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?

  • $\begingroup$ 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.) $\endgroup$ – Elliot G May 28 '19 at 7:04
  • $\begingroup$ 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. $\endgroup$ – Marc van Leeuwen May 28 '19 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!

  • 1
    $\begingroup$ Indeed, the full derivation of the uniqueness of the Jordan canonical form took up much of the second half of MA2101 Linear Algebra II at NUS. $\endgroup$ – Parcly Taxel May 28 '19 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.