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I want to show that if $\lambda$ is a real eigenvalue of a symplectic matrix $A$ then its char poly is of the form $\det(A-\mu id) = (\lambda-\mu)(\frac{1}{\lambda}-\mu) \det(\hat{A}- \mu id) $ where $\hat{A}$ is again a symplectic matrix.

My proof goes like this: First, we take the eigenvector $e_1$ to the eigenvalue $\lambda$. We can extend this vector to a symplectic basis $e_1,f_1,e_2,f_2,\ldots,f_n$ satisfying the canonical conditions on such a basis with symplectic form $\omega.$

With respect to this basis, the first column of $A$ is $(\lambda,0,\ldots,0)^T$ by the eigenvector property.

The entry $(2,2)$ of the matrix is given by $\omega(e_1,f_1) = \frac{1}{\lambda} \omega(e_1,Af_1) = \frac{1}{\lambda} \omega(e_1,f_1) = \frac{1}{\lambda}.$

Moreover, all other entries in the second row are zero( besides the entry $(2,2)$) as for any $k \in \operatorname{span}\{{f_1}\}^{\perp}.$

$$\omega(e_1,Ak) = \frac{1}{\lambda} \omega(Ae_1,Ak) = \frac{1}{\lambda} \omega(e_1,k) = 0.$$

Thus, we can calculate the determinant of $A-\mu id$ in this basis by first expanding with respect to the first column. Since the only non zero entry is $(\lambda-\mu)$ we only get one term. Then we calculate the determinant of the submatrix with entries $(2,\ldots,n)\times(2,\ldots,n).$ We expand this determinant with respect to the second row. Here, the only non-zero entry is $(2,2).$ This one is $\frac{1}{\lambda-\mu}.$ So we indeed end up with something of the form $\det(A-\mu id) = (\lambda-\mu)(\frac{1}{\lambda}-\mu) \det(\hat{A}- \mu id) .$

But now the big question is: Why is the submatrix $\hat{A}$ corresponding to the entries $(3,\ldots,n)\times(3,\ldots,n)$ of $A$ again symplectic?

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Perhaps you want to prove that if $\lambda\in \mathbb{C}$ is an eigenvalue of $A$ with multiplicity $\alpha$, then the same is true for $1/\lambda$.

Since $A$ satisfies $A^{-1}=\Omega^{-1}A^T\Omega$, $A^{-1}$ is similar to $A^T$; consequently $A^{-1}$ is similar to $A$ and we are done.

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