Linear independence in Finite Fields How can we define linear independence in vectors over $\mathbb{F_{2^m}}$ ?
Let vectors $v_1,v_2,v_3$ $\in$ $\mathbb{F_{2^m}}$,
If $v_1,v_2,v_3$ are linearly independent,then
$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3$=0 if and only if $\alpha_1=\alpha_2=\alpha_3=0$  and
$\alpha_1,\alpha_2,\alpha_3 \in \mathbb{F_2}$ ? or $\mathbb{F_{2^m}}$ ?
Thanks in advance

Linear independence is defined the same way in every vector space:
$\{v_i\mid i\in I\}$ is a linearly independent subset of $V$ if $\sum_{i=1}^n \lambda_i v_i=0$ implies all the $\lambda_i=0$ for all $i$, where the $\lambda_i$ are in the field.
In short, you definitely would not take the $\lambda_i$ from $F^m$. You are probably thinking of multiplying coordinate-wise. The definition of a linear combination, though, takes coefficents from the field (and $F^m$ is not a field).

To address the edits, which radically changed the question:
Linear independence depends on the field (no pun intended.) If you want them to be linearly independent over $F$, then $\lambda_i$ can only come from $F$. If you want it to be linearly independent over $F_{2^k}$, then the $\lambda_i$ are all from $F_{2^k}$.
For a simple example, look at $F_2$ and $F_8$. If $x\in F_8\setminus F_2$, then $\{1,x\}$ is linearly independent over $F_2$, but it is linearly dependent over $F_8$.
