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

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

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. ∎

Title | characteristic polynomial of a symplectic matrix is a reciprocal polynomial |
---|---|

Canonical name | CharacteristicPolynomialOfASymplecticMatrixIsAReciprocalPolynomial |

Date of creation | 2013-03-22 15:33:18 |

Last modified on | 2013-03-22 15:33:18 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 7 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 53D05 |

Related topic | ReciprocalPolynomial |

Related topic | CharacteristicPolynomialOfAOrthogonalMatrixIsAReciprocalPolynomial |