DEGENERATE COCYCLE WITH INDEX-1 AND LYAPUNOV EXPONENTS

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Stochastics and Dynamics, Vol. 7, No. 2 (2007) 229–245c© World Scientific Publishing Company

DEGENERATE COCYCLE WITH INDEX-1 AND LYAPUNOVEXPONENTS

NGUYEN HUU DU∗, TRINH KHANH DUY and VU TIEN VIET

Faculty of Mathematics, Mechanics, and InformaticsHanoi National University, 334 Nguyen Trai

Thanh Xuan, Hanoi, Vietnam∗[email protected]

Received 7 February 2007Revised 23 March 2007

This paper deals with the solvability of initial-value problem and with Lyapunov expo-nents for linear implicit random difference equations, i.e. the difference equations wherethe leading term cannot be solved. An index-1 concept for linear implicit random differ-ence equations is introduced and a formula of solutions is given. Paper is also concernedwith a version of the multiplicative theorem of Oseledets type.

Keywords: Random dynamical systems; linear implicit equation Lyapunov exponent;index-1 tractable.

1. Introduction

Difference equations might define the simplest dynamical systems, but neverthe-less, they play an important role in the investigation of a dynamical system. Thedifference equations arise naturally when we want to study the evolution of bio-logical population or economic models on a fixed period of time. They can also beillustrated as discretization of continuous time systems in computing process.

An important class of difference equations is the linear one which leads to prod-ucts of matrices. The linear difference equations can be found when we handlelinearizations of nonlinear system

f(Xn+1, Xn, n) = 0, n ≥ 0 (1.1)

along solution. This leads to the difference equation

AnXn+1 = BnXn + qn, n ≥ 0, (1.2)

where An = ∂f∂xn+1

and Bn = ∂f∂xn

.If in (1.2), the matrix An is invertible for all n ∈ N, we can multiply both sides

of (1.2) by A−1n to obtain

Xn+1 = A−1n BnXn + A−1

n qn, (1.3)

which has been studied for a long time by many authors both in theory and practice.

229

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230 N. H. Du, T. K. Duy & V. T. Viet

However, in case where the assumption for solvability of implicit functions (1.1)cannot be fulfilled (for example when ∂f

∂xn+1is degenerate), the matrix An may be

degenerate for some n and it is unable to solve explicitly the leading term Xn+1

to obtain classical difference equations (1.3). This situation also occurs when weconsider the backward form of (1.2), i.e.

BnXn = AnXn+1 − qn, n = −1,−2, . . . ,

and in general, there is no reason to assume that (Bn) is invertible.The difference equation of the form (1.2) also appears in discretizing the linear

differential algebraic equation (see [7])

A(t)X(t) = B(t)X(t) + q(t),

by explicit Euler method.In that cases, solving the system (1.1) and (1.2) becomes more complicated. In

fact, we are faced with an ill-posed problem where the solution of an equation mayexist only on a submanifold or even, without any further assumption, the solutionof Cauchy problem does not exist.

In this paper we are concerned with the solvability of initial value problem,dynamic property of the solution and with the Lyapunov spectrum for a real noiseequation

AnXn+1 = BnXn,

X0 = x ∈ Rm,

n = 0,±1,±2, . . . , (1.4)

where (An, Bn) is a stationary process.When all matrices An and Bn are invertible, the solutions of (1.4) form a ran-

dom dynamic system which is well investigated (see Refs. [1 and 5] for example).However, without the assumption of invertibility of An and Bn, most of works haveto consider as Eq. (1.4) separately for n ≤ 0 or for n ≥ 0 and as far as we know,there is no work dealing with the dynamics of (1.4) over Z.

Therefore, the aim of this paper is to study dynamic property of (1.4). Wedevelop techniques in [7] to difference equations. First, we introduce a conceptof index-1 for Eq. (1.4). By means of this concept, we give an expression of thesolutions. Hence, we can prove the dynamic property and consider their Lyapunovexponents.

We are focused only on the case of index-1 tractable and we have to assumethat rankA0 is a nonrandom constant which implies that the sequence of matri-ces (An)n∈Z has constant rank with probability one. So far there is no availabletechnique to solve this problem with rankAn varying in n. Moreover, a higherindex concept of Eq. (1.3) is not defined and it requires some further works. In ouropinion, this work can be considered as a pioneer one dealt with implicit randomdynamical systems.

The paper is organized as follows: In Sec. 2 we deal with the index-1 tractableconcept of the pencil of matrices An, Bn. By this concept of index-1, we set

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Degenerate Cocycle with Index-1 and Lyapunov Exponents 231

out the condition of solvability for Cauchy problem and give an explicit formulaof solutions. In Sec. 3 we prove the cocycle property of solutions over randomdynamics. Section 4 gives the proof of multiplicative ergodic theorem MET forimplicit difference equation (1.4). Here, we show that the space R

m can be splitinto measurable subspaces R

m = ⊕τi=0Wi where for any i, the space W0 ⊕ Wi is

invariant and on the set Wi, the solution (Xn) of (1.4) has the same Lyapunovexponent.

2. Solutions of Implicit Linear Difference Equation

Henceforth, Z denotes the set of all integers. Assume that (Ω,F , P ) is a probabilityspace satisfying the normal conditions (see [8]). Let θ : (Ω,F , P ) → (Ω,F , P ) be aninvertible, P -preserving transformation. Let A(·) and B(·) be two random variablesvalued in the space of m × m-matrices. We consider the equation

A(θnω)Xn+1(ω) = B(θnω)Xn(ω),

X0 = x ∈ Rm a.s, n ∈ Z,

(2.1)

where θn = θ θn−1. Equation (2.1) is called real noise linear implicit differenceequation.

2.1. Some surveys on linear algebra

In this section, we survey some basic properties of linear algebra. Let (A, A, B) bea triple of matrices. Assume that rankA = rankA = r and let T ∈ Gl(Rm) suchthat T |kerA is an isomorphism between kerA and kerA. We can obtain such anoperator T as follows: let Q (resp. Q) be a projector onto kerA (resp. onto kerA);find nonsingular matrices V and V such that Q = V Q(0)V −1 and Q = V Q(0)V

−1

where Q(0) = diag(0, Im−r) and finally we obtain T by putting T = V V −1.Denote S = x : Bx ∈ im A and let Q be a projector onto kerA.

Lemma 2.1. The following assertions are equivalent.

(a) S ∩ kerA = 0,(b) the matrix G = A + BTQ is nonsingular,(c) R

m = S ⊕ kerA.

Proof. (a) ⇒ (b): Let x ∈ Rm such that (A+BTQ)x = 0 ⇐⇒ B(TQx) = A(−x).

This equation implies TQx ∈ S. Since S ∩ kerA = 0 and TQx ∈ kerA, it followsthat TQx = 0. Hence, Qx = 0 which implies Ax = 0. This mean that x ∈ kerA.Thus, x = Qx = 0, i.e. the matrix G = A + BTQ is nonsingular.

(b) ⇒ (c): It is obvious that x = (I − TQG−1B)x + TQG−1Bx. We see thatTQG−1Bx ∈ kerA and B(I −TQG−1B)x = B − (A+ BTQ)G−1Bx+ AG−1Bx =AG−1Bx ∈ im A. Thus, (I − TQG−1B)x ∈ S and we have R

m = S + kerA.Let x ∈ S ∩ kerA, i.e. x ∈ S and x ∈ kerA. Since x ∈ S, there is a z ∈ R

m

such that Bx = Az = APz and since x ∈ kerA, T−1x ∈ kerA. Therefore, T−1x =

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QT−1x. Hence, (A + BTQ)T−1x = (A + BTQ)Pz which follows that T−1x = Pz.Thus, T−1x = 0 and then x = 0. So, we have (c).

(c) ⇒ (a) is obvious.The lemma is proved.

Lemma 2.2. Assume that the matrix G is nonsingular. Then, the following rela-tions hold:

(i) P = G−1A where P = I − Q. (2.2)(ii) G−1BTQ = Q. (2.3)(iii) Q := TQG−1B is the projector onto kerA along S. (2.4)(iv) If Q is a projector onto kerA, then

PG−1B = PG−1BP, (2.5)

QG−1B = QG−1BP + T−1Q, (2.6)

with P = I − Q.

Proof. (i) Note that GP = (A + BTQ)P = AP = A we get (2.2).

(ii) From BTQ = (G − A), it follows that G−1BTQ = (I − P ) = Q. Thus, wehave (2.3).

(iii) By virtue of (2.3), Q2 =TQG−1BTQG−1Bby (2.3)

= TQQG−1B =TQG−1B = Q and AQ = ATQG−1B = 0. This means that Q is a projectoronto kerA. From the proof of (c), Lemma 2.1, we see that Q is the projectoronto kerA along S.

(iv) Since T−1Qx ∈ kerA for any x,

PG−1BQ = PG−1BTT−1Q = PG−1(A + BTQ)QT−1Q = 0.

Therefore, PG−1B = PG−1BP so we have (2.5). Finally,

QG−1B = QG−1BP + QG−1BTT−1Q = QG−1BP + QG−1(A + BTQ)QT−1Q

= QG−1BP + QT−1Q = QG−1BP + T−1Q.

Lemma 2.2 is proved.

2.2. Existence of solutions of (2.1) for n ≥ 0

Henceforth, we assume that rank A(ω) = k for P -a.s ω ∈ Ω where k (0 < k ≤ m)is a nonrandom constant. Let T be a random variable with values in Gl(Rm) suchthat T (ω)|kerA(ω) is an isomorphism between kerA(ω) and kerA(θ−1ω). Let Q(ω)be a measurable projector onto kerA(ω). Denote

S(ω) = z : B(ω)z ∈ im A(ω).

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Definition 2.3. The implicit difference equation (2.1) is said to be index-1tractable if

S(ω) ∩ kerA(θ−1ω) = 0 for a.s. ω ∈ Ω. (2.7)

By virtue of Lemma 2.1 we see that Eq. (2.1) is index-1 tractable if and only ifthe matrices

G(ω) := A(ω) + B(ω)T (ω)Q(ω)

are nonsingular with probability one.

Remark 2.4. (1) In fact, what we are doing below is true if the space Ω can bedivided into θ-invariant subsets Ωi, i = 1, 2, . . . , q such that rankA is constant onevery Ωi and (2.7) is satisfied for P -almost sure ω.

(2) In general, since A(ω) is degenerate, the dimension of the space of solutionsvaries in n. If the dimension of the space of solutions at step n + 1 is greaterthan one at the step n, it may cause bifurcations or multi-valued functions. Upto now, there has been no systematic analysis of such systems. As we can seebelow, S(ω) is in fact the space of solutions. Therefore, with the index-1 tractableassumption (condition (2.7)), dimS(ω) is constant. That implies the regularity ofinherent ordinary difference equation.

For the sake of simplicity we put

Qn(ω) = Q(θnω), Pn = I − Qn, An(ω) = A(θnω), Bn(ω) = B(θnω),

Gn(ω) = G(θnω), Tn(ω) = T (θnω).

Throughout this paper, if there is no confusion, we will omit ω in formulas.Assume that Eq. (2.1) is an index-1 tractable. By multiplying both sides of (2.1)by PnG−1

n and QnG−1n respectively and applying (2.3)–(2.6) for A = An, A = An−1

and B = Bn we come to the following equation:PnXn+1 = PnG−1

n BnPn−1Xn,

0 = QnG−1n BnPn−1Xn + T−1

n Qn−1Xn.

Putting Yn = Pn−1Xn; Zn = Qn−1Xn, (2.1) is equivalent toYn+1 = PnG−1

n BnYn,

Zn = −TnQnG−1n BnYn,

Xn = Yn + Zn, n = 0, 1, 2, . . . .

(2.8)

From (2.8) it follows that the solution with initial condition X0 = x exists if

Q0G−10 B0x = 0

with probability 1. By (2.4) we see that the random space Jf(ω) = ξ ∈ Rm :

(Q0G−10 B0)(ω)ξ = 0 does not depend on the choice of the projector Q and the

transformation T . Moreover, by Lemma 2.2

Qn−1 := TnQnG−1n Bn, (2.9)

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is the projector onto ker An−1 along Sn = ξ ∈ Rm : Bnξ ∈ im An. Hence, the

matrices

Gn = An + BnTnQn

are also nonsingular with probability one.

Lemma 2.5. Qn−1(ω), called canonical projector, and the matrix (PnG−1n Bn)(ω)

are independent from the choice of Q(ω) and T (ω), where Pn = I − Qn.

Proof. Since Qn−1 is the projector onto kerAn−1 along the space Sn, it is inde-pendent from the choice of Q and T .

Let T ′ be another linear transformation from Rm onto R

m satisfying T ′ |ker A0

to be an isomorphism from kerA0 onto kerA−1 and Q′ be another measurableprojector onto kerA. Denote T ′

n(ω) = T ′(θn(ω)), Q′n(ω) = Q′(θn(ω)) and G′

n(ω) =An(ω) + Bn(ω)T ′

n(ω)Q′n(ω). It is easy to see that

PnG−1n Bn = PnG−1

n G′nG ′−1

n Bn = PnG−1n (An + BnT ′

nQ′n)G ′−1

n Bn

= PnPnG ′−1n Bn + PnG−1

n BnT ′nQ′

nG ′−1n Bn = PnG ′−1

n Bn + PnG−1n BnQn−1

= PnG ′−1n Bn + PnPnG−1

n BnQn−1see (2.5)

= PnG ′−1n Bn.

The lemma is proved.

Using the canonical projectors we have Qn−1(ω)Xn(ω) = 0 which impliesXn(ω) = Pn−1(ω)Xn(ω) for a.s. ω ∈ Ω. Therefore, the forward equation of (2.1)for n ≥ 0 with the initial condition X0 = x ∈ Jf is reduced to a classical differenceequation

Xn+1 = PnG−1n BnPn−1Xn,

X0 = x, n = 0, 1, 2, . . . ,(2.10)

which follows that

Xn(ω) =

(0∏

i=n−1

PiG−1i BiPi−1

)(ω)x, n ∈ N, X0(ω) = x. (2.11)

Summing up we see that the initial value problem of (2.1) for n ≥ 0 has a uniquesolution given by (2.11) provided x ∈ Jf .

Remark 2.6. Since (PnG−1n Bn)(ω) = (PnG−1

n Bn)(ω), Eq. (2.10) is rewrittenunder the form

Xn+1 = PnG−1n BnPn−1Xn,

X0 = P−1x, (x ∈ Rm), n = 0, 1, 2, . . . .

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2.3. Solutions (2.1) for n < 0

For n < 0, Eq. (2.1) turns into the implicit difference equationBnXn = AnXn+1,

X0 = x; n = −1,−2, . . . .(2.12)

We assume that rank B(ω) = r for P -a.s. ω ∈ Ω where r (0 < r ≤ m) is anonrandom constant. Let T be a random variable with values in Gl(Rm) such thatT (ω)|ker B(ω) is an isomorphism between kerB(ω) and kerB(θω). Let Q(ω) be ameasurable projector onto kerB(ω). Put

G(ω) := B(ω) + A(ω)T (ω)Q(ω).

Assume that G is nonsingular with probability 1. This assumption implies thatEq. (2.12) is index-1 tractable. Let

Qn(ω) = Q(θnω); Gn(ω) = G(θnω); Pn = I − Qn; T n(ω) = T (θnω). (2.13)

For any n < 0, denote Yn = PnXn; Zn = QnXn, Eq. (2.12) leads toYn = PnG

−1

n AnYn+1,

Zn+1 = −TnQnG−1

n AnYn+1,

Xn = Yn + Zn, n = −1,−2, . . . .

We find the canonical projector. Denote Qn := Tn−1 Qn−1 G−1

n−1An−1 which isthe projector onto kerBn along the space Sn−1 = ξ : An−1ξ ∈ im Bn−1. LetGn(ω) = Bn(ω) + An(ω)TnQn. By using a similar argument as above we obtain

Lemma 2.7. The canonical projector Qn and the matrix PnG−1

n An are indepen-dent of the choice of Q and T , where Pn = I − Qn.

With the canonical projector Qn we have QnXn = 0 for any n < 0. Therefore,Xn =−1∏i=n

PiG−1i AiPi+1x, n = −1,−2, . . . ,

X0 = x ∈ Jb,

(2.14)

where Jb = ξ ∈ Rm : Q0ξ = 0. Similarly as in Sec. 2.2 we see that the initial

value problem for n ≤ 0 has a unique solution given by (2.14) provided that x ∈ Jb.

Remark 2.8. (1) Let us give some comments on the expressions of solutions (2.11)and (2.14). To obtain (2.11) for n ≥ 0 we assume the initial condition X0 = x ∈ Jf

(equivalently X0 = P−1x, x ∈ Rm) to be satisfied and obtain (2.14) for n ≤ 0 the

equality X0 = P0x is required. Thus, there exists uniquely the solution of (2.1) forn ∈ Z with the initial condition X0 = x if and only if x ∈ Jf ∩ Jb.

(2) Unlike random ordinary difference equations, in general, the existence of solutionof random implicit difference equations implies that the initial condition must be ameasurable selection of the corresponding ω → Jf(ω) ∩ Jb(ω).

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3. Dynamic Property

Since Q0 is the projector onto kerB0, Q−1Q0 = T0Q0G−10 B0Q0 = 0. Similarly,

Q0Q−1 = 0. Hence, the projectors P−1 and P0 commute, i.e.

P−1P0 = P0P−1 (3.1)

with probability one.Put

Φ(n, ω) =

0∏i=n−1

(PiG−1i BiPi−1)(ω) =

0∏i=n−1

(PiG−1i Bi)(ω) if n > 0,

(P−1P0)(ω) if n = 0,

−1∏i=n

(PiG−1i AiPi+1)(ω) =

−1∏i=n

(PiG−1

i Ai)(ω) if n < 0.

(3.2)

We are now in a position to give a fundamental expression in random dynamictheory, called cocycle property (cf. L. Arnold [1]):

Theorem 3.1. For any m, n ∈ Z the following relation holds:

Φ(n + m, ω) = Φ(n, θmω) · Φ(m, ω). (3.3)

Proof. If m, n < 0 or m, n > 0 the relation (3.3) follows from properties of randommatrix products.

We consider the case m = 0. Since Q0 is a projector onto kerB0,

Φ(1, ω)Φ(0, ω) = P0G−10 B0P−1P0 = P0G

−10 B0P0

= P0G−10 B0(I − Q0) = P0G

−10 B0 = Φ(1, ω).

Hence Φ(n, ω)Φ(0, ω) = Φ(n, ω) for all n > 0. The case n < 0 is similar becauseQ−1 is the projector onto kerA−1.

On the other hand, A0Φ(1, ω) = B0Φ(0, ω). Therefore, by multiplying both sidesof this relation with T 0Q0G

−10 we obtain

T 0Q0G−10 A0Φ(1, ω) = T 0Q0G

−10 B0Φ(0, ω) = T 0Q0P0Φ(0, ω) = 0.

Noting that Q1 = T 0Q0G−10 A0 we have Q1Φ(1, ω) = 0, which leads to Φ(0, θω)·

Φ(1, ω) = Φ(1, ω). This means that (3.3) is true for n = 0, m = 1. Hence

Φ(0, θmω)Φ(m, ω) = Φ(0, θ(θm−1ω))Φ(1, θm−1ω)Φ(m − 1, ω) = Φ(m, ω)

for n = 0 and m ≥ 1. The case n = 0, m < 0 is similar.For n = −1, m = 1 we have

Φ(−1, θω)Φ(1, ω) = P0G−10 A0Φ(1, ω) = P0G

−10 B0Φ(0, ω)

= P0P0Φ(0, ω) = Φ(0, ω).

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Suppose that n < 0 < m.

Φ(n, θmω)Φ(m, ω) = Φ(n + 1, θm−1ω)Φ(−1, θmω) · Φ(1, θm−1ω)Φ(m − 1, ω)

= Φ(n + 1, θm−1)Φ(0, θm−1ω)Φ(m − 1, ω)

= Φ(n + 1, θm−1ω)Φ(m − 1, ω).

By induction we obtain

Φ(n, θmω) · Φ(m, ω) =

Φ(0, θn+mω)Φ(n + m, ω) = Φ(n + m, ω) if n + m > 0,

Φ(0, ω)Φ(0, ω) = Φ(0, ω) if n + m = 0,

Φ(n + m, ω)Φ(0, ω) = Φ(n + m, ω) if n + m < 0.

The case n > 0 > m is proved similarly by noting that P−1P0 = P0P−1, thusTheorem 3.1 is proved.

Denote by Xn(z(ω), ω) the solution of (2.1) satisfying X0(z(ω), ω) = z(ω). It isclear that

Xn(x(ω), ω) = Φ(n, ω)x with x(ω) = (P−1P0)(ω)x.

Example 3.2. Consider the initial value problemXn+1 = B(θnω)Xn,

X0 = x ∈ Rm, n ∈ Z.

Assume that rankB = r is nonrandom constant. Let Q(0) = diag(0, Im−r). PutQ0 = 0, Q0 = V Q(0)V

and T = V (θ)V

where B = U

(Σ 0

0 0

)V

is a singular

value decomposition of B. Suppose that G0 = B + T Q0 = B + V (θ)Q(0)V

isnonsingular with probability 1. We see that Q0 = V Q(0)V (θ−1)G

−1

−1; P0 = I − Q0

and then Xn(ω, x) = Φ(n, ω)x with

Φ(n, ω) =

(Bn−1 · Bn−2 · · ·B0)(ω) if n > 0,

P0(ω) if n = 0,

−1∏i=n

(PiG−1

i )(ω) if n < 0.

4. Lyapunov Exponents — Multiplicative Ergodic Theorem

Firstly, we recall some notions of generalized inverse of matrices. Let M be anm × m- matrix. We say a matrix X is generalized inverse of M if the followingrelations hold:

(i) MXM = M, (ii) XMX = X, (iii) (MX) = MX, (iv) (XM) = XM.

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In [2] it is proved that such a matrix X , denoted by M+, exists uniquely. If M

is nonsingular, then M+ = M−1.

For any number α we write

α+ =

0 if α = 0,

α−1 if α = 0.

By this notation we have[diag(α1, α2, . . . , αm)

]+ = diag(α+1 , α+

2 , . . . , α+m).

Moreover, if M is a diagonal-block matrix then[diag(M1, M2, . . . , Mm)

]+ = diag(M+1 , M+

2 , . . . , M+τ ).

Using the Jordan form of matrices we see that

Lemma 4.1. λ is an eigenvalue with multiplier d of the matrix M iff λ+ is aneigenvalue with multiplier d of the generalized inverse M+.

Lemma 4.2. For any matrix M we have (MM)+ =(M)+M+ and ker(MM)+

coincides with ker MM.

We are now in position to study Lyapunov exponents of solutions (Xn) of (2.1).Suppose that θ is an ergodic transformation on (Ω,F , P ) and the following conditionis satisfied.

Hypothesis 4.3.

ln ‖P0G−10 B0‖ ∈ L1(Ω,F , P ) and ln ‖P0G

−10 A0‖ ∈ L1(Ω,F , P ). (4.1)

We note that this assumption is independent of the choice of T and Q.Since (Φn) is the product of ergodic stationary matrices PnG−1

n Bn for n > 0and PnG−1

n An for n < 0, by [6] we get

(i) Under the assumption (4.1), there exist the limits

limn→∞(Φ(n, ω)Φ(n, ω))1/2n =: ∆(ω) (4.2)

and

limn→−∞(Φ(n, ω)Φ(n, ω))1/2|n| =: ∆(ω).

(ii) Let 0 <eλ1 < eλ2 < · · · <eλτ be the different nonzero eigenvalues of ∆ andλ0 =−∞. We denote U0 = ker ∆(ω), and Ui, i = 1, 2, . . . , τ the eigenspacewith multipliers di = dimUi corresponding to the eigenvalue eλi . Then,τ ; di, λi, i = 1, 2, . . . , τ are nonrandom constants. Let Vk = U0 ⊕ U1 ⊕ · · · ⊕Uk, k = 0, 1, 2, . . . , τ such that

0 ⊂ V0 ⊂ V1 ⊂ · · · ⊂ Vτ = Rm

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defines a filtration of Rm. For each x ∈ R

m the Lyapunov exponent

λ(ω, x) = limn→∞

1n

ln ‖Φ(n, ω)x‖ (4.3)

exists and

λ(ω, x) = λk ⇐⇒ x ∈ (Vk\Vk−1)(ω)

for k = 1, 2, . . . , τ . Furthermore, λ(ω, x) = −∞ ⇔ x ∈ V0 and the spaces Vi

are invariant in the sense

Φ(n, ω)Vi(ω) ⊂ Vi(θnω) (4.4)

for any n ≥ 0; i = 0, 1, 2, . . . , τ .(iii) A similar result can be formulated for the limits with n → −∞ . That is, let

0 < eλτ < eλτ−1 < · · · < eλ1 be the different nonzero eigenvalues of ∆, λτ+1 =−∞. Denote Uτ+1 = ker ∆(ω) and Uτ , . . . ,U1 the corresponding eigenspaceswith multipliers di = dimU i, i = 1, 2, . . . , τ . Then τ ; di, λi, i = 1, 2, . . . , τ arenonrandom constants. Let Vk = Uτ+1 ⊕ Uτ ⊕ · · · ⊕ Uk, k = τ + 1, τ , . . . , 1,

such that

0 ⊂ Vτ+1 ⊂ Vτ ⊂ · · · ⊂ V1 = Rm (4.5)

defines another filtration of Rm. For each x ∈ R

m the Lyapunov exponent

λ(ω, x) = limn→−∞

1|n| ln ‖Φ(n, ω)x‖

exists and

λ(ω, x) = λk ⇔ x ∈ (Vk \ Vk+1)(ω)

for k = 1, 2, . . . , τ , and λ(ω, x) = −∞ ⇔ x ∈ Vτ+1. Moreover, the spaces V i isinvariant in the sense

Φ(n, ω)Vi(ω) ⊂ V i(θnω), ∀ n < 0 (4.6)

for i = 1, 2, . . . , τ + 1.

Denote by π0 the projector P−1P0 and πn = π0(θn). We note that under theassumption (4.1), the ith nonzero eigenvalue (in decreasing order for n < 0 andincreasing order for n > 0) of (Φ(n, ω)Φ(n, ω))1/2n converges to the ith one of∆(ω)’s. Moreover, for any n, from (3.3) we see that (Φ(n, ω)Φ(n, ω))x = 0 if andonly if π0x = 0. Hence, ker π0 = ker ∆ = ker ∆ which implies that V0 = Vτ+1.Furthermore,

πnΦ(n, ω) = Φ(n, ω)π0(ω).

Indeed, from (3.3) we have π0(ω) = Φ(−n, θnω)Φ(n, ω) which follows that

Φ(n, ω)π0(ω) = Φ(n, ω)Φ(−n, θnω)Φ(n, ω)

= Φ(n, θ−nθnω)Φ(−n, θnω)Φ(n, ω) = πn(ω)Φ(n, ω).

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Therefore, by (4.4) we conclude that Φ(n, ω)π0(ω)Vi(ω) = πn(ω)Φ(n, ω)Vi(ω) ⊂πn(ω)Vi(θnω) for P -almost ω ∈ Ω. Further, dim Φ(n, ω)Vi(ω) = dimπn(ω)Vi(θnω)which implies

Φ(n, ω)π0(ω)Vi(ω) = πn(ω)Vi(θnω), ∀ n ≥ 0.

Hence, for i = 1, 2, . . . , τ ,

Φ(n, ω)π0(ω)(Vi\Vi−1)(ω) = πn(ω)(Vi(θnω)\Vi−1(θnω)) ∀ n ≥ 0, (4.7)

λ(x, ω) = λi ⇔ x ∈ Vi(ω)\Vi−1(ω), a.s. ω.

For n < 0, from (4.7) we have

Φ(−n, θnω)π0(θn(ω))(Vi(θnω)\Vi−1(θnω)) = π0(ω)(Vi(ω)\Vi−1(ω)).

Therefore,

Φ(n, ω)π0(ω)(Vi(ω)\Vi−1(ω)) = Φ(n, ω)Φ(−n, θnω)πn(ω)(Vi(θnω)\Vi−1(θnω))

= πn(ω)(Vi(θnω)\Vi−1(θnω)).

Thus, (4.7) is true for n ∈ Z.Similarly,

Φ(n, ω)π0(V i\V i+1)(ω) = πn(V i(θnω)\V i+1(θnω)) ∀ n ∈ Z, (4.8)

λ(x, ω) = λi ⇔ x ∈ Vi(ω)\Vi+1(ω).

In the case where π0 is symmetric, by (3.3) and the relations

Φ(−n, θnω)Φ(n, ω)Φ(−n, θnω) = Φ(−n, θnω)πn = Φ(−n, θnω),

Φ(n, ω)Φ(−n, θnω)Φ(n, ω) = Φ(n, ω)π0 = Φ(n, ω),

it follows that

Φ(−n, θnω) = (Φ(n, ω))+.

Therefore, by virtue of Lemma 4.2 we have

Φ(−n, θnω)Φ(−n, θnω) =(Φ(n, ω)Φ(n, ω)

)+.

Hence, by Lemma 4.1 we can use the same argument as in [1] to obtain

τ = τ, λi = −λi, di = di, i = 1, 2, . . . , τ. (4.9)

In fact, this property is valid without the symmetry of π0. Since π0 is a pro-jector, we can find a random variable V (ω) with values in Gl(Rm) such thatV (ω)π0V

−1(ω) = diag(I, 0). Put

Ψ(n, ω) = V (θnω)Φ(n, ω)V −1(ω), n ∈ Z, ω ∈ Ω.

It is easy to see that

Ψ(n + m, ω) = Ψ(n, θm(ω))Ψ(m, ω), Ψ(0, ω) = diag(I, 0).

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This means that Ψ(n, ω) is a cocycle with symmetric projector diag(I, 0). Moreover,

Ψ(−n, ω) = Ψ(n, θ−nω)+. (4.10)

However, the integrable conditions of this cocycle are not satisfied. Therefore, weare unable to immediately deduce the conclusion (4.9).

It notes that rankΦ(n, ω) =: d is a nonrandom constant. Indeed, from the rela-tions rank (MN) ≤ minrankM, rankN and Φ(n, ω) = Φ(n−m, θmω)Φ(m, ω), itfollows that rankΦ(n, ω) ≤ rankΦ(m, ω) for all m, n ∈ Z. Therefore rankΦ(n, ω) =rankΦ(0, ω) for all n ∈ Z. On the other hand, Φ(0, ω) = Φ(1, θ−1ω)Φ(−1, ω)which implies that rankΦ(0, ω) ≤ rankΦ(1, θ−1ω) = rankΦ(0, θ−1ω). Similarly,rankΦ(0, ω) ≤ rankΦ(0, θω). Hence, rankΦ(0, ω) = rankΦ(0, θω). This means thatrankΦ(0, ω) is an invariant function. By ergodicity of θ it follows that rankΦ(0, ω)is constant.

Remark 4.4. It is easy to see that dimJf = dimSn = rankAn for all n > 0 anddimJb = dimSn = rankBn for all n < 0. However, there is no relation betweenrankΦ(0, ω) and rankAn, rankBn.

We denote by Λk the measurable function successively defined by

Λ1(ω) + · · · + Λk(ω) = limn→∞

1n

ln ‖ ∧k Φ(n, ω)‖, k = 1, . . . , d.

From [1] we get

Λk(ω) = limn→∞

1n

ln δk(Φ(n, ω)), k = 1, . . . , d,

where δk(Φ(n, ω)) are the nonzero singular values of Φ(n, ω), and

Λ1(ω) ≥ Λ2(ω) ≥ · · · ≥ Λd(ω).

We see that λ1 < · · · < λτ are the different numbers in the sequence Λ1 ≥ Λ2 ≥· · · ≥ Λd (Λk = const in the ergodic case) and di is the frequence of appearanceof λi in this sequence. Similarly, λτ < · · · < λ1 are the different numbers in thesequence Λ−

1 ≥ Λ−2 ≥ · · · ≥ Λ−

d , where

Λ−k (ω) = lim

n→∞1n

ln δk(Φ(−n, ω)), k = 1, . . . , d.

We have

limn→∞

1n

ln ‖Φ(n, ω)‖ = Λ1(ω) P -a.s.

Therefore, due to P -preserving property of the transformation θ we get

1n

ln ‖V (θnω)‖ +1n

ln ‖Φ(n, ω)‖ +1n

ln ‖V −1(ω)‖ −→ Λ1(ω) in probability,

− 1n

ln ‖V −1(θnω)‖ +1n

ln ‖Φ(n, ω)‖ − 1n

ln ‖V (ω)‖ −→ Λ1(ω) in probability.

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Hence, by using the inequalities

− ln ‖V −1(θnω)‖ + ln ‖Φ(n, ω)‖ − ln ‖V (ω)‖≤ ln ‖Ψ(n, ω)‖≤ ln ‖V (θnω)‖ + ln ‖Φ(n, ω)‖ + ln ‖V −1(ω)‖,

it follows that1n

ln ‖Ψ(n, ω)‖ → Λ1(ω) in probability as n → +∞. (4.11)

Similarly, we can also prove that for k = 1, . . . , d

1n

ln ‖ ∧k Ψ(n, ω)‖ → Λ1(ω) + · · · + Λk(ω) in probability as n → +∞. (4.12)

Hence,

1n

ln δi(Ψ(n, ω)) −→ Λi(ω) in probability as n → +∞, i = 1, . . . , d.

This relation implies that

1n

ln δi(Ψ(n, θ−nω)) −→ Λi(ω) in probability as n → +∞, i = 1, . . . , d. (4.13)

By the same argument as above we obtain

1n

ln δi(Ψ(−n, ω)) −→ Λ−i (ω) in probability as n → +∞, i = 1, . . . , d. (4.14)

From (4.10) we get

δi(Ψ(−n, ω)) = 1/δd+1−i(Ψ(n, θ−nω)).

Using (4.13) and (4.14) we obtain

Λ−i = −Λd+1−i.

In particular λi = −λi, di = di, τ = τ .

We now show that

Pω : V i(ω) ∩ Vi−1(ω) = V0(ω) = 1

for any i = 1, 2, . . . , τ . Suppose on the contrary,

K1 := ω : V i ∩ (Vi−1(ω) \ V0) = ∅ then PK1 = β > 0. (4.15)

By definition of Vi−1 and Vi we see that for n > 0 the sets

K2(n) := ω : ‖Φ(n, ω)x‖ ≤ exp((λi−1 + δ)n) ‖x‖, ∀ x ∈ Vi−1,K3(n) := ω : ‖Φ(−n, ω)y‖ ≤ exp((−λi + δ)n) ‖y‖, ∀ y ∈ V i,

(4.16)

have the following properties

PK2(n) ≥ 1 − β

4, PK3(n) ≥ 1 − β

4(4.17)

if δ = 4−1|λi − λi−1| for n > N(β).

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On the other hand, θ is P -preserving transformation then

Pθ−nK3(n) = Pω : ‖Φ(−n, θnω)y‖ ≤ exp((−λi + δ)n).‖y‖∀y ∈ V i(θnω) = PK3(n) ≥ 1 − β

4.

It is clear that for all n > N(β)

PK2(n) ∩ θ−nK3(n) ≥ 1 − β

2.

We now take ω ∈ K1(n) ∩ K2(n) ∩ θ−nK3(n) (it is evident PK1(n) ∩ K2(n) ∩θ−nK3(n) > β

2 ). If y ∈ (Vi−1 ∩ Vi) \ V0), then π0y = 0. By the definition of K2(n)and θ−nK3(n) we have

‖π0y‖ = ‖Φ(−n, θnω)Φ(n, ω)y‖ ≤ exp((λi−1 − λi + 2δ)n) ‖π0y‖,that is a contradiction because this relation is impossible.

Summing up we obtain

Theorem 4.5. (MET for implicit linear difference equation) Suppose that

ln ‖P0G−10 B0‖, ln ‖P0G

−10 A0‖ ∈ L1.

Then,

(i) λ1, λ2, . . . , λτ and d1, d2, . . . , dτ are nonrandom numbers.(ii) For any i = 1, 2, . . . , τ, the set Vi ∩ V i is invariant.(iii) There exist subspaces W0 = V0, W1, . . . , Wτ such that

Rm = ⊕τ

i=0Wi, Wi ⊕ V0 = Vi ∩ V i, dim Wi = di, i = 1, 2, . . . , τ, (4.18)

and for any x ∈ Wi\0lim

n→±∞1n

ln ‖Φ(n, ω)x(ω)‖ = λi = limn→±∞

1n

ln ‖Xn(x(ω), ω)‖,

where x = P−1P0x.

Proof. Only the relation (4.18) is to be proved. Assume that Wi⊕V0 = Vi∩V i. Bydefinition, dimVi−1 +dimV i = m+dimV0. Therefore, dimWi = dimVi +dimV i −dimV0 − m = di. Thus we have (4.18).

Example 4.6. We give a simple example with A(·) and B(·) depending on a sta-tionary process ξ(n) = ξ(θn). Let A(ω) = A(ξ(ω)) and B(ω) = B(ξ(ω)) (there isan abuse of only notations used here), where,

A(+) :=

1 0 00 1 10 0 0

, A(−) =

−1 0 −10 1 00 0 0

,

B(+) :=

0 0 00 1 −1

−1 0 1

, B(−) :=

0 1 10 1 −10 0 1

.

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By calculations we obtain

Q0(++) =

0 0 0−1 0 −1

1 0 1

, Q0(+−) =

0 0 0−1 0 −1

1 0 1

,

Q0(−+) =

0 0 −10 0 00 0 1

, Q0(−−) =

0 0 −10 0 00 0 1

.

Therefore,

Jf(ω) =

expand(0, 1, 0); (1, 0, 1) if ξ(ω) = +,

expand(0, 1, 0); (1, 0, 0) if ξ(ω) = −.

Similarly,

Q0(++) =

1 0 0−1 0 0−1 0 0

, Q0(−+) =

−1 −1 −11 1 11 1 1

,

Q0(+−) =

1 0 00 0 00 0 0

, Q0(−−) =

1 1 10 0 00 0 0

.

Hence,

Jb(ω) =

expand(0, 1, 0); (0, 0, 1) if ξ(ω) = +,

expand(−1, 0, 1); (−1, 1, 0) if ξ(ω) = −.

Thus, the solution Xn of Eq. (2.1) exists on the whole Z if X0 ∈ expand(0, 1, 0)with ξ(ω)= + and X0 ∈ expand(−1, 1, 0) with ξ(ω) = −.

It is easy to see that the matrix PnG−1n Bn := Hθn+1ω,θnω,θn−1ω, where

H+++ =

0 0 0−2 1 10 0 0

, H++− =

0 0 0−1/2 1 −1/2

0 0 0

,

H−+− =

0 0 0−1/2 1 −1/2

0 0 0

, H+−+ =

0 −1/2 −1/20 1 10 −1/2 −1/2

,

H+−− =

0 −1/2 00 1 00 −1/2 0

, H−++ =

0 0 0−2 1 10 0 0

,

H−−− =

0 −1 00 1 00 0 0

, H−−+ =

0 −1 −10 1 10 0 0

.

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Since rank (H)= 1, it is clear that for any n, the largest eigenvalue λ(n)1 of

Φ(n)Φ(n) equals to trace(Φ(n)Φ(n)) but it is difficult to compute trace(Φ(n)

Φ(n)). Suppose that (ξ(n)) is an i.i.d. sequence with P (ξ(n)= +)=0.4, P (ξ(n) =−) = 0.6. A simulation program on the Maple software allows us to estimate theunique finite Lyapunov exponent of this system λ[π0x] ≈ 0.098 for any x ∈ R

3 suchthat π0x = 0.

5. Discussion

So far we are able to solve the initial value problem for implicit difference equationsonly in case where rankA(ω) and rankB(ω) are nonrandom. Moreover, an index-1tractable of the pencil matrices (A, B) is required. It is interesting if we can freethese assumptions or how can we define a higher index of (2.1).

Acknowledgments

The authors would like to extend their appreciations to the anonymous referee(s)for his very helpful suggestions which greatly improve this paper.

References

1. L. Arnold, Random Dynamical Systems (Springer-Verlag, 1998).2. L. S. Campbell and D. Meyer, Generalized Inverses of Linear Transformations (Dover,

1991).3. P. K. Anh, N. H. Du and L. C. Loi, On linear implicit non-autonomous systems of

difference equations, J. Diff. Eqns. Appl. 8 (2002) 1085–1105.4. H. Furstenberg and Y. Kiffer, Random matrix products and measures on projective

spaces, Isr. J. Math. 46 (1983) 12–32.5. I. Ya. Goldsheid and G. A. Margulis, Lyapunov exponents of random matrices product,

Usp. Mat. Nauk 44 (1989) 13–60.6. V. M. Gundlach and O. Steinkamp, Product of random rectangular matrices, Math.

Nachr. 212 (2000) 54–76.7. R. Marz, On linear differential algebraic equations and linearization, Appl. Numer.

Math. 18 (1995).8. A. N. Shiryaev, Probability, 2nd edn. (Springer-Verlag, 1995).

DEGENERATE COCYCLE WITH INDEX-1 AND LYAPUNOV EXPONENTS

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