Relaxing the eigenvector equation
An eigenvector satisfies (T - lambda*I)v = 0. A generalized eigenvector only needs (T - lambda*I)^k v = 0 for some positive power k. So instead of being killed in one stroke of (T - lambda*I), it is killed after a few. The set of all such v for a fixed lambda is the generalized eigenspace, written G(lambda, T).
T = [ 5 1 0 ;
0 5 0 ;
0 0 5 ] (only eigenvalue lambda = 5)
N = T - 5*I = [ 0 1 0 ;
0 0 0 ;
0 0 0 ]
N e1 = 0 -> e1 is an ordinary eigenvector
N e3 = 0 -> e3 is an ordinary eigenvector
N e2 = e1 (not 0) -> e2 is NOT an eigenvector ...
N^2 e2 = N e1 = 0 -> ... but e2 IS a generalized eigenvector (k=2)
So G(5, T) = span{e1, e2, e3} = all of R^3,
even though there are only 2 independent ordinary eigenvectors.Fitting: split into kernel-part and image-part
Take any single operator N (think of N = T - lambda*I). The chains ker N ⊆ ker N^2 ⊆ ... and im N ⊇ im N^2 ⊇ ... both stabilize at the same power m. The Fitting decomposition then states V = ker N^m (+) im N^m, and both summands are N-invariant.
On the first summand ker N^m, the operator N is nilpotent (some power is zero). On the second summand im N^m, the operator N is invertible. So Fitting cleanly separates the part where T behaves like lambda plus a nilpotent shift from the part where T - lambda*I is harmless. This is the single-eigenvalue heart of the bigger theorem coming next.
Notice the connection back to Guide 2: the generalized eigenspace G(lambda, T) is exactly ker N^m, the kernel summand of Fitting for N = T - lambda*I. It is the largest invariant subspace on which T - lambda*I is nilpotent, and it always contains the cyclic subspaces generated by the eigenvectors of lambda.
Why this matters for canonical forms
Generalized eigenspaces never fall short the way ordinary eigenspaces do: the dimension of G(lambda, T) always equals the algebraic multiplicity of lambda. That guarantee is what lets the whole space be rebuilt from generalized eigenspaces — the topic of the next guide — and it is why Jordan form exists when diagonal form does not.