Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Well-known and useful facts are:

  • any symmetric matrix over $\mathbb R$ is semi-simple (i.e. diagonalizable), and
  • any hermitean matrix over $\mathbb C$ is semi-simple.

I will loosely speak about the shape of a matrix and mean the existence of some (linear) relations between matrix-entries (or functions of the matrix-entries).

Question: Let $k$ be an algebraically closed field of characteristic $p$. Is there any result whatsoever, which says that a rich class of matrices of a given shape consists only of semi-simple matrices.

Since I am more interested in positive results, the notion of shape is kept flexible. However, if it could be proved that semi-simplicity is not implied by any shape in some reasonable class of shapes, this would be interesting as well.

share|improve this question
1  
I doubt any shape (other than diagonal) implies semisimplicity, as all finite Chevalley groups in characteristic $p$ have order divisible by $p$, hence have nontrivial unipotent elements. –  BS. Jan 13 '11 at 12:01
    
@Andreas: It seems unlikely that your loosely formulated question has an interesting answer. What's true is that the semisimple matrices form a Zariski-dense subset in the space of all square matrices; so there are plenty of them. But you are probably looking for a set of semisimple matrices given by polynomial conditions on entries, which would give a (proper) closed set in the Zariski topology. Still, "rich" is a flexible term. –  Jim Humphreys Jan 13 '11 at 14:06
    
Jim, since I am stuck with this problem, I fear that any concrete question will have a negative answer. That is why I am trying to open up the realm of possible answers. Anything non-trivial and positive would certainly count as an answer. (Rich is only supposed to exclude the class of diagonal matrices.) –  Andreas Thom Jan 13 '11 at 14:11
    
@BS: Wouldn't that only be an issue if the "shape" was required to be closed under multiplication? –  ndkrempel Jan 13 '11 at 15:22
    
@ndkrempel : indeed this is more a hint than a proof. I was thinking to the cases of (not necessarily definite) orthogonal and unitary groups over $R$, which have only semisimple elements exactly when this is the case for the corresponding (pseudo-) symmetric or hermitian matrices. There must be a link, but I don't know enough of algebraic group theory to elaborate on this. –  BS. Jan 13 '11 at 15:42
show 7 more comments

4 Answers 4

up vote 3 down vote accepted

This is only a hint, not an answer.

There is a simple characterization of semisimple matrices over finite fields. Namely, if $A\in M_n(F_q)$, its eigenvalues lie in $F_{q^m}$, $m=lcm(2,\dots,n)$, and there is $P\in GL_n(F_{q^m})$ such that $P^{-1}AP$ is a diagonal of Jordan blocks $\lambda_i I + N_i$, $i=1,\dots,s$. But it is easy to see that $(\lambda_i I+N_i)^{q^m}=\lambda_i I$ (note that $q^m \gt n$), so that $A$ is semisimple if and only if $A^{q^m}=A$.

Now you might want to start to study the possibilities for vector spaces $V\subset M_n(F_q)$ (or other subvarieties) such that $A^{q^m}=A$ for all $A\in V$.

share|improve this answer
    
Thanks a lot for this answer. –  Andreas Thom Jan 14 '11 at 19:37
add comment

A basic observation: If $K$ is a field where $0$ is a sum of nonzero squares, say $0=\sum_{i=1}^n x_i^2$, then $\left( x_i x_j \right)_{1 \leq i,j \leq n}$ is a symmetric, nonzero, matrix with square $0$. Such a matrix cannot be semisimple.

So the implication "symmetric implies semisimple" only works over formally real fields.

share|improve this answer
add comment

Well, distinct eigenvalues over the algebraic closure is enough to ensure semisimplicity, so the discriminant of the characteristic polynomial is a polynomial in the matrix entries whose non-vanishing ensures semisimplicity. If you prefer a closed set, you could require it to have a specific non-0 value, say 1. Whether this qualifies as a "shape" is another matter.

share|improve this answer
    
What if 90% of the eigenvalues are pairwise different. Can you "almost-diagonalize" the matrix then, for a suitable definition of "almost-diagonalize"? –  Łukasz Grabowski Jan 15 '11 at 17:37
add comment

There is a kind of spectral theorem describing a class of linear operators on Banach spaces over non-Archimedean fields possessing orthogonal (in the non-Archimedean sense) spectral decompositions. See A. N. Kochubei, Non-Archimedean normal operators, J. Math. Phys. 51 (2010), article 023526 (or ArXiv: 0908.4381).

share|improve this answer
    
imho it'd be more helpful if you describe the theorem roughly, and reference for the details to the paper. Or at least if you write which exactly theorem from the paper you are referring to. –  Łukasz Grabowski Jan 15 '11 at 17:55
    
For a matrix $A$ with the maximum of absolute values of elements equal to 1, form the matrix $A'$ of the images of elements in the residue field. If the field, over which the matrix $A$ is defined, contains all its eigenvalues, $A'$ is diagonalizable and not scalar, then $A$ has a spectral decomposition of the kind resembling classical Hermitian or normal matrices over complex numbers. The infinite-dimensional case is similar though it is formulated in a more complicated way. - Anatoly Kochubei –  Anatoly Kochubei Jan 15 '11 at 20:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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