The element $E$ is nilpotent; if the representation is irreducible, it has the maximal possible rank. Denote the rank $k$ so that the dimension of the underlying vector space $V$ will be $k+1$. There is a basis $(v_0,\ldots,v_k)$ of $V$ such that

1. $F$ will be a Jordan cell i.e. $Ev_i=v_{i-1}, Ev_1=0$. $v_{k}$ is the highest weight vector.

2. $Hv_i=(2i-k)v_i$.

3. $Ev_i=i(k-i+1)v_{i+1},Ev_k=0$.

These formulae in some form or another can be found in many sources (e.g in Serre's Lie algebra book).

The element $E$ is nilpotent; if the representation is irreducible, it has the maximal possible rank. Denote the rank $k$ so that the dimension of the underlying vector space $V$ will be $k+1$. There is a basis $(v_0,\ldots,v_k)$ of $V$ such that

1. $F$ will be a Jordan cell i.e. $Fv_i=v_{i-1}, Fv_0=0$. $v_{k}$ is the highest weight vector.

2. $Hv_i=(2i-k)v_i$.

3. $Ev_i=i(k-i+1)v_{i+1},Ev_k=0$.

These formulae in some form or another can be found in many sources (e.g in Serre's Lie algebra book).

[upd: one can reconstruct all this from $F$ given in some basis by setting $v_0$ to be any nonzero vector in $Im F^k=\ker F$ and then defining $v_1,\ldots, v_k$ by formula 3 above. With this definition formulae 1 and 2 will be true as well.]