characteristic polynomial of a symplectic matrix is a reciprocal polynomial

Theorem 1.

The characteristic polynomial of a symplectic matrix is a reciprocal polynomial.

Proof.

Let $A$ be the symplectic matrix, and let$p(\lambda )=det(A-\lambda I)$ beits characteristic polynomial. We wish to prove that

$$p(\lambda )=\pm {\lambda }^{n}p(1/\lambda ).$$

By definition, $AJ{A}^T=J$ where $J$ is the matrix

Since $A$ and $J$ are symplectic matrices, their determinants are $1$, and

$p(\lambda )$$=$$det(AJ-\lambda J)$

$=$$det(AJ-\lambda AJ{A}^T)$

$=$$det(-\lambda A)det(J)det(-{\displaystyle \frac{1}{\lambda }}J+J{A}^T)$

$=$$\pm {\lambda }^{n}det(A-{\displaystyle \frac{1}{\lambda }}I).$

as claimed.∎