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.



Number of Jordan canonical forms for an nxn matrix
[<https://oeis.org/A001970>](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 $\pi(A(\zeta))=\zeta$.


