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.

  • 1
    $\begingroup$ The number of Jordan forms for a nilpotent $n \times n$ matrix is the number of partitions of $n,$ just because the sizes of its Jordan blocks must add up to $n.$ $\endgroup$ Jun 23, 2012 at 14:42
  • $\begingroup$ It's true only if matrix has a single eugenvalue. $\endgroup$
    – Alexander
    Jun 23, 2012 at 15:23

2 Answers 2


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 $\pi(A(\zeta))=\zeta$.

  • $\begingroup$ I have found confirmation in the article Peter J. CAMERON "Some sequences of integers" Discrete Mathematics 75 (1989) 89-102 p. 93. Thank you for your help. $\endgroup$
    – Alexander
    Jun 24, 2012 at 16:57

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.