vector space dimension of linear varieties Suppose $Y=Z(f_1,\cdots ,f_r)\subseteq \mathbb{A}^n_k$ where each $f_i$'s are linear homogeneous polynomials which are $k$-linear independent. Then $Y$ is also a vector space over $k$. My question: Is the vector space dimension of $Y$ is same as the dimension of $Y$ as an affine variety? If yes how?  
Thank you.

Yes.
Note that $Y$ is given as $Ax=0$, where $A$ is a $n \times r$-matrix containing the coefficients of the $f_i$.
Thus $Y= \ker A$.
This should now be clear, maybe depending upon your definition of "dimension".
EDIT Now that we have a definition of dimension to work with: we can at least immediately bound the dimension: any linear subspace has a filtration $0 \subset V_1 \subset V_2 \subset \ldots \subset Y$. Here $0$ is a point, $V_1$ is a line, $V_2$ is a plane and so on. Then the Zariski dimension must be greater than or equal to $\dim \ker A$. 
Too see that we must equality, suppose that we could fit an irreducible algebraic subset inbetween $V_k$ and $V_{k+1}$. Without loss of generality, we could assume that $V_k$ is defined by $x_0=x_1=\ldots=x_{n-k}=0$, and $V_{k+1}$ is defined similarly (with one condition less).
If $V$ is an irreducible algebraic subset such that $V_k \subset V \subset V_{k+1}$, then $V$ must be defind by some ideal $I_V$ such that $I_{V_{k+1}} \subset I_V \subset I_{V_k}$. But this means that $I_V$ is  stricly contained in and non-zero in $I_{V_k}/I_{V_{k+1}} = (\overline{x_{n_k}}) \simeq k$. But this is impossible, since the latter is a 1-dimensional vector space.
