# Equivalence of two definitions of Lyapunov exponents

I saw in articles two different definitions for Lyapunov exponents of a discrete dynamical system.

Let's consider a discrete dynamical system $$x_{k+1}=f(x_{k}),\quad x_{k}\in\mathbb{R}^{n},\quad k=0,1,2,\ldots,$$ $x_{0}$ -- given, where $f:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is a smooth map.

Let $f'(x)$ be Jacobian of $f$ at $x$, $T$ denote the matrix transpose, $\lambda_{i}(\cdot)$ be the $i$th eigenvalue of a matrix. Let $\Phi_{m}=f'(x_{m-1})\ldots f'(x_{1})f'(x_{0})$.

Definition 1. The $i$th Lyapunov exponent at $x_{0}$ is given by $$l_{i}(x_{0})=\lim_{m\to\infty}\frac{1}{2m}\ln|\lambda_{i}(\Phi_{m}^{T}\Phi_{m})|,\quad i=1,2,\ldots,n.$$

Definition 2. The $i$th Lyapunov exponent at $x_{0}$ is given by $$l_{i}(x_{0})=\lim_{m\to\infty}\frac{1}{m}\ln|\lambda_{i}(\Phi_{m})|,\quad i=1,2,\ldots,n.$$

Are these definitions equivalent? If the answer is no, could you give a counterexample of $f$? And what is the correct definition of Lyapunov exponents?

• Another website: math.stackexchange.com is perhaps more suitable for such questions
– Ilya
Mar 15 '13 at 13:44
• Definition 2 is wrong, but can be corrected by replacing the $i$th eigenvalue of $\Phi_m$ with the $i$th singular value (after which the two definitions agree). To construct a counterexample, you could construct a dynamical system where the $\Phi_m$'s are conjugate to rotations (so have eigenvalues on the unit circle), but the entries of $\Phi_m$ are large, so the top singular value is large. Mar 15 '13 at 14:28

Formally speaking, both definitions are wrong, because Lyapunov exponents are not about the growth of matrices $\Phi_m$ as such (be it in terms of eigenvalues or of singular values), but about their growth along directions in the vector space $V=\mathbb R^n$. More precisely, given a vector $v\in V$, the Lyapunov exponent of the sequence $\Phi_m$ along $v$ is defined as $$\chi(v) = \limsup_m \frac1m \log \| \Phi_m v \| \;.$$ The a priori difference between this definition and your definitions 1 or 2 should be obvious.
However, under the additional condition that $$\tag{\ast} \log\|A_m\|,\log\|A_m^{-1}\|=o(m)$$ for the increments $A_m=\Phi_m\Phi^{-1}_{m-1}$ (in your notation $A_m=f'(x_{m-1})$ all these definitions are equivalent.
The key notion here is that of Lyapunov regularity, which means that the limit $\chi = \lim_m \log |\det\Phi_m|$ exists and equals to the sum of the Lyapunov exponents (taken with their multiplicities). One can pretty easily show that Lyapunov regularity is equivalent to existence of a positive definite matrix $\Lambda$ such that $\log \|\Phi_m\Lambda^{-m}\|,\log\| (\Phi_m\Lambda^{-m})^{-1} \|=o(m)$. Then the Lyapunov exponents of the sequence $\Phi_n$ are just the logarithms of the eigenvalues of $\Lambda$.
Under condition $(\ast)$ Lyapunov regularity is equivalent to existence of the limit $\Lambda =\lim (\Phi^*_m\Phi_m)^{1/2m}$ (actually it is enough to require just existence of the limits of logarithms of the norms of $\Phi_m$ is all external powers of $V$), and the Lyapunov exponents are precisely the logarithms of the eigenvalues of $\Lambda$. Without condition $(\ast)$ this may fail; the simplest example is provided by the matrices $\Phi_m=\left( 1\;\; e^m\atop 0 \;\; 1 \right)$.
To expand on Anthony's comment somewhat, the two definitions differ in that the second limit can fail to exist even when the first limit does exist. For example, consider a case in which the sequence of Jacobian matrices alternates between the following two matrices: $$A_1:=\left$$\begin{array}{cc}0&2\\\2&0\end{array}\right$$,\qquad A_2:=\left$$\begin{array}{cc}0&1\\\4&0\end{array}\right$$.$$ (This could arise if, for example, $f$ is a transformation of the disjoint union of two 2-tori which acts by first applying to each torus the toral automorphism defined by one of the above matrices, and then subsequently interchanging the two tori.) Now, products of the form $(A_2A_1)^n$ or $(A_1A_2)^n$ have $\lambda_1=8^n$ and $\lambda_2=2^n$, but products of the form $A_1(A_2A_1)^n$ or $A_2(A_1A_2)^n$ have $\lambda_1=\lambda_2=2^{2n+1}$. The second limit therefore fails to exist. On the other hand, RW's condition (*) is clearly satisfied, and it is not difficult to show that the first and second singular values which appear in the first definition are respectively $8^n$ and $2^n$ for all of the above products and so the first limit exists.