On the spectra of generalized Fibonacci and Fibonacci-like operators
aa r X i v : . [ m a t h . SP ] A p r On the spectra of generalized Fibonacci and Fibonacci-likeoperators
Ivan Slapniˇcar ∗ April 24, 2017
Abstract
We analyze the spectra of generalized Fibonacci and Fibonacci-like operators inBanach space l . Some of the results have application in population dynamics. Let l denote the Banach space of all real sequences x def = ( x , x , x , · · · ) such that k x k def = P | x k | < ∞ . Let H : l → l be a linear operator on l . The resolvent set of H , ρ ( H ) isthe set of all complex numbers λ such that the operator λI − H has a bounded inverse,where I : l → l is the identity operator. The set σ ( H ) def = C \ ρ ( H ) is the spectrumof H . The spectrum is further subdivided into three mutually disjoint parts, the pointspectrum σ p ( H ), the continuous spectrum σ c ( H ) and the residual spectrum σ r ( H ). Thepoint spectrum is the set of all λ ∈ C such that λI − H has no inverse. As in the finitedimensional case, such λ are also called eigenvalues and the corresponding non-zero vectors x ∈ l , such that ( λI − H ) x = 0 are called eigenvectors. The continuous spectrum is theset of all λ not in ρ ( H ) or σ p ( H ) for which the range of λI − H is dense in l . The residualspectrum is the set of all λ in σ ( H ) which are not in σ p ( H ) or σ c ( H ). The spectral radiusof H is r σ ( H ) def = sup λ ∈ σ ( H ) | λ | . (1)The operator H has a matrix representation H in the standard basis e ik def = δ ik , where δ ik is the Kronecker symbol.We shall also use two standard results: first, if the operator H is bounded or closed andhas a matrix representation H , then the transpose matrix H t is the matrix representationof the operator H t : l ∞ → l ∞ and (see e.g. [TaLa80], [Gol66, Corollary II.5.3] or [Hal55,Theorems 3.2 and 3.3]) σ p ( H t ) ⊆ σ p ( H ) ∪ σ r ( H ) , σ r ( H ) ⊆ σ p ( H t ) . (2) ∗ University of Split, Faculty of Electrical Engineering, Mechanical Engineering and Naval Architecture,R. Boskovica b.b., 21000 Split, Croatia, e-mail: [email protected]. Author acknowledges the grantnumber 023-0372783-1289 of the Ministry of Science, Education and Sports of the Republic of Croatia andthe grant FP7 People IEF “MATLAN” of the European Commission. H is bounded, then (see for example [TaLa80, (3-5)]) r σ ( H ) = lim k →∞ k H k k /k . (3)Our aim is to classify spectra of two classes of generalized Fibonacci and Fibonacci-like operators. For the first class of operators their spectral radii are expressed in termsof largest real positive roots of certain polynomials and the coefficients of their powersbehave like generalized Fibonacci sequences, as we shall see in section 2.The second class of operators, which also has applications in mathematical biology, isanalyzed in a similar manner in section 3. Let the linear operator F n : l → l be defined by( x , x , x , · · · ) → ∞ X k = n +1 x k , x , x , x , · · · ! , n = 1 , , , . . . (4)Each F n is bounded and its matrix representation in the standard basis is F n = n z }| { · · · · · · · · · · · · ... ... . . . ... ... ... ... ... ... · · · · · · · · · ... ... ... ... ... . . . ... ... ... . . . , (5)Following the analysis of the spectrum of F by Halberg [Hal55], the spectrum of F n is classified in several steps which are summarized as follows:1. first, by solving the equation ( λI − F n ) x = 0 , x = 0 , (6)we show that the point spectrum is σ p ( F n ) = { λ ∈ C : λ n +1 − λ n − , | λ | > } , (7)2. second, by solving the equation( λI − F n ) x = y, x = 0 , (8)we compute the inverse ( λI − F n ) − and show that the resolvent set consists of all λ such that | λ | > σ p ( F n ), that is, ρ ( F n ) = { λ ∈ C : | λ | > , λ n +1 − λ n − = 0 } , (9)2. third, we analyze the transposed operator F tn and show that σ p ( F tn ) = { λ ∈ C : | λ | ≤ , λ = 1 } , (10)which, together with (2), implies that the residual spectrum of F n is σ r ( F n ) = { λ ∈ C : | λ | ≤ , λ = 1 } . (11)4. Finally, since the spectrum of F n is closed, is also contains the point λ = 1. Sincethis point is neither in the point spectrum nor in the residual spectrum, it must bein the continuous spectrum, that is σ c ( F n ) = { } . (12)We proceed with the detailed analysis of each step. Step 1.
The equation (6) can be written as0 = λx − x n +1 − x n +2 − x n +3 − · · · ,x = λx ,x = λx , ... (13) x k = λx k +1 , ...Since λ = 0 implies x = 0, zero is not an element of σ p ( F n ). If λ = 0, by applying (13)recursively, we have x k +1 = 1 λ x k = 1 λ x k − = 1 λ x k − = · · · = 1 λ k x , k ≥ . (14)Thus x = x (cid:0) λ λ · · · λ k · · · (cid:1) t (15)and k x k = | x | X | λ | k . (16)If | λ | ≤
1, then k x k = ∞ , so x / ∈ l . If | λ | >
1, then k x k = | x | | λ | / ( | λ | − λx − x n +1 − x n +2 − x n +3 − · · · = λ x − λ n x − λ n +1 x − λ n +2 x − · · · = x (cid:20) λ − λ n (cid:18) λ + 1 λ + 1 λ + · · · (cid:19)(cid:21) = x (cid:18) λ − λ n − λ (cid:19) = x λ n +1 − λ n − λ n − ( λ − . x = 0, we conclude that σ p ( F n ) consists of those roots of the polynomial p n +1 ( λ ) def = λ n +1 − λ n − | λ | >
1, as stated in (7). Since p n +1 (1) = − < p ′ n +1 ( λ ) > λ ∈ R , λ ≥
1, that is, p n +1 is strictlyincreasing for λ >
1, we conclude that F n has exactly one real eigenvalue larger than one.Let us denote this eigenvalue by λ max ( F n ). By Ostrovsky’s theorem [Pra04, Theorem1.1.4, p. 3], λ max ( F n ) is the unique positive root of p n +1 ( λ ) and the absolute values of allother roots are strictly smaller. Consequently, all other eigenvalues of F n are in absolutevalue strictly smaller than λ max ( F n ) which, in turn, implies r σ ( F n ) = λ max ( F n ) . (18)Figure 1 shows σ p ( F n ) for various values of n . Step 2.
The equation (8) can be written as y = λx − x n +1 − x n +2 − x n +3 − · · · ,x = 1 λ ( x + y ) ,x = 1 λ ( x + y ) , (19)... x k +1 = 1 λ ( x k + y k +1 ) , ...By setting u = ∞ X k = n +1 x k , v = ∞ X k = n +1 y k , and using (19), we have u = 1 λ x n + 1 λ u + 1 λ v, y = λx − u. After rearranging, we have u = 1 λ − x n + v ) . Thus, y = λx − λ − x n + v ) . (20) These roots are the eigenvalues and the vectors x defined by (15) are the corresponding eigenvectors. Figure 1: The point spectra σ p ( F n ) for various values of n .By recursively applying (19), we have x = 1 λ ( x + y ) ,x = 1 λ ( x + y ) = 1 λ x + 1 λ y + 1 λ y , ... x k +1 = 1 λ ( x k + y k +1 ) = 1 λ k x + 1 λ k y + 1 λ k − y + 1 λ k − y + · · · + 1 λ y k +1 , (21)...Inserting x n into (20) gives y = λ x − λ − (cid:18) λ n − x + 1 λ n − y + 1 λ n − y + · · · + 1 λ y n + v (cid:19) , x gives x = 1 λ n +1 − λ n − (cid:18) λ n − ( λ − y + y + λ y + λ y + · · · + λ n − y n + λ n − v (cid:19) . By inserting this into (21) we have x = ( λI − F n ) − y = 1 λ n +1 − λ n − A + B ) y where the matrix representations of A and B are given by A = ( λ − λ n − λ λ λ · · · λ n − λ n − λ n − · · · ( λ − λ n − λ λ λ · · · λ n − λ n − λ n − · · · ... ... ... ... ... ... ... ... ... · · · ( λ − λ λ n − λ n − λ n − · · · λ λ λ · · · ( λ − λ n − λ n − λ n − · · · λ λ · · · ( λ − λ λ n λ n − λ n − · · · λ λ λ λ · · · ... ... ... ... ... . . . ... ... ... . . . , and B = · · · λ · · · λ λ · · · λ λ λ · · · ... ... ... ... ... . . . , respectively. Obviously, for | λ | > k A k < ∞ and k B k < ∞ . Thus, for | λ | > λ not being the root of λ n +1 − λ n −
1, the operator λI − F n has a bounded inverse,so the resolvent set of F n is given by (9). Step 3.
The point spectrum of the transposed operator F tn consists of all λ ∈ R suchthat ( λI − F tn ) x = 0 , x = 0 , k x k ∞ < ∞ . This is equivalent to x = λ x ,x = λ x = λ x , ... x n +1 = λ x n = λ n x ,x n +2 = λ x n +1 − x = ( λ n +1 − x ,x n +3 = λ x n +2 − x = ( λ n +2 − λ − x , ... x k = ( λ k − − λ k − n − − λ k − n − − · · · − λ − x , ... Next row of A is obtained by dividing the previous row by λ . x k = (cid:18) λ k − − λ k − n − − λ − (cid:19) x . For | λ | ≤ λ = 1 we have | x k | < (cid:18) | λ − | (cid:19) | x | , which implies k x k ∞ < ∞ . For λ = 1 we have x = x ,x = x , ... x n +1 = x ,x n +2 = 0 ,x n +3 = − x ,x n +4 = − x , ... x k = − ( k − n − x , ...so k x k ∞ = ∞ . We conclude that the point spectrum of F tn is given by (10). This, in turn,implies (11) and (12) as described before. In this section we describe the relationship between operators F n and generalized Fibonaccisequences. A generalized Fibonacci sequence { f ( n ) } is defined by f ( n )1 = 1 , f ( n )2 = 1 , · · · , f ( n ) n +1 = 1 , f ( n ) k = f ( n ) k − + f ( n ) k − n − , k > n + 1 . (22)For n = 1 this definition yields the classical Fibonacci sequence f = 1 , f = 1 , f k = f k − + f k − , k > . (23)7y induction we can prove that the k -th power of the matrix F n from (5) for k > n hasthe form F kn = f ( n ) k − n f ( n ) k − n +1 f ( n ) k − n +2 · · · f ( n ) k − f ( n ) k f ( n ) k f ( n ) k · · · f ( n ) k − n − f ( n ) k − n f ( n ) k − n +1 · · · f ( n ) k − f ( n ) k − f ( n ) k − f ( n ) k − · · · ... ... ... ... ... ... ... ... · · · f ( n )1 f ( n )2 f ( n )3 · · · f ( n ) n f ( n ) n +1 f ( n ) n +1 f ( n ) n +1 · · · · · · · · · · · · · · · ... ... ... ... ... ... ... ... · · · · · · · · · · · · · · · · · · · · · · · · · · · ... ... ... . . . ... ... ... ... · · · . We conclude that k F kn k = 1 + k X i =1 f ( n ) i . (24)By applying (22) to the terms in parentheses we have2 k X i =1 f ( n ) i = f ( n )1 + · · · + f ( n ) n + (cid:0) f ( n ) n +1 + f ( n ) n +2 + · · · + f ( n ) k − (cid:1) + f ( n ) k + (cid:0) f ( n )1 + f ( n )2 + · · · + f ( n ) k − n − (cid:1) + f ( n ) k − n + f ( n ) k − n +1 + · · · + f ( n ) k = f ( n )1 + f ( n )2 + · · · + f ( n ) n + f ( n ) n +2 + · · · + f ( n ) k + f ( n ) k + f ( n ) k − n + f ( n ) k − n +1 + · · · + f ( n ) k = k X i =1 f ( n ) i − f ( n ) n +1 + f ( n ) k − n + f ( n ) k − n +1 + · · · + f ( n ) k + f ( n ) k . From this, by applying (22) again recursively, we obtain k X i =1 f ( n ) i = f ( n ) k − n + f ( n ) k − n +1 + · · · + f ( n ) k + f ( n ) k − f ( n ) k − n +1 + · · · + f ( n ) k + f ( n ) k +1 − f ( n ) k − n +2 + · · · + f ( n ) k +1 + f ( n ) k +2 − f ( n ) k + n +1 − . Inserting this into (24) gives k F kn k = f ( n ) k + n +1 (25)8nd from (18) it follows that lim k →∞ (cid:0) f ( n ) k + n +1 (cid:1) /k = λ max ( F n ) . Also, by using standard techniques in analyzing linear recurrence relations with constantcoefficients, we can prove that for all i, j lim k →∞ [ F kn ] i,j [ F kn ] i +1 ,j ≡ lim m →∞ f ( n ) m +1 f ( n ) m = λ max ( F n ) . For example, by setting n = 1 we have for the Fibonacci sequence (23)lim k →∞ ( f k +2 ) /k = r σ ( F ) = 1 + √ , lim k →∞ f k +1 f k = 1 + √ . Now we would like to consider the family of linear operators Γ n : l → l defined by( x , x , x , · · · ) → ρ ∞ X k = n +1 ( k − n ) x k , x , x , x , · · · ! , n = 1 , , , . . . (26)for some real positive ρ . The domain of Γ n isDom Γ n = ( x ∈ l : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k = n +1 ( k − n ) x k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ∞ ) , and its matrix representation in the standard basis is Γ n = n z }| { · · · ρ ρ ρ ρ ρ · · · · · · · · · ... ... . . . ... ... ... ... ... ... · · · · · · · · · ... ... ... ... ... . . . ... ... ... . . . . (27)However, the operator Γ n is not closed as illustrated by the following example. The proof is derived using the fact that f ( n ) l has the form f ( n ) l = α λ l max ( F n ) + P ni =1 α i λ li , where λ max ( F n ) and λ i are the roots of the characteristic polynomial (17), and | λ max ( F n ) | > | λ i | . xample 1 Let us define the sequence { x ( m ) } of vectors in l by m + n − z }| { x ( m ) = (cid:18) · · · m · · · (cid:19) t . Then x ( m ) → (cid:0) · · · (cid:1) t , while Γ n x ( m ) = (cid:0) ρ · · · m · · · (cid:1) t → (cid:0) ρ , · · · (cid:1) t . Although the point spectrum of Γ n is defined and can be computed in a standardmanner (see later), the resolvent set of Γ n is empty, which makes the analysis of Γ n lessinteresting. Instead, we shall consider the family of operators G n : l → l formally definedby G n = D n Γ n D − n , where D n = diag (cid:0) n z }| { , · · · , , ρ, ρ, ρ, ρ, · · · (cid:1) . That is, for n ∈ N the operator G n is defined by( x , x , x , · · · ) → ∞ X k = n +1 x k , x , x , · · · , x n − , ρ x n , x n +1 , x n +2 , x n +3 , x n +4 , · · · ! , and its matrix representation in the standard basis is G n = n z }| { · · · · · · · · · · · · · · · · · · ... ... . . . ... ... ... ... ... ... · · · · · · ρ · · · · · · · · · · · · · · · · · · · · · · · · · · · ... ... ... ... ... ... ... ... . . . ... . (28)Let us define the polynomial q n +1 ( λ ) by q n +1 ( λ ) = λ n +1 − λ n + λ n − − ρ. (29)Similarly as in section 2, the spectrum of G n is classified in several steps which are sum-marized as follows: 10. first, by solving the equation ( λI − G n ) x = 0 , x = 0 , (30)we show that the point spectrum is σ p ( G n ) = { λ ∈ C : q n +1 ( λ ) = 0 , | λ | > } , n ≥ . (31)2. second, by solving the equation( λI − G n ) x = y, x = 0 , (32)we compute the inverse ( λI − G n ) − and show that the resolvent set consists of all λ such that | λ | > σ p ( G n ), ρ ( G n ) = { λ ∈ C : | λ | > , λ / ∈ σ p ( G n ) } , (33)3. third, we analyze the transposed operator G tn and show that σ p ( G tn ) = { λ ∈ C : | λ | ≤ , λ = 1 } , (34)which, together with (2), implies that the residual spectrum of G n is σ r ( G n ) = { λ ∈ C : | λ | ≤ , λ = 1 } . (35)4. Finally, since the spectrum of G n is closed, is also contains the point λ = 1. Sincethis point is neither in the point spectrum nor in the residual spectrum, it must bein the continuous spectrum, that is σ c ( G n ) = { } . (36)The proofs are similar to the ones from section 2, but more tedious. Step 1.
The equation (30) can be written as0 = λx − x n +1 − x n +2 − x n +3 − · · · , (37) x k = λx k +1 , k = 1 , · · · , n − ,ρ x n = λ x n +1 ,k − n + 1 k − n x k = λ x k +1 , k = n + 1 , n + 2 , · · · . Since λ = 0 implies x = 0, zero is not an element of σ p ( G n ). If λ = 0, by applying (37)recursively, we obtain x k = 1 λ k − x , k = 2 , , · · · , n,x k = ρ k − nλ k − x , k = n + 1 , n + 2 , · · · . (38)11his, in turn, implies x = x (cid:0) λ λ · · · λ n − ρλ n ρλ n +1 ρλ n +2 · · · ( k − n ) ρλ k − · · · (cid:1) t , so that k x k = | x | (cid:18) − ( λ ) n − λ + ρλ n (1 + 2 λ + 3 λ + · · · ) (cid:19) . If | λ | ≤
1, then k x k = ∞ , so x / ∈ l . If | λ | >
1, then, by using differentiation of thegeometric series, we have k x k = | x | (cid:18) λ n − λ n − ( λ −
1) + ρλ n λ ( λ − (cid:19) < ∞ , thus, x ∈ l . By inserting (38) into the first equality of (37) and using differentiation ofthe geometric series we have0 = λ x − ρ (cid:18) λ n x + 2 λ n +1 x + 3 λ n +2 x + · · · (cid:19) = x (cid:20) λ − ρλ n (cid:18) λ + 3 λ + 4 λ + · · · (cid:19)(cid:21) = x (cid:18) λ − ρλ n − λ ) (cid:19) . Finally, solving this equation with x = 0 gives (31).We shall now prove that σ p ( G n ) consists of λ max ( G n ), a unique simple real eigenvaluelarger than one and all other eigenvalues have modulus smaller than λ max ( G n ). This alsoimplies r σ ( G n ) = λ max ( G n ) . (39)The proof is based on the ideas from the proof of [Pra04, Theorem 1.1.4, pp. 3-4]. Indeed,if n = 1 then the roots of q ( λ ) are 1 ± √ ρ and the statement holds. For n ≥ q n +1 ( λ ) = λ n − ( λ − − ρ, (40) q ′ n +1 ( λ ) = λ n − [( n + 1) λ − nλ + ( n − . (41)Since q n +1 (1) = − ρ < q ′ n +1 ( λ ) > λ ∈ R , λ >
1, that is, q n +1 ( λ ) is strictlyincreasing for λ >
1, we conclude that q n +1 ( λ ) has exactly one real root larger than oneor, equivalently, that G n has exactly one real eigenvalue larger than one. Let us denotethis eigenvalue by λ max ( G n ). Let z = λ max ( G n ) be some other real or complex eigenvalueof G n and let ζ = | z | >
1. Since z is also the root of q n +1 ( λ ), the relation (40) implies z n − ( z − = ρ, which, in turn, implies | z | n − | z − | = ρ. (42)Since ζ >
1, this implies ζ n − ( ζ − ≤ ρ, q n +1 ( ζ ) = ζ n − ( ζ − − ρ ≤ . (43)Since q n +1 ( λ ) is strictly increasing for λ >
1, and q n +1 ( λ max ( G n )) = 0, we conclude that ζ ≤ λ max ( G n ) and that the equality in (43) holds only if ζ = λ max ( G n ). But, the equalityin (43) and (42) imply | z − | = ζ − , that is, z ∈ R and z = ± ζ = ± λ max ( G n ). The choice z = − λ max ( G n ) is impossible since q n +1 ( − λ max ( G n )) = 0, and the second choice contradicts the assumption z = λ max ( G n ).Therefore, ζ < λ max ( G n ) as desired. Remark 1
Although the above analysis is sufficient for our purposes, by inspecting thepolynomial q n +1 ( λ ) and its derivative from (40) and (41), respectively, we can establishfurther facts about its roots. From (41) we see that the derivative q ′ n +1 ( λ ) has exactlytwo real positive simple roots λ = n − n +1 and λ = 1 and, if n >
2, also the root λ = 0. If n > λ is multiple. Let ρ = 4 λ n − / ( n + 1) . The number of real roots of q n +1 ( λ )in the open interval (0 , λ max ( G n )) is governed by ρ as follows: if ρ > ρ , then there are nosuch roots, if ρ = ρ there is exactly one root equal to λ and if ρ < ρ there are exactlytwo roots, one smaller and one larger than λ . Finally, if n is odd, then q n +1 ( λ ) also hasa simple negative real root. As we have already proved, λ max ( G n ) is the root with strictlymaximal absolute value. Remark 2
It is easy to see that the point spectrum of Γ n from (26) and (27) is equal tothe point spectrum of G n . Step 2.
The equation (32) can be written as y = λx − x n +1 − x n +2 − x n +3 − · · · , (44) x k +1 = 1 λ ( x k + y k +1 ) , k = 1 , · · · , n − ,x n +1 = 1 λ ( ρ x n + y n +1 ) ,x k +1 = 1 λ (cid:18) k − n + 1 k − n x k + y k +1 (cid:19) , k = n + 1 , n + 2 , · · · . By recursively applying the above three equalities, we obtain x k = ( k − n ) ρλ k − x + ( k − n ) ρλ k − y + ( k − n ) ρλ k − y + · · · + ( k − n ) ρλ k − n +1 y n + (45)+ k − n λ k − n y n +1 + k − n λ k − n − y n +2 + k − n λ k − n − y n +3 + · · · ++ k − nk − n − λ y k − + 1 λ y k , k = n + 1 , n + 2 , · · · . For | λ | >
1, by inserting (45) into (44), rearranging, and using differentiation of thegeometric series, we have y = λ x − λ ( λ − (cid:18) ρλ n x + ρλ n y + ρλ n − y + · · · + ρλ y n + 1 λ y n +1 (cid:19) −− α n +2 y n +2 − α n +3 y n +3 − α n +4 y n +4 − · · · , (46)13here α n + k = 1 λ + k + 1 k λ + k + 2 k λ + k + 3 k λ + · · · , k = 2 , , , · · · , and | α n + k | ≤ | λ | (cid:18) ∞ X i =1 k + ik | λ | i (cid:19) ≤ | λ | (cid:18) ∞ X i =1 ( i + 1) 1 | λ | i (cid:19) = | λ | ( | λ | − . (47)Solving for (46) for x gives x = 1 q n +1 ( λ ) (cid:20) λ n − ( λ − y + ρ y + ρ λ y + ρ λ y + · · · + ρ λ n − y n + λ n − y n +1 ++ ¯ α n +2 y n +2 + ¯ α n +3 y n +3 + ¯ α n +4 y n +4 + · · · (cid:21) , (48)where ¯ α n + k = λ n − ( λ − α n + k , and | ¯ α n + k | ≤ | λ | n − ( | λ | + 1) ( | λ | − . (49)By inserting (48) into (44) and (45), we have x = ( λI − G n ) − y = 1 q n +1 ( λ ) ( A + B ) y where the matrix representations of A and B are given by A = ( λ − λ n − ρ ρ λ ρ λ · · · ρ λ n − λ n − ¯ α n +2 ¯ α n +3 · · · ( λ − λ n − ρλ ρ ρ λ · · · ρ λ n − λ n − α n +2 λ ¯ α n +3 λ · · · ( λ − λ n − ρλ ρλ ρ · · · ρ λ n − λ n − α n +2 λ ¯ α n +3 λ · · · ( λ − λ n − ρλ ρλ ρλ · · · ρ λ n − λ n − α n +2 λ ¯ α n +3 λ · · · ... ... ... ... · · · ... ... ... ... · · · ( λ − λ ρλ n − ρλ n − ρλ n − · · · ρ λ λ α n +2 λ n − ¯ α n +3 λ n − · · · ( λ − ρλ n − ρλ n − ρλ n − · · · ρ λ ¯ α n +2 λ n − ¯ α n +3 λ n − · · · ( λ − λ ρλ n − ρλ n − ρλ n − · · · ρλ ¯ α n +2 λ n − ¯ α n +3 λ n − · · · ( λ − ρλ ρ λ n ρ λ n − ρ λ n − · · · ρ λ ρλ ρ ¯ α n +2 λ n ρ ¯ α n +3 λ n · · · ( λ − ρλ ρ λ n +1 ρ λ n ρ λ n − · · · ρ λ ρλ ρ ¯ α n +2 λ n +1 ρ ¯ α n +3 λ n +1 · · · ( λ − ρλ ρ λ n +2 ρ λ n +1 ρ λ n · · · ρ λ ρλ ρ ¯ α n +2 λ n +2 ρ ¯ α n +3 λ n +2 · · · ( λ − ρλ ρ λ n +3 ρ λ n +2 ρ λ n +1 · · · ρ λ ρλ ρ ¯ α n +2 λ n +3 ρ ¯ α n +3 λ n +3 · · · ... ... ... ... · · · ... ... ... ... · · · , B = · · · · · · λ · · · · · · λ λ · · · · · · λ λ λ · · · · · · ... ... ... ... ... · · · ... ... ... ... ... ... ... · · · λ n − λ n − λ n − λ n − · · · λ · · · ρλ n ρλ n − ρλ n − ρλ n − · · · ρλ λ · · · ρλ n +1 ρλ n ρλ n − ρλ n − · · · ρλ λ λ · · · ρλ n +2 ρλ n +1 ρλ n ρλ n − · · · ρλ λ λ λ · · · ρλ n +3 ρλ n +2 ρλ n +1 ρλ n · · · ρλ λ λ λ λ · · · ρλ n +4 ρλ n +3 ρλ n +2 ρλ n +1 · · · ρλ λ λ λ λ λ · · · ... ... ... ... ... · · · ... ... ... ... ... ... ... · · · , respectively. For | λ | >
1, by using differentiation of the geometric series, (49), and theargument used in (47), we have k A k ≤ ∞ and k B k ≤ ∞ . Thus, for | λ | > q n +1 ( λ ) = 0, the operator λI − G n has a bounded inverse and its resolvent set is given by(33). Step 3.
The point spectrum of the transposed operator G tn consists of all λ ∈ R , | λ | ≤
1, such that ( λI − G tn ) x = 0 , x = 0 , k x k ∞ < ∞ . This is equivalent to x k = λ k − x , k = 2 , · · · , n,x n +1 = λ n ρ x ,x n + k = 1 k (cid:18) λ n + k − ρ − λ k − − λ k − − λ k − − · · · − ( k − λ − ( k − (cid:19) x , k ≥ . If λ = 1, we have x n + k = 1 k (cid:20) λ n + k − ρ − (cid:0) λ k − + 2 λ k − + · · · + ( k − λ + ( k − (cid:1) ( λ − ( λ − (cid:21) x = 1 k (cid:18) λ n + k − ρ − λ k − k λ + ( k − λ − (cid:19) x , k ≥ . Therefore | x n + k | ≤ k (cid:18) ρ + 1 + k + k − | λ − | (cid:19) x ≤ (cid:18) ρ + 2 | λ − | (cid:19) x , k ≥ , which implies k x k ∞ < ∞ . For λ = 1 we have x n + k = 1 k (cid:18) ρ − − − − · · · − ( k − − ( k − (cid:19) x = 1 k (cid:18) ρ − ( k − k (cid:19) x , k ≥ , k x k ∞ = ∞ . We conclude that the point spectrum of G tn is given by (34). This, in turn,implies (35) and (36) as described before. Example 2
The lesion forming plant pathogen potato late blight (phytophthora infes-tans) grows radially on a leaf with a constant daily rate. The latency period for a lesionto become infectious is five days, and the sporulating area is infectious for one day. In[PSV05] the epidemic spread of such pathogen is modeled with the infinite dimensionalLeslie matrix of the form of Γ as defined in (27). Further, the upper bound for thespeed of invasion in computed via minimization of the largest eigenvalue λ max (Γ ( s )).From Remark 2 it follows that this eigenvalue is the largest unique positive root of q ( λ )from (29). Here the parameter ρ has the form ρ ( s ) = const × M ( s ) where M ( s ) is somemoment-generating function (for example, M ( s ) = exp( σ s /
2) for the Gaussian kernelor M ( s ) = 1 / (1 − σ s ) for the Laplace kernel). Here Γ ( s ) appears naturally due to thefact that the considered pathogen has a latency period of five days. It is interesting that λ max (Γ ( s )) can be computed analytically: λ max (Γ ) = 13 + 2 / (cid:0) √ ρ + p √ ρ + 729 ρ (cid:1) / + (cid:0) √ ρ + p √ ρ + 729 ρ (cid:1) / · / . The speed of invasion is bounded by v ∗ = min
The author wishes to thank the anonymous referee for the valuablecomments and remarks, in particular for correcting an error in the previous version of themanuscript and for providing Example 1. The author also wishes to thank Jussi Behrndtfrom the Technische Universit¨at Berlin for helpful discussions.
References [Gol66] S. Goldberg, Unbounded linear operators, McGraw-Hill, New York, 1966.[Hal55] C. J. Halberg, Spectral Theory of Linked Operators in l pp