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