For which $n, k$ is $S_{n,k}$ a basis? Fun algebra problem 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 $V$?
 A: I guess the answer is that they are a basis iff $k$ and $n$ are coprime. Here's one half of the proof.
For the sake of clarity, I'll assume that $V = \mathbb R^n$ and $(v_i)$ is the canonical basis
If $d$ divides both $n$ and $k$, let's $\zeta = \exp(2i\pi/d)$ and $f : \mathbb C^n \to \mathbb C$ the complex linear form defined by $f(x_1, \ldots, x_n) = \sum_{k=1}^n x_i \zeta^i$. Seen as vectors of $\mathbb C^n$, the elements of $S_{k,n}$ all belong to $\ker f$, so the determinant of this system of vectors is 0 and they are not a basis of $\mathbb R^n$. (That's one of the superpowers of the determinant: when you compute the determinant of a system of vectors, the formulas are the same in every field, so even a complex linear relation prevents real vectors to form a basis).
Added later: the other half of the proof.
If $n$ and $k$ are coprime, $k$ has an inverse modulo $n$. So you can find $p > 0$ and $q$ such that $pk = qn + 1$.
Now, let's $V$ be the (real) vector space generate by our vectors. In particular $V$ contains the sum of all the vectors of $S_{n,k}$, which is just $k\cdot(1,1, \ldots, 1)$. So $(1, 1, \ldots, 1) \in V$. Now, add $p$ vectors of $S_{n,k}$: the one beginning with $v_1$, the one beginning with $v_{k+1}$, the one beginning with $v_{2k+1}$ and so on (the indices are to be understood modulo n). Simply put, you take the first vector in $S_{n,k}$, then the one which begins just after the last one stops, and so on. It's just like tiling a floor! (except that the room is something like a 1-dimensional analogue of the world of Snake and that you don't stop when the floor is entirely tiled.)
The sum of this $p$ vectors has all its components equal to $q$, except for one which is $q+1$. So if you take the difference with $q\cdot(1,1, \ldots, 1)$, you get one of the vectors $(v_i)$. 
It's now easy to get the other vectors $(v_i)$: either you copy this "tiling" procedure with another first vector or you argue that $V$ is stable under cyclical permutation of the coordinates.
P.S.: actually, it's not to hard to use this tiling procedure to prove the necessary condition as well. Indeed if you start to make sums of vectors in the same way, with $d = \mathrm{gcd}(n,k) > 1$, the sum of the $n/d$ first vectors will be the constant vector $(k/d, \ldots, k/d)$. It's a novel way to prove that $(1,1, \ldots, 1)$ is in $V$, and together, these two ways give a (real) linear relation between the vectors. But I like my complex proof more, so I will not delete it.
A: 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.
