3
$\begingroup$

I asked this question in https://math.stackexchange.com/questions/204115/decomposition-of-matrices-in-semisimple-and-nilpotent-parts ​​but remains unanswered.

For any matrix $A\in M_n(\mathbb F)$, where $\mathbb F$ is an algebraically closed field, there is a matrix $S\in M_n(\mathbb F)$ such that

$$SAS^{-1}=D+N,$$ where $D$ is diagonal and $N$ nilpotent. Moreover, this decomposition is unique.

Suppose now that $A\in M_n(\mathbb K)$, but $\mathbb K$ is not necessarily algebraically closed. It is also true that there is a matrix $L\in M_n(\mathbb K)$ such that

$$LAL^{-1}=R+M,$$

where $M$ is nilpotent and $R$ is diagonalizable in the algebraic closure of $\mathbb K$? Moreover when we consider the decomposition in $\mathbb K$ and in the algebraic closure of $\mathbb K$ the nilpotent part is the same?

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Also $L$ is redundant. You may take $L=E$ (the identity matrix). $\endgroup$
    – Anton Klyachko
    Commented Sep 29, 2012 at 12:24
  • 3
    $\begingroup$ There is no uniqueness if you don't require that the two matrices in the decomposition commute. $\endgroup$
    – BS.
    Commented Sep 30, 2012 at 10:00

2 Answers 2

Reset to default
6
$\begingroup$

  1. You don't need to conjugate if you want $R$ to be diagonalizable (as opposed to diagonal).

  2. I assume you want $R$ and $M$ to have coefficients in $K$, otherwise just work in the algebraic closure.

  3. The statement is then true if $K$ is perfect and possibly false otherwise, as you can see by taking $A=[[0, 1], [t, 0]]$ in $K=F_2(t)$.

EDIT : see http://en.wikipedia.org/wiki/Jordan%E2%80%93Chevalley_decomposition

Improve this answer
$\endgroup$
3
  • $\begingroup$ We should demand that $R$ and $M$ commute. $\endgroup$
    – Anton Klyachko
    Commented Sep 29, 2012 at 15:58
  • 2
    $\begingroup$ The example in #3 (which readily adapts to any imperfect field $k$ using the $k$-linear multiplication by $a^{1/p}$ on $V = k(a^{1/p})$) isn't an entirely satisfactory counterexample because it is semisimple over $k$ (though not "geometrically semisimple"; i.e., not diagonalizable over $\overline{k}$). The Wikipedia entry has now been updated to give an example over any imperfect field $k$ in which the operator isn't a sum of two commuting $k$-linear operators that are respectively semisimple (just over $k$!) and nilpotent. $\endgroup$
    – grp
    Commented Oct 1, 2012 at 11:23
  • $\begingroup$ @grp, what is the definition of semisimplicity that is distinct from geometric semisimplicity? (For those who learned the Jordan decomposition from Borel, they are the same by definition.) Is it that the minimal polynomial is irreducible over $k$? $\endgroup$
    – LSpice
    Commented Aug 1, 2015 at 19:06
5
$\begingroup$

To amplify the points made by Laurent Berger, the literature I've seen (dating from around 1950) always specifies perfect fields; so I believe it was understood very early that Jordan-type decomposition breaks down for imperfect fields. I'm not sure whether this is discussed in the modern textbooks on theoretical linear algebra such as those by Curtis and Hoffman-Kunze. (In basic linear algebra it's rare to mention imperfect fields, though this becomes a real issue in Lie theory.)

Beyond the Jordan normal form for a matrix (originally developed over a field of characteristic 0 containing all the eigenvalues), the work of Chevalley has been essential for the more flexible notion of "Jordan decomposition" and related matrix polynomials over a perfect field not containing the eigenvalues. Of course he was motivated especially by the theory of linear algebraic groups, but even for computational linear algebra his viewpoint is historically important and justifies the term Jordan-Chevalley decomposition Much of the history has been written down in a joint paper by Danielle Couty and colleagues: see the arXiv preprint here.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .