# The defining matrix of a symplectic matrix

Just a beginner in symplectic geometry, and the definition of symplectic matrix bothers me. A $$2n\times 2n$$ real matrix $$M$$ is said to be symplectic if it satisfies the following condition: $$M^T\Omega M=\Omega$$ where $$\Omega$$ is a fixed $$2n\times 2n$$ real, invertible and skew-symmetric matrix.

My question is: since $$\Omega$$ can be arbitrary, so if $$\Omega,\Delta$$ are both satisfy the condition, then the following statement must be true: $$M^T\Omega M=\Omega \Rightarrow M^T\Delta M=\Delta.$$

But I don't know how to prove this. Can anyone help me? Thanks.

There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $$f\circ\Delta =\Omega\circ f$$ where $$\Omega$$ and $$\Delta$$ are the linear map associated to the corresponding matrices.
Actually, what you are trying to prove is not strictly true since the definition of the symplectic matrix is basis dependent. To demonstrate this, consider the similarity transformation $$M' = D^{-1} M D$$ and define $$\Delta = D^T \Omega D$$. Then the following proposition can be shown with some matrix algebra:
$$M^T\Omega M=\Omega \Rightarrow M'^T\Delta M'=\Delta.$$
In fact, choosing $$D$$ appropriately allows you to write the symplectic bilinear form in the symplectic basis:
$$\Delta = \begin{bmatrix} 0 & \mathbb{1}_{n \times n} \\ -\mathbb{1}_{n \times n} & 0 \end{bmatrix}.$$