Timeline for Integer matrices with no integer eigenvalues
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2010 at 21:18 | comment | added | Hej | @Bjorn, thanks; at least now I am convinced that there is a possibility that this problem is undecidable. | |
| May 16, 2010 at 12:03 | comment | added | KConrad | Bjorn, indeed you are right. Vesa, if you read this I apologize. | |
| May 16, 2010 at 6:20 | comment | added | Bjorn Poonen | Wow, I didn't expect my answer to be so controversial! My proof proves what it claims to prove, nothing more. I answered only the question with a question mark: "Is it possible that a problem like this is undecidable?" I agree with all of you that this has no bearing on the answer for the two particular 2 by 2 matrices. @KConrad: I think that Vesa Halava is male. | |
| May 16, 2010 at 5:11 | comment | added | KConrad | Bjorn, I just started to look at the link you posted and in the introduction the author says she uses Paterson's method but also that in Chapter 3 (the very one you direct us to look at) she shows Paterson's method "does not suit for problems of 2 x 2 matrices". Hej's problem of course is specifically about 2 x 2 matrices, so do you really think there is a way to modify your argument to really work in the 2 x 2 case? | |
| May 16, 2010 at 5:06 | comment | added | Andrey Rekalo | Exactly! It seems the result of the paper is not applicable here. Besides, the authors claim that their particular problem for integer matrices is undecidable for $n\times n$ matrices, when $n>2$ and decidable for $2\times 2$-matrices. | |
| May 16, 2010 at 4:58 | comment | added | Wadim Zudilin | @Bjorn: But is it an answer to the question? In general, yes, it's undecidable but in this particular case... Almost every real number has irrationality exponent 2 (metric NT) but to find the irrationality exponent of a given number, like $\pi$, is a different task (diophantine NT). It looks like the original problem is quite diophantine, but it's really hard to write it "in symbols". | |
| May 16, 2010 at 4:53 | comment | added | Victor Protsak | But, but, but... how can there be a zero matrix in <A,B> is A and B are invertible? Maybe your general problem is too general! | |
| May 16, 2010 at 4:34 | history | answered | Bjorn Poonen | CC BY-SA 2.5 |