Matrices commutable with a given Jordan canonical form

Question

Let $K$ be a field and $A\in M_n(K)$ be a Jordan canonical form. I tried to find the dimension of $C(A)$ where $C(A)$ is the space of all matrices that commutes with $A$.
I think I solved this question, but I'm not confident. Could someone see my answer below and tell me if it is correct? If this is correct, what can be said about $\dim C(A)$ when $A$ is not necessarily triangularizable?

Let $a$ be an eigenvalue of $A$. Then, the generalized eigenspace corresponding to $a$ is stable under the transformation of $B$. Actually, if $(A-a)^nx=0$, $(A-a)^nBx=B(A-a)^nx=0$.
Because of this, we only have to consider the case where $A$ has only one eigenvalue. Moreover, we can assume that the only eigenvalue is zero, because $C(A)=C(A+a)$ for $a\in K$.
Now, $A$ is the form $\operatorname{diag}\{J_{d_1}, \ldots, J_{d_k}\}$ where $J_p$ is the nilpotent Jordan block of dimension $p$. Each $J_{d_s}(s=1,\ldots, k)$ has a corresponding subspace $V_s=\operatorname{span}\{e_{d_1+\ldots+d_{s-1}+1},\ldots, e_{d_1+\ldots+d_{s}}\}\subset K^n$ of dimension $d_s$ and obviously, $K^n=\oplus V_s$. Let $i_s$ be an inclusion $V_s\rightarrow K^n$ and $p_s$ be a projection $K^n\rightarrow V_s$.
$B\in C(A)$ is equivalent to $p_s\circ B\circ i_t\circ J_{d_t}$ and $J_{d_s}\circ p_s\circ B\circ i_t$ for all $s, t=1,\ldots, k$. Since the space of possible $p_s\circ B\circ i_t$ has a dimension $\min\{d_s, d_t\}$, $\dim C(A)=\Pi \min\{d_s, d_t\}$

I used these two pages as references: How to find the set of all matrices that commute with a given matrix? and What commutes with a matrix in Jordan canonical form?.

Answer 1

I vaguely remember that someone has mentioned in a comment on this site that there is a known formula for $\dim C(A)$ in terms of some combinatorial function (partition function?), but I cannot locate the relevant comment.

Anyway, suppose $X$ is a $p\times q$ rectangular matrix with $p,q>1$ such that $J_p(\lambda_1)X=XJ_q(\lambda_2)$. When $\lambda_1\ne\lambda_2$, $X$ has to be zero. When $\lambda_1=\lambda_2$ instead, let $d=|p-q|$. If we adopt the convention that Jordan blocks are upper triangular, then $X$ must take the form of $$ X = \begin{cases} T & p=q,\\ \pmatrix{T\\ 0_{d\times q}} & p>q,\\ \pmatrix{0_{p\times d}& T} & p<q \end{cases} $$ where $T$ is a $\min(p,q)\times\min(p,q)$ upper triangular Toeplitz matrix. Therefore the dimension of the matrix subspace containing all feasible choices of $X$ is given by $\min(p,q)$.

Now let $\lambda_1,\ldots,\lambda_r$ be the distinct eigenvalues of $A$. For each $\lambda_k$, let there be $b_k$ corresponding Jordan blocks in $A$, with sizes $s_{k1}\ge\cdots\ge s_{kb_k}$. It follows from our previous discussion that $$ \dim C(A) =\sum_{k=1}^r \sum_{i=1}^{b_k}\sum_{j=1}^{b_k}\min(s_{ki},s_{kj}) =\sum_{k=1}^r \sum_{i=1}^{b_k}(2i-1)s_{ki} =2\left(\sum_{k=1}^r \sum_{i=1}^{b_k}is_{ki}\right)-n. $$