1.Since $[,]$ is non-degenerate, there should be such an eigenvector $\eta$ corresponding to the eigenvalue $\lambda'$ for the eigenvector $\xi$ corresponding to the eigenvalue $\lambda$ that $[\xi, \eta] \neq 0$, which is possible only if $\lambda' = \bar{\lambda}$. Thus, the two-dimensional invariant plane $\pi_\lambda$ is nonnull.
2.Since $\xi$ is a real vector, there is a number $a$ such that $\xi = a\xi_1 + \bar{a}\xi_1$, where $\xi_1$ is the eigenvector with the eigenvalue $\lambda$. Then
$$ [S(a\xi_1 + \bar{a}\bar{\xi}_1), a\xi_1 + \bar{a}\bar{\xi}_1] = (\lambda - \bar{\lambda}) |a|^2 [\xi_1, \bar{\xi}_1] = (2|a|^2 Im \lambda) i(I\xi_1, \xi_1), $$ the sign of which is independent of $a$, i.e. independent of $\xi$. Note that $i(I\xi_1, \xi_1)$ is a real number.
3.Denote $(G\xi,\xi) = i(I\xi_1, \xi_1)$, and denote $S$ by $M$. Assume $M$ is stable and all its eigenvalues are Krein-definite. If $M$ is not strongly stable, there would be ${M_n}$ of unstable symplectic matrices converging to $M$. Either $M_n$ has eigenvalue outside the unit circle, or $M_n$ has an eigenvalue on the unit circle which is not semi-simple. 66
Thus, there is a $G$-isotropic unit eigenvector $x_n$: $$M_n x_n = \lambda_n x_n,~(Gx_n, x_n) = 0.$$ Since $\lambda_n$ is a root of $|M_n - z I|$, we can extract from the sequence $\lambda_n$ a subsequence converging to a root of $M - zI$, i.e. $x_n \rightarrow x, \lambda_n \rightarrow \lambda$. Then $(Gx,x) = 0$ which is impossible since all eigenvalues of $M$ is Krein-definite.
Conversely, assume $M$ is strongly stable. Then all the eigenvalues of $M$ lie on the circle and are semi-simple. For every eigenvalue $\lambda$ with positive imaginary part, we can choose in the eigenspace $Ker(M-\lambda I)$ a $G$-orthogonal basis, say $[\xi_1, \cdots, \xi_m]$ with $(G\xi_k,\xi_k) = \pm 1$. We can take $[\bar{\xi_1},\cdots,\bar{\xi_m}]$ as a basis for the conjugate eigenspace $Ker(M-\bar{\lambda} I)$. If $\pm 1$ is an eigenvalue, the corresponding eigenspace is real and even-dimensional; we can choose its $G$-orthogonal basis to be $[\xi_1,\cdots,\xi_m,\bar{\xi_1},\cdots,\bar{\xi_m}]$ with $(G\xi_k,\xi_k)=1=-(G\bar{\xi}_k,\bar{\xi}_k)$. Putting everything together, we get a $G$-orthonogonal basis of eigenvectors of $[\xi_1,\cdots,\xi_n,\bar{\xi}_1,\cdots,\bar{\xi}_n]$ for $C^{2n}$. We have $(G\xi_k,\xi_k)=-(G\bar{\xi}_k,\bar{\xi}_k)$, aand by rearranging the basis, we can assume that $(G\xi_k,\xi_k)=1$ for $1 \leq k \leq n$.
Assume there is an eigenvalue $\lambda$ which is not definite. 66