Jordan form of a nilpotent

A nilpotent operator N satisfies N^m = 0 for some m. Its only eigenvalue is 0, so its Jordan form has lambda = 0 in every block: pure shift blocks with 0s on the diagonal and 1s on the superdiagonal. This is the cleanest case, and the whole Jordan theorem is built on top of it.

The structure theorem for a nilpotent N on an n-dimensional space says: the space decomposes as a direct sum of cyclic subspaces, each spanned by a chain v, Nv, N^2 v, ..., on which N acts as a single nilpotent Jordan block. The block sizes are a partition of n, determined by the kernels of N, N^2, ... exactly as in the rank-jump formula (here lambda = 0).

Why it is the core case: for a general matrix A, restrict to the generalized eigenspace of an eigenvalue lambda; there A - lambda*I is nilpotent. Apply the nilpotent theorem to A - lambda*I, then add lambda*I back, and you get Jordan blocks for lambda. Doing this for every eigenvalue and stacking the results proves the full Jordan canonical form. So 'understand nilpotents' really means 'understand everything.'

A clean way to picture it: a nilpotent's chains are dots arranged in rows (one row per chain, longest at top); N shifts every dot one step toward its eigenvector and pushes the bottom row to zero. The column heights of that diagram are the Weyr characteristic.