A theorem states that every nilpotent linear transformation $f\in\mathrm{End}_F(V)$ has a basis such that the matrix of $f$ with respect to said basis is a Jordan matrix.
Is the converse also true or what would be a counterexample for if a matrix $A$ is a Jordan matrix, then the linear transformation corresponding to $A$ is nilpotent?
I know $\mathrm{dim}(V)=n,\mathrm{End}_F(V)\cong \mathrm{Mat}_n (F)$, but is it enough?
Or in other words, if I take any Jordan matrix and keep multiplying it by itself, does there always exist some integer at which point the power of the matrix becomes the 0 matrix?
In general no! For example, consider $I$, the identity matrix. It is in Jordan normal form, but it is clearly not nilpotent.
The converse you are looking for is probably that if a matrix is in Jordan normal form and all its eigenvalues are 0 then it is nilpotent.