If your representation $V$ is irreducible, it carries a unique up to scalars non-degenerate "contravariant form", with the property that $E$ and $F$ are adjoint.  If you have chosen a self-dual basis (so the form looks like dot product) then the matrix form of $F$ is therefore the transpose of $E$.  The matrix $H$ is determined by $H=[E,F]$.

In the general case there are of course many "contravariant forms" on a reducible finite dimensional representation $V$, but perhaps there is a natural choice in whatever situation you are interested in.