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

We learned in linear algebra that an $N \times N$ matrices can be placed into jordan normal forms with "blocks": $$ \left[ \begin{array}{cccc} \lambda & 1 & 0 & 0 \\\\ 0 & \lambda & 1 & 0 \\\\ 0 & 0 & \lambda & 1 \\\\ 0 & 0 & 0 & \lambda \end{array} \right] \hspace{0.25in}\text{or }\hspace{0.25in} \left[ \begin{array}{cccc} \lambda & 0 & 0 & 0 \\\\ 0 & \lambda & 0 & 0 \\\\ 0 & 0 & \lambda & 0 \\\\ 0 & 0 & 0 & \lambda \end{array} \right]$$

What is the analogue of "blocks" for a pair of $N \times N$ matrices, $\phi,\psi:V \to V$ ? and for a triple of matrices. $\phi, \psi,\rho: V \to V $ ?

share|improve this question
3  
I think that this is a "wild" problem (this has a precise technical meaning, but it means roughly that it is not possible to parametrize in a reasonable fashion pairs of matrices up to equivalence, ie pairs up to simultaneous conjugation in teh general linear group). –  Geoff Robinson Aug 28 '11 at 22:02
    
i mean... i knew this was a quiver problem to begin with and that the problem had no answer but i wasn't sure what to ask –  john mangual Aug 28 '11 at 22:19
    
if you restrict to certain quivers ("tame" or "finite type") or if you restrict to certain kinds of representations you get a complete classification. I was reading ams.org/notices/200502/fea-weyman.pdf and I had trouble following the distinctions they were making between "real", "imaginary" and "Schur" roots. –  john mangual Aug 28 '11 at 22:30
    
Are you sure the first sentence is correct? I guess you meant "an $N \times N$ matrix..." –  Qiaochu Yuan Aug 29 '11 at 2:06

2 Answers 2

I think you mean an $N\times N$ matrix in the first sentence.

Then the answer is the following: Then if you are interested in the most general case the answer is that there is no such normal form (if you are interested in one valid for all $N$). You can translate this problem by asking for representations of the algebra $k\langle X,Y\rangle$, the polynomial ring in two non-commuting variables. The representation theory is undecidable. Look also for the definition of a wild algebra.

If your algebra has special properties, then there might be an answer. For example, if you ask for pairs of commuting 2-nilpotent matrices, you can transfer this to the problem of the representation theory of $k[x,y]/(x^2,y^2)$, which is well-known, look for special biserial algebras.

With three variables, it does not even matter, what kind of additional conditions you impose (except maybe for trivial ones like two matrices to be equal or restriction of dimension). You will never be able to find a normal form (this is also related to wild algebras, see the $3$-Kronecker quiver)

If you need additional references, please ask in the comment.

share|improve this answer
    
What about pairs of arbitrary commuting matrices? I guess this should follow from the nilpotent case using some form of Jordan-Chevalley. –  Mark Aug 29 '11 at 8:16
    
@Geoff: I'm not an expert, but I think this is not true. See this article of Gel'fand and Ponomarev: springerlink.com/content/h24716h1462442v0 –  Julian Kuelshammer Aug 29 '11 at 12:21
2  
@Geoff: I think that your comment just allows you to reduce to the study of modules over $k[x,y]/(x^N,y^N)$, but that is still hard. –  Neil Strickland Aug 29 '11 at 15:16
    
@Neil: By hard you mean wild in general I guess. –  Julian Kuelshammer Aug 29 '11 at 15:35
    
That's unfortunate. Another question, though: in the cases where classification is feasible, can everything be made to work (under suitable hypotheses) in the infinite-dimensional setting as well? For example, an algebra of bounded linear operators on a separable Hilbert space, generated by two self-adjoint 2-nilpotent operators. –  Mark Aug 29 '11 at 21:06

For pairs of matrices $(A,B)$ (or rather, we prefer to think about the pencil $Ax-B$), there is the Weierstraß canonical form: there are $P,Q$ such that $PAQ$ and $PBQ$ are direct sums of blocks of four easy-to-describe types. I think this is described in the book by Gantmacher. There surely is a short description in Handbook of linear algebra by Hogben.

For triples of matrices under the same equivalence class, there should be no canonical form.

share|improve this answer
1  
I'm not sure how this answers the question, which (I assume) is about simultaneous conjugation. –  Qiaochu Yuan Aug 29 '11 at 20:28

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.