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.

  • $\begingroup$ Some remarks: 1) $\mathbf{F}_q$ is canonically identified with a subset of $\mathbf{F}_{q^n}$ - it is the set of elements satisfying $x^q=x$. 2) Your displayed equation makes no sense to me. 3) Note that the rank of $M$ equals the dimension of the space spanned by the $\alpha_i$'s, by considering a sub-matrix corresponding to a maximal independent subset of those. $\endgroup$
    – Uri Bader
    Jun 27, 2017 at 8:56
  • $\begingroup$ Looking again at your displayed equation, probably you just had a typo there. If you meant to write $v\in (\mathbb{F}_q)^m$, then you got it right. In that case, it seems that what you have missed is the observation about the rank of $M$ (see above) and the resulting comparison of dimensions. $\endgroup$
    – Uri Bader
    Jun 27, 2017 at 9:10
  • $\begingroup$ Thanks for pointing out that finite fields embed canonically into each other! And sorry I didn't write $(\mathbb F_q)^n$ very clearly, edited! $\endgroup$ Jun 27, 2017 at 16:24

1 Answer 1


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.

  • $\begingroup$ Oh yeah that works! Perfect, thanks $\endgroup$ Jun 27, 2017 at 16:46

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