Positive determinant of a matrix

Let $A \in \mathbb{R}$ be a matrix  such that $A^3=A +\mathbb{I}$. Prove that $\det A >0$.

Solution

Since $A^3=A+\mathbb{I}$ this means that $A$ is a root of the polynomial $x^3-x-1$. All eigenvalues of the matrix are roots of this polynomial. However, there is only one real root satisfying the equation and two conjugate complex numbers. But, the product of these two eigenvalues is real and positive.


Now, the determinant of the matrix is given by the product of the eigenvalues. Trivially, the real root of the polynomial is positive. Hence the determinant of the matrix is positive.