Generalized Eigenspaces are Disjoint
We know that eigenspaces of distinct eigenvalues are disjoint. That is, no vector is an eigenvector of two different eigenvalues. This extends to generalized eigenvectors as well. If we have a nonzero then we know for sure that for .
To see this, let’s say that we have a that’s in two generalized eigenspaces, and start hitting it with over and over again. Eventually, it disappears, but before it does, it’s a nonzero eigenvector with eigenvalue . I say that this (not generalized) eigenvector is also in the generalized eigenspace of . Indeed, each time we hit with we write it as the sum . Since , the same is true of the scalar multiple . And since is a polynomial in , its kernel is invariant. Thus we have as well. And thus is invariant under the transformation , and we can assume (without loss of generality) that is an actual eigenvector with eigenvalue , as well as a generalized eigenvector with eigenvalue .
But now let’s start hitting with . Applying it once we find . So no matter how many times we apply the transformation, we just keep multiplying by , and we’ve assumed that this is nonzero. So cannot possibly be in the generalized eigenspace of . And so generalized eigenspaces corresponding to distinct eigenvalues can only intersect trivially.