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 $\mathbb{F}_{q^{m}}$ with $m=GCD(h_1,h_2,\ldots,h_{k-1},n)$? I mean, it is obvious that if they are dependent, then that determinant is 0, but is the converse true as well?

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?