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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](http://books.google.com/books?id=4dVCaFCcExcC&pg=PA262)...2011-09-25
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    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.2011-09-26
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    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.2014-05-06
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    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.2014-05-06

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