Let $\mathrm{F}_q$ be a subfield of $\mathrm{F}_{q^m}$. $\mathrm{F}_{q^m}$ can be seen as an $m$-dimensional vector space over $\mathrm{F}_q$. Let $v_1,\ldots, v_k \in \mathrm{F}_{q^m}^n$ be linearly independent over $\mathrm{F}_{q^m}$. So this set spans a $k$-dimensional subspace of $\mathrm{F}_{q^m}^n$. Clearly this set of vectors are also linearly independent over $\mathrm{F}_q$.
Question: How can we make an $mk$-dimensional subspace of $\mathrm{F}_q^n$ out of this set of linearly independent?
note: to make the question more specific, we can consider $v_1=u_1,u_2,\ldots,u_n$ where $u_1,u_2,\ldots, u_n$ are linearly independent elements in $\mathrm{F}_{q^m}$ ($n\leq m$) and $v_i=(u_1^{q^{i-1}},\ldots,u_n^{q^{i-1}})$ for $1\leq i\leq k$. We already know the set of $n$-vectors $v_1,\ldots,v_k$ with this form are linearly independent over $\mathrm{F}_{q^m}$ and generate a $k$-dimensional subspace of $\mathrm{F}_{q^m}^n$.
I have this answer to my own question but I am not quite sure:
Let $a_1,\ldots, a_m$ be a basis of $\mathrm{F}_{q^m}$ over $\mathrm{F}_{q}$. So the vector multiplication $(a_1,\ldots,a_m)^T\times (v_1,\ldots, v_k)=(b_{11},b_{12},\ldots, b_{mk})$ is the basis of our $mk$-dimensional vector space over $\mathrm{F}_q$.
Is it always true to make an $mk$ linearly independent vectors over $\mathrm{F}_q$ in such a way?