Prove that $\det(A) > 0$ Let $A \in \mathcal{M}_{n}(\mathbb{R})$ be a real $n \times n$ matrix such that : $A^{3} = A + \mathrm{I}_{n}$. Prove that $\det(A) > 0$. 
Here is what I tried : $X^{3}-X-1$ is a null polynomial of $A$. As a consequence : $\mathrm{Sp}(A) \subset \left\{ \text{roots of } X^3-X-1 \right\}$. Let $\alpha$ be the only real root of $X^{3}-X-1$. By studying the variations of $x \, \mapsto \, x^{3}-x-1$, we can easily prove that $\alpha > 0$. Since $X^{3}-X-1$ has no other real root and is a real polynomial, if $z \in \mathbb{C}$ is a root of $X^3-X-1$, then $\overline{z}$ is a root too. As a consequence, $\left\{ \text{roots of } X^3 -X-1 \right\} = \left\{ \alpha, z, \overline{z} \right\}$. The determinant of $A$ is equal to the product of the eigenvalues of $A$ (with multiplicity).
If $\mathrm{Sp}(A) = \left\{ \alpha \right\}$, then $\det(A) = \alpha^{n} > 0$. 
If $\mathrm{Sp}(A) = \left\{ \alpha, z, \overline{z} \right\}$, we must prove that $\overline{z}$ has the same multiplicity than $z$ (I'm not sure about this) so that : $\det(A) = \alpha^{m} \vert z \vert^{2p} > 0$ (where $m$ is the multiplicity of $\alpha$ in the characteristic polynomial and $p$ is the multiplicity of $z$). 
If $\mathrm{Sp}(A) = \left\{ z, \overline{z} \right\}$, then $\det(A) = \vert z \vert^{n} > 0$.  
Is this correct ?
 A: You have the right idea.
To prove multiplicity: it is a theorem that for any real polynomial $p(t)$, if $z$ is a root with multiplicity $p$, then so is $\overline{z}$.  Apply this to the polynomial $\det(A - tI)$.
A: From the eq. $A^3=A+I_n$ it readily follows that $A^{-1}=A^2-I_n$. That allows us to show first that $\det(A) \neq 0$. Moreover, from a straightforward application of the properties of determinants, we end up with the equality
$$\det(A-I_n)\det(A+I_n)=\frac{1}{\det(A)}.$$
To prove that $det(A)>0$, one needs to employ a contradiction argument: If one has that $det(A)<0$, by direct application of Bolzano theorem the auxiliar function $f(\lambda)=\det(A-\lambda I_n)$ (monic polynomial of degree $n$) has at least one [real] root, say $c$, on the open interval $]-1,1[$, since $f(1)f(-1)<0$. Moveover, for a given eigenpair $(c,v)$, one obtains the eigenvalue equation
$$ c^3 v =A^3v=(A+I_n)v=(c+1)v,$$
and whence, the equation $c^3=c+1$. This is turn allows us to show that $c^3$ is always positive, since $0<c+1<2$ (contradiction, because the inequality $-1<c<1$ implies $-1<c^3<1$?!).
