What Vol I already gave you
Recall from Vol I that an eigenvector v of an operator T satisfies Tv = lambda*v. The line spanned by v — the set {c*v : c scalar} — has a quiet but crucial property: T never sends any vector on that line off of it. Apply T and you stay on the same line. That stability is the seed of everything in this track.
Diagonalization, the high point of Vol I, was really just the lucky case where the whole space splits into a direct sum of such lines. But many operators have no eigenbasis. So we keep the *stability* idea and drop the *one-dimensional* restriction.
The definition that generalizes the eigenline
A subspace U is an invariant subspace for T when T(U) is contained in U: every vector of U is mapped back inside U. The whole space V and the zero subspace {0} are always invariant — trivially. Eigenlines are the 1-dimensional invariant subspaces. The interesting question is what lives in between.
Let T act on R^3 by the matrix (in the standard basis)
T = [ 2 1 0 ;
0 2 0 ;
0 0 5 ]
Check a candidate subspace U = span{ e1, e2 } (the xy-plane).
T(e1) = 2*e1 -> in U
T(e2) = 1*e1 + 2*e2 -> in U (a combination of e1, e2)
So T(U) is contained in U: U is INVARIANT.
But e2 is NOT an eigenvector (T(e2) is not a multiple of e2).
The 2-dim invariant subspace exists even though only ONE
independent eigenvector lives inside it.Why invariance is the right lens
If U is invariant, then T restricts to a well-defined operator on U alone — written T|U. You can study that smaller operator in isolation. Decomposing V into invariant pieces is exactly the strategy of breaking a hard operator into smaller, understandable ones. The full collection of invariant subspaces forms the lattice of invariant subspaces, a structure that records every way T can be sliced.
When an invariant U comes paired with another invariant subspace W so that V = U (+) W, we call W an invariant complement of U. That pairing is the dream: it would let us study T on U and on W completely separately. The hard truth — explored next guide — is that the complement need not exist, and that obstruction is precisely what makes some operators refuse to diagonalize.