Number of perturbations of the Jordan form

Question

I am looking for information about the number of Jordan forms that can be obtained from a given Jordan form of a small perturbation.

For example, if a Jordan form consists of a single cell 2x2 $$J=\begin{pmatrix} \lambda_0 &1\\\ 0&\lambda_0\end{pmatrix},$$ by small perturbation $\begin{pmatrix} 0&0\\\ 0&\varepsilon\end{pmatrix}$, we can only get the matrix $$\begin{pmatrix} \lambda_0 &0\\\ 0&\lambda_1\end{pmatrix}$$ i.e. only two variants are possible.

If the Jordan form consists of a single cell 3x3, there may be such cases: $$J=\begin{pmatrix} \lambda_0 &1&0\\\ 0&\lambda_0&1\\\ 0&0&\lambda_0\end{pmatrix},$$ $$J+ \begin{pmatrix} 0 &0&0\\\ 0&0&0\\\ 0&0&\varepsilon\end{pmatrix}\sim\begin{pmatrix} \lambda_0 &1&0\\\ 0&\lambda_0&0\\\ 0&0&\lambda_1\end{pmatrix},$$ $$J+ \begin{pmatrix} 0 &0&0\\\ 0&\varepsilon_1&0\\\ 0&0&\varepsilon_2\end{pmatrix}\sim\begin{pmatrix} \lambda_0 &0&0\\\ 0&\lambda_1&0\\\ 0&0&\lambda_2\end{pmatrix}.$$

i.e. only three variants are possible.

I think I proved that if the Jordan form consists of a single cell mxm, then the number of variants equal to $p(m)$ (see http://en.wikipedia.org/wiki/Partition_%28number_theory%29).

It seems to me that these results have been obtained by someone, but I can not find them.

25.11 We are working over $\mathbb{C}$.

When we have a Jordan form $$J=\begin{pmatrix}\lambda_0&0\\\0&\lambda_0\end{pmatrix}\ \ \ \mbox{denote by}\ \ \ 1(\lambda_0)+1(\lambda_0),$$ we may obtained $$2(\lambda_0) \ \mbox{ or} \ \ 1(\lambda_0)+1(\lambda_1).$$ There are three variants.

If we have a Jordan form $2(\lambda_0)+1(\lambda_0)$ we may obtained $$3(\lambda_0), 2(\lambda_0)+1(\lambda_1), 1(\lambda_0)+1(\lambda_1)+1(\lambda_1),$$ $$1(\lambda_0)+1(\lambda_1)+1(\lambda_2).$$ There are five variants.

If we have a Jordan form $3(\lambda_0)+1(\lambda_0)$ we may obtained

$$4(\lambda_0), \ \ 3(\lambda_0)+1(\lambda_1), \ \ 2(\lambda_0)+2(\lambda_1),$$ $$2(\lambda_0)+1(\lambda_1)+1(\lambda_1), \ \ 2(\lambda_0)+1(\lambda_1)+1(\lambda_2),$$ $$1(\lambda_0)+2(\lambda_1)+1(\lambda_1)=2(\lambda_1)+1(\lambda_1)+1(\lambda_0),$$ $$1(\lambda_0)+1(\lambda_1)+1(\lambda_2)+1(\lambda_3).$$ There are eight variants. And so on.

Answer 1

We may suppose that all the eigenvalues of the $n\times n$ matrix $M$ are equal, say 0, and the same is true of the perturbed matrices (if we are only interested in the possible sizes of the Jordan blocks). Thus the Jordan form of $M$ is described by a partition $\lambda=(\lambda_1,\lambda_2,\dots)$ of $n$, where the $\lambda_i$'s are the block sizes. Then a partition $\mu=(\mu_1,\mu_2,\dots)$ describes the Jordan form of a perturbation if and only if $\mu\leq \lambda$ (dominance order), i.e., $\mu_1+\cdots+\mu_i\leq \lambda_1+\cdots+\lambda_i$ for all $i$. See http://en.wikipedia.org/wiki/Nilpotent_orbit. (If we are working over $\mathbb{C}$ then the Zariski closure of a nilpotent orbit will coincide with the closure under the "standard" topology on $n\times n$ matrices.)

Answer 2

Your question is strange in that it considers both cases where the characteristic polynomial changes and where it does not; see my comments to Richard Stanley's answer. It seems like if you break up a multiple root of the characteristic polynomial into smaller groups, each corresponding to a new root, then you can distribute each Jordan block size for the original eigenvalue into at most one block for each new eigenvalue. Then for any partitions into blocks so obtained for an eigenvalue (old or new), also take everything above it in the dominance order, to reflect the closure relation among nilpotent orbits. Just how many cases this gives is not so clear to me; you do not seem to care which newly created eigenvalue is close to which old one in distinguishing patterns, and this might affect the outcome.