For any $x\neq0$ in a Lie algebra, is there always a matrix representation $\rho$ such that $\rho(x)^2\neq0$?

Question

For any $x\neq0$ in a Lie algebra $L$, is there always a matrix representation $\rho:L\to\mathfrak{gl}(V)$ such that $\rho(x)^2\neq0$ ?

(Of course $\rho(x)^2$ means ordinary multiplication/composition, not the commutator.)

All the spaces involved are finite-dimensional, over $\mathbb R$ (or some field with characteristic $0$, or not $2$; but maybe this is irrelevant).

This question generalizes to $\rho(x)^k\neq0$ for various $k$, and further to $\det\!\big(\rho(x)\big)\neq0$.

Answer 1

Yes (for arbitrary finite-dimensional Lie algebras over a field with $2\neq 0$).

Indeed, first using Ado's theorem, it is enough to assume that $\mathfrak{g}=\mathfrak{gl}_n(K)$. For a nonzero $x$ therein, if $x^2\neq 0$ we are done. Otherwise, we can suppose that $x=E_{1n}$ ($xe_n=e_1$, $xe_i=0$ for $i<n$). Then in the second symmetric power $S^2(K^n)$, we have $x(e_n\odot e_n)=2(e_1\odot e_n)$ and $x(e_1\odot e_n)=e_1\odot e_1$, so the corresponding operator doesn't square to $0$.

For $\rho(x)^k\neq 0$ you can argue similarly (probably excluding positive characteristic $\le k$).

For $\det\neq 0$, it does not work: for $E_{12}\in\mathfrak{sl}_2(K)$ in characteristic $0$ is mapped to a nilpotent operator by every finite-dimensional representation. (I don't know about positive characteristic.)