Please note: Below, I keep trying to put [ capital B ] but it gets turned into !! In Dennery and Krzywicki, they give an example of how to put a matrix in Jordan canonical form (pp. 167-170). They start with a 4x4 matrix [A] that looks kind of messy and transform it to a quasi-diagonal form [A]' = [B^(-1)] [A] that is less full (I'm distinguishing between operators A and their matrix representations [A]). So far so good. They go on to say that vectors in the null space of (A-2E)^3 are also in the null space of ([A]' - 2[E])^2. This, they say, shows that A has no generalized eigenvectors of rank 3. They seem to be assuming that if ([A]'-2[E])^2 [v]' = 0 then (A-2E)^2 |v> = 0 (|v> is a vector without reference to a particular basis, and [v]' = [B^(-1)] [v]). Then they would have (A-2E)^3 |v> = 0 implies that (A-2E)^2 |v> = 0 and this statement does indeed seem incompatible with the existence of a generalized eigenvector of rank 3 for A. Now for my big confusion: If ([A]'-2[E])^2 [v]' = 0 is just a particular representation of (A-2E)^2 |v> = 0, then it would seem that so is ([A]-2[E])^2 [v] = 0. But ([A]'-2[E])^2 [v]' = ([B^(-1)] [A] - 2[E])^2 [B^(-1)] [v] = [B^(-1)]^2 ([A] - 2[E])^2 ^2 [B^(-1)] [v] = [B^(-1)]^2 ([A] - 2[E])^2 [v] = 0. The operators ([A]'-2[E])^2 and ([A] - 2[E])^2 do not seem to have to same vectors in their nullspaces. There is a factor of that would have to be the identity matrix there in the last equation. So what, then, does it mean to talk about *the* nullspace of (A-2E)^2? What would particular representations of the operator look like?