This question has come out while reading J. Moser "*New Aspects in the Theory of Stability 
of Hamiltonian Systems*". I'm particularly interested to the Appendix, where one investigates the stability of elliptic fixed points of Hamiltonian dynamical systems, in the time independent case. I start presenting the **framework**.


Let us consider the Hamiltonian dynamical system
$$
 \dot{x}\_\nu = H\_{y\_\nu}(x,y), \qquad \dot{y}\_\nu=-H\_{x\_\nu}(x,y) \qquad \qquad (1)
$$


where $\nu=1,2,\dots,n$. Hamiltonian $H$ does not depend on time $t$ and it is assumed to be a real analytic function of $x\_\nu,\,y\_\nu$, with $\nu=1,2,\dots,n$ in the neighboorhood of $x=y=0$, the expansion of which starts with quadratic terms. Then $x=y=0$ is an equilibrium solution. 


One can construct a fundamental system of solutions of exponential form 
$$
 w^{(\nu)}=e^{\gamma\_\nu t}p^{(\nu)}
$$
where $p^{(\nu)}$ are constant vectors or, in the case of multiple eigenvalues $\gamma\_{\nu}$, possibly polinomials in $t$. The numbers $\gamma\_\nu$ are obtained as the eigenvalues of the matrix determined by the linear terms of the right-hand side of (1). Suppose that all eigenvalues are distinct and purely imaginary, i.e. of the type $\gamma\_\nu=i\beta\_\nu$, with $\beta\_\nu$ real. So the spectrum has the form 
$$
 \pm i \beta\_1, \quad \pm i \beta\_2, \,\dots,\, \pm i \beta\_n.
$$
So one obtains a collection of distinct numbers $\beta\_\nu$, $-\beta\_\nu$ with $\nu=1,2,\dots,n$


So far, so good. 


For later purposes, one needs to define the sign of $\beta\_\nu$. The Author says that the sign of $\beta\_\nu$ is taken in such a way that 
$$
\mathcal{Im}\left[w^{(\nu)},\overline{w^{(\nu)}}\right]<0.
$$ 
Square brackets are defined as Lagrange Brackets (an outdated therminology, nowadays called symplectic form [see comments]). More precisely, given any two $2n$-dimensional vectors, $x$ and $\tilde{x}$, with components $x\_\nu$ and $\tilde{x}\_\nu$, their Lagrange Bracket (read: symplectic form) is defined as 
$$
 [x,\tilde{x}]=\sum\_{\nu=1}^n (x\_\nu\tilde{x}\_{\nu+n}-x\_{\nu+n}\tilde{x}\_\nu)
$$
In passing, one has to remember that an Hamiltonian sysytem is marked by the fact that, for any two solutions $x$ and $\tilde{x}$, the Lagrange bracket (read: symplectic form) $[x,\tilde{x}]$ is $t$-independent.


**Questions:** can someone please show 


1) How to explicitly compute:


$$
 [w^{(\nu)},w^{(\mu)}] = ?
$$


$$
 [w^{(\nu)},\overline{w^{(\mu)}}] = ?
$$


2) How to prove that
$$
 [w^{(\nu)},\overline{w^{(\nu)}}] = [p^{(\nu)},\overline{p^{(\nu)}}]
$$


$$
 [p^{(\nu)},\overline{p^{(\nu)}}] \text{ is purely imaginary}
$$
3) How to define the sign $\beta\_{\nu}$ in such a way that
$$
\mathcal{Im}\left[w^{(\nu)},\overline{w^{(\nu)}}\right]<0.
$$ 
I am from the Physics community, so I kindly ask to display all important passages. 



Symplectic forms and sign of eigenvalues
I have found the solution myself. 


I write the system of linear ordinary differential equations in matrix form:


$$
\begin{pmatrix}
\dot{\vec{x}}(t)\\
\dot{\vec{y}}(t)
\end{pmatrix}
=\mathbb{J}
\begin{pmatrix}
\vec{x}(t)\\
\vec{y}(t)
\end{pmatrix}
$$
where $\mathbb{J}$ is the Jacobian matrix associated to the linear system and $\vec{x}$, $\vec{y}$ are $n$-dimensional vectors.


What Moser calls "fundamental system of solutions" is simply a set of $2n$ functions $w^{(j)}(t)$ having the form
$$
\vec{w}^{(j)}(t)=e^{i\beta\_jt}
\begin{pmatrix}
P\_{1,j}\\
P\_{2,j}\\
\dots \\
P\_{2n,j}
\end{pmatrix}
$$
for $j=1,2,...,n$. The remaining $n$ fundamental functions are simply the complex conjugates: 
$$
\vec{w}^{(j+n)}(t)=\overline{\vec{w}^{(j)}}(t)=e^{-i\beta\_jt} \begin{pmatrix}
P\_{1,j}^\*\\
P\_{2,j}^\*\\
\dots \\
P\_{2n,j}^\*
\end{pmatrix}
$$
for $j=1,2,\dots,n$. Notice that numbers $i\beta\_j$ are the eigenvalues of $\mathbb{J}$ while $\vec{P}\_j$ are the corresponding eigenvectors.


**Answers:**


1)Applying the definition of symplectic form given by Moser, one can write that:
$$
 \left[\vec{w}^{(\nu)}(t),\vec{w}^{(\mu)}(t)\right]=e^{it(\nu+\mu)} \left[\vec{P}\_\nu,\vec{P}\_{\mu} \right]
$$
From the Theory of Hamiltonian system, it is known that the symplectic form of two solutions must be $t$-independent. This implies that 
$$
 \left[\vec{P}\_\nu,\vec{P}\_{\mu} \right]=0
$$
(Probably there exist a proof which exploits the property of eigenvectors $\vec{P}\_j$'s but I was not able to find it.)


Similarly, one can prove that
 $$
 \left[\vec{w}^{(\nu)}(t),\overline{\vec{w}^{(\mu)}}(t)\right]=e^{it(\nu-\mu)} \left[\vec{P}\_\nu,\vec{P}\_{\mu}^\* \right]=0
$$


2) If one computes the symplectic form between two complex conjugate solutions, one discovers that the exponential term cancel:
 $$
 \left[\vec{w}^{(\nu)}(t),\overline{\vec{w}^{(\nu)}}(t)\right]=e^{it(\nu-\nu)} \left[\vec{P}\_\nu,\vec{P}\_{\nu}^\* \right]= \left[\vec{P}\_\nu,\vec{P}\_{\nu}^\* \right]
$$
One can see that $\left[\vec{P}\_\nu,\vec{P}\_{\nu}^\* \right]$ is purely imaginary, in fact:
$$
\left[\vec{P}\_\nu,\vec{P}\_{\nu}^\* \right]=\sum\_{k=1}^n \left[P\_{k,\nu}P\_{k+n,\nu}^\* -P\_{k+n,\nu}P\_{k,\nu}^\*\right] = \sum\_{k=1}^n 2\mathcal{I}\{P\_{k,\nu}P\_{k+n,\nu}^\* \}
$$


3) First of all notice that:
$$
 \left[\vec{P}\_\nu,\vec{P}\_{\nu}^\* \right]=-\left[\vec{P}\_\nu^\*,\vec{P}\_{\nu} \right]
$$
Notice also that, if eigenvalue $+i\beta\_\nu$ is associated to eigenvector $\vec{P}\_\nu$, eigenvalue $-i\beta\_\nu$ is associated to eigenvector $\vec{P}\_\nu^\*$. So if the computation of the imaginary part of symplectic form gives a positive result, one just needs to exchange the roles of $+i\beta\_\nu$ and $-i\beta\_\nu$.


