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A symplectic matrix is a $2n \times 2n$ matrix $M$ satisfying the condition

$$ M^\mathsf{T} \Omega M = \Omega \tag{1} $$

where the matrix $\Omega$ is usually chosen as the block matrix

$$ \Omega = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix} $$

which satisfies $\Omega^{-1} = \Omega^\mathsf{T} = -\Omega$. The inverse of $M$ exists, and is found from Eq. (1) as

$$ M^{-1} = \Omega^{-1} M^\mathsf{T} \Omega = -\Omega M^\mathsf{T} \Omega \tag{2} $$

However, I am having trouble in showing that $M^{-1}$ is also symplectic, i.e. also satisfy the defining property Eq. (1):

$$ (M^{-1})^\mathsf{T} \Omega M^{-1} = \Omega \tag{3} $$

  • Route 1: Multiply Eq. (1) on the right by $M^{-1}$, and on the left by $(M^\mathsf{T})^{-1} = (M^{-1})^\mathsf{T}$, I can produce Eq. (3)

    $$ \Omega = (M^{-1})^\mathsf{T} \Omega M^{-1} $$

  • Route 2: Taking inverse of Eq. (1), I get

    $$ \begin{align*} M^{-1} \Omega^{-1} (M^\mathsf{T})^{-1} &= \Omega^{-1} \\ \Rightarrow \quad M^{-1} \Omega (M^{-1})^\mathsf{T} &= \Omega \end{align*} $$

    which does not seem to lead to Eq. (3).

Question: How can I arrive at Eq. (3) using Route 2? Alternatively, can symplectic matrices be equivalently defined using the following condition

$$ M \Omega M^\mathsf{T} = \Omega \quad ? \tag{4} $$

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  • $\begingroup$ For Route 2, apply both inversion and transpose to get equ (2). $\endgroup$
    – coiso
    May 21, 2023 at 3:54

1 Answer 1

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$M$ is invertible because $\det(M^T)\det(\Omega)\det(M)=\det(\Omega)\implies\det(M^T)\det(M)=1$ since $\det(\Omega)=\pm 1$.

Now, $\Omega$ is also invertible. So, taking inverses of Eq. (1) you get,

$(M^T\Omega M)^{-1}=\Omega^{-1}$.

$M^{-1}\Omega^{-1}(M^T)^{-1}=\Omega^{-1}$.

Since $\Omega^{-1}=-\Omega$ and $(M^T)^{-1}=(M^{-1})^T$, we obtain

$(M^{-1})\Omega (M^{-1})^T=\Omega$.

Now premultiply by $M$ and postmultiply by $M^T$.

$\Omega=M\Omega M^T$.

So $M^T$ is symplectic. But $(M^{-1})\Omega (M^{-1})^T=\Omega$ shows that $(M^{-1})^T$ is symplectic. Thus, $M^{-1}$ is also symplectic as your matric is still symplectic after taking tranposes.

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  • $\begingroup$ Lots of transposition in Route 2... Thank you! $\endgroup$ May 21, 2023 at 4:59

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