Number of Jordan forms from given characteristic polynomial and partitions

Question

Suppose I have a characteristic polynomial $$f(x) = (x-a_1)^{r_1}(x-a_2)^{r_2}\cdots(x-a_n)^{r_n}$$ of a matrix whose size is $m×m$, I began to think that how many Jordan forms are there corresponding to the above char. polynomial.

For $$f(x)= (x+1)^3 (5-x)^5$$ I have manually calculated that there are 21 such forms. (This was our home work).

But what I noticed that $21 = 7 × 3$ and $p(5)=7$ and $p(3)=3$ where $p$ is the partition function.

Then I found here Given the characteristic (and minimal) polynomial of $T:V\to V$, how many distinct Jordan forms are possible? that the number of Jordan forms is $p(r_1)p(r_2)\cdots p(r_n)$

What I think is that the number of Jordan forms corresponding to a particular $a_i$ is $p(r_i)$ and so total number is their product.

I am not able to prove the above observation rigorously,how to see a bijective relation between number of partitions and a Jordan block?

Answer 1

The point is that every linear term of the characteristic polynomial corresponds to the block of one of the eigenvalues, but each one of these eigenvalue blocks is broken down into subblocks corresponding to the eigevectors and their corresponding generalized eigenvectors.

To put an example, if our characteristic polynomial is $(x - 1)^4$ then we are talking about a $4\times 4$ matrix. The partitions of $4$ are the following: $$ 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1, $$ which would correspond to the canonical forms (the empty spaces are 0's) $$ \begin{pmatrix} 1&1&&\\ &1&1&\\ &&1&1\\ &&&1\\ \end{pmatrix}, \begin{pmatrix} 1&1&&\\ &1&1&\\ &&1&\\ &&&1\\ \end{pmatrix}, \begin{pmatrix} 1&1&&\\ &1&&\\ &&1&1\\ &&&1\\ \end{pmatrix}, \begin{pmatrix} 1&1&&\\ &1&&\\ &&1&\\ &&&1\\ \end{pmatrix}, \begin{pmatrix} 1&&&\\ &1&&\\ &&1&\\ &&&1\\ \end{pmatrix}, $$ Which ultimately correspond to saying about the transformation the following (respectively):

• It has exactly one eigenvector of eigenvalue 1,
• It has exactly two eigenvectors of eigenvalue 1, but one of them has a cyclic basis of length 3,
• It has exactly two eigenvectors of eigenvalue 1, each part of a cyclic basis of lenght 2,
• Three eigenvectors, and one of them is part of a cyclic basis of length 2,
• Four eigenvectors of eigenvalue 1, which is the diagonalizable case.

What we saw here can be repeated for every characteristic polynomial by analizying each one of its linear factors. The factor $(x - a_i)^{r_i}$ corresponds to a block of size $r_i\times r_i$, whose diagonal consists of $a_i$. The partition enters the picture in establishing in which positions of the superdiagonal there are $1$'s and where are $0$'s, precisely to divide the $r_i\times r_i$ big block into smaller blocks of the sizes given by the partitions and corresponding to the cyclic subspaces of the generalized eigenvectors.