Consider the matrix

$$ M=\begin{bmatrix} \alpha_1 & \alpha_2 & \dots & \alpha_m \\ \alpha_1^q & \alpha_2^q & \dots & \alpha_m^q \\ \alpha_1^{q^2} & \alpha_2^{q^2} & \dots & \alpha_m^{q^2} \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_1^{q^{m-1}} & \alpha_2^{q^{m-1}} & \dots & \alpha_m^{q^{m-1}} \\ \end{bmatrix} $$ over $\mathbb F_{q^n}$ for some prime power $q$. Such a matrix is sometimes called a Moore Matrix. It is a classically know result that $M$ is nonsingular if and only if $a_1, \ldots, a_m$ is linearly independent over $\mathbb F_q$ (see, for instance, [Lidl and Niederreiter][1] lemma 3.51).

My question is a strengthening of this result: do the linear dependencies of $\{a_1, \ldots, a_m\}$ over $\mathbb F_q$ completely determine the kernel of $M$ (i.e. the linear dependencies of the columns of $M$ over $\mathbb F_{q^n}$)? More precicely, do we have the equality $$ \mathrm{ker}M = \mathrm{span}_{\mathbb F_{q^n}} \{ v \in {\mathbb (F_q)}^n | Mv = 0 \} $$ Note that this requires identifying $\mathbb F_q$ as a subset of $\mathbb F_{q^n}$ in some way, which may introduce ambiguities/issues, but I am primarily interested in a case where $q$ is prime where this is not a problem. Of course, one inclusion is trivial, but I have been unable to arrive at the other.

For my purposes I am specifically interested in the case of $q=2$, for which I have been unable to find a counterexample with a brute force search in sage. In this case the hypothesis is equivalent to asking whether there exists some basis for $\mathrm{ker}M$ whose entries are all either $0$ or $1$.

Any leads are appreciated.

[1]: Rudolf Lidl and Harald Niederreiter. Introduction to finite fields and their applications. Cambridge university press. 1986.