I'm trying to get a hold of it with various examples and simplify the process as much as possible. For this one: $A=\left(\begin{matrix}2 & 1\\ -1 & 0 \end{matrix}\right)$ it's quite obvious that the JNF is: $J=J_{2}\left(1\right)=\left(\begin{matrix}1 & 1\\ 0 & 1 \end{matrix}\right)$
Now, to find the basis which consists $P$ s.t. $J=P^{-1}AP$, I'll denote $N=A-I$ and consecutively take the differnce between the kernels, meaning:
$\lambda=1\Rightarrow\ker N\\=\ker\left(\begin{matrix}1 & 1\\ -1 & -1 \end{matrix}\right)=\text{Span}\left\{ v=\left(-1,1\right)\right\} \\\Rightarrow\ker N^{2}\setminus\ker N=\\=\mathbb{R}^{2}\setminus\text{Span}\left\{ v=\left(-1,1\right)\right\} \\(**)=\text{Span}\left\{ u=\left(-1,0\right)\right\} $
The thing is, in the stage denoted (**) I could have taken another 2 or 3 options of $u\in \mathbb R ^{2} \setminus \text{Sp}{(-1,1)}$ but only $(-1,0)$ gives me the desired $P$.
Right now I'm guessing and checking, but isn't there a systematic way to choose the right $u$? Also, supposed $n\geq3$ (still, a single eigenvalue), could I have generalized such method to any $n$?
You have to make sure that the second vector $v$ you find satisfies $(A-\lambda I)^2v = 0$ but $(A-\lambda I)v \ne 0$