1.5. The Discrete-Time Case: Linear Difference Equations 21 Let Ek = E(μk),k = 1,...,r, be the real generalized eigenspaces, and suppose that there are 1 ≤ ≤ d distinct moduli |μk| of the eigenval- ues μk. Let λ1 . . . λ be the logarithms of these moduli and order them as λ1 . . . λ , 1 ≤ ≤ r. Define the Lyapunov space of λj as Lj = L(λj) := Ek, where the direct sum is taken over all real generalized eigenspaces Ek associated to eigenvalues μk with λj = log |μk|. Note that j=1 L(λj) = Rd. A matrix A is invertible if and only if 0 is not an eigenvalue. Hence log |μ| is finite for every eigenvalue of A ∈ Gl(d, R). As in the continuous time case, it is helpful to look at the scalar case first: For xn+1 = λxn, 0 = λ ∈ R, the solutions are ϕ(n, x0) = λnx0. Hence the Lyapunov exponent is a limit (not just a limit superior) and is given by lim n→±∞ 1 t log |λnx0| = lim n→±∞ 1 n log (|λ|n) + lim n→±∞ 1 n log |x0| = |λ| . Furthermore, the Lyapunov exponent of the time-reversed equation xn+1 = λ−1xn is |λ|−1. Remark 1.5.5. Consider a solution xn,n ∈ Z, of xn+1 = Axn. Then the function yn := x−n,n ∈ Z, satisfies yn+1 = x −(n+1) = A−1x−n = A−1yn,n ∈ Z. Hence we call xn+1 = A−1xn,n ∈ Z, the time-reversed equation. One finds that the Lyapunov exponents do not depend on the norm and that they remain unchanged under similarity transformations of the matrix. For arbitrary dimension d, the following result clarifies the relation- ship between the Lyapunov exponents of xn+1 = Axn and the moduli of the eigenvalues of A ∈ Gl(d, R). Furthermore, it shows that they are associated with the decomposition of the state space into the Lyapunov spaces. Theorem 1.5.6. Consider the linear difference equation xn+1 = Axn with A ∈ Gl(d, R). Then the state space Rd can be decomposed into the Lyapunov spaces Rd = L(λ1) ⊕ . . . ⊕ L(λ ), and the Lyapunov exponents λ(x0),x0 ∈ Rd, are given by the logarithms λj of the moduli of the eigenvalues of A. For a solution ϕ(·,x0) (with x0 = 0) one has λ(x0) = limn→±∞ 1 n log ϕ(n, x0) = λj if and only if x0 ∈ L(λj). Proof. For any matrix A there is a matrix T ∈ Gl(d, R) such that A = T −1 JRT , where JR is the real Jordan form of A. Hence we may assume that A is given in real Jordan form. Then the assertions of the theorem can

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