Nguyen's singularity conjecture for random matrices with fixed row sums

Let QnQ_n be a random matrix whose independent rows are chosen uniformly from the vectors in {0,1}n\{0,1\}^n having sum exactly n/2\lfloor n/2\rfloor. Nguyen's conjecture.

P[Qn is singular]=(12+on(1))n.\mathbb{P}[Q_n\emph{ is singular}]=\left(\frac{1}{2}+o_n(1)\right)^n.

This model is a dependent-entry analogue of random adjacency matrices of regular directed graphs and serves as a test bed for inverse Littlewood–Offord questions. The source states that the paper resolves Nguyen’s conjecture.

Sources & referencesView supporting material

Primary source

Vishesh Jain, Ashwin Sah and Mehtaab Sawhney, “Sharp invertibility of random Bernoulli matrices”, arXiv:2010.06553 (2021).

Additional references

2 papers in this index state this conjecture (2020). The statement above is taken from the most recent of them; the others are arXiv:2007.06318.

Progress summary

Refreshed
Open

The conjecture remains open: known work shows singular matrices are very rare, but not the precise exponential rate predicted.

Nguyen conjectured that for the fixed-row-sum random matrix model, the singularity probability is asymptotic to (1/2)n(1/2)^n. The conjecture was posed in Nguyen’s work after identifying equal rows as the apparent dominant source of singularity.

Known results

  • Nguyen proved, for even nn, that P(Qn is singular)=O(nC)\mathbb{P}(Q_n\text{ is singular})=O(n^{-C}) for every C>0C>0.
  • Equal rows give the lower bound P(Qn is singular)(1/2+o(1))n\mathbb{P}(Q_n\text{ is singular})\geq (1/2+o(1))^n.
  • Subsequent work proved exponential upper bounds, including P(Qn is singular)2ecn\mathbb{P}(Q_n\text{ is singular})\leq 2e^{-cn}, but not the conjectured asymptotic.

Current status (as of August 2026): The conjectured asymptotic remains unproved; only polynomial and exponential-scale nonsingularity estimates are established.

Sources

Solutions 0

No solutions have been posted yet.