# 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. $\Box$

Title eigenvalues of an involution EigenvaluesOfAnInvolution 2013-03-22 13:38:57 2013-03-22 13:38:57 Koro (127) Koro (127) 4 Koro (127) Proof msc 15A21