eigenvalues of an involution

Proof. For the first claim suppose $\lambda$ is an eigenvaluecorresponding 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 isthe 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 bothsides, 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.$\mathrm{\square }$