Growing a subspace from one seed vector
Pick any nonzero vector v and watch its orbit under T: v, Tv, T^2 v, T^3 v, ... The span of this orbit is the cyclic subspace generated by v. By construction it is invariant — applying T to any orbit element just gives the next one, which is already in the span. It is the smallest invariant subspace containing v.
The orbit can't keep growing forever inside a finite-dimensional space. At some step T^k v becomes a linear combination of the earlier vectors. The first such k is the dimension of the cyclic subspace, and the relation it produces is the operator's minimal polynomial restricted to v.
T = [ 2 1 ;
0 2 ] acting on R^2, seed v = e1 = (1,0)
T v = (2,0) = 2*v -> already a multiple of v
So span{v} alone is invariant: a 1-dim cyclic subspace.
Try instead the seed w = e2 = (0,1):
w = (0,1)
T w = (1,2) -> NOT a multiple of w
T^2 w = (4,4) = 4*w + ... -> a combination of w and T w
Cyclic subspace of w = span{ w, Tw } = all of R^2 (dimension 2).The shear that cannot be split
Return to the shear T = [2, 1; 0, 2]. Its only eigenvalue is 2, and up to scaling its only eigenvector is e1. So the only 1-dimensional invariant subspace is the line U = span{e1}. Now ask: is there an invariant complement — another invariant line W with R^2 = U (+) W?
- Any invariant line W must be spanned by an eigenvector (a 1-dim invariant subspace is an eigenline).
- But the only eigenline is U itself, so no second invariant line exists.
- Therefore U has NO invariant complement: R^2 cannot be split into two T-invariant lines.
What the obstruction is telling us
Diagonalization wanted V to be a direct sum of eigenlines, each with an invariant complement. The shear shows that even when invariant lines exist, complements can vanish. We have two ways forward: enlarge what counts as a 'good' vector beyond strict eigenvectors, or settle for triangular instead of diagonal. The next guide takes the first road; the last guide takes the second.