In a exercise on Humphreys's book on Lie algebras, the reader is asked to view $M = \mathfrak{sl}(3,F)$ as an $L$-module via the adjoint representation. I'm having trouble on how the action of $L$ on $M$ can be defined.
Well, I know that given a representation $\phi$ of $L$ we can view any vector space $V$ as a $L$-module via the action $x\cdot v = \phi(x)(v).$ So in the case that $\phi: L \rightarrow \mathfrak{gl}(L)$ is the adjoint representation and $V =M$, the action of $x\in L$ on any $v\in M$ must be like: $x\cdot v = \mbox{ad} x(v)$. But how can this be defined? Since $M$ contains $3\times 3$ matrices and $L$ just $2\times 2$ matrices, it's not possible to define the bracket of $[x,v]$. So i bet the action via the adjoint representation must be something different. How is it possible to view $M$ as a $L$-module?
Hint : find matrices $E,F,H \in M$ such that the Lie algebra generated by $E,F,H$ is isomorphic to $L$.