When can two linear operators on a finite-dimensional space be simultaneously Jordanized?

IN a comment to Qiaochu's answer here it is mentioned that two commuting matrices can be simultaneously Jordanized (sorry that this sounds less appealing then "diagonalized" :P ), i.e. can be brought to a Jordan normal form by the same similarity transformation. I was wondering about the converse - when can two linear operators acting on a finite-dimensional vector space (over an algebraically closed field) be simultaneously Jordanized? Unlike the case of simultaneous diagonalization, I don't think commutativity is forced on the transformations in this case, and I'm interested in other natural conditions which guarantee that this is possible.

EDIT: as Georges pointed out, the statements that two commuting matrices are simultaneously Jordanizable is in fact wrong. Nevertheless, I am still interested in interesting conditions on a pair of operators which ensures a simultaneous Jordanization (of course, there are some obvious sufficient conditions, i.e. that the two matrices are actually diagonalizable and commute, but this is not very appealing...)

Solution 1:

The comment you mention does not seem to be correct: in reality, over any field there exist commuting matrices which cannot be simultaneously Jordanized. Here is an example.

Let $n\geq 3$ and let $J_n$ be the $n$-th Jordan block, the $n\times n$ matrix whose entries are all $0$ except just above the diagonal where the $n-1$ entries equal 1.
I claim that although $J_n$ obviously commute with its square $J_n^2$, these matrices cannot be simultaneously Jordanized.
Indeed any matrix $P$ Jordanizing $J_n$ will satisfy $$P^{-1}J_nP=J_n$$ because of the uniqueness of Jordan forms.
But this will force (by squaring that equality) $$P^{-1}J_n^2P=J_n^2$$ which is not in Jordan form.
Hence no matrix $P$ can simultaneously Jordanize both $J_n$ and $J_n^2$.

Solution 2:

I am 2 years late, but I would like to leave a comment, because for matrices of order 2 exists a very simple criterion.

Thm: If $A,B$ are complex matrices of order 2 and not diagonalizable then $A$ and $B$ can be simultaneously Jordanized if and only if $A-B$ is a multiple of the identity.

Proof: Suppose $A-B=aId$.

Since $B$ is not diagonalizable then $B=RJR^{-1}$, where $J=\left(\begin{array}{cc} b & 1 \\ 0 & b\end{array}\right)$

Thus, $A= RJR^{-1}+aId=R(J+aId)R^{-1}=R\left(\begin{array}{cc} b+a & 1 \\ 0 & b+a\end{array}\right)R^{-1}$. Therefore $A$ and $B$ can be simultaneously Jordanized.

For the converse, let us suppose that $A$ and $B$ can be simultaneously Jordanized.

Since $A$ and $B$ are not diagonalizable then $A=RJ_AR^{-1}$ and $B=RJ_BR^{-1}$, where $J_A=\left(\begin{array}{cc} a & 1 \\ 0 & a\end{array}\right)$ and $J_B=\left(\begin{array}{cc} b & 1 \\ 0 & b\end{array}\right)$.

Therefore, $A-B=RJ_AR^{-1}-RJ_BR^{-1}=R(J_A-J_B)R^{-1}=R\left(\begin{array}{cc} a-b & 0 \\ 0 & a-b\end{array}\right)R^{-1}=(a-b)Id$. $\ \square$

Now, we can find many examples of matrices that commute and can not be simultaneously Jordanized.

Example: The matrices $\left(\begin{array}{cc} a & 1 \\ 0 & a\end{array}\right), \left(\begin{array}{cc} b & -1 \\ 0 & b\end{array}\right)$ are not diagonalizable and their difference is not a multiple of the identity, therefore they can not be simultaneously Jordanized. Notice that these matrices commute.