Let $E$ be a finite dimensional vector space over a field $K$. We say that an endomorphism $u:E\rightarrow E$ is simple if and only if the only subspaces of $E$ that are stable by $u$ are $\{0\}$ and $E$.
Let us denote by $\chi_u$ and by $\mu_u$ respectively the characteristic and the minimal polynomials of $u$.
1) Prove that if $u$ is simple, then $\mu_u = \chi_u$.
2) Prove that $u$ is simple if and only if $\mu_u$ is irreducible over $K$.
I was able to prove 1) by considering the subspace $\operatorname{Span}(x,u(x),\ldots ,u^{d-1}(x))$ of $E$ where $d$ is the degree of $\mu_u$ and $x$ is any nonzero vector of $E$.
I could also prove the direct implication of 2). If $u$ is simple and $\mu_u$ is written as a product of two polynomials $AB$, then the kernels and images of $A(u)$ and of $B(u)$ are stable by $u$, so must be either $\{0\}$ either the whole space $E$. It follows that $A(u)$ and $B(u)$ either are $0$, either are invertible. Because $AB(u)=0$, we then see that $\mu_u$ must divide $A$ or $B$, so that it is irreducible.
However, I have some doubts about the reverse implication. I suspect it may not be true. Have a look at the following conter-example. Take $E = \mathbb R^4$ with its usual basis $e=(e_1,e_2,e_3,e_4)$. Consider the endomorphism $u$ having matrix representation
$$ \operatorname{Mat}_e(u)= \left[ {\begin{array}{cccc} 0 & 0 & 0 & -1\\ 0 & 0 & -1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{array} } \right] $$
The minimal polynomial of $u$ is $X^2+1$ which is irreducible over $\mathbb R$. However, the subspace $\operatorname{Span}(e_1,e_4)$ is stable by $u$ since $u(e_1)=e_4$ and $u(e_4)=-e_1$.
Unless I am mistaken somewhere, there must be an error in this exercise. However, I couldn't come up with a correction for it. Would somebody know another characterization of simple endomorphisms ?
Thank you very much.