# How many different structures of Jordan forms in $M(n,\mathbb{C})$?

The possible Jordan forms for $n \times n$ matrices in $\mathbb{C}$ can be found if we know that the characteristic polynomial is: $$p(x)=(x-\lambda_1)^{k_1}(x-\lambda_2)^{k_2}\cdots (x-\lambda_m)^{k_m}$$ for $1\le k_i\le n$ and with $k_1+k_2+\cdots+k_m=n$.

So, taking all possible partitions of $n$ we have different normal forms with different possible Jordan blocks, whose structure depends on the minimal polynomial and from the dimension of the generalized eigenspaces.

My problem is to find the number of such different forms of Jordan matrices for a fixed $n$ (note that I'm not interested on the actual value of the eigenvalues but only whether the eigenvalues in two blocks are equal or not).

I found many questions that asks essentially the same for special or general cases, but not a complete answer.

With a bit of work, using the possible forms of the characteristic and minimal polynomials, I foun $6$ possible forms for $n=3$ and $15$ forms for $n=4$ (if my work is correct), but I don't see a method to extend this result to a general case and the work of count all possibilites becomes really tedious if $n$ increase (clearly the result depends by the multiplicity of the eigenvalues in the characteristic and minimal polynomial).

There is some general result?

I found a possible answer here, but I don't agree with the result, so I submit my work to the scrutiny of Math-stackexchange. The idea is that the number of different Jordan blocks is ''the partitions of the partitions''.

Fo $4\times 4$ matrices this means that we start from the paritions of the integer $n=4$ that are the $p(4)=5$ partitions: $$4,\quad 3+1, \quad 2+2, \quad 2+1+1, \quad 1+1+1+1+1$$ These are the possible factorizations of the characteristic polynomial:

the partition $4$ correspond to the polynomial $(x-a)^4$,

$3+1$ corresponds to the polynomial $(x-a)^3(x-b)$

$2+2$ correspond to $(x-a)^2(x-b)^2$

$2+1+1$ corresponds to $(x-a)^2(x-b)(x-c)$

$1+1+1+1$ corresponds to $(x-a)(x-b)(x-c)(x-d)$

Now we take a second partition for any of such partitions that indicate the possible dimensions of the generalized eigenspace corresponding to an eigenvalue. So:

for $(x-a)^4$ we have $p(4)=5$ possibilities, i.e. the eigenspaces of dimension $4, \quad 3+1,\quad 2+2,\quad 2+1+1,\quad 1+1+1+1$

these numbers are the dimension of the Jordan blocks and correspond to the matrices: $$\begin{bmatrix} a&1&0&0\\ 0&a&1&0\\ 0&0&a&1\\ 0&0&0&a \end{bmatrix} \quad \begin{bmatrix} a&1&0&0\\ 0&a&1&0\\ 0&0&a&0\\ 0&0&0&a \end{bmatrix} \quad \begin{bmatrix} a&1&0&0\\ 0&a&0&0\\ 0&0&a&1\\ 0&0&0&a \end{bmatrix} \quad \begin{bmatrix} a&1&0&0\\ 0&a&0&0\\ 0&0&a&0\\ 0&0&0&a \end{bmatrix} \quad \begin{bmatrix} a&0&0&0\\ 0&a&0&0\\ 0&0&a&0\\ 0&0&0&a \end{bmatrix} \quad$$

for $(x-a)^3(x-b)$ we have the $p(3)\cdot p(1)=3$ possibilities : $(3,1), \quad (2+1,1), \quad (1+1+1,1)$ with the matrices:

$$\begin{bmatrix} a&1&0&0\\ 0&a&1&0\\ 0&0&a&0\\ 0&0&0&b \end{bmatrix} \quad \begin{bmatrix} a&1&0&0\\ 0&a&0&0\\ 0&0&a&0\\ 0&0&0&b \end{bmatrix} \quad \begin{bmatrix} a&0&0&0\\ 0&a&0&0\\ 0&0&a&0\\ 0&0&0&b \end{bmatrix}$$

for $(x-a)^2(x-b)^2$ we have $p(2)\cdot p(2)=4$ possibilities: $(2,2),\quad (2,1+1),\quad (1+1,2),\quad (1+1,1+1)$, with matrices $$\begin{bmatrix} a&1&0&0\\ 0&a&0&0\\ 0&0&b&1\\ 0&0&0&b \end{bmatrix} \quad \begin{bmatrix} a&1&0&0\\ 0&a&0&0\\ 0&0&b&0\\ 0&0&0&b \end{bmatrix} \quad \begin{bmatrix} a&0&0&0\\ 0&a&0&0\\ 0&0&b&1\\ 0&0&0&b \end{bmatrix} \quad \begin{bmatrix} a&0&0&0\\ 0&a&0&0\\ 0&0&b&0\\ 0&0&0&b \end{bmatrix} \quad$$

for $(x-a)^2(x-b)(x-c)$ we have $p(2)\cdot p(1) \cdot p(1)=2$ possibilities : $(2,1,1),\quad (1+1,1,1)$, with matrices $$\begin{bmatrix} a&1&0&0\\ 0&a&0&0\\ 0&0&b&0\\ 0&0&0&c \end{bmatrix} \quad \begin{bmatrix} a&0&0&0\\ 0&a&0&0\\ 0&0&b&0\\ 0&0&0&c \end{bmatrix}$$

for $(x-a)(x-b)(x-c)(x-d)$ we have only one possible matrix $$\begin{bmatrix} a&0&0&0\\ 0&b&0&0\\ 0&0&c&0\\ 0&0&0&d \end{bmatrix}$$

So I find a total of $15$ possible Jordan block forms. And this is the problem, because the reference gives , for $n=4$ a total of $14$ forms, and this is the number indicated in the referred A001970 sequence. Where is my mistake?

• I hope that now the question is well posed. – Emilio Novati Apr 15 '18 at 13:53
• Are your $2,1+1$ and $1+1,2$ really different? – ancient mathematician Apr 17 '18 at 6:51
• I think @ancient has it right. There's no way to distinguish $a$ from $b$, so those two are indistinguishable, so that's how to get 14. – Gerry Myerson Apr 17 '18 at 7:13
• I think it's one of these cases where one has to be very very careful about what counts as different. The way @EmilioNovati poses the question (the actual set of distinct ev is a given) perhaps leads to 15, but the canonical count (only worrying about ev being same or different) leads to 14. – ancient mathematician Apr 17 '18 at 7:53
• If $a$ and $b$ are different, then the Jordan blocks $[a]$ and $[b]$ are not similar, so there's an uncountable infinity of Jordan forms for $n=1$. I don't think we want to go there, Emilio. – Gerry Myerson Apr 19 '18 at 2:54

The short answer is that your counting is not really consistent. You distinguish the cases $(2,1+1)$ and $(1+1,2)$ but consider cases like $(3,1)$ and $(1,3)$ as the same. Because $4$ is so small this is the only "exception", but for larger dimensions it becomes a problem.

A slightly longer answer is this. There are at least two things which one can count.

(A) Given some $n$ and $k\leqslant n$, and distinct $\lambda_1,\dots,\lambda_k$ one can count how many conjugacy classes of $n\times n$ matrices there are which have eigenvalues $\lambda_1,\dots,\lambda_k$ (and no others).

I think this is close to your original question.

If one does this one has to distinguish in the $n=4$, $k=2$ case, between $(1+1,2)$ and $(2,1+1)$. But one also has to distinguish between $(3,1)$ and $(1,3)$, etc.

(B) More generally one can also count the number of "shapes" of Jordan forms. It's usually more useful, I think, and one can get the answers to (A) from this.

To make precise what it is that we are counting, again fix $n$, and consider the set $\mathcal{J}$ of all Jordan forms of $n\times n$ matrices. Now $\text{Symm}(\mathbb{C})$, the group of all permutations of $\mathbb{C}$, acts on $\mathcal{J}$ in the obvious way, by acting on each diagonal element of a Jordan matrix. Formally what we count is the number of orbits of $\text{Symm}(\mathbb{C})$ on $\mathcal{J}$.

In this case, since we can swap any pair $a\not= b$, we must not distinguish between $(2,1+1)$ and $(1+1,2)$.