Here is the full statement of the problem:
Let $V$ be a vector space over a field $F$, $X$ be a basis for $V$, and $V$* $=Hom_{F}(V,F)$, which is a vector space over $F$. For each $x\in X$, denote $f_{x}$ the unique element of $V$* such that
$$f_{x}(y) = \left\{\begin{array}{11} 1_{F} & x=y\\ 0_{F} & x\ne y \\ \end{array}\right.$$
Let $X$* $=\left\{f_{x} | x\in X\right\}$. Prove that if $X$ is finite, then $X$* is a basis for the vector space $V$* over $F$.
So I know that I need to prove that $X$* is a linearly independent set and that $X$* generates $V$* but I am at a loss as to how this could be done. Any suggestions to get me started would be very helpful and appreciated.
Hint: Let $f \in V^* $. How could we write $f$ as a linear combination of $f_x$. First notice that if $v \in V$, since $X$ is a basis (lets say $X = \{x_1,x_2,...,x_n\}$), $v = v_1x_1+...+v_n x_n$, where $v_i \in F$. First notice that $$ f_{x_i}(v) = v_i $$ This means that $$ f(v) = f( v_1x_1+...+v_1x_n) = v_1f(x_1)+...+v_nf(x_n) = f_{x_1}(v)f(x_1)+...+f_{x_n}(v)f(x_n) $$ What does this say about the funnctios $f_{x_i}$ spanning $V^*$?
For the linear independence think about what happens if $\lambda_i \in F$ and $$ \lambda_1 f_{x_1}+...+\lambda_n f_{x_n} =0 $$ This would need to hold for every $w \in V$... $$ \lambda_1 f_{x_1}(w)+...+\lambda_n f_{x_n}(w) =0 $$