# Why is Jordan Normal Form unique?

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?

• My (handwaving) answer is that you can sort of picture how the generalized eigenvectors fully determine the conjugacy class of the matrix. Once you have a set of $n$ linearly independent (generalized) eigenvectors, the action on the entire space is determined. (At least up to conjugation.) – Elliot G May 28 at 7:04
• Important point: the Jordan form is unique, but the same form is obtained for many different (Jordan) bases. This is even true for the diagonalisable case. – Marc van Leeuwen May 28 at 13:19

There's a lot to say about this if you want the full story, so the following is a very brief answer tailored specifically to the example in your second paragraph.$$\def\vec#1{{\bf#1}}$$
To find Jordan bases for $$\lambda$$ you need to consider the generalised eigenspace \eqalign{GE_\lambda &=\{\vec v\mid (A-\lambda I)^k\vec v=\vec 0\ \hbox{for some k\ge1}\,\}\cr &=\ker(A-\lambda I)\cup \ker(A-\lambda I)^2\cup \ker(A-\lambda I)^3\cup\cdots\ .\cr} In the example you have given you will find that $$\ker(A-2I)^2=\ker(A-2I)$$ so $$GE_2=\ker(A-2I)\cup\ker(A-2I)\cup\ker(A-2I)\cup\cdots=\ker(A-2I)$$ and this is $$1$$-dimensional. This means that the only vectors $$\vec x$$ satisfying $$(A-2I)\vec x=\vec 0$$ will be multiples of the eigenvector you have already, and your hypothetical vector $$\vec v$$ will not exist.
For the other eigenvalue you will have $$\ker(A-3I)\subsetneq \ker(A-3I)^2\quad\hbox{and}\quad \ker(A-3I)^2=\ker(A-3I)^3$$ (you can tell this from the dimensions) and so $$GE_3=\ker(A-3I)\cup\ker(A-3I)^2\cup\ker(A-3I)^2\cup\cdots=\ker(A-3I)^2$$ which is $$2$$-dimensional, so you will be able to find your $$\vec w$$.
Exercise. Try it with $$A=\pmatrix{5&1&1\cr -2&2&-1\cr -3&-1&1\cr}\ .$$ Confirm that the eigenvalues are $$2,3,3$$, find eigenvectors $$\vec x$$ and $$\vec y$$ for $$2$$ and $$3$$ respectively, confirm that $$(A-2I)\vec v=\vec x$$ has no solution but $$(A-3I)\vec w=\vec y$$ does. Good luck!