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Cyclic Subspaces and the Failure of Complements

Build an invariant subspace from a single vector by repeatedly applying T, then meet the shear operator whose only invariant line has no invariant complement — the first sign that diagonalization is not guaranteed.

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 same operator gives a 1-dim cyclic subspace from one seed and a 2-dim one from another.

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?

  1. Any invariant line W must be spanned by an eigenvector (a 1-dim invariant subspace is an eigenline).
  2. But the only eigenline is U itself, so no second invariant line exists.
  3. 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.