Timeline for Is the set of power matrices decidable?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2018 at 21:22 | comment | added | François Brunault | It is not hard to bound the coefficients of the characteristic polynomial $\chi_A$ in terms of $M$ alone, so there are only a finite number of possibilities for $\chi_A$. Also, if the logarithmic Mahler measure $m(\chi_M)$ is $>0$ (which amounts to say $\chi_M$ is not a product of cyclotomic polynomials), then we can also bound $k$, because $m(\chi_M) = k m(\chi_A)$ and there is a lower bound for the Mahler measure of a polynomial of degree $n$ with integer coefficients. | |
| Dec 19, 2018 at 16:45 | comment | added | YCor | How close? this is just finding a much harder problem, that is undecidable. Linear algebra is simpler than algebraic geometry. | |
| Dec 19, 2018 at 15:03 | comment | added | Andrej Bauer | How close is this to asking whether solvability of diophantine equations in homogeneous polynomials is decidable? | |
| Dec 19, 2018 at 10:03 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |