dimension of a vector space seen as a variety

It is true that the dimension of $V$ as a vector space is equal to it's dimension as a variety. This is because $I(V)$ can be generated by $n-\dim V$ linear forms, where we mean dimension as a linear space.

To prove this, select a basis $v_1,\cdots,v_m$ for $V$ and complete it to a basis of $k^n$ with the vectors $w_{m+1},\cdots,w_n$. Writing these vectors as the columns of a matrix $M$, we see that this matrix is invertible and takes the standard basis vectors $e_i\mapsto v_i$ if $i\leq m$ and $e_i\mapsto w_i$ if $i>m$. So $M$ and $M^{-1}$ define an isomorphism between $V$ and the subspace $V'$ consisting of vectors with final $n-m$ coordinates zero (this is an isomorphism as vector spaces and also as varieties). This means that as the coordinate functions $x_{m+1},\cdots,x_n$ generate the ideal of $V'$, their pullbacks under $M^{-1}$ generate the ideal of $V$, and we see the claim.