Suppose $H$ is Hamiltonian. Show that $e^H$ must be symplectic.

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Definition/Facts: Let $K$ denote the skew-symmetric matrix $$\left[ \begin{matrix} 0 & I_n \\ -I_n & 0 \end{matrix} \right]. $$ A matrix $A\in\mathbb R^{2n\times 2n}$ is said to be Hamiltonian if $K^{-1}A^TK=-A$ and symplectic if $K^{-1}A^TK = A^{-1}$.

A possible solution is: $$K^{-1}(e^H)^TK = K^{-1}e^{H^T}K = e^{K^{-1}H^TK} = e^{-H} = (e^H)^{-1}.$$

Thus, $e^H $ is symplectic.

Question: How does
$$K^{-1}e^{H^T}K = e^{K^{-1}H^TK}$$