I know that there's a theorem stating that if a linear operator A in complex space V has a characteristical polynomial p(λ) = $(λ - λ_1)^q_1*...*(λ- λ_r)^q_r$ where $ q_i $ is a multiplicity of eigenvalue $ λ_i $ then there is a basis in V in which operator matrix has a Jordan normal form.
But does each linear operator has such a basis? Maybe there are some limitations?
As you said, this is true for any linear operator that acts on a finite dimensional vector space over the complex field. This is not true in general. You don't have to search too hard for a counter example - just choose a matrix with complex eigenvalues over a real vector space - it cannot have a Jordan basis over $\mathbb{R}$, since this would mean that the diagonal would have complex entries.
The key fact here is that $\mathbb{C}$ is algebraically closed - any polynomial can be decomposed into linear factors. Without this property, you will have trouble finding a Jordan basis.