I know that the Jordan Normal Form of a matrix is unique (up to reordering the Jordan blocks), but I don't really see why. Say we're looking at a 3x3 case. Now, all we need to do to compute the Jordan Normal form of the matrix is find eigenvalue(s) and get eigenvectors. Sure, we may not always get 3 linearly independent eigenvectors (which would then make the matrix diagolizable), but we can certainly find 3 generalized eigenvectors to form a Jordan basis. My question, however, is that can we only find 3 such vectors? Can't we find several other different generalized eigenvectors (by perhaps finding them for a different eigenvalue), giving us a different Jordan basis? Does rank-nullity somehow prevent this? I'd appreciate if someone could explain.
If my question is unclear, let me give an example: Say we find eigenvalues 2, 3, 3 for a matrix A, and find that ker$(A-2 I)$ and ker$(A-3 I)$ are both one-dimensional (and so we only get one 2 eigenvectors). Couldn't I now get different vectors v and w such that $(A-2 I)$v and ker$(A-3 I)$w give me each eigenvector respectively, in which case {both eigenvectors and v} and {both eigenvectors and w} would form Jordan bases. Does rank-nullity theorem somehow prevent me from finding two such distinct v and w?