How does the size of Jordan blocks in the Jordan form of matrix A relate to exponents in the factorization of the minimal polynomial?

Question

How does the size of Jordan blocks in the Jordan canonical form of matrix A relate to exponents $d_i$ in the factorization of the minimal polynomial $m_A(x) = (x-\lambda_1)^{d_1}(x-\lambda_2)^{d_2}\ldots(x-\lambda_m)^{d_m}$?

What does it say about the diagonalizability of A?

The reason I'm posting is that I'm out of my depth completely. What I can surmise is there will be at least one Jordan block the size of a particular exponent $d_i$. Please, if anyone decides to answer, provide at least a short explanation as I really want to understand this better. Thank you very much.

Answer 1

$d_i$ is the size of the largest Jordan block associated with $\lambda_i$. $A$ is diagonalizable if and only if all $d_i$ are equal to $1$ (i.e. there are no Jordan blocks with size larger than $1$).

Answer 2

The multiplicity $d_i$ of $\lambda_i$ as root of the minimal polynomial equals the maximal size of a Jordan block for $\lambda_i$ (and its multiplicity as root of the characteristic polynomial is the sum of the sizes of those Jordan blocks, as you probably know). In particular all roots of the minimal polynomial are simple, in other words all $d_i$ are $1$, if and only if all Jordan blocks are of size$~1$, or equivalently the matrix is diagonalisable.