Man, I am really lost with what you are asking now. I think you need to review some of your definitions, it is getting to the point that these posts are becoming incoherent to me.
You do not find a basis for a matrix.
A matrix represents a linear transformation, a basis is for a vector space, not a function. Matrices have ranks, it does not make sense to talk about the rank of a row.
The dual space to $\displaystyle \mathbb{R}^n$ does not have a "standard" basis. It is not canonically isomorphic to $\displaystyle \mathbb{R}^n$. It is isomorphic, but only in terms of a basis. If you give me a basis for $\displaystyle \mathbb{R}^n$, I will give you a basis for $\displaystyle (\mathbb{R}^n)^*$.
I already wrote out the basis for the dual space in a previous post on here titled "bassis(I can't type apparently) for the Dual Space".
Look at that. If you want the basis for the dual space for $\displaystyle \mathbb{R}^n$ subordinant to the standard basis, then $\displaystyle e_1=(1,0,0,...,0)$
$\displaystyle e_2=(0,1,0,...,0)$
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$\displaystyle e_n=(0,0,0,...,1)$
and this yields the basis for $\displaystyle (\mathbb{R}^n)^*$
$\displaystyle \epsilon_1(a_1e_1+a_2e_2+...+a_ne_n)=a_1$
$\displaystyle \epsilon_2(a_1e_1+a_2e_2+...+a_ne_n)=a_2$
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$\displaystyle \epsilon_n(a_1e_1+a_2e_2+...+a_ne_n)=a_n$