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.



Is the kernel of a Moore Matrix in $\mathbb F_{q^n}$ determined by the linear dependencies of the generators over $\mathbb{F}_q$?
Consider the vector $(\alpha\_1,\ldots,\alpha\_m)$ as the $\mathbf{F}\_q$-linear transformation $T:(\mathbf{F}\_q)^m\to \mathbf{F}\_{q^n}$ given by 
$$ (\beta\_1,\ldots,\beta\_m) \mapsto \sum \beta\_i\alpha\_i. $$
Then $\ker(M)$ is the $\mathbf{F}\_{q^n}$-span of $\ker(T)$ in $(\mathbf{F}\_{q^n})^m$.
To see this, note that $\ker(T)<\ker(M)$ and compare dimensions.


