Analytic proof that Lyapunov exponents in Hamiltonian systems pairwise sum to zero I have read that in Hamiltonian systems, Lyapunov exponents come in pairs $(\lambda_i, \lambda_{2N-i+1})$ such that their sum is equal to zero.
Is there a way of proving this analytically?
EDIT: 
Saw this here.

In symplectic systems, LEs come in pairs $(\lambda_i, \lambda_{2N-i+1})$ such that their sum is equal to zero. This means that the Lyapunov spectrum is symmetric. It is a way of emphasizing the invariance of Hamiltonian dynamics under change of the time arrow. 

 A: *

*We are considering a discrete time evolution $$x_{n}~=~f(x_{n-1})~=~f^{n\circ}(x_0), \qquad n~\in~\mathbb{N},$$ in a $2N$-dimensional symplectic manifold $(M,\omega)$, where $f$ is a symplectomorphism. 

*Let us for simplicity work in local coordinates. Define the Jacobian matrix as
$$\tag{1} A(x,n)^{i}{}_{j}~:=~\frac{\partial (f^{n\circ} (x))^i}{\partial x^j}. $$

*In local Darboux coordinates, the Jacobian matrix (1) is a symplectic matrix 
$$\tag{2}  A^T\Omega A~=~ \Omega, \qquad \Omega ~:=~\begin{bmatrix} 0_N & -I_N \cr I_N & 0_N \end{bmatrix}.$$ 

*Note that the transposed $A^T$ is also a symplectic matrix. Note that $A^TA$ is a positive definite symplectic matrix. 

*Symplectic quartet mechanism: For a diagonalizable$^1$ symplectic matrix, the eigenvalues form quartets $$\tag{3} \{\lambda,\bar{\lambda}, \lambda^{-1},\bar{\lambda}^{-1}\}$$
in the complex plane $\mathbb{C}$. 
A quartet becomes a doublet on the real axis and on the unit circle.

*Define the Lyapunov exponents $$\tag{4} \left\{\lambda_1(x,n), \ldots, \lambda_{2N}(x,n)\right\}~\subset~\mathbb{R} $$ as the eigenvalues of the Hermitian matrix
$$\tag{5} \Lambda(x,n)~:=~\frac{1}{2n}\ln \left(A(x,n)^TA(x,n)\right).$$

*It follows from the symplectic doublet mechanism (3), that the eigenvalues (4) are distributed symmetrically around 0 on the real axis $\mathbb{R}$.
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$^1$Not all symplectic matrices are diagonalizable. 2D Counterexample: $$\tag{6} A~=~\begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix}.$$
