I would like to know about the properties of the diagonal elements of a triangular matrix $A$ reduced from a symplectic matrix $S$, size $2n\times 2n$ for $n\in\mathbb N$.
The symplectic matrix $S$ can be decomposed as
$$S=O\left(\begin{matrix}D&0\\0&D^{-1}\end{matrix}\right)O'$$
where $O$ and $O'$ are two orthogonal matricies and $D$ is a diagonal matrix of real numbers.
The determinant of this matrix is $1$.
If we triangulize this matrix into a form
$$A=\left(\begin{matrix}a_{1,1}&a_{2,2}&\dots& a_{1,2n}\\ 0&a_{2,2}&\dots& a_{2,2n}\\ \vdots&\ddots&\ddots& \vdots\\ 0&\dots&0& a_{2n,2n}\end{matrix}\right)$$
we preserve the determinant.
Clearly, the determinant leads us to see that the product of all diagonal elements of $A$ is $1$:
$$\det A=a_{1,1}a_{2,2}\dots a_{2n,2n}=1$$
However, I also have a conjecture that the diagonal elements of $A$ also satisfy a system of equations
$$a_{1,1}a_{n+1,n+1}=1\\ \vdots\\ a_{n,n}a_{2n,2n}=1$$
i.e. it somehow preserves the properties of the diagonal $D$ in the original symplectic matrix.
Is this a known property?