Question

How many Jordan canonical forms may have an nxn matrix?

In the article https://oeis.org/A000219 states that the number of Jordan canonical forms for an nxn matrix is the Number of planar partitions of n. But calculating the normal form of $4\times4$ matrix I'm obtained 14 distinct Jordan forms, and for $5\times5$ I'm obtained 27 distinct Jordan forms. Author of the article (https://oeis.org/A000219) states that the normal form 2 11 can't be obtained. I do not understand this. For example $$ J=\begin{pmatrix} \lambda&1&0&0\\ 0&\lambda&0&0\\ 0&0&\lambda'&0\\ 0&0&0&\lambda'\ \end{pmatrix}. $$

I think that the number of Jordan canonical forms for an nxn matrix is the number of partitions of n. (https://oeis.org/A001970). But how to prove it, I do not know.

I would like to hear your opinion on this matter. Sorry for my english.

Answer 1

https://oeis.org/A001970 is the number of partitions of partitions (not the number of partitions). I agree this seems right. I also agree that in your notation 2 11 is a valid arrangement.

I think (as you suggest) the correct claim should be that the collection of Jordan normal forms for an $n\times n$ matrix is in bijection with the set $\mathcal P\mathcal P(n)$ defined as follows.

Let $\mathcal P(n)$ denote the collection of partitions of $n$. Equip $\mathcal P(n)$ with an arbitrary total order $\le_n$ and define $\mathcal P=\bigcup\mathcal P(n)$. For $\lambda\in\mathcal P$, let $n(\lambda)$ be the number of elements of which $\lambda$ is a partition and let $t(\lambda)$ be the number of pieces of the partition. Extend the orders defined on $\mathcal P(n)$ to a total order defined on $\mathcal P$ by $\lambda\le\lambda'$ if $n(\lambda) < n(\lambda')$ or $\lambda\le_k \lambda'$ if $n(\lambda)=n(\lambda')=k$

Let $\mathcal P\mathcal P(n)$ denote the collection of partitions of partitions of $n$. That is the collection of sequences $(\lambda_1,\lambda_2,\ldots,\lambda_j)$ in $\mathcal P^*$ such that $\lambda_1\ge \lambda_2\ge\ldots \lambda_j$ and $n(\lambda_1)+\ldots+n(\lambda_j)=n$.

Given an $n\times n$ matrix $A$, it has some number $j$ of distinct eigenvalues, $\alpha_1,\ldots,\alpha_j$. For each eigenvalue, let the dimension of the generalized eigenspace be $d_j$. The generalized eigenspace may then be expressed as a direct sum of Jordan blocks, giving a partition of $d_j$. We may assume that the $\alpha_i$ are ordered in such a way that the $d_j$ that the partitions are decreasing in the above sense. Hence we obtain from $A$ a unique element of $\mathcal P\mathcal P(n)$. Denote this mapping from a matrix to an element of $\mathcal P\mathcal P(n)$ by $\pi$.

Conversely, given an element $\zeta$ of $\mathcal P\mathcal P(n)$, let $\zeta=(\lambda_1,\ldots,\lambda_j)$ be the decreasing sequence of partitions such that $n(\lambda_1)+\ldots+n(\lambda_j)=n$. For each $\lambda_i$, we can find an $n(\lambda_i)\times n(\lambda_i)$ matrix $B_i$ with all generalized eigenvalues being $\alpha_i$, where the block structure of the Jordan blocks matches $\lambda_i$. Then let $A(\zeta)$ be the block-diagonal matrix with these blocks on the diagonal. We see that $\pi(A(\zeta))=\zeta$.

Answer 2

It looks like you found an error in a note in the OEIS.

Jordan canonical forms give you a partition of $n$ into the algebraic multiplicities of the set of eigenvalues, and the number of ways each multiplicity $m$ can be split into blocks is the number of partitions of $m$, so the number of Jordan canonical forms is the number of "partitions of partitions," A001970, or the Euler transform (#3) of the number of partitions.