Canonical Transformation and Symplectic Conditions I have one question regarding Canonical transformation and symplectic matrix. 
I have read some notions from the following note: 
http://www.chim.unifi.it/orac/MAN/node6.html
For me it is not clear how it is obtained the following  relation: 
$$\dot y = MJM^t \frac{\partial H}{\partial y}.$$
Can you explain me, please. 
Update1
What I have tried. 
I'll consider the  case when $n=2$. 
We try to make a transformation $y \to y(x)$.
relation 1(a): 
$$\dot y = \frac{\partial y}{\partial  x} \dot x $$
relation 2(a):
$$\dot x = J \frac{\partial H}{\partial x}$$
BUT
relation 3(a):
$$\frac{\partial H}{\partial x}=\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}$$
So, we have: 
$$\dot y= \frac{\partial y}{\partial x} \dot x = \frac{\partial y}{\partial  x} J \frac{\partial H}{\partial x}=\frac{\partial y}{\partial  x} J\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}.$$
From here I get stuck.
Looks similar but I don't know how to continue to obtain same form. 
I don't know why there appear the transpose of that matrix. 
Can you help me to write in coordinates $\displaystyle \frac{\partial y}{\partial x}$.
UPDATE2
from relation 1(a):
$$\dot y = \frac{\partial y }{\partial x} \dot x $$
In matrix notation, for $n=2$, I'll have:
relation (1b)
$$\begin{pmatrix}
\dot y_{1}\\ \dot y_{2}
\end{pmatrix}=\begin{pmatrix}
\frac{\partial y_{1}}{\partial x_{1}}& \frac{\partial y_{1}}{\partial x_{2}}\\
\frac{\partial y_{2}}{\partial x_{1}}&\frac{\partial y_{2}}{\partial x_{2}}
\end{pmatrix} \begin{pmatrix}
\dot x_1\\\dot x_2
\end{pmatrix}.$$
Now, From relation 3(a) I want to obtain in a matrix form relation 3(b):
$$\begin{pmatrix}
\frac{\partial H}{\partial x_1}\\
\frac{\partial H}{\partial x_2}
\end{pmatrix}=\begin{pmatrix}
\frac{\partial H}{\partial y_1}\\
\frac{\partial H}{\partial y_2}
\end{pmatrix} \begin{pmatrix}
\frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2}\\
\frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2}
\end{pmatrix}.$$ 
So,the relation from UPDATE1 
$$\dot y=\frac{\partial y}{\partial  x} J\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}.$$
will become: 
$$\begin{pmatrix}
\dot y_{1}\\ \dot y_{2}
\end{pmatrix}=\begin{pmatrix}
\frac{\partial y_{1}}{\partial x_{1}}& \frac{\partial y_{1}}{\partial x_{2}}\\
\frac{\partial y_{2}}{\partial x_{1}}&\frac{\partial y_{2}}{\partial x_{2}}
\end{pmatrix} \begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix}\begin{pmatrix}
\frac{\partial H}{\partial y_1}\\
\frac{\partial H}{\partial y_2}
\end{pmatrix} \begin{pmatrix}
\frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2}\\
\frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2}
\end{pmatrix} $$
And from here, again, it seems impossible for me... more and more, the matrix multiplication is not possible... 
Thanks!
 A: Don't know if you're still interested in this, but here goes.
Take the coordinate transformation $\vec{x}\to\vec{y}(\vec{x})$, i.e. we consider $\vec{y}$ to be a function of $\vec{x}$.
Let the Hamiltonian function in terms of the new coordinates be denoted $\tilde{H}$, that is, $\tilde{H}(\vec{y}(\vec{x}))=H(\vec{x})$.
Rather than doing the calculation in terms of matrices, it is easier to do it in terms of components (tensor notation). Then you don't have to worry about tracking rows/columns, and also the notation is a lot more compact. Let $y_i$ denote the $i$'th component of $\vec{y}$, etc. Also, $M_{ij}$ denotes the $ij$ component of the $M$ matrix, which is $\frac{\partial y_i}{\partial x_j}$ (there is a typo in the definition of $M$ in your link).
Our goal is to show
$$\dot{y}_i=\left(M J M^T \frac{\partial H}{\partial \vec{y}}\right)_i.$$
Consider the LHS above, $\frac{d}{dt}\left(y_i (\vec{x}(t))\right)$. By the chain rule, this is equal to 
$$\frac{\partial y_i}{\partial x_j}\frac{dx_j}{dt}$$ 
with implicit summation over $j$. But $\frac{\partial y_i}{\partial x_j}=M_{ij}$, so
$$\dot{y}_i=\frac{\partial y_i}{\partial x_j}\frac{dx_j}{dt}=M_{ij}\dot{x}_j=(M\dot{\vec{x}})_i$$
$$\implies \dot{\vec{y}}=M\dot{\vec{x}}=MJ\frac{\partial H}{\partial \vec{x}}.$$
Now recalling $H(\vec{x})=\tilde{H}(\vec{y}(\vec{x}))$, apply the chain rule to get
$$\frac{\partial H}{\partial x_i}=\frac{\partial \tilde{H}(\vec{y})}{\partial y_j}\frac{\partial y_j}{\partial x_i}=M_{ji}\frac{\partial \tilde{H}}{\partial y_j}=(M^T)_{ij}\frac{\partial \tilde{H}}{\partial y_j}=\left(M^T \frac{\partial \tilde{H}}{\partial \vec{y}}\right)_i$$
$$\frac{\partial H}{\partial \vec{x}}=M^T \frac{\partial \tilde{H}}{\partial \vec{y}}.$$
Substitute this into the above gives
$$\dot{\vec{y}}=MJM^T\frac{\partial \tilde{H}}{\partial\vec{y}}$$
as required.
