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Mar
17 |
comment |
Can a closed disc in the plane be partitioned into three disjoint sets which are pair-wise isometric?
This is probably stupid, but why not cutting the cake in three equal parts (taking only half of the boundary)? |
Mar
9 |
comment |
Applications of the Cayley-Hamilton theorem
@LSpice: well as commutator have zero trace, C-H (in dimension 2) says that $[A,B]^2$ is a scalar matrix, which therefore commutes with every other matrix $C$. One can also see this without mentionning C-H of course... |
Mar
4 |
awarded | Nice Question |
Mar
2 |
comment |
Applications of the Cayley-Hamilton theorem
One can use C-H to show some "universal" identities for matrices. For example, for every $2 \times 2$ matrices $A,B,C$, one has :$ [[A,B]^2,C]=0$ (where $[,]$ is the commutator). |
Jan
22 |
comment |
Exact determinant of a circulant matrix
This determinant is also equal to the resultant of $X^n-1$ and $P:=c0+...+c_{n-1}X^{n-1}$. This resultant can be computed efficiently by modular algorithms (as mentionned on wikipedia for example, en.wikipedia.org/wiki/Resultants#Computation). There are also so-called subresultant algorithms (variation on using Euclid algorithm to compute the resultant but avoiding to work in the rationals). |
Jan
5 |
comment |
Zeros of polynomials modulo non-prime
@user83633 : yes of course! I was not in a good day... |
Dec
18 |
comment |
Zeros of polynomials modulo non-prime
My previous comment is wrong: over Z/n, one root of a polynomial gives a linear factor, but two distinct roots don't give a factorization by a product of two linear factors (due to the non integrity of the ring). |
Dec
18 |
comment |
Zeros of polynomials modulo non-prime
Since roots give factorisation of polynomials even over the ring $\mathbf Z/n$, it is equivalent to search a characterisation of subsets $S \subset \mathbf Z/n$ s. t. for every $x$ not in $S$, $\prod_{s \in S} (x-s) \neq 0$. |
Dec
18 |
comment |
Zeros of polynomials modulo non-prime
@user83633: as the OP suggests with her example mod 6, this is not as simple (due to the condition that $p$ must be non zero outside of $S$). |
Dec
8 |
revised |
Concrete solution to the (oriented) Oberwolfach problem with one table
added 364 characters in body |
Dec
8 |
comment |
Concrete solution to the (oriented) Oberwolfach problem with one table
Actually, there is a concrete solution for $n=8$ on page 3 of Bermond, J.-C.; Faber, V., Decomposition of the complete directed graph into k-circuits.J. Combinatorial Theory Ser. B 21 (1976), no. 2, 146–155. But they say it was found on a computer so it does not help that much. |
Dec
8 |
comment |
Concrete solution to the (oriented) Oberwolfach problem with one table
One reference is : Alspach, Brian; Gavlas, Heather; Šajna, Mateja; Verrall, Helen Cycle decompositions. IV. Complete directed graphs and fixed length directed cycles. J. Combin. Theory Ser. A 103 (2003), no. 1, 165–208. But then one has to remount to older and older papers, treating special cases or equivalent problems. |
Dec
7 |
comment |
Concrete solution to the (oriented) Oberwolfach problem with one table
There is of course a non-oriented variant of the problem. I'll be glad to hear about it as well. |
Dec
7 |
revised |
Concrete solution to the (oriented) Oberwolfach problem with one table
added 119 characters in body |
Dec
7 |
asked | Concrete solution to the (oriented) Oberwolfach problem with one table |
Nov
19 |
revised |
When is a submanifold of $\mathbf R^n$ given by global equations?
edited tags |
Nov
18 |
revised |
Gysin exact sequence for a singular subvariety
added 7 characters in body |
Nov
18 |
revised |
Gysin exact sequence for a singular subvariety
added 40 characters in body |
Nov
18 |
revised |
Gysin exact sequence for a singular subvariety
edited tags |
Nov
17 |
awarded | Yearling |