How $L$ acts trivially on any one-dimensional $L$-module

Question

My question is from Humphreys Introduction to Lie Algebras and Representation Theory. There is a lemma in section 6.3 which says, if $\phi: L \to \mathfrak{gl}(V)$ is a representation of a semisimple Lie algebra $L$ then $\phi(L) \subset \mathfrak{sl}(V)$ and in particular $L$ acts trivially on any one-dimensional $L$-module.
I understand how the first part of the statement follows but not why $L$ acts trivially on any one-dimensional $L$-module. First, "acts trivially" is never defined. Does this mean $x \cdot v = \phi(x)v = 0$ for all $x \in L$ and $v \in V$? If so how do we get from $\phi(x)$ is a trace zero matrix to $\phi(L)(V) = 0$ where $V$ is a one-dimensional $L$-module?

Answer 1

Because a linear map $\psi$ from a $1$-dimensional vector space $V$ into itself can only be of the form $v\mapsto\lambda v$ and then $\lambda=\operatorname{tr}\psi$.

Answer 2

I like the answer above but I just realized this is very clear: $\mathfrak{sl}(V)$ are exactly the trace zero matrices and when dim$V=1$ we just have $\mathfrak{sl}(V)=0$.