# eigenvalues of an involution

*Proof.* For the first claim suppose $\lambda $ is an eigenvalue^{}
corresponding to an eigenvector^{} $x$ of $A$. That is, $Ax=\lambda x$.
Then ${A}^{2}x=\lambda Ax$, so $x={\lambda}^{2}x$. As an eigenvector, $x$ is non-zero, and
$\lambda =\pm 1$. Now property (1) follows since the determinant^{} is
the product of the eigenvalues. For property (2), suppose that
$A-\lambda I=-\lambda A(A-1/\lambda I)$, where $A$ and $\lambda $ are as above.
Taking the determinant of both
sides, and using part (1), and the properties of the determinant, yields

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

Property (2) follows. $\mathrm{\square}$

Title | eigenvalues of an involution |
---|---|

Canonical name | EigenvaluesOfAnInvolution |

Date of creation | 2013-03-22 13:38:57 |

Last modified on | 2013-03-22 13:38:57 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 4 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 15A21 |