lattice of invariant subspaces
Collect every T-invariant subspace of V into one set, written Lat T. It is not just a list — it has structure: given two invariant subspaces, their intersection and their sum are again invariant. So Lat T is a lattice, ordered by inclusion, with meet = intersection and join = sum, bottom {0} and top V.
The shape of this lattice is a fingerprint of the operator. A scalar operator (T = c I) is the wildest: every subspace is invariant, so Lat T is the full subspace lattice. At the other extreme, a single Jordan block is the tamest: its invariant subspaces form a single chain, one of each dimension, totally ordered. Diagonalizable operators with distinct eigenvalues sit in between, with a Boolean lattice of 2^k subspaces from k eigenlines.
Here is the surprising part: non-diagonalizable operators have RICHER, more interesting lattices in the structural sense, not poorer ones. A nilpotent Jordan block forces its invariant subspaces into one rigid chain, encoding the whole Jordan structure. Studying Lat T abstractly is the gateway to the famous invariant subspace problem in infinite dimensions — does every bounded operator on a Hilbert space have a nontrivial closed invariant subspace? In finite dimensions over C the answer is always yes, by Schur.
The extremes: a scalar matrix makes everything invariant; a single Jordan block leaves only a single nested chain.
A single Jordan block has a TOTALLY ordered invariant-subspace lattice — exactly one invariant subspace of each dimension — which is the cleanest way to see why it is indecomposable.