All Questions
14 questions
7
votes
2
answers
347
views
Matrices over $\mathbb{F}_p$ that have nonzero determinant under any element permutation
$\DeclareMathOperator\GL{GL}$A few months ago, the following discussion took place on AoPS, concerning matrices that have nonzero determinant under any permutation of their entries: https://...
8
votes
0
answers
170
views
Random walk on matrix until singularity
Consider a random walk on matrices, where one starts with the matrix $M=I_n$ and at each step randomly chooses an entry of $M$ to increase by $1$.
I’m interested in two things about this walk:
What’s ...
0
votes
0
answers
45
views
On full rank submatrices of a construction
Take two matrices $T_1$ and $T_2$ in $\mathbb Z^{n\times n}$ with entries uniformly in $[-b,b]\cap\mathbb Z$ at some $b>0$. The matrices will be of rank $n$ each with probability at least $1-\frac1{...
1
vote
1
answer
519
views
How typical are integer isometries on a hypercube? Littlewood-Offord problem for Bernoulli Gram matrices
Let $m\geq 3$ be fixed and $n\to\infty$. Consider $v=(v_j)_{j\leq m}$ with $v_1,\ldots,v_m\in \{-1,+1\}^n$. Let:
$N_I(v)$ be the number of sequences $u_1,\ldots,u_m\in \{-1,+1\}^n$ isometric to $v$ ...
2
votes
1
answer
150
views
Intersection of a lower dimensional space and a discrete set
Let $H\subset \mathbb{R}^n$ with dimension ${\rm dim}(H)=\ell<n$; let $S$ be a finite subset of reals.
My question is the following. Is it correct to say,
$$
{\rm card}(H \cap V)\leqslant |S|^\...
1
vote
1
answer
394
views
On rank of random $0/1$ matrices
It is known that a $0/1$ matrix picked from uniform distribution from $\{0,1\}^{n\times n}$ is non-singular with probability $1-o(1)$.
Fix an integer $t$.
Consider a random matrix formed the ...
1
vote
1
answer
121
views
Probability for high mutual coherence on all subsets of a Gaussian vector set
We examine as set of independent normal vectors:
$$ \forall i \in [N]\triangleq \{1,\dots,N\}:\,\mathbf{x}_{i}\sim\mathcal{N}\left(0,\mathbf{I}_{d}\right)$$
For any $\epsilon>0$ and $K\leq N$, we ...
5
votes
0
answers
235
views
Riemann theta function inequality for a class of large random matrices
The following is essentially the same question as in this previous post, but since I have completely re-formulated it (hopefully for the better ;-), I decided to post a new question instead of an edit....
3
votes
0
answers
202
views
Difficult Gaussian-sum inequality for large random Bernoulli-Toeplitz matrices
I have come across the following problem in an attempt to prove an entropy bound for large random Bernoulli-Toeplitz matrices (Conjecture 1 on p. 16 of this preprint by Clifford et al. 2015), which is ...
1
vote
2
answers
478
views
Worst case difference in rank by column-row swapping
Given a matrix $m\in\{-1,+1\}^{n\times n}$. Consider $m^\sigma$ to be collection of all matrices obtained from $m$ by permuting rows and columns.
Consider $\mathscr{M}[m^\sigma]$ to be collection of ...
3
votes
1
answer
85
views
Vanishing Restricted Isometric Constant
In compressed sensing, we are interested in the restricted isometry property. Suppose the design matrix is $n$ by $p$, consisting of $np$ iid $\mathcal{N}(0, 1/n)$ entries. Assume both $n$ and $p$ are ...
1
vote
1
answer
220
views
Probabilistic statement on matrix ranks
Given $A\in\{0,1\}^{n\times n}$ with $\operatorname{rank}(A)=r=2^{O((\log_2n)^{\frac{1}{c+1}})}$.
Denote $\mathsf{1_n}\in\{0,1\}^n$ as vector with $1$s.
Does
$$\lim_{n\rightarrow\infty}\mathsf{P_{A\...
1
vote
0
answers
55
views
Find a common expression (parameterized by y = 1/2, 1 and 2) uniting three (rational) polynomials in k [closed]
I am seeking a common "simple" expression (preferably/presumably a sum of products of linear factors, or sum of products of low-degree factors) uniting the three polynomials (parameterized by $y=\...
6
votes
0
answers
375
views
Kasteleyn, Gessel-Viennot and eigenvalues
The Kasteleyn matrix (for counting perfect matchings) and the Lindström-Gessel-Viennot matrix (for counting families of nonintersecting lattice paths) are tightly related, as observed many times by ...