Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Questions about the properties of vector spaces and linear transformations, including linear systems in general.
12
votes
Accepted
Vector of integers such that almost all dot products are positive
Sure, this is true. By the comment of Joseph Van Name we basically want a hyperplane passing through $0$ separating all the permutations of our numbers viewed as vectors in $\mathbb{R}^n$ from $(x_1, …
2
votes
Positive linear recurrent sequence
Even more is true: for all linear recurrences either $|a_k| \le xr^k$ for some $x>0, r < 1$ or $\limsup |a_k| > 0$.
Indeed, for any linear recurrence we have $a_k = \sum_{m = 1}^N b_m k^{c_m} d_m^k$ f …
4
votes
Accepted
An upper estimate for $|\det(A+B)|$
Okay, here is the proof that $C(n) = \frac{2^{n-1}}{\sqrt{n}^n}$.
Firstly, it is enough to find best constant $c(n)$ in $|\det(A)| \le c(n)||A||^n$. Indeed, we have
$$|\det(A+B)| \le c(n)||A+B||^n \l …