Let \(A \in M_n(\mathbb{C})\) (or an algebraically closed field). How do you determine the number and sizes of the blocks in the Jordan Canonical Form of \(A\)?
Solution
Fix an eigenvalue \(\lambda\) and set \[ d_k=\dim\ker(A-\lambda I)^k, \qquad d_0=0. \] If the Jordan blocks for \(\lambda\) have sizes \(m_1,\dots,m_r\), then \[ d_k=\sum_{j=1}^r\min(k,m_j). \] Therefore \[ d_k-d_{k-1} \] is exactly the number of Jordan blocks of size at least \(k\).
Consequently: \[ \#\{\text{blocks for }\lambda\}=d_1, \] \[ \#\{\text{blocks of size exactly }k\} =(d_k-d_{k-1})-(d_{k+1}-d_k) =2d_k-d_{k-1}-d_{k+1}. \] The largest block size is the least \(m\) for which \[ d_m=d_{m+1}, \] and equivalently it is the exponent of \((x-\lambda)\) in the minimal polynomial of \(A\).
Thus the nullities of the powers \((A-\lambda I)^k\), for each eigenvalue \(\lambda\), determine all Jordan block numbers and sizes.