Proof. For the first claim suppose $\lambda$ is an eigenvalue corresponding to an eigenvector $x$ of $A$ . That is, $Ax = \lambda x$ . Then $A^2x = \lambda Ax$ , so $x=\lambda^2x$ . 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.