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Let $n\geq2$ be a natural number and let $x_1,\ldots,x_n$ be $n$ real numbers. Is there a general sufficient condition to guarantee that the set of $n$ "cyclicly permuted" vectors $\left\{(x_1,x_2,\ldots,x_n), (x_n, x_1,\ldots,x_{n-1}), \ldots, (x_2,x_3,\ldots,x_1)\right\}$ is linearly independent?

When $n=2$ it is sufficient and necessary that $(x_1+x_2)(x_1-x_2)\ne0$. When $n=3$, the determinant of the matrix whose rows are those vectors is $(x_1+x_2+x_3)\left(\frac{{(x_1-x_2)}^2+{(x_1-x_3)}^2+{(x_2-x_3)}^2}{2}\right)$, so the necessary and sufficient condition is that the sum of those $3$ numbers is not zero, and they are not all the same number.

In general, the determinant of the matrix is $$ \prod_{k=0}^{n-1}\left(\sum_{\ell=0}^{n-1}e^{\frac{2\pi ik\ell}{n}}\cdot x_{\ell+1}\right). $$ But I am unable to deduce some intuitive condition for that to be non-zero.

Since I think that determinant is a product of discrete Fourier transforms, I also tag this as related to Fourier transform.

If this is inappropriate, I will remove that tag.

P.S.

The determinant can be found as Lemma 5.26 in Washington's Introduction to cyclotomic fields.

Any help is greatly appreciated.

awllower
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    $\sum x_k\ne0$ is always necessary; if $n$ is even then $\sum (-1)^k x_k\ne0$ is also necessary. It's also necessary that $(x_1,\dots,x_n)$ is not a nontrivial cyclic permutation of itself, in other words, that there does not exist $d\mid n$ such that $x_{k+d}=x_k$ always. One might speculate that these are the only necessary conditions. (The answer is probably simpler over the complex numbers, $-1$ is the only nontrivial root of unity in $\Bbb R$, which explains its appearance.) – Greg Martin Oct 21 '22 at 07:36
  • I think that when $n$ is prime then your answer (but not proof?) for $n=3$ generalises. This is a special case of @GregMartin's conjecture. I think one can come at this by looking at the action of the GG of the rou over the field $\mathbb{Q}(x_1,\dots, x_n)$. – ancient mathematician Oct 21 '22 at 07:48
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    I don't think there is any "intuitive" condition. Let's take $n=5$, and take $x_3,x_4,x_5$ arbitrary reals. Let $\beta=e^{2\pi i/5}$. Since ${1,\beta}$ is a linearly independent set over the reals, it's a basis for the complex numbers over the reals, so there exist reals $x_1,x_2$ such that $x_1+x_2\beta=-(x_3\beta^2+x_4\beta^3+x_5\beta^4)$, and you get determinant zero. – Gerry Myerson Oct 21 '22 at 08:01
  • @GerryMyerson Your argument shows that for any three numbers there is a unique pair of numbers such that the determinant is zero, but we don't know about the relation between these five numbers, except that the "Fourier transform" is zero. What I am asking is if there is a condition to determine when that "Fourier transform" is zero. Thanks for the comment! – awllower Oct 21 '22 at 08:25
  • @ancientmathematician I will happily accept it as an answer if a proof of GregMartin's conjecture is provided. :) – awllower Oct 21 '22 at 08:27
  • @GregMartin This is an interesting conjecture! Thanks for that! – awllower Oct 21 '22 at 08:28
  • We know there is no relation between the three numbers $x_3,x_4,x_5$, as they are arbitrary, and doesn't that mean there's nothing relating the five numbers other than $x_1+x_2\beta+x_3\beta^2+x_4\beta^3+x_5\beta^4=0$? I think this argument shows that it's not enough to avoid the situations given by @Greg. – Gerry Myerson Oct 21 '22 at 08:28
  • @GerryMyerson I see what you mean now. But your argument also works for $n=3$, and in that case there is a good condition, so I still have hope. ;P – awllower Oct 21 '22 at 08:32
  • @awllower For the alternating sum case: Let $(v_j)_j$ be the $n$ vectors considered. Then if $n$ is even, consider $\sum_j (-1)^jv_j$. Each coordinate of that becomes $\pm\sum_k (-1)^k x_k$. – Milten Oct 21 '22 at 08:32
  • @Milten Thanks for the help. I now see how to see the alternating sum being zero: take $k=n/2$. – awllower Oct 21 '22 at 08:34
  • When $n=3$, there's only one degree of freedom. Once you fix $x_1$, that fixes $x_2$ and $x_3$, so there's a formula for them in terms of $x_1$. That argument fails badly for $n=5$. – Gerry Myerson Oct 21 '22 at 08:48
  • @GerryMyerson Surely (at least when the $x_i$ are real transcendentals) from one relation $x_1+x_2\beta+\dots=0$ we get (by applying the automorphisms $\beta\mapsto\beta^j$) three other relations, and then see that the $x_i$ are all equal? – ancient mathematician Oct 21 '22 at 08:48
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    @anc, no, you can see that $x_3,x_4,x_5$ are arbitrary reals from the argument I gave. Your automorphisms are over the rationals, that is, they fix the rationals, but there's no way to extend them to general reals. – Gerry Myerson Oct 21 '22 at 08:51
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    @GerryMyerson. Thanks. Confused myself there. My argument only works over the rationals for $n=p$. – ancient mathematician Oct 21 '22 at 09:06
  • @GregMartin Have you any idea whether your "conjecture" (if that's not too strong a word) holds over the rationals in the case $n=105$? – ancient mathematician Oct 21 '22 at 09:09
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    @GerryMyerson has correctly broken my "conjecture"! One simple counterexample that comes from his construction is the vector $(\phi,-1,\phi,0,0)$ where $\phi=\frac12(\sqrt5+1)$ is the golden ratio; its set of 5 cyclic permutations spans a real vector space of dimension only 3. – Greg Martin Oct 21 '22 at 16:14
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    I'm guessing it might be helpful to look at the irreducible real representations of $\mathbb{Z}/n\mathbb{Z}$, which would probably be the trivial representation, the sign representation if $n$ is even, and then direct sums of the complex $k$ and $-k$ representations otherwise. I guess the corresponding basis vectors of the decompositions would be $[ \cos(2\pi k \ell/n) ]{\ell=0}^{n-1}$ and $[ \sin(2\pi k \ell/n) ]{\ell=0}^{n-1}$. – Daniel Schepler Oct 22 '22 at 03:29

1 Answers1

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A sufficient condition is that one of the elements dominates the others. Namely, for some $i\in[n]$, $|x_i|> \sum_{j\neq i}|x_{j}|$. Note that we can choose the matrix to have only $x_i$ on its diagonal entries. Then, our matrix is strictly diagonally dominant, making it non-singular.

Idontgetit
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  • Thanks for the answer. I will wait to see if there are better answers for a while, before accepting the answer. Thanks again. :) – awllower Oct 22 '22 at 05:12