Yesterday I asked a question on eigenvalues of $A^{T}A$.
The reason why I asked this question was that I'd seen in articles two different definitions for Lyapunov exponents of a discrete dynamical system. Below I will give these definitions. As I see now, these definitions are likely not equivalent. So, I have another question. What is the correct definition of Lyapunov exponents? Or, perhaps, these two definitions are really equivalent, since there is the operation of limit in their formulations.
The definitions are following.
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 $$ l_{i}(x_{0})=\lim_{m\to\infty}\frac{1}{m}\ln|\lambda_{i}(\Phi_{m})|,\quad i=1,2,\ldots,n. $$
If these two definitions are not equivalent could you give a counterexample of $f$.
Thank you very much for your attention to my question. It is very important for me.
