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.