# Prove the inverse of a symplectic matrix is also symplectic

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

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

• For Route 2, apply both inversion and transpose to get equ (2). May 21, 2023 at 3:54

$$M$$ 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.

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