Let us denote by $\boldsymbol{v}_i$ the columns of an $m \times n$ GF($8$) matrix. The field elements are enumerated $\{0,1,2,...,q-1\}$. To define the arithmetic operations between field elements, we consider each each element as a polynomial of degree at most $2$ with GF($2$) coefficients that are its binary representation. For example, $$0 \leftrightarrow 0,1 \leftrightarrow 1,2 \leftrightarrow z,3 \leftrightarrow z + 1,4 \leftrightarrow {z^2},5 \leftrightarrow {z^2} + 1,6 \leftrightarrow {z^2} + z,7 \leftrightarrow {z^2} + z + 1$$
The GF($8$) operations are then carried out as addition/multiplication of polynomials, modulo an irreducible polynomial of degree $3$.
I would like to find the number of matrices with columns such that $$\sum\limits_{i = 1}^n {{\alpha _i} \cdot {\boldsymbol{v}_i}} = 0,$$ $\alpha_i \in \{0,1,2,3\}$, only if $\alpha_i$ are all zero. Note that the set $\{0,1,2,3\}$ is a subgroup of the additive group of GF($8$). In addition, if instead we conidered $\tilde \alpha_i \in \{0,1,2,...,q-1\}$, the question reduces to determining the number of full-rank matrices, which is $\prod\limits_{i = 1}^n {\left( {{q^m} - {q^{i - 1}}} \right)}$.
Taking the concrete example $m=2, n=3$, I encountered a problem when trying to mimic the calculation of the number of full-rank matrices. For the first column, any column except the zero vector can be used. Thus, the first column can be chosen in $q^m-1$ ways. The second column cannot be chosen such that $\alpha_1 \cdot \boldsymbol{v}_1 + \alpha_2 \cdot \boldsymbol{v}_2 = 0$, or equivalently $(\alpha_1 / \alpha_2) \cdot \boldsymbol{v}_1 = \boldsymbol{v}_2$. This means that the second column can be chosen in $q^m - 8$ ways, since $| {\{ {{\alpha _1}/{\alpha _2}} \}} | = 8$ ($\alpha_2 \ne 0$). Continuing in this fashion, the $i^{\rm th}$ column can be chosen in $q^m-8^{i-1}$ ways. However, this leads to $\left( {64 - 1} \right)\left( {64 - 8} \right)\left( {64 - 64} \right) = 0$ ways in the example above, instead of the correct number which is 84,672 (found by a computer exhaustive search).
I suspect that the inclusion-exclusion principle should be used, but I am not sure how.