Sorry for the delay; it’s getting crowded around here again.
Anyway, an irreducible module for a Lie algebra is a pretty straightforward concept: it’s a module such that its only submodules are and . As usual, Schur’s lemma tells us that any morphism between two irreducible modules is either or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the identity transformation. This, of course, doesn’t depend on any choice of basis.
A one-dimensional module will always be irreducible, if it exists. And a unique — up to isomorphism, of course — one-dimensional module will always exist for simple Lie algebras. Indeed, if is simple then we know that . Any one-dimensional representation must have its image in . But the only traceless matrix is the zero matrix. Setting for all does indeed give a valid representation of .