# characteristic polynomial of a symplectic matrix is a reciprocal polynomial

###### Proof.

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

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

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

 $J=\left(\begin{array}[]{cc}0&I\\ -I&0\end{array}\right).$

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

 $\displaystyle p(\lambda)$ $\displaystyle=$ $\displaystyle\det(AJ-\lambda J)$ $\displaystyle=$ $\displaystyle\det(AJ-\lambda AJA^{T})$ $\displaystyle=$ $\displaystyle\det(-\lambda A)\det(J)\det(-\frac{1}{\lambda}J+JA^{T})$ $\displaystyle=$ $\displaystyle\pm\lambda^{n}\det(A-\frac{1}{\lambda}I).$

as claimed. ∎

Title characteristic polynomial of a symplectic matrix is a reciprocal polynomial CharacteristicPolynomialOfASymplecticMatrixIsAReciprocalPolynomial 2013-03-22 15:33:18 2013-03-22 15:33:18 matte (1858) matte (1858) 7 matte (1858) Theorem msc 53D05 ReciprocalPolynomial CharacteristicPolynomialOfAOrthogonalMatrixIsAReciprocalPolynomial