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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

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One comment: when making definitions, you shouldn't use the biconditional "iff". The first "if" there is also superfluous. – rschwieb Jan 24 '13 at 18:33
The edit seems to have made the formulation of the definition worse. In particular, an "iff" became "if" where what's really needed is "only if". – Andreas Blass Jan 24 '13 at 18:40
When you talk about linear independence, you need to mention what is your vector space, is it $F_{2^m}$ over $Z_2$? – mez May 10 '13 at 23:31

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 $F_8$.

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Then the answer is $\alpha_1,\alpha_2,\alpha_3 \in $\mathbb{F_{2^m}}$ ? Sorry I could not understand – elif Jan 24 '13 at 18:37
Thank you for your answer. But if I have $k \times k$ matrix, named $A$, such that each element is selected from $\mathbb{F_{2^m}}$. In which case $A$ is invertible? if $A$'s rows are linearly independent over $\mathbb{F_{2^m}}$ or over $\mathbb{F_2}$ – elif Jan 24 '13 at 19:19
@elif In that case, if they are independent over $F_{2^m}$. There is another way to see if it's invertible: check to see if its determinant is nonzero. – rschwieb Jan 24 '13 at 19:23

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