Let $G(R^n)$ be the general linear group, and $g(R^n)$ the set of all linear transformations on $R^n$, and $\mathscr g$ the Lie algebra of $G(R^n)$. We define a function $J:\mathscr g \to g(R^n)$ by $<J(X)v,w>=X(e)f_{v,w}$, where $X\in \mathscr g$ and $e$ is the identity in $G(R^n)$ and $v,w\in \mathbb R^n$ and $f_{v,w}:G(R^n)\to \mathbb R$ is given by $T\mapsto <Tv,w>$.
My problem is I strongly believe that $<AJ(X)v,w>=X(A)f_{v,w}$ for all $A\in G(R^n)$ but I couldn’t prove it even using the fact that $X$ is invariant. How should I show that?
If I show that, in fact I will have established an isomorphism between the two Lie algebras above. By the way I know a more natural isomorphism between them so my question is not really about isomorphism.