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Given a $n$ dimensional vector space over the finite field $F_q$, called $V(F_q)$, and a set of $M$ vectors $\vec c_m=(c_0,c_1,...c_{n-1})^T$ that fulfill $(\sum c_k )=0$. Further the complete set of $\{\vec c_m\}$ is symmetric under all permutations of elements of a given $\vec c_m$.

I'm searching $\vec s\in V(F_q)$, the smallest (in the sense of the taxicab norm) element of $V(F_q)$ which has:

$$ \vec s \cdot \vec c_m \neq 0 \bmod q, \forall m$$

If each $\vec c_m$ makes up an $n-1$ dimensional subspace, orthogonal to $\vec c_m$, my feeling says that $\vec s$ lies in the complement of union of all these subspaces. If and how is this related to Grassmanians? I ask because the application of Gaussian Binomial Coefficients tells us that

the Gaussian binomial coefficient ${\displaystyle {n \choose k}_{q}}$ counts the number of $k$-dimensional vector subspaces of an $n$-dimensional vector space over $F_q$ (a Grassmannian).

Is the union of all these subspaces a Grassmanian?

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  • $\begingroup$ "If each $\vec{c}_m$ makes up an $n−1$ dimensional subspace". What do you mean by this? The vector $\vec{c}_m$ belongs to an $n$-dimensional vector space, which you denote by $V(F_q)$. Which $n - 1$ dimensional subspace of $V(F_q)$ are you referring to? $\endgroup$ – Michael Albanese Jun 19 at 22:09
  • $\begingroup$ @MichaelAlbanese I clarified: If each $\vec c_m$ makes up an $n-1$ dimensional subspace, orthogonal to $\vec c_m$, my feeling says that $\vec s$ lies in the complement of union of all these subspaces. $\endgroup$ – draks ... Jun 20 at 13:41
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    $\begingroup$ I agree with that statement. $\endgroup$ – Michael Albanese Jun 20 at 19:10
  • $\begingroup$ What makes you think that the problem you are trying to solve has anything to do with grassmannians? $\endgroup$ – Michael Albanese Jun 22 at 14:25
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    $\begingroup$ I don't think that there is any reasonable way to frame this problem in terms of grassmannians. $\endgroup$ – Michael Albanese Jun 28 at 19:04

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