The gap Volume I left open
In Volume I you learned that an operator is diagonalizable exactly when it has an eigenbasis — a full basis of eigenvectors. That happens precisely when, for every eigenvalue, the geometric multiplicity (dimension of the eigenspace) equals the algebraic multiplicity (its multiplicity as a root of the characteristic polynomial). Most matrices you met behaved.
But consider A = [2, 1; 0, 2]. Its only eigenvalue is lambda = 2 with algebraic multiplicity 2, yet (A - 2I) = [0, 1; 0, 0] has a one-dimensional null space spanned by e1. There is just one eigenvector direction, not two. No eigenbasis exists; A is defective. Volume I's machinery stalls here.
Widening the kernel: powers of (A - lambda I)
An eigenvector satisfies (A - lambda I)v = 0. A generalized eigenvector of rank k satisfies the weaker condition (A - lambda I)^k v = 0 while (A - lambda I)^(k-1) v != 0. Eigenvectors are exactly the rank-1 case. The set of all of them for a fixed lambda is the generalized eigenspace.
The kernels nest and grow: ker(A - lambda I) is contained in ker(A - lambda I)^2 contained in ker(A - lambda I)^3, and so on. Each power can only enlarge the null space until it stabilizes. The chain of null spaces stops growing at some power, and when it does, the stable kernel is the entire generalized eigenspace — and its dimension equals the algebraic multiplicity. That is the key promise that fills Volume I's gap.
A = [2, 1;
0, 2] lambda = 2
(A - 2I) = [0, 1; 0, 0] ker has dim 1 -> { e1 }
(A - 2I)^2 = [0, 0; 0, 0] ker has dim 2 -> all of R^2
rank-1 (true eigenvector): v1 = e1, (A-2I) e1 = 0
rank-2 (generalized) : v2 = e2, (A-2I) e2 = e1 != 0
(A-2I)^2 e2 = 0Why this always works over C
Over the complex numbers the characteristic polynomial always splits into linear factors, so the algebraic multiplicities sum to n. Summing the generalized eigenspaces over all distinct eigenvalues gives a direct-sum decomposition of the whole space C^n. This is the primary decomposition in disguise: even when no eigenbasis exists, a generalized eigenbasis always does.