# Given a Jordan canonical basis, how to find out to which generalized eigenspace picked generalized eigenvector belongs

Suppose we have finite-dimensional linear operator $$A:V\to V$$ , that has eigenvalues $$\lambda_1 ,\lambda_2, ... \lambda_n$$ . It is known that we can decompose $$V$$ into direct sum of generalized eigenspaces $$\Lambda_i$$, corresponding to each $$\lambda_i$$, with respect to $$A$$. $$V = \bigoplus_{i=1}^n \Lambda_i$$ Usually we do this decomposition by solving $$B_i^kv=0,$$ i.e finding $$\operatorname{Ker} B^k_i$$, where $$B^k_i = (A-\lambda_iI)^k$$. This is simple and gives us the complete structure of decomposition, that can be visualized as a diagram with arrows showing application of corresponding $$B_i$$ to basis vectors of each generalized eigenspace until all of them turn into $$0$$

$$\require{AMScd}$$ $$\begin{CD} e_5 @>B_1>> e_4 @>B_1>> e_3 @>B_1>>0\\ \\ @. e_6 @>B_1>> e_2 @>B_1>>0\\ \\ e_8 @>B_2>> e_7 @>B_2>> e_1 @>B_2>>0\\ \end{CD}$$ In the example above, $$V$$ is split into two generalized eigenspaces: the first is spanned by $$(e_5, e_4, e_3, e_6, e_2)$$, and the second is spanned by $$(e_8, e_7, e_1)$$. All of this is very good, but is it possible, given a canonical Jordan basis, i.e. set of generalized eigenvectors $$(e_i)$$, and a set of eigenvalues, to reconstruct which generalized eigenvectors belong to which generalized eigenspace $$\Lambda_i$$ ? In other words, is it possible to draw arrows in the diagram above, if we don't know them?

It is obvious that we can try each $$B_i$$ to see which of them sends (after some number of applications) the vector to $$0$$, but I'm keen if there is other approach, that doesn't require us to do such a bruteforcing. Every book I had a look at before asking this question doesn't ever mention any problem similar to this, or maybe I haven't read them thoroughly enough. Sorry if this question is trivial and I just can't see the solution, but I stuck with it really hard.