[Paper Review] Singularity of Bernoulli matrices in the sparse regime $pn = O(\log(n))$

Han Huang|arXiv (Cornell University)|Sep 29, 2020

Random Matrices and Applications19 references4 citations

TL;DR

This paper establishes the singularity of sparse Bernoulli matrices when $ pn = O(\log n) $, proving that singularity is asymptotically equivalent to the presence of zero rows or columns. Using probabilistic and combinatorial techniques, it extends the Litvak-Tikhomirov conjecture to the sparse regime, showing that the probability of singularity is dominated by zero rows/columns even as $ p $ decays logarithmically with $ n $.

ABSTRACT

Consider an $n imes n$ random matrix $A_n$ with i.i.d Bernoulli($p$) entries. In a recent result of Litvak-Tikhomirov, they proved the conjecture $$ \mathbb{P}\{\mbox{$A_n$ is singular}\}=(1+o_n(1)) \mathbb{P}\big\{\mbox{either a row or a column of $A_n$ equals zero}\big\}. $$ for $ C\frac{\log(n)}{n} \le p \le \frac{1}{C}$ for some large constant $C>1$. In this paper, we setted this conjecture in the sparse regime when $p$ satisfies $$ 1 \le \liminf_{n ightarrow \infty} \frac{pn}{\log(n)} \le \limsup_{n ightarrow \infty} \frac{pn}{\log(n)} < + \infty. $$

Motivation & Objective

• To extend the Litvak-Tikhomirov conjecture on Bernoulli matrix singularity to the sparse regime where $ pn = O(\log n) $.
• To determine whether the singularity of $ A_n $ is still primarily driven by the presence of zero rows or columns in the sparse setting.
• To analyze the asymptotic behavior of the singularity probability when $ p $ decays logarithmically with $ n $, beyond the dense regime.
• To establish that the dominant contribution to singularity comes from zero rows or columns, even in the sparse regime.

Proposed method

• Analyzes the probability that an $ n \times n $ random matrix with i.i.d. Bernoulli($ p $) entries is singular.
• Employs probabilistic bounds and combinatorial estimates to control the contribution of structured dependencies beyond zero rows/columns.
• Uses the second moment method and moment estimates to show that the probability of singularity is asymptotically equivalent to the probability of having a zero row or column.
• Applies concentration inequalities and tail bounds for binomial random variables to handle the sparse regime where $ p \to 0 $ as $ n \to \infty $.
• Considers the regime $ 1 \le \liminf \frac{pn}{\log n} \le \limsup \frac{pn}{\log n} < \infty $, ensuring $ p $ decays slowly enough to maintain non-trivial structure.
• Relies on symmetry and exchangeability of entries to simplify the analysis of rare events like singularity.

Experimental results

Research questions

• RQ1 In the sparse regime where $ pn = O(\log n) $, is the singularity of a Bernoulli matrix still primarily caused by zero rows or columns?
• RQ2 Does the Litvak-Tikhomirov conjecture on matrix singularity hold when $ p $ decays logarithmically with $ n $?
• RQ3 What is the asymptotic behavior of the probability that a sparse Bernoulli matrix is singular?
• RQ4 Are there non-zero row/column configurations that contribute significantly to singularity in the sparse regime?
• RQ5 How does the threshold $ pn = O(\log n) $ affect the likelihood of structural dependencies causing singularity?

Key findings

• The probability that an $ n \times n $ Bernoulli matrix with $ p $ satisfying $ pn = O(\log n) $ is singular is asymptotically equivalent to the probability that it has at least one zero row or column.
• The contribution of non-zero row/column dependencies to singularity is negligible in the sparse regime, confirming that zero rows/columns dominate.
• The result extends the Litvak-Tikhomirov conjecture to the sparse regime, validating its robustness under decaying $ p $.
• The analysis confirms that $ \mathbb{P}(\text{singularity}) = (1 + o_n(1)) \cdot \mathbb{P}(\text{zero row or column}) $ holds under the given sparse condition.
• The proof relies on controlling higher-order dependencies and showing their probability decays faster than the zero row/column event.
• The regime $ 1 \le \liminf \frac{pn}{\log n} \le \limsup \frac{pn}{\log n} < \infty $ is sufficient to maintain the dominance of zero rows/columns in singularity.