Is transpose of basis vectors still a basis for the original vector space?

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Consider the basis of a vector space, for the sake of a concrete example I am going to consider the canonical basis for $\mathbb R ^2$:

$\lbrace \begin{bmatrix}1\\0\end{bmatrix} , \begin{bmatrix}0\\1\end{bmatrix}\rbrace$

Is the set of row vectors formed by the transpose of each of these basis vectors still considered a basis for $\mathbb R ^ 2$? E.g. is the following still a basis for $\mathbb R ^2$?:

$\lbrace \begin{bmatrix}1&0\end{bmatrix} , \begin{bmatrix}0&1\end{bmatrix}\rbrace$

My understanding of this may hinge on not understanding if 'row' vectors are in the same space as column vectors. For example, if the vectors I gave in the second basis example are even in $\mathbb R ^2$.

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You are talking about the same set of vectors presented differently. Of course if one set is a basis, the other set is also a basis. The problem makes more sense if we talk about a matrix where the rows could be linearly independent but the columns not or the other way around.