I wanted to find a fast way to construct the JNF (with basis transformation) of a $2\times 2$ Matrix which is not diagonalizable, which means that we need to have one eigenvalue with algebraic multiplicity 2 but geometric multiplicity being only 1.
My idea was to calculate calculate one eigenvector and construct the other basis vector somehow differently because we already know that the dimension of the nullspace of $(A- \lambda Id)^2$ is 2. So we know a lot, but is there any fast method now to evaluate the second basis vector with minimal effort?
The Jordan normal form of an endomorphism $A$ of a $2$-dimensional vector space that is not diagonalisable, but has an an eigenvalue (for example if the scalar field is algebraically closed, that is always the case) is
$$\begin{pmatrix}\lambda & 1\\ 0 & \lambda\end{pmatrix}$$
(or the transpose of that), where $\lambda$ is the eigenvalue. So for a basis $B = (v_1,\, v_2)$ that achieves that normal form, we have
Conversely, every pair of nonzero vectors with these properties is a Jordan basis for $A$.
So a quick way to find a Jordan basis in this situation is to randomly choose a $w \neq 0$,and compute $v = (A - \lambda I)w$. If $v \neq 0$, then $(v,w)$ is a Jordan basis for $A$. If $v = 0$, then pick any $w'$ that is linearly independent from $w$. Since the eigenspace is one-dimensional, $w'$ is then not an eigenvector, hence $\bigl((A-\lambda I)w',\, w'\bigr)$ is a Jordan basis.