AI & ML






A probabilistic construction of maximum $h$-scattered subspaces
arXiv math.IT0 views
arXiv:2610.32737v1 Announce Type: cross Abstract: Let $1\le h\le n-2$ and $g=\gcd(n,h+1)$. We prove that, for every prime power $q$, a maximum $h$-scattered $\mathbb{F}_q$-subspace of $V(r,q^n)$ exists whenever $(h+1)\mid rn$ and $$r\ge\frac{h+1}{g}\bigl(h(n-h-2)+1\bigr).$$ In particular, for fixed $n$ and minimum distance $\delta\in\{2,\ldots,n-1\}$, $\mathbb{F}_{q^n}$-linear MRD codes of length $N$ exist for every $q$ and every sufficiently large $N$ such that $n$ divides $(n-\delta+1)N$, with a bound on $N$ depending only on $n$ and $\delta$. We take the $\mathbb{F}_{q^s}$-span of the columns of a random matrix $A$ over the algebra $R=\mathbb{F}_{q^n}\otimes_{\mathbb{F}_q}\mathbb{F}_{q^s}$, for a suitable $s$. Such a subspace is maximum $h$-scattered if and only if the kernel of $A$ contains no nonzero vector whose expansion with respect to an $\mathbb{F}_q$-basis of $\mathbb{F}_{q^s}$ has rank weight at most $h+1$. When $\gcd(n,s)=1$, $R$ is a field, and a first-moment argument, counting these vectors up to multiplication by $\mathbb{F}_{q^n}^*\mathbb{F}_{q^s}^*$, shows that with positive probability the kernel contains none of them if $s>h(n-h-2)$. When $d=\gcd(n,s)>1$, $R$ is a product of $d$ fields, and vectors vanishing in some factors lie in the kernel with larger probability. We control them by proving that, for an $(h+1)$-dimensional $\mathbb{F}_q$-subspace $D$ of $\mathbb{F}_{q^n}$, the elements of $D\otimes_{\mathbb{F}_q}\mathbb{F}_{q^s}$ vanishing outside $k$ given factors, $0Source:https://arxiv.org/abs/2610.32737
PreviousA counterexample to Tokareva's conjecture on sums of two bent functionsNextVariance of relative entropy over random states
Related posts
Introducing the Grid Protectors Program
0OpenAI Blog
The world has a new plan to save nature, but it still needs the money
0Ars Technica

Could data center backlash flip Florida?
0The Verge

Vivo’s first Pro Max phone proves its Ultra is still in a league of its own
0The Verge

Why Peacock is making a huge bet on YouTube
0The Verge
Ninja’s first-ever microwave just replaced two of my kitchen appliances at once
0TechCrunch
show = false, 2600)"
x-show="show" x-cloak x-transition.opacity.duration.200ms
class="fixed bottom-8 left-1/2 z-[80] -translate-x-1/2 rounded-lg border border-line bg-card px-4 py-2.5 text-sm text-heading shadow-card-hover"
role="status" aria-live="polite">