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 ?
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 $\det(A - tI)$.