Show that $W$ is a symplectic matrix i.e. $W^T J W=J$ Good day,
This is my first question, I hope all information is given. If not, feel free to ask.
Currently I am reading "New Trends for Hamiltonian Systems and Celestial Mechanics" by E. A. Lacomba & J. Libre, 1994. On page 325 it says the following:
$\rightarrow$ Let $I$ be the identity matrix and  $J=\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$.
Further we have
$W=\frac{1}{\sqrt{N}} \left[ \omega^{i \cdot j} \right]_{0 \leq i,j \leq N-1} =\frac{1}{\sqrt{N}}\begin{pmatrix}
 1    & 1    & \dots  & 1      \\
 1    & \omega    & \dots   & \omega^{N-1}     \\
 \vdots & \vdots     & \ddots  & \vdots  \\
 1     & \omega^{N-1} & \dots      & \omega^{(N-1)^2} \end{pmatrix} $ with $ \omega=e^{2 \pi i / N} $ the nth root of unitary.
Using the complex quantity $ \omega $ makes it easier to write down and perform this transformation (here not necessary), but one should not forget that in the real space $\omega$ corresponds to a 2 $ \times $ 2 submatrix of the form 
$ \omega \leftrightarrow \begin{pmatrix} 
\cos 2 \pi /N & -\sin 2 \pi /N  \\
\sin 2 \pi /N & \cos 2 \pi /N \end{pmatrix} $
Keeping this relationship in mind one sees that for $W^T$ we have to use the conjugate transposed matrix, i.e.  $W^T=\overline{W}^t.$ Thus we have a unitary matrix with
$ W^{-1}=W^T $ (...) Due to the structure of $W$ we have in $ \mathbb{R}^{2N} $ 
$$W^T J W = J$$ $\leftarrow$
Okay, now my question: How do I get $W^T J W = J$? There is no proof given, maybe it's trivial.
Here are my thoughts:


*

*Maybe I could prove it by using the proportions $W^T=W^{-1}$ and $J^T=J^{-1}=-J$ but I couldn't get it, I think I have to multi plicate it formally. 

*Okay, it's mentioned it should be in $\mathbb{R}^{2N}$, but W itself is in $\mathbb{C}^{N \times N}$, so I should get it in its real form. But how? 
a. By using the correspondence of $\omega$ to its rotating matrix. Like that:
$$ \frac{1}{\sqrt{N}} \begin{pmatrix} 
    1 & 0 & 1 & 0 & \dots & 1 & 0 \\
    0 & 1 & 0 & 1 & \dots & 0 & 1 \\
    1 & 0 & \cos\left( \frac{2 \pi}{N} \right)  & -\sin\left( \frac{2 \pi}{N}\right) & \dots & \cos\left( \frac{2 \pi (N-1)}{N} \right)  & -\sin\left( \frac{2 \pi (N-1)}{N} \right) \\
    0 & 1 & \sin\left( \frac{2 \pi}{N} \right) & \cos\left( \frac{2 \pi}{N} \right)  & \dots & \sin\left( \frac{2 \pi (N-1)}{N} \right) & \cos\left( \frac{2 \pi (N-1)}{N} \right)  \\
    \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
    1 & 0 & \cos\left( \frac{2 \pi (N-1)}{N} \right)  & -\sin\left( \frac{2 \pi (N-1)}{N} \right) & \dots & \cos\left( \frac{2 \pi (N-1)^2}{N} \right)  & -\sin\left( \frac{2 \pi (N-1)^2}{N} \right)  \\
    0 & 1 & \sin\left( \frac{2 \pi (N-1)}{N} \right) & \cos\left( \frac{2 \pi (N-1)}{N} \right) & \dots & \sin\left( \frac{2 \pi (N-1)^2}{N} \right) & \cos\left( \frac{2 \pi (N-1)^2}{N} \right)  \end{pmatrix} $$
    It would be unitary. But I believe this approach is wrong, not sure, but I tried to compute the case $N=4$ to get $W^T J W = J$ but without success. This form can be simplifed by using $\omega^N=1$ e.g. $\omega^{N-1}=\omega^{-1}$ and so on.
b. Compute the eigenvalues and try to compute a real Jordan Normal Form, but this could be tiring and at $N=4$ this lead me to failure.
Okay. These were my approaches. I'm thankful for every help, let it be another approach/proof or a corrections to mine.
Thanks a lot,
Marvin

Preliminaries: Let us see how the product of two complex $n\times n$ matrices $A$ and $B$ looks like in terms of their real and imaginary parts
$$
A=A_r+iA_i,\quad B=B_r+iB_i\quad\Rightarrow\quad AB=A_rB_r-A_iB_i+i(A_rB_i+A_iB_r).
$$
It means that the complex multiplication for matrices works the same as for numbers as long as one respects the order.
Similar to numbers one can construct the $2n\times 2n$ real matrices
$$
A\sim\left[\matrix{A_r & A_i\\-A_i & A_r}\right], \quad 
B\sim\left[\matrix{B_r & B_i\\-B_i & B_r}\right]\quad\Rightarrow\quad
$$
$$
\quad\Rightarrow\quad AB\sim \left[\matrix{A_rB_r-A_iB_i & A_rB_i+A_iB_r\\-(A_rB_i+A_iB_r) & A_rB_r-A_iB_i}\right]=\left[\matrix{A_r & A_i\\-A_i & A_r}\right]\left[\matrix{B_r & B_i\\-B_i & B_r}\right].
$$
So the operations of matrix multiplication and taking the real counterpart are perfectly commuting.
Answer: The matrix $J=\left[\matrix{0 & I\\-I & 0}\right]$ is the real counterpart of the complex $n\times n$ matrix $iI$. The complex matrix $W$ is unitary, i.e. $W^*W=I$ (or $W^TW=I$ in your notations, but I prefer to keep ${^T}$ for real transpose to avoid confusions). It means that
$$
iI=W^*W(iI)=W^*(iI)W=(W_r^T-iW_i^T)(iI)(W_r+iW_i).
$$
Note here that $iI$ is a scalar multiple of identity, thus commutes with any matrix. Now taking the real counterparts for each matrix on both sides gives
$$
J=\left[\matrix{W_r^T & -W_i^T\\W_i^T & W_r^T}\right]J
\left[\matrix{W_r & W_i\\-W_i & W_r}\right]=
\left[\matrix{W_r & W_i\\-W_i & W_r}\right]^TJ\left[\matrix{W_r & W_i\\-W_i & W_r}\right].
$$
