I'm reading Introduction to Lie Algebras and Representation Theory from James Humphreys and I do not understand the statement made at the top of page 27.
Given a vectorspace $V$ (finite dimensional) over a field $F$ (char 0), there is a standard isomorphism of vector spaces $V^* \otimes V \rightarrow End(V)$ degined by $(f \otimes v)(w) = f(w)v$. Now if $V$ is an $L$-module ($L$ a lie algebra) then $End(V)$ can be viewed as an $L$-module by $(x.f)(v) = x.f(v) - f(x.v)$.
I want to verify this using the isomorphism of $V^* \otimes V$ and $End(V)$. However, I got stuck in the calculations. This is what I did:
Let $f(v) = \sum \lambda_{ij} e_i^*(v) e_j$ be an endomorphism of $V$ and $x \in L$. Then $(x, f)$ in $L \times End(V)$ can be identified with $ (x, \sum \lambda_{ij} e_i^* \otimes e_j) $ in $L \times (V^* \otimes V)$. "Differentiating" this element gives $ \sum \lambda_{ij} \left( (x.e_i^*) \otimes e_j + e_i^* \otimes (x.e_j) \right) $ in $V^* \otimes V$. Then this isomorphism discribed above evaluated in a vector $v \in V$ gives $ \begin{align} \sum \lambda_{ij} (x.e_i^*(v)e_j + e_i^*(v)(x.e_j)) &= - \sum \lambda_{ij} e_i^*(x.v)e_j + \sum \lambda_{ij} e_i^*(v)(x.e_j) \\ &= \sum \lambda_{ij} e_i^*(v)(x.e_j) - f(x.v). \end{align} $
I cannot manage to simplify this anymore. So I probably made a mistake somewhere. Could someone show me the correct calculations?
Edit: I think I found the answer already! Since $e_i^*(v) \in F$ it can be put trough the action $x$, hence $ \sum \lambda_{ij} e_i^*(v)(x.e_j) - f(x.v) = x.(\sum \lambda_{ij} e_i^*(v)e_j) - f(x.v) = x.f(v) - f(x.v). $ I think this is correct?