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.