Consider the Lie algebra $sl\_2$ 
with the standard basis $(e,f,h),$ where
 \begin{equation\*}\label{sl2}
[h,e]=2\,e, [h,f]=-2\,f,[e,f]=h.
\end{equation\*}


Let $V$ be finite-dimensional $sl\_2$-module and let we know that element $e$ acts on $V$ as linear operator with a matrix $E$ (in some fixed basis).


**Question**. How one may find the matrix $F,H$ which correspond to elements $f$ and $h.?$


It is posible to recover $F,H$ from $E?$



Representation of Lie algebra sl_2.
Edited in light of clarifications made by OP:


Given a nilpotent matrix $E$ acting on a finite dimensional vector space $V$, it is always possible to extend it to a representation of $sl\_2$ in such a way that it represents $e$. The extension is almost never unique: conjugating the representing matrices $F$ and $H$ by anything in the centralizer of $E$ gives a new extension. 


The existence statement is the Jacobson-Morozov lemma (part of whose standard proof is reproduced in another answer) applied to the semisimple Lie algebra $sl(V)$. See Proposition 2 of section 2 of paragraph 11 of Bourbaki's "Lie Groups and Lie Algebras", Chapter VIII (see the Corollary following the Proposition for the extent to which uniqueness is true: basically, up to conjugacy).


On the other hand, if you have some additional rigid structure, there might be a unique extension. For instance, if you know a contravariant form and have an orthonormal basis at your disposal then $F$ is the transpose of $E$ (written in terms of the given orthonormal basis) and $H$ is determined by $H=[E,F]$.


