Consider an $n \times n$ matrix $M$ over field $F$. Over any field (not necessarily algebraically closed) any matrix can be conjugated to the Frobenius normal form (e.g. Wikipedia). It becomes block-diagonal and so sizes of blocks define some partition of $n = \sum_i b_i $.
Consider centralizer of $M$, i.e. set: $C(M) = \{ B \in Mat(n,F) : BM=MB \} $ of all matrices which commute with $M$. It is clearly a linear subspace of $Mat(n,F)$ and so it has some dimension $d$.
Question 1: Is it true that dimension of the centralizer of a matrix depends only on the partition $n = \sum_i b_i $ (i.e. on integers $b_i$) appearing in the Frobenius normal form ?
If true:
Question 2: How to express the dimension of centralizer in terms of the partition $b_i$ ?
Question 2b: Is it true that the dimension of centralizer given by the formula $\sum t_i^2$, where $t_i$ comes from the transpose Young diagram to $b_i$ ?
If 1 is true:
Question 3: If $M$ is in Frobenius normal form, what can be the said about the form of matrices commuting with it ? For example if there are two blocks of unequal sizes, then seems commuting matrix will also have same block-diagonal form ?
Examples: if $M$ is a matrix at general position i.e. all eigenvalues are different (may be in some algebraic extension of $F$), then all above is true and simple: Frobenius normal form consists of just one block of maximal size $n\times n$, the space of commuting matrices has dimension $n$, which is $\sum 1^2$ (since $1+1+...+1$ is transpose Young diagram $n$), question 3 is tautologically true, since any matrix is block diagonal for block size is $n\times n$.
Other extreme is $M$ - scalar matrix (i.e. diagonal with all values on the diagonal are equal), its Frobenius normal form is $M$ itself, and we have $n$ blocks of sizes $1\times 1$ , everything commute with it - so $n^2$ is commutant dimension, which is $n^2$ since transpose to $n=1+1+...+1$ is partition of one term $n$.
In case matrix is diagonalizable and has eigenvalues $\lambda_i$ with multiplicities $c_i$, then dimension of the centralizer is clearly $\sum c_i^2$, so probably these $c_i$ corresponds to $t_i$ in the question ? (See also For non-diagonalizable matrices, the dimension of centralizer can be different from $\sum\limits_{j=1}^k d_j^2$ )
Example - more complicated: The sizes of blocks in FNF are exactly the degrees of polynomials $f_i$ entering the FNF theorem. They satisfy : $f_i|f_{i+1}$. That means:
if there are two blocks of the same size in FNF, then they are identically the same !
So consider example of matrix with FNF when all blocks of the same size, say $k$, so $k|n$, say $n=k*l$. Now it is
easy to describe what is commuting with a matrix which consists of several identical blocks: because it can be represented as tensor product $Block \otimes Id_{l} $, and hence $Com(Block) \otimes Mat(l) $ is commuting with such matrix.
$Com(Frobenius~ block)$ are just polynoms from block itself, so it has dimension as vector space equal to the size of the block i.e. to $k$. And hence dimension of the vector space $Com(Block) \otimes Mat(l) $ is equal to $k*l^2$. That is exactly was in required in questions 1,2,2b. Indeed Young diagram for our matrix is just rectangular $k \times l$ , so its transpose is $l\times k$, so summing up heights of columns we get $\sum_i l^2 = k*l^2$.
Remark: The top degree polynomial of $f_i$ coincides with the minimal polynomial. The product of all $f_i$ is equal to the characteristic polynomial.
PS
Nice discussion of centralizers is here: The dimension of centralizer of a Matrix. And also see answers here: https://math.stackexchange.com/a/1950688/21498
and here: https://mathoverflow.net/a/105076/10446
PSPS
All that should be standard and simple for experts, but I think worth to be discussed here.
