# Simultaneous Jordanization?

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...)

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I have to wonder if anybody's ever seen the Kronecker canonical form... – J. M. Sep 25 '11 at 22:18
What we see here is that certain definitions of the Jordan normal form are inherently "flawed". The suggestion "upper triangularizable" probably inherits the same "flaws", because why should we prefer "upper triangularizable" over "lower triangularizable"? The "correct" way would probably just be a rank condition ($N_k^k=0$, $rank(N_k)=k-1$). Would be interesting to see what this means for the "Kronecker canonical form", which actually has important practical applications for "differential algebraic equations", whereas I'm not aware of any important applications of the Jordan normal form. – Thomas Klimpel Sep 26 '11 at 7:37
As the answer by Daniel shows, simultaneous Jordanisation is just too restrictive a condition: even if matrices have Jordan normal forms on bases that only differ by scaling of the basis vectors, it is unlikely that any scaling can be found that makes the above-diagonal entries $1$ simultaneaously. The simplest examples are pairs such as $(N,\lambda N)$ where $N=(\begin{smallmatrix}0&1\\0&0\end{smallmatrix})$ and $\lambda\notin\{0,1\}$: then $(e_1,e_2)$ being a Jordan basis for $N$ requires $e_1=Ne_2$, and for $\lambda N$ it requires $e_1=\lambda Ne_2$, obviously contradictory conditions. – Marc van Leeuwen May 6 '14 at 9:22
Actually I could have said that simpler: if $N$ is in Jordan Normal Form on some basis, then on that same basis $\lambda N$ has an off-diagonal coefficient $\lambda$, so it is not in JNF. – Marc van Leeuwen May 6 '14 at 15:27

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 $P^{-1}J_n^2P=J_n^2$, which is not in Jordan form. Hence no matrix $P$ can simultaneously Jordanize both $J_n$ and $J_n^2$.

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+1. A nice argument. I dare guess that the comment was meant to say that a set of commuting matrices is simultaneously upper triangularizable (over an algebraically closed base field). IIRC this follows from the Lie-Kolchin theorem. – Jyrki Lahtonen Jul 4 '11 at 12:40
Thanks! A very nice and clean argument, I might add. However, part of my question remains (I edited it accordingly). – Mark Jul 5 '11 at 2:07

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.

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This is nicely observed, +1. However, I think you should start your answer with with the example, for the case $a=b=0$, which shows that "simultaneously Jordanisable" is an extremely strict condition that even fails in very easy cases. The rest is just to show just how restrictive the condition is. – Marc van Leeuwen May 6 '14 at 9:09

There are many theorems about symmetric matrices in Prasolov's book if you are interested. But also, there are results such as

## Theorem 20.2.4

Let $A$ and $B$ be arbitrary complex square matrices and there is no nonzero column $x$ such that $x^*Ax = x^*Bx = 0$. Then there exists an invertible matrix $T$ such that $T^*AT$ and $T^*BT$ are triangular matrices.

I hope that counts as the closest next to Jordanization. Since the triangular hermitian matrices are the diagonal ones, many Hermitian matrix related results follow. You can find most of them classified in Horn and Johnson's book.

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If $T$ is not unitary (and the theorem does not seem to says so) then the fact that $T^*AT$ is triangular has no relation to triangularising $A$, which requires conjugation by an invertible matrix. IMHO this answer is completely unrelated to the question. – Marc van Leeuwen May 6 '14 at 9:01