are permutation coordinate of a point linear independent under cyclic group with odd length?

Question

suppose we have a cyclic group $C_{k}$ with odd length . if this group act on a point in $R^{k}$ with permutation coordinate, how I can show this points are linearly independent? if they are not can you give me an example?

for example if $C_{3}=<(1,2,3),e,(1,3,2)>$ acts on $(10,12,0)\in R^{3}$ then $(0,10,12)$ , $(10,12,0)$ and $(12,0,10)$ are linearly independent

Answer 1

What happens if you cyclically permute the entries in $(1, 0, -1)$?

If you start with any $k$ dimensional vector whose coordinates sum to $0$ then the sum of the $k$ permuted vectors will be $0$, so they won't be independent.

I suspect that they will be independent if the entries are positive. For linear independence that's equivalent to the sum being nonzero since you can add a constant vector to everything and not change the conclusion.

I'll leave it to someone else to provide a proof.

Edit. This turns out to be deeper and more interesting problem than I thought. The wikipedia page on circulant matrices provides this:

The polynomial $f(x)=c_{0}+c_{1}x+\dots +c_{{n-1}}x^{{n-1}}$ is called the associated polynomial of matrix $C$.

The rank of a circulant matrix $C$ is equal to $n − d$ , where $d$ is the degree of $\gcd (f(x),x^{n} -1) $.

https://en.wikipedia.org/wiki/Circulant_matrix https://en.wikipedia.org/wiki/Circulant_matrix