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.


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