I worked out with symplectic matrices that the transpose is also symplectic for the $2\times 2$ case since the algebra was easy and the determinant of the matrix just needed to equal $1$. [The expression for determinant is easy for $2\times 2$ matrices.] and the transpose has equal determinant. But what about the $n\times n$ case. Is it also true?
Linear Algebra – Prove Transpose of a Symplectic Matrix is Symplectic
linear algebramatricessymplectic-linear-algebra
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I am not sure what your view is on what the transpose of a matrix is, so let me give my point of view first.
Suppose that $V$ is a real (finite dimensional) vector space with an inner product, $\langle \cdot, \cdot \rangle: V \times V \rightarrow \mathbb{R}$.
Now given two vectors $v,w \in V$, we can compute the inner product $\langle v, Aw \rangle$. One can now ask if there is a matrix $B$ (dependent on $A$), such that $\langle Bv , w \rangle = \langle v, Aw \rangle$ for all $v,w \in V$. This turns out to be the transpose of the matrix $A$, that is $B = A^{T}$. (Exercise: verify this...)
Given a vector $v \in V$, its length squared is given by $|v|^{2} =\langle v, v \rangle$. Given two vectors $v,w$, there is a relation between the cosine of the angle between them, $\theta$, and their inner product \begin{equation} \cos(\theta) = \frac{\langle v, w \rangle}{|v||w|}. \end{equation}
Now, a rotation is a linear map $A:V \rightarrow V$, that preserves angles and preserves lengths. That is, for any $v \in V$ we want \begin{equation} |Av|^{2} = |v|^{2}, \quad \Longleftrightarrow \quad \langle Av, Av \rangle = \langle v, v \rangle. \end{equation} Now if $v,w \in V$ let $\theta$ be the angle between $v$ and $w$, then we want the angle between $Av$ and $Aw$ to be $\theta$ as well, or in other words \begin{equation} \frac{\langle v, w \rangle}{|v||w|} = \cos(\theta) = \frac{\langle Av, Aw\rangle}{|Av||Aw|} = \frac{\langle Av, Aw \rangle}{|v||w|}. \end{equation} We conclude that we want $\langle v, w \rangle = \langle Av, Aw \rangle$ to hold for all $v$ and $w$. We can rewrite this using the transpose \begin{equation} \langle v, w \rangle = \langle Av, Aw \rangle = \langle v, A^{T} A w \rangle. \end{equation} Since this is supposed to be true for all $v,w \in V$ we may conclude that $A^{T}A = \text{Id}$.
No, not all symplectic $(0,1)$-matrices are block-triangular. The following matrix is symplectic: $$\begin{pmatrix}0&1&0&0\\1&0&1&0\\0&1&0&1\\0&0&1&0\end{pmatrix}$$
A computer search found 44 counterexamples in dimension $4\times 4$, so here is another less symmetric one: $$\begin{pmatrix}1&1&0&1\\1&1&1&0\\1&0&1&0\\0&1&0&1\end{pmatrix}$$
WolframAlpha code to check it's symplectic:
[[1,1,0,1],[1,1,1,0],[1,0,1,0],[0,1,0,1]]^T*[[0,0,1,0],[0,0,0,1],[-1,0,0,0],[0,-1,0,0]]*[[1,1,0,1],[1,1,1,0],[1,0,1,0],[0,1,0,1]]
Best Answer
If $S$ is symplectic of size $2n\times 2n$, i.e., satisfies $S^TJS=J$, so is $S^T$. To see this, take the inverse of the equality $(S^{−1})^T JS^{-1}= J$ (because if $S$ is symplectic, then $S^{-1}$ is symplectic too).