the results in Section 3 with those of B.-S. Tam on nonnegative matrices [8, Theorems 2.4 and 2.5]. We shall use the graph theoretic concepts dened in [7,x2], and [8,x1].

In the complex Banach latticeCj `

each ideal is of the form I

=

i(= i-th component of

x) = 0 ifi2=gfor somef1;:::;`g; see [6, p. 2]. Let

and be nonempty subsets off1;:::;`g. Forx2 j

C` we denote by x

the subvector

of x with indices from. For an ``-matrix A we denote by A

the submatrix

of A with row indices from and column indices from. We writeA

instead of

A

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dened in Section 2 and the ideals Ifor a class of A (i.e., is a strongly connected

component in the directed graph associated with A [8,x 1].

Lemma 4.1. LetAbe a nonnegative square matrix, and let be a class of A.

(I)If I is an A-invariant ideal andI\I 6=f0g, then II.

(II)If II 0

Ij, thenis nonbasic (i.e. r(A) < r(A)).

(III)Letbe a basic class. Thenis distinguished iII 1iI

IhiI= Jk

for somek2f1;:::;mg.

(IV) If has access to a distinguished basic class of A, then either is this distin-guished class of AorII

0.

Proof. (I) The ideal I\I is A-invariant. Since A is an irreducible matrix,

I\I 6=f0gimplies I\I = I.

(II) We have r(A)maxfr(A

0);r(Aj)

g< r(A).

(III) Let be a distinguished basic class of A. Then there exists a nonnegative eigenvector x of A corresponding to r, such that xis strictly positive; see [7, Theorem

3.1]. By Proposition 3.2(I), we have that x2I

1. Therefore I I

1. If is a basic

class of A with II

1, then I

J Ih, by (II). Let be a basic class of A with

I Ih. Then I J, since r(Aj) < r(A). Ai is the direct sum of the irreducible

operators Aik= AijJk (k = 1;:::;m), so we get A = Aik for some k2f1;:::;mg.

Then I = Jk for this k. If I = Jk for some k 2 f1;:::;mg, then r(A) = r(A)

and Ais irreducible. Therefore is a strongly connected set in the directed graph

associated with A. From the triangular structure of A1with respect to the direct sum

I1 = Ij J

1

:::JmI

0 it follows that is a strongly connected component in

this graph.

(IV) Let have access to a distinguished basic class of A. Assume 6= . Then

is nonbasic. Furthermore, there exists a nonnegative eigenvector x of A corresponding to r, such that x ist strictly positive; see [7, Theorem 3.1]. Proposition 3.2(I)

implies x 2I

1. Therefore x

2 N(r,A

1). From Proposition 2.2(II) it follows that

N(r,A

1)

J I

0. Therefore I

JI

0. From the last part of the proof of (III)

we see that I\J6=f0gwould imply r(A) = r(A). Hence we obtain II 0, since

is nonbasic.

B.-S. Tam [8, Lemma 2.1] showed that A has fully cyclic peripheral spectrum (see [6, Denition 2.5]), i for all 2

j

C withjj= r(A)

fjuj: u2N(,A)gN(r,A):

Thus, if A has a fully cyclic peripheral spectrum, then N(r,A) is a sublattice of j

C`. By Proposition 3.2, the latter assertion implies statement (a) in [8, Theorems 2.4 and 2.5]. In the matrix case we have the following result as a supplement to Theorem 3.4.

Theorem 4.2. For a nonnegative square matrix the following assertions are equivalent.

(I) N(r,A

1)is a sublattice, wherer = r(A),

(II) there does not exist a class ofA which has access to two dierent distinguished basic classes.

Proof. We prove that the assertion (II) is equivalent to Theorem 3.4(II).

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distinguished basic classes 1 and

2. Then

has to be a nonbasic class, since a

basic class does not have access to a distinguished basic class. By Lemma 4.1(IV) this impliesI

I

0. From [7, 3.1(i)] it follows that there are two nonnegative eigenvectors w

i)). By Proposition 3.2(II), w

are disjoint sets, we get that q h(

2) are disjoint nonnegative vectors in N(r,A

h). But w

1and w

2 are not disjoint, and this contradicts Theorem 3.4(II).

(II))(I) Ifz is inN(r,A

h), then

jzjbelongs toN(r,A

h); see Proposition 2.2(II).

Then supp(z) = supp(jzj) is the union of some distinguished basic classes. From

[7, 3.1(i) and (ii)] it follows that for the unique nonnegative w 2 N(r,A 1) with q

h(

w) =jzj its support is the union of all classes which have access to distinguished

basic classes contained in supp(z). Since w is a nonnegative eigenvector of A, the

ideal generated by w is A-invariant; this ideal is I

, where

= supp(w). Now let

z 1 and

z

2 be disjoint vectors in

N(r,A

h), and assume that the unique nonnegative w

distinguished basic class 1in

i). Therefore I

i are distinguished. Thus

has access to two

dierent distinguished basic classes of A. This contradicts (II), so w 1 and

w 2 are

disjoint. Thus Theorem 3.4(II) holds.

Theorem 4.2 and Proposition 3.2 show that statements (a) and (b) in [8, Theorem 2.4] together are equivalent to the statement thatN(r,A) is a sublattice of

j

C`.

The next theorem will show, how conditions (c) in Theorems 2.4 and 2.5 of [8] are related to our results.

Theorem 4.3. Consider for a nonnegative square matrix A and a peripheral

eigenvalueof Asatisfying the following statements.

(I)If is an eigenvalue ofA

for some class

of A, thenis distinguished.

(II)N(,A)I 1.

(III) If is an eigenvalue ofA

for some class

of A, then all initial classes of the

for some class

of A, thenis also an eigenvalue

of A

for some distinguished class

of A, which has access to .

(V)is an eigenvalue of A

for some distinguished class

of A.

assumption, all (nal) classes in supp(u) have access to a distinguished basic class of A. By Lemma 4.1(III) and (IV), we haveu2I

1.

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belongs to it. Then is basic, since jj= r(A). Let ! be the union of all classes of

A, which have access to and are dierent from . Set =f1;:::;`gn(![). Note

that one or both of the sets ! and can be empty. Then the corresponding matrices in the decomposition below do not appear. With respect to the decompositionCj`=

I II! the matrix A has the block triangular form 2

4

A 0 0

A A 0

A! A! A!

3

5:

By the choice of , is not an eigenvalue of A!. Take an eigenvector u of A

corre-sponding to . Dene u!= (,A!)

,1A

!u and u = [o>;u>

;u>

!]>. Then u is an

eigenvector of A corresponding to . Now u2N(,A)I

1 by assumption. Then

o6= [o >;u>

;o>]>

2I\I

1implies I \I

1

6

=f0g. By Lemma 4.1(I), we have II 1.

Since is basic, we get from Lemma 4.1 (III) that it is distinguished. (III)) (IV) is clear.

(IV))(III) Take an initial class of A in F(). Then is an eigenvalue of A. By

assumption, there exists a distinguished basic class of A with >= and is an

eigenvalue of A. Then >= >= , thus 2F(). Therefore = , since is

initial in F().

(IV)) (V) is clear, since each eigenvalue of A has to be an eigenvalue of A for at

least one class of A.

(V) ) (VI) For a distinguished basic class let u be an eigenvector of A

cor-responding to . By Lemma 4.1(III), A = Aik for some k 2 f1;:::;mg. Let

uh= [u> 1;:::;u

>

m]> with u

k = u and ui= 0 if i6= k. The vector uhis an eigenvector

of Ahcorresponding to , since Ah= Ai1

:::Aim. Dene u

0= ( ,A

0) ,1A

0huh,

and u1 = [u >

h;u> 0]

>. Then u

1 is an eigenvector of A1 corresponding to . Thus

f0g6= N(,A

1) = N(

,A)\I 1.

(VI) ) (V) Let u 2 I

1 be an eigenvector of A corresponding to . If is a nal

class of A in supp(u), then is a basic class [8, Lemma 2.3]. Since I I 1, is

distinguished (see Lemma 4.1(III).

The following examples will show that the converses of the three one-way impli-cations in Theorem 4.3 are not true in general.

Example 4.4. Let

A =

2

6 6 6 6 6 6 6 6 4

0 1 0 0 0

1 0 0 0 0

0 1 1 0 0

a b 0 0 1

c d 0 1 0

3

7 7 7 7 7 7 7 7 5

with nonnegative numbers a;b;c and d:

Then =f1;2g, =f3gand =f4;5gare the classes of A. The matrices A, A

and A have eigenvalues 1 and,1. The matrix A has the eigenvalue 1. Therefore all

classes of A are basic. The classes and are distinguished and is not distinguished. Further, here I1=

fx2 j

C5

: x1= x2= 0

g= II and I

0=

(10)

ELA

22 K.-H. Forster and B. Nagy

(1) a =d= 1 and b =c = 0. Then statement (I) is not true for =,1, but

N(,1,A) =fx2 j

C5: x

1= x

2= x

3= 0 ;x

4+ x

5= 0 gI

1. For

=r(A) statement

(I) in Theorem 4.3 is equivalent to: \All basic classes ofA are distinguished." This

is equivalent to p = 1, where p is the order of the pole r(A) of the resolvent, cf.

Section 2. The example also shows that even for = r(A) (II) does not imply (I),

since N(1,A) =fx2 j

C5

:x 1=

x 2= 0

;x 4

,x 5= 0

gI 1 and

p= 2.

(2) a = b = c = d = 1. Then statement (II) is not true for = ,1, since

N(,1,A) =fx2 j

C5: x

1+ x

2= 0 ;x

2+ 2 x

3= 0 ;x

4+ x

5= 0

g. But statement (III)

is true for=,1, sinceF() =f;gwith >| andF() =fg.

(3)a=b=c=d= 0. Then statement (IV) is not true for=,1, since>6=.

But statement (V) is true, since,1 is an eigenvalue ofA .

Part (b) of [8, Theorem 2.5], which states that each distinguished basic class ofA

is initial (in the familyof all classes ofA), is equivalent toN(r, A 1) =

N(r, A h)

f0g.

This follows immediately from [7, 3.1(i) and (ii)]. Combining this observation with Proposition 3.8 we obtain the following result.

Corollary 4.5. Let A be a nonnegative square matrix, such that each

distin-guished class of A is initial (in the family of all classes of A). Then A has a fully

cyclic peripheral spectrum i N(,A)I

1 for all peripheral eigenvalues of A.

This corollary is stronger than [8, Theorem 2.5] as Theorem 4.3 (I) )(II) and

Example 4.4(1) show.

REFERENCES

[1] K.-H. Forster and B. Nagy, On the local spectral radius of a nonnegative element with respect to an irreducible operator,Acta Sci. Math.55:401{423, 1991.

[2] G. Greiner, Zur Perron-Frobenius Theorie stark stetiger Halbgruppen,Math. Z.177:155{166, 1981.

[3] P. Meyer-Nieberg,Banach lattices, Springer Verlag, Berlin{Heidelberg{New York, 1991. [4] R. Nagel (ed.),One-parameter semigroups of Positive Operators, Lecture Notes in

Mathemat-ics 1184, Springer Verlag, Berlin 1986.

[5] U.G. Rothblum, Algebraic eigenspaces of non-negative matrices,Linear Algebra Appl.12:281{ 292, 1975.

[6] H.H. Schaefer,Banach Lattices and Positive Operators, Springer Verlag, New York, 1974. [7] H. Schneider, The inuence of the marked reduced graph of a nonnegative matrix on the

Jordan form and on related properties: a survey, Linear Algebra Appl.84:161{189, 1986. [8] B.-S. Tam, On nonnegative matrices with a fully cyclic peripheral spectrum, Tamkang