the symplectic version of Gram-Schmidt process and one application of it. I am trying to understand the following proposition:


cf. McDuff and Salamon, Introduction to Symplectic Topology(second
  edition), Corollary 2.4,page 40

One can assume $\omega_0 =\sum_i dx_i \wedge dy_i$. Here the theorem 2.3 is just the existence of a symplectic basis. What confused me is how to apply the Gram-Schmidt process to a symplectic linear space rather than an inner product space. 
 A: You are also in an inner product space as well. 
Since we know $\omega_t$ is a nondegenerate skew symmetric bilinear form (a symplectic form since we are not over a field of characteristic 2) then 
$\omega_t(u,v) = u^T A v$
for some skew symmetric full rank $A \in M_{2n}(\mathbb{R})$.
Now the pullback condition is that 
$\Psi_t^\ast \omega_t(u,v) = \omega_t(\Psi_tu,\Psi_tv) = (\Psi_tu)^T A \Psi_tv = u^T (\Psi_t^T A \Psi_t) v = u^T J v = \omega_0(u,v)$
So the question amounts to if any $A$ described above is congruent to $J = \pmatrix{ 0 & -\mathbb{I} \\ \mathbb{I} & 0 }$, with the answer providing the $\Psi_t$. You also must show that such a family would vary smoothly along with $\omega_t$, but by fixing a basis this is very simple to prove.
As far as how Graham-Schmidt is used, think of how you would construct $\Psi_t$ knowing there exists a basis $(e_1,\dots,e_n,f_1,\dots,f_n)$ which is a standard symplectic basis with respect to $\omega_t$. Immediately one could consider the change of basis matrix $K$ from the standard symplectic basis $(x_i,y_i)$ for $\omega_0$ and the $(e_i,f_i)$ mentioned above so that 
$K^{-1}AK = J$.
Now think of what condition is needed to turn the above similarity into congruency while preserving the requirements of a symplectic basis. 
We want $K^{-1} = K^T$ so we want $K \in O(2n)$ which means its columns are an orthonormal basis, hence Graham-Schmidt (of course you also must show that Graham-Schmidt preserves the behavior of the basis elements with respect to $\omega_t$).
