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Let $V$ be a finite-dimensional real vector space, and let $C\subset V$ be a closed convex cone, not contained in a hyperplane, and such that $C\cap(-C)=\{0\} $. Let $n$ be a nilpotent endomorphism of $V$, with $\ n^r\neq 0$, $\ n^{r+1}=0\ (r>0)$ , and let $u=I+n$. If $r$ is odd, I claim that $u(C)\neq C$ (proof below). Does this still hold for $r$ even?

Proof for $r$ odd: assume $u(C)=C$. Let $v\in C$. Expanding $(1+n)^k$ we see that $\binom{k}{r}^{-1}u^k(v)\rightarrow n^r(v) $ as $k\rightarrow \infty$, hence $n^r(v)\in C$. Similarly taking $k\rightarrow -\infty $ gives $-n^r(v)\in C$, hence $n^r(v)=0$. Since $C$ spans $V$ this implies $n^r=0$, contradiction.

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  • $\begingroup$ Assume $u(C)=C$. Apply Jordan Normal Form Theorem (which holds over any field, if the characteristic polynomial is split) to get a pair of vectors $e_1,e_2$ with $u e_1=e_1$ and $ue_2=e_1+e_2$. Then for any $k\in\mathbb Z$ you have $u^kr_2=ke_1+e_2$. The affine line $\mathbb{R}e_1+e_2$ meets the cone, hence must be contained in the cone which is impossible. $\endgroup$
    – Antonius
    Commented Nov 8 at 10:26
  • $\begingroup$ @Nandor: "The affine line meets the cone, hence must be contained in the cone". Why??? $\endgroup$
    – abx
    Commented Nov 8 at 11:34
  • $\begingroup$ The line is invariant and each orbit on it leaves the cone. $\endgroup$
    – Antonius
    Commented Nov 8 at 12:36
  • $\begingroup$ @Nandor: I don't see at all why your line meets the cone. Think of a generatrix of the cone that you translate outside. $\endgroup$
    – abx
    Commented Nov 8 at 13:21
  • $\begingroup$ Ok, here is another proof: Let $A$ be the cone intersect with the sphere. Then $n$ maps the cone to itself, then we scale every vector and obtain a continuous self-map on $A$. By the Brouwer fixed-point theorem this map has a fixed point, hence $n$ has a positive eigenvalue, which it cant, being nilpotent. $\endgroup$
    – Antonius
    Commented Nov 8 at 13:54

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This answer gives a counter-example (with $r=2$).

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