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I've read in my Algebra book what seems to be like an easy way to find a canonical basis given a nilpotent mapping $N:V\to V$, and the corresponding cycle tableau. But I'm not sure I completely understand the method or if it can apply to the general case as well, since it is not given so clearly in the general case but rather is illustrated using an example.

A "cycle tableau", as I take it, is basically just understood in terms of columns of connected boxes. Where the number of boxes in the first column equals $\dim(\ker(N))$, in the second column $\dim(\ker(N^2))-\dim(\ker(N))$, in the third $\dim(\ker(N^3))-\dim(\ker(N^2))$, and so on until the final column contains only $1$ box. For an example, check here, which shows that for example $\dim(\ker(N^2))=5$, and $\dim(\ker(N^4))=8$.

The method goes like this: For each cycle of length $k$ corresponding to a row of length $k$ in the tableau, first take some non-zero vector $y\in \ker(N)\cap \operatorname{Img}(N^{k-1})$. Then some vector $x$ s.t. $N^{k-1}(x)=y$. Then take some eigenvectors of $N$, $y_2,y_3,...,y_n$ such that $\{y,y_2,y_3,...,y_n\}$ forms a linearly independent set, until you have the number of vectors in the cycle. Or in other parts of the text it says that "the remaining vectors in the cycle apart from $x,y$ will be $N(x),N^2(x),...,N^{k-2}(x)$? How is this the same? Then when you do this process for all cycles a canonical basis for $N$ would be the union of all the vectors you found.

Am I understanding the method correctly? And how could I apply it to the general problem of finding a canonical basis for nilpotent mappings?

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  • $\begingroup$ The notion of "cycle tableau" is not standard, and should be defined. I, for one, have no idea what it is. $\endgroup$ Apr 18, 2016 at 5:27
  • $\begingroup$ Edited with an explanation of what I mean. $\endgroup$ Apr 18, 2016 at 5:58

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Yes, sounds like you are just computing a cyclic basis.

Just compute the chain of kernels.

Then select vectors from each level of kernel so that you have an independent set.

E.g., be careful, vectors in the first kernel are (of course) again in the second kernel, third kernel,etc. That is, if you kill the vector on a first application of the operator $N$, then $N^2$ applied to the vector leaves it dead :)

With this basis, you will have carved out a cyclic subspace.

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  • $\begingroup$ Ok, now that you mention it I see that it is just computing a cyclic basis. But how does that give me a canonical basis for a nilpotent mapping? $\endgroup$ Apr 18, 2016 at 6:27
  • $\begingroup$ That is a canonical basis - and with this basis, suitably ordered, you can represent your nilpotent matrix in Jordan Canonical Form. Regarding your cyclic basis, what you really have is a Jordan Canonical Basis. Yours is an especially fundamental and simple computation of a Jordan form; since your mapping is nilpotent, the spectrum is just {0}. $\endgroup$
    – User001
    Apr 18, 2016 at 6:38

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