2010
DOI: 10.1016/j.jfa.2009.04.016
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On the singularity probability of discrete random matrices

Abstract: Let n be a large integer and M n be an n by n complex matrix whose entries are independent (but not necessarily identically distributed) discrete random variables. The main goal of this paper is to prove a general upper bound for the probability that M n is singular. For a constant 0 < p < 1 and a constant positive integer r, we will define a property p-bounded of exponent r. Our main result shows that if the entries of M n satisfy this property, then the probability that M n is singular is at most (p 1/r + o(… Show more

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Cited by 136 publications
(180 citation statements)
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“…A motivating example is for random Bernoulli matrices B whose entries are ±1 valued symmetric random variables. If all entries are independent, it is conjectured that the singularity probability of B is (12+o(1))n, while the best current bound (12+o(1))n is due to Bourgain, Vu and Wood . The typical norm of the inverse in this case is ||B1||=O(n) , see .…”
Section: Introductionmentioning
confidence: 99%
“…A motivating example is for random Bernoulli matrices B whose entries are ±1 valued symmetric random variables. If all entries are independent, it is conjectured that the singularity probability of B is (12+o(1))n, while the best current bound (12+o(1))n is due to Bourgain, Vu and Wood . The typical norm of the inverse in this case is ||B1||=O(n) , see .…”
Section: Introductionmentioning
confidence: 99%
“…taking values −1 or 1 with probability 1/2 each, showing that the probability of singularity is bounded above by θ n for θ = .999. The value of θ has been improved by Tao and Vu [25], [26] to θ = 3/4 + o(1) and by Bourgain, Vu and Wood [5] to θ = 1/ √ 2 + o(1). Slinko [22] considered Ginibre random matrices whose entries have the same uniform distribution taking values in a finite set, proving also that the probability of singularity is O(n −1/2 ) as n → ∞.…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…Denote the SVD of the block rating matrix B by B = UBΣBV ⊤ B , then the SVD of R is simply R = U KΣB V ⊤ , where U = UC UB and V = VCVB. When r → ∞, B has full rank almost surely [7]. We will assume B is full rank in the following proofs, which implies that UBU ⊤ B = I and VBV ⊤ B = I.…”
Section: Proof Of Theoremmentioning
confidence: 99%