Consider the finite field $\mathbb{F}_{2^q}$ for some $q$. Let $\mathbb{F}_{2^q}$ is constructed by a primitive polynomial $\bf f$ of degree $q$ over $\mathbb{F}_{2}$. Consider $\alpha$ is a root of $\bf f$.
My question: Is there a systematic method to obtain $k$ elements $\alpha^{i_1}$, $\alpha^{i_2}$, $\cdots$, $\alpha^{i_k}$ with $0<i_1<i_2<\cdots <i_k<2^q-1$, such that the elements $\alpha^{i_j}$, $1\leq j \leq k$, satisfy the following two conditions: (For simplicity, we can assume that there are these elements in $\mathbb{F}_{2^q}$)
All linear combinatorics of $\alpha^{i_j}$, $1\leq j \leq k$, are not equal to zero or the unit element. In the other words, there is no binary elements $b_j$ and $c_j$ with $1\leq j \leq k$, such that ($b_j,c_j\in \{0,1\}$) $$ b_1 \,\alpha^{i_1}+b_2 \,\alpha^{i_2}+\cdots +b_k \,\alpha^{i_k}=0,\\ c_1 \,\alpha^{i_1}+c_2 \,\alpha^{i_2}+\cdots +c_k \,\alpha^{i_k}=1. $$
Example: Consider $\mathbb{F}_{2^8}$ that is constructed with ${\bf f}={x}^{8}+{x}^{7}+{x}^{6}+{x}^{3}+{x}^{2}+x+1$. Let $\alpha$ is a root of $\bf f$. By search approach, elements $\alpha^{9}$, $\alpha^{10}$, $\alpha^{143}$, $\alpha^{147}$ and $\alpha^{192}$ hold on the given conditions.
I do not know, this question is related to the cyclotomic coset subject or not? In addition, when we fix $q$ and $k$, then how many $k$--tuples $(\alpha^{i_1},\alpha^{i_2},\cdots , \alpha^{i_k})$, there are which satisfy the mentioned conditions in $\mathbb{F}_{2^q}$.
Thanks for any suggestion.
The requirement is equivalent to the set $$S=\{\alpha^{i_1},\alpha^{i_2},\ldots,\alpha^{i_k},1\}$$ being linearly independent over the subfield $\Bbb{F}_2$.
Because there are $k+1$ elements in $S$, a necessary condition is that $k+1\le q$, or $k\le q-1$. Assuming that inequality it is easy to find examples. To mention the most obvious: because $\alpha$ is a zero of a primitive polynomial (of degree $q$) the elements $1,\alpha,\alpha^2,\ldots,\alpha^{q-1}$ form a basis of $\Bbb{F}_{2^q}$ as a vector space over $\Bbb{F}_2$. Consequently if $\{1\le i_1<i_2<i_3<\ldots<i_k\le q-1\}$ the conclusion is immediate. More generally, if the rows of $(k+1)\times q$ matrix $A$ (entries in $\Bbb{F}_2$) $$ A=\left(\begin{array}{ccccc} 1&0&0&\cdots&0\\ a_{11}&a_{12}&a_{13}&\cdots&a_{1q}\\ a_{21}&a_{22}&a_{23}&\cdots&a_{2q}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a_{k1}&a_{k2}&a_{k3}&\cdots&a_{kq}\\ \end{array}\right) $$ are linearly independent, then the choices $$\alpha^{i_j}=\sum_{t=1}^qa_{jt}\alpha^{t-1},\qquad j=1,2,\ldots,k,$$ will work.
However, if $q$ is a bit larger, it is difficult to extract the exponents $i_j$ from this (an instance of the discrete logarithm problem). If $q$ is large you may not even have a primitive polynomial! For example, the field may be given by having a multiplication table of a suitable normal basis $x,x^2,x^4,x^8,\ldots,x^{2^{q-1}}$. In that case you can use any proper subset of a normal basis. This is because any normal basis of $\Bbb{F}_{2^q}$ has the property that $1$ is the sum of all the basis elements.