On complex irreducible representations of Lie algebras

In my lecture notes I have stumbled upon the following:

"Indeed, the statement [Lie's theorem] immediately implies that any such representation [complex representation of a solvable Lie algebra] contains an invariant subspace of dimension one, which in particular implies that irreducible complex representations are automatically one-dimensional and trivial."

Two questions here:

1. Why is the prerequisite "solvable" no longer needed in the last statement? Or is it just true for any representation of a solvable Lie algebra?

2. I get that it must be one-dimensional (cf. for example Complex irreducible representation of solvable lie algebra). But why does it follow that such representation must be trivial?

1. Solvability is still needed, because it is false otherwise. Take $L=\mathfrak{sl}(2)$. It has an irreducible representation in every dimension. Moreover, the field has to be algebraically closed of characteristic zero; otherwise there are counterexamples.
2. Sometimes "trivial" just means $1$-dimensional, even though the action is not by zero. Compare with the group case, which was answered here: Why does the trivial representation have degree 1?
• In my definition, the trivial representation is defined as the representation $\rho \colon \mathfrak{g} \rightarrow \mathfrak{gl}(V)$ where any $X \in \mathfrak{g}$ sends every element of $V$ to zero (i.e. $X$ acts by the zero map). How do I understand this in the context of your reference which is about representations of groups? – Jakob Elias Jan 20 '17 at 16:05
• @JakobElias Indeed, it does not follow. For example, abelian Lie algebras have plenty of non-trivial $1$-dimensional representations (I removed my previous comment since it made some wrong claims about nilpotent Lie algebras).. – Tobias Kildetoft Jan 20 '17 at 18:32