A nowhere-zero point in a linear mapping

Importance: High ✭✭✭
Author(s): Jaeger, Francois
Subject: Combinatorics
» Matrices
Recomm. for undergrads: no
Prize: none
Posted by: mdevos
on: March 8th, 2007
Conjecture   If $ {\mathbb F} $ is a finite field with at least 4 elements and $ A $ is an invertible $ n \times n $ matrix with entries in $ {\mathbb F} $, then there are column vectors $ x,y \in {\mathbb F}^n $ which have no coordinates equal to zero such that $ Ax=y $.

The motivation for this problem comes from the study of nowhere-zero flows on graphs. If $ A $ is the directed incidence matrix of a graph $ G $, then a nowhere-zero $ {\mathbb F} $-flow on $ G $ is precisely a vector $ x $ so that $ x $ has all entries nonzero, and $ Ax=0 $. The above conjecture is similar, but is for general (invertible) matrices. Alon and Tarsi have resolved this conjecture for all fields not of prime order using their polynomial technique.

Definition: Say that a $ m \times n $ matrix $ A $ is $ (a,b) $-choosable if for all $ X_1,X_2,\ldots,X_m \subseteq {\mathbb F} $ with $ |X_i|=a $ and for all $ Y_1,Y_2,\ldots,Y_n \subseteq {\mathbb F} $ with $ |Y_j|=b $, there exists a vector $ x \in X_1 \times X_2 \ldots \times X_m $ and a vector $ y \in Y_1 \times Y_2 \ldots \times Y_n $ so that $ Ax=y $. Note that every matrix is $ (1,|{\mathbb F}|) $-choosable, but that an $ n \times n $ matrix is $ (|{\mathbb F}|,1) $-choosable if and only if it is invertible.

Alon and Tarsi actually prove a stronger property than Jaeger conjectured for fields not of prime order. They prove that if $ {\mathbb F} $ has characteristic $ p $, then every invertible matrix over $ {\mathbb F} $ is $ (p,|{\mathbb F}|-1) $-choosable. This result has been extended by DeVos [D] who showed that every such matrix is $ (p,|{\mathbb F}|-p+1) $-choosable. Yang Yu [Y] has verified that the conjecture holds for $ n \times n $ matrices with entries in $ {\mathbb Z}_p $ when $ n < 2^{p-2} $.

Jaeger's conjecture is true in a very strong sense for fields of characteristic 2. DeVos [D] proved that every invertible matrix over such a field is $ (k+1,|{\mathbb F}|-k) $-choosable for every $ k $. The following conjecture asserts that invertible matrices over fields of prime order have choosability properties nearly as strong.

Conjecture   [The choosability in $ {\mathbb Z}_p $ conjecture (DeVos)] Every invertible matrix with entries in $ {\mathbb Z}_p $ for a prime $ p $ is $ (k+2,p-k) $-choosable for every $ k $.

This is essentially the strongest choosability conjecture one might hope to be true over fields of prime order. I (M. DeVos) don't have any experimental evidence for this at all, so it could be false already for some small examples. However, I suspect that if The permanent conjecture is true, that this conjecture should also be true. In any case, I (M. DeVos) am offering a bottle of wine for this conjecture.


[A] N. Alon, Combinatorial Nullstellensatz, Combinatorics Probability and Computing 8 (1999) no. 1-2, 7-29. MathSciNet

[AT] N. Alon, M. Tarsi, A Nowhere-Zero Point in Linear Mappings, Combinatorica 9 (1989), 393-395. MathSciNet

[BBLS] R. Baker, J. Bonin, F. Lazebnik, and E. Shustin, On the number of nowhere-zero points in linear mappings, Combinatorica 14 (2) (1994), 149-157. MathSciNet

[D] M. DeVos, Matrix Choosability, J. Combinatorial Theory, Ser. A 90 (2000), 197-209. MathSciNet

[Y] Y. Yu, The Permanent Rank of a Matrix, J. Combinatorial Theory Ser. A 85 (1999), 237-242. MathSciNet

* indicates original appearance(s) of problem.