Linear Algebra and its Applications

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Linear Algebra its Applications 432 21) 394 41 Contents lists available at ScienceDirect Linear Algebra its Applications journal homepage: wwwelseviercom/locate/laa On the Perron exponents of

4 November 28 Accepted 23 August 29 Available online 19 September 29 Submitted by H Schneider Keywords: Time-varying linear systems Characteristic exponents This note studies properties of Perron or

1 Linear Algebra its Applications ) Contents lists available at ScienceDirect Linear Algebra its Applications journal homepage: wwwelseviercom/locate/laa On the Perron exponents of discrete linear systems Adam Czorni Department of Automatic Control, Silesian Technical University, ul Aademica 16, Gliwice, Pol A R T I C L E I N F O A B S T R A C T Article history: Received 4 November 28 Accepted 23 August 29 Available online 19 September 29 Submitted by H Schneider Keywords: Time-varying linear systems Characteristic exponents This note studies properties of Perron or lower Lyapunov exponents for discrete time-varying system It is shown that for diagonal systemoforders there are at most 2 s 1 lower Lyapunov exponents By example it is demonstrated that in non-diagonal case it is possible to have arbitrarily many different Perron exponents Finally it is shown that the exponent is almost everywhere equal to the lower Lyapunov exponent of the matrices coefficient sequence 29 Elsevier Inc All rights reserved 1 Introduction Consider the linear discrete time-varying system xn + 1) = An)xn), n, 1) where An) are s-by-s real matrices For the coefficient matrices denote Φm) = Am 1)A) Φ) = I, where I is the identity matrix For an initial condition x the solution of 1) is denoted by xn, x ) so xn, x ) = Φn)x By denote the Euclidean norm in R s the induced operator norm For initial condition x /= Lyapunov exponent λx ) Perron exponent πx ) or the lower Lyapunov exponent of solution xn, x ) are defined as λx ) = lim sup n xn, x ) 1/n address: AdamCzorni@polslpl /$ - see front matter 29 Elsevier Inc All rights reserved doi:1116/jlaa29821

2 A Czorni / Linear Algebra its Applications ) πx ) = lim inf n xn, x ) 1/n, respectively For the matrix sequence Φn) =[φ ij n)] i,j=1,,s the Lyapunov λφ) Perron πφ) exponents are defined as follows λφ) = lim sup n Φn) 1/n πφ) = lim inf n Φn) 1/n, where is any matrix norm The Lyapunov exponents have been investigated in many papers see [1,2] the reference therein) The Perron exponents [6] are defined similarly to Lyapunov exponents but, as we show in this paper, have very different properties Some properties of Perron exponents for continuous time systems have been established in [3,5] 2 Numbers of Perron exponents It is well nown [2] that the set of all Lyapunov exponents of system 1) contains at most s elements, but as it is shown in the next example, that the set of Perron exponents may have more elements Example 1 Consider two dimensional system 1) with [ 5 n ] 3 n 1 2 n for even n An) = [ 4 n 1 3 n ] 5 n 1 4 n for odd n 2 n 1 Then [ 5 n ] 2 Φn) = [ n 3 n ] 4 n for even n for odd n for x 1 = [1 ] T, x 2 = [ 1] T, x 3 = [1 1] T,wehaveπx 1 ) = 3, πx 2 ) = 2, πx 3 ) = 4 The next Theorem shows that the situation from the Example is typical for diagonal system namely we show that diagonal system 1) has at most 2 s 1 Perron exponents Theorem 1 For the diagonal system the set {πx ) : x R s, x /= } has at most 2 s 1 elements In the Proof we will use the following Lemma 2 If a i) ) N, i = 1,, s are sequences of nonnegative numbers such that for certain positive constants c i, i = 1,, s exists the limit lim s c i a i), then the limit exists for each nonnegative constants c i, i = 1,, s, it does not depend on these constants is equal to max lim sup a i),,s

3 396 A Czorni / Linear Algebra its Applications ) Proof Denote α = max lim sup a i) let i be such that the maximum is achieved Fix ε>,,s select such that for all i = 1,, s > we have a i) α + ε or equivalently Then a i) α + ε) m 1 a i ) s c i a i) m 2s α + ε), where m 1 = min c i, m 2 = max c i consequently,,l,,l m1 s that is a i ) α lim sup s c i a i) m 2 s α + ε) c i a i) α + ε By virtue of arbitrary choice of ε> the proof of the Lemma is complete Proof of Theorem 2 If the matrices An) are diagonal so are Φn) Letb i) n, i = 1,, s be the elements on diagonal of Φn) consider the initial condition x =[y 1,, y s ] T /= Then s xn, x ) 2 = y i b i) ) 2 n Let n ) N be an increasing sequence of natural numbers such that πx ) = lim xn, x ) 1/n Consider another initial condition x =[y 1,, y s ]T such that y i = if only if y i = Then we have πx s ) lim inf y ) 2 1/n i bi) n, s but according to Lemma 3 the limit of y ) ) 2 1/n i bi) n exists is equal to the limit of ) ) 2 1/n therefore πx ) πx ) In the same way one can show that πx ) πx ) s y i b i) n It implies that πx ) = πx ) The last fact means that πx ) /= πx ) only if y i = y i /= or y = i y i /= for at least one i = 1,, s Therefore the number of different values of πx ) is less or equal 2 s 1 In the next theorem we show that for general case there are no bounds for the number of different values of Perron exponents

4 A Czorni / Linear Algebra its Applications ) Theorem 3 For each natural number p there exists system 1) with s = 2 p different Perron exponents Proof Fix a natural number p denote by rn) the remainder of division of nby p Define two sequences of real numbers a n = p + 1) n b n = rn) + 1) n rn) + 1)p + 1) n It is easy to calculate that if [ an+1 An) = a n a ] n+1 b a n n + b n+1, 1 then Φn) = [ ] an b n 1 Moreover [ Φm + 1) 1 ] 1 m+ 1 = 2m+ 1) ) 1 2m+ 1) + 1 m [ ] 1 Φn) n, 1 therefore π [ 1] T ) = for = 1,, p 1, π [ 1] T ) = p Properties of Perron exponents For the Lyapunov exponents we have the following well-nown relation max λx ) = λφ) x R s In this paragraph we show a similar relation for the Perron exponents In that purpose we will use ideas proposed by Izobov in [5] for continuous time systems Let us introduce certain additional notation For the matrix sequence Φn) =[φ ij n)] i,j=1,,s we define sets M l,, l = 1,, s of natural numbers as follows { M l = n : max φ ij n) } = φ l n) i,j=1,,s If Φn) /= for all n then for each infinite set M l natural number p we define the set A p) l R s consisting of vectors x =[c 1,, c s ] T for which there exists an infinite sequence m j ) j=1,2,, m j, m j M l such that s c iφ i m j ) φ l m j ) dc)e mj p, for certain positive constant dc) R all j N Lemma 4 The set s s p=1 =1 l=1 Ap) l is of zero Lebesgue measure

5 398 A Czorni / Linear Algebra its Applications ) Proof Fix, l {1,, s} such that M l is infinite for r > natural p introduce additional set B r = { [c 1,, c s ] T A p) l : c i r, i /= l} Moreover for m M l r > denote by M r m) the set of all [c 1,, c s ] T such that s c + φ i m) l c i e m p, c φ l m) i r, i /= l, i /= l The Lebesgue measure of a set K we denote by mesk) We calculate mesm r m)) as follows r r bm) mesm r m)) dc 1 dc s dc l 22r) s 1 e m/p, r r am) where am) = s i /= l c i φ i m) φ l m) e m p, bm) = For the introduced sets we have B r M r m) q<m M l mesm r m)) 2r) s e m p q<m M l m>q s i /= l q c i φ i m) φ l m) + e m p for all r > natural q Therefore the set B r is measurable mesb r ) = for all r > consequently mesa p) l ) mes s ) s p=1 =1 l=1 Ap) l = The next theorem contains the main results of this section Theorem 5 The function π : R s R is almost everywhere constant, there exists x R s such that sup πx) = πx ) x R s πx ) = πφ) 2) 3) Proof If for certain n we have that max i,j=1,,s max i,j=1,,s φ ij n) > for all natural n Denote A = p=1 s =1 for all x 1, x 2 R s \A x 3 R s we have πx 1 ) = πx 2 ) πx 3 ) πx 2 ) φ ij n ) = then the theorem is trivially satisfied Assume s l=1 Ap) l First we will show that For any square s by s matrix Y =[y ij ] i,j=1,,s we have see [4]) Y s max yij, 4) i,j=1,,s therefore from the relation

6 A Czorni / Linear Algebra its Applications ) xn, x 1 ) = xn, x 2 ) + Φn) x 1 x 2 ), we obtain that xn, x 1 ) xn, x 2 ) + s x 1 x 2 max φ ij n) 5) i,j=1,,s By the definition of πx 2 ) there is an increasing sequence n ) N of natural numbers such that πx 2 ) = lim xn, x 2 ) 1/n Without loss of generality, we may assume that the sequence n ) N entirely lies in some M l otherwise, we can always extract a subsequence with this property) Then from 5) weget xn, x 1 ) xn, x 2 ) 1 + s x 1 x 2 φ l n ) ) 6) xn, x 2 ) We also have s xn, x 2 ) φ in )x i) 2, where x 2 = [ x 1) ] T 2,, xs) 2 therefore φ l n ) xn, x 2 ) φ l n ) s φ in )x i) 2 The last inequality together with the definition of the set A the fact x 2 / A imply lim inf by 6) we obtain φ l 1 + s x 1 x 2 n ) ) 1/n = 1, xn, x 2 ) πx 1 ) πx 2 ) 7) Because in the proof of the last inequality we have used only the fact that x 2 / A we also proved πx 3 ) πx 2 ) 8) Moreover we can repeat the same argument as in the proof changing the role of x 1 x 2, therefore πx 2 ) πx 1 ) 9) From 7) 9) we obtain the first two statements of this theorem It remains to show 3) From the obvious relation xn, x ) Φn) x we obtain the first desired inequality max πx) πφ) x Rs We will show that for any initial condition x R s \A, x =[c 1,, c s ] T we have πx ) πφ) 1) By the definition of πx ) there is an increasing sequence n ) N of natural numbers such that πx ) = lim xn, x 2 ) 1/n

7 4 A Czorni / Linear Algebra its Applications ) Because we see that xn, x ) = Φn ) xn, x ) Φn ), πx ) πφ) lim inf = πφ) lim xn, x ) ) 1/n Φn ) ) 1/m, 11) xm, x ) Φm ) where m ) N is a subsequence of n ) N which implements the last liminf Without loss of generality, we may assume that the sequence m ) N entirely lies in some M l otherwise, we can always extract a subsequence with this property) Then, taing into account 4) we obtain lim xm, x ) Φm ) ) 1/m = lim inf s c ) 1/m iφ im ) φ l m ) Suppose that the last lower limit is less then 1 Then there exists a subsequenceq ) N ofm ) N q ) N implements the last lower limit) a positive integer p 1 such that s lim c ) 1/q iφ iq ) < e 1 p 1 φ l q ) Consequently, for any given ε>the inequality s c ) 1/q iφ iq ) < e 1 p 1 + ε 12) φ l q ) is valid starting from some 1 Suppose that it is satisfied for all otherwise we may eliminate finitely many terms of q )Ifε is sufficiently small, then there exists a natural number p 2 such that e 1 p 1 + ε<e 1 p 2 The last inequality together with 12) implies that x A p 2) l, which contradicts the choice of x Therefore s lim inf c ) 1/m iφ im ) 1 φ l m ) This together with 11), implies 1) the proof of the Theorem is complete 4 Conclusions It is shown that for discrete time-varying system with diagonal coefficients of order s there are at most 2 s 1 Perron exponents By example it is demonstrated that for any natural number p there is two dimensional system with exactly p Perron exponents Finally it is shown that the Perron exponent, considered as a function of initial conditions, is almost everywhere, with respect to the Lebesgue measure, equal to the Perron exponent of the sequence of matrices coefficients Acnowledgment This research has been supported by State Committee for Scientific Research under Grant N

8 A Czorni / Linear Algebra its Applications ) References [1] L Arnold, H Crauel, JP Ecmann Eds), Lyapunov Exponents, Lecture Notes in Mathematics, vol 1486, Springer-Verlag, Berlin, Heidelberg, 1991 [2] L Barreira, Ya Pesin, Lyapunov Exponents Smooth Ergodic Theory, Univ Lecture Ser, vol 23, Amer Math Soc, Providence, RI, 22 [3] BF Bylov, RE Vinograd, DM Grobman, VV Nemytsii, Theory of Lyapunov Exponents its Applications to Stability Theory, Moscow, Naua, 1966 [4] RA Horn, CR Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1985 [5] NA Izobov, Asymptotic Characteristic of Linear Wealy Nonlinear Systems, C Sci Phys-Math) Dissertation, Mins, 1967 [6] O Perron, Die ordnungszahlen linearer differentialgleichungssysteme, Math Z 31 4) 193)

Linear Algebra and its Applications

Linear Algebra and its Applications 433 (200) 867 875 Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: www.elsevier.com/locate/laa On the exponential exponents