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In a finite field $\{0,1,2\}^2$, given a set of vectors $[0\:1],[1\:0],[1\:1],[2\:2]$, we can have the linear combination, $c_1[1\:0]+c_2[0\:1]+c_3[1\:1]+c_4[2\:2] = [s_1\:s_2]\in\{0,1,2\}^2$, where $c_1,c_2,c_3,c_4\in \{0,1,2\}$.

The basis of this set of vectors is $\{[0\:1],[1\:0]\}$. The dimension of this basis is 2, hence this set of vectors span the entire finite field $\{0,1,2\}^2$.

However if the value each constant can take now changes to $c_1\in\{0,1\}, c_2\in\{0,1,2\}, c_3\in\{0,1\},c_4\in\{0,1\}$, is there a more efficient method to determine if these set of vectors still span the finite field $\{0,1,2\}^2$, without rigorously checking if each vector from the finite field $\{0,1,2\}^2$ can be achieved from the linear combination,

$c_1[1\:0]+c_2[0\:1]+c_3[1\:1]+c_4[2\:2] = [s_1\:s_2]\in\{0,1,2\}^2$, where $c_1\in\{0,1\}, c_2\in\{0,1,2\}, c_3\in\{0,1\}, c_4\in\{0,1\}$ ?

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    How is $\{0,1,2\}^2$ a finite field? Do you mean a 2-dimensional vector space over the field $F=\mathbb{Z}_3$ of three elements? For it to be a field you need to specify, how you multiply vectors. A minimal requirement is a rule $$[0\ 1][1\ 0]=[s_1\ s_2]$$ for some elements $s_1,s_2\in F$. The rest of the multiplication table then follows from the field axioms (but some care is needed in selecting $[s_1\ s_2]$ for otherwise you may get zero divisors or some such atrocities).2012-08-31
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    @JyrkiLahtonen: I think that indeed the first sentence should read "In a vector space $\{0,1,2\}^2$ over a finite field $\{0,1,2\}$ ...", and that $\{0,1,2\}$ denotes $\mathbb Z/3\mathbb Z$.2012-08-31

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