Suppose $A(p, n)=(a_{ij}(p))_{i, j \leq n}$ is an $n\times n$ random matrix over $\mathbb{F_2}$, with all its entries being i.i.d. and such that $P(a_{ij}(p) = 1) = p$, where $p$ is some real number from $[0; 1]$. What is the largest possible probability, that $A(p, n)$ is non-singular and with what $p$ is it reached?

Note, that $A(p, n)$ in non-singular iff $det(A(p, n)) = 1$.

Solution for $n=1$:

$det(A(p, 1)) = 1$ with probability $p$. The maximum of $det(A(p, 1))$ is $1$ and it is reached with $p = 1$.

Solution for $n = 2$:

$det(A(p, 2)) = 1$ with probability $2p^2(1 - p^2)$. The maximum of $P(det(A(p, 2))=1)$ is $\frac{1}{2}$ and it is reached with $p = \frac{1}{\sqrt{2}}$.

However, I would like to know some sort of general formula (or at least asymptotics).


After I failed to solve this problem using determinants, I tried to prove this using the fact that asquare matrix is non-singular iff its rows are linearly dependent. As there exists only one non-zero element in $\mathbb{F_2}$, we can write linear dependence of the vector system $\{v_i\}_{i \leq n}$ in $\mathbb{F}_2^n$ as $\forall S \subset \{1, ... , n\}$ such that $S \neq \emptyset$ we have $\sum_{i \in S} v_i \neq \overline{0}$. I know the probability that a given set of vectors with i.i.d. random entries Bernoulli distributed with parameter $p$ $\{v_i\}_{i \leq k}$ over $\mathbb{F}_2^n$ satisfy $\sum_{i = 1}^k v_i \neq \overline{0}$ is $(1 - \frac{p((1 - 2p)^k - 1)}{1 - 2p})$. However, I do not know how to proceed further in this direction.

  • 1
    $\begingroup$ (1) math.stackexchange.com/questions/54246/… $\endgroup$
    – d.k.o.
    Nov 30 '19 at 11:55
  • 1
    $\begingroup$ (2) math.stackexchange.com/questions/71288/… $\endgroup$
    – d.k.o.
    Nov 30 '19 at 11:55
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    $\begingroup$ Doesn't the question you marked as a duplicate (and also the other one in a comment above) only address the case $p=\frac12$? I think your question is considerably harder, since we can't use the symmetry in the way we used it in those other questions. Thus I don't think it's a duplicate? $\endgroup$
    – joriki
    Jan 18 '20 at 8:38
  • $\begingroup$ @joriki, when I marked this question as duplicate, I thought that the answers for $p = \frac{1}{2}$ can be easily generalised to the case of arbitrary $p$. Now I see, that it is not the case... I voted to reopen the question. $\endgroup$ Jan 18 '20 at 9:32
  • $\begingroup$ OK -- I've reopened the question. $\endgroup$
    – joriki
    Jan 18 '20 at 17:32

Random tests give the following results:












100;;plateau for $p\in [0.2,0.3]$;;0.290

It seems that the lower bound is $\approx 0.290$.

EDIT. In fact, it seems that the above lower bound is $\Pi_{k=1}^{\infty}(1-2^{-k})\approx 0.288788$, that is $lim_{n\rightarrow\infty} Prob(det(A)=1)$ when $p=1/2$.


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