TITLE: For which $n, k$ is $S_{n,k}$ a basis? Fun algebra problem
QUESTION [8 upvotes]: Here it is a nice algebra problem I had some fun with
Let $V$ be a vector space over $\mathbb R$ of finite dimension $\dim V = n$. Let $v = \{ v_1, \dots, v_n\}$ be a basis for $V$. Let $$S_{n,k} = \{ v_1 + v_2 + \dots + v_k, v_2 + \dots + v_{k+1}, \dots, v_n + v_1 + \dots + v_{k-1}\}$$ 
For which $n, k$ we have that $S_{n,k}$ is still a basis for $V$?

REPLY [2 votes]: Hint:  The set $S_{n,k}$  is a basis if  the  corresponding  determinant is not zero. But  this determinant is the Circulant.
The  rank of the circulant matrix equals $n-d$ where $d$ is degree of the polynomial $(1+x+\cdots+x^{k-1},x^n-1).$ Thus the cirulant is not zero iff the polinomials are coprime. Since  $1+x+\cdots+x^{k-1}=\dfrac{x^k-1}{x-1}$ we get that $(n,k)=1.$
So $S_{n,k}$  is a basis if $n$  and $k$ are coprime.