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Let $a_1,a_2,\ldots,a_k$ be elements of the Galois field of order $q^n$, say $\mathbb{F}_{q^n}$, with $n \geq k$.

By Lemma 3.51 of Lidl, Niederreiter-Finite Fields,

$\left| \begin{array}{cccc} a_1 & a_1^{q} & \cdots & a_1^{q^{k}} \\ a_2& a_2^{q} & \cdots & a_2^{q^{k}} \\ \cdots & \cdots& \cdots & \cdots \\ a_k & a_k^{q} & \cdots & a_k^{q^{k}} \\ \end{array} \right| =0$ if and only if $\{a_1,a_2,\ldots,a_k\}$ are linearly dependent over $\mathbb{F}_q$.

My question is:

is it true that:

$\left| \begin{array}{cccc} a_1 & a_1^{q^{h_1}} & \cdots & a_1^{q^{h_{k-1}}} \\ a_2 & a_2^{q^{h_1}} & \cdots & a_2^{q^{h_{k-1}}} \\ \cdots & \cdots& \cdots & \cdots \\ a_k & a_k^{q^{h_1}} & \cdots & a_k^{q^{h_{k-1}}} \\ \end{array} \right| =0$

if and only if $\{a_1,a_2,\ldots,a_k\}$ are linearly dependent over $\mathbb{F}_{q^{m}}$ where $$m=GCD(h_1,h_2,\ldots,h_{k-1},n)?$$

Question. It is obvious that if they are dependent, then that determinant is $0$. But, is the converse true as well?

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  • $\begingroup$ No, take $h_2=h_1$ for example. $\endgroup$ – Fedor Petrov May 16 '17 at 15:05
  • $\begingroup$ yes, obviusly they are all different. $\endgroup$ – user46071 May 16 '17 at 15:06
  • $\begingroup$ Take them equal modulo $n$. Well, I am sure that even if they are different modulo $n$ this does not imply the linear dependence. $\endgroup$ – Fedor Petrov May 16 '17 at 15:09

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