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# 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 $p(\lambda)=\det(A-\lambda I)$ 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. ∎

Related:

ReciprocalPolynomial, CharacteristicPolynomialOfAOrthogonalMatrixIsAReciprocalPolynomial

Major Section:

Reference

Type of Math Object:

Theorem

Parent:

## Mathematics Subject Classification

53D05*no label found*

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