Number of perturbations of the Jordan form

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.

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your result is equivalent to the following: a small petrubation of a single jordan block is a sum of jordan blocks with different eigenvalues. This follows form the fact that rank of $J - \lambda I$ is $> n-1$ and and the "norm" of the operator is large, i.e., a small pertrubation will not decrease the rank. – kassabov Nov 25 '11 at 16:14
It's clear in case of a single block. The problem with counting variants in cases where several blocks. – Alexander Nov 25 '11 at 16:48

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.)
We are working over $\mathbb{C}$. – Alexander Nov 25 '11 at 5:33
@Richard: Your reply needs two corrections. First, since we say orbit $O_2$ is a perturbation of $o_1$ iff $O_1\subseteq\overline{O_2}$, you should say $\mu\geq\lambda$ in the dominance order (this takes care of OP's comment). But second, one cannot assume nilpotent matrices (or any other constant characteristic polynomial): since diagonalizable matrices are dense, one always finds the partition (1,1,\ldots,1)\$ in a neigbourhood, although within the nilpotent cone it describes the smallest, closed, orbit. One can go up inside the cone, or down outside it. – Marc van Leeuwen Nov 26 '11 at 14:51