A basis of dual vector space. Let $V$ be a vector space over field $K$, $V^*$ is a dual vector space, i.e., the set of all linear transformations from $V$ to $K$.
Let $(x_1, x_2,...,x_n)$ be a basis of $V$. I know the fact that the set of linear transformations $(x^*_1,x^*_2,...,x^*_n)$ forms a basis of $V^*$, it's easy and straightforward.

Conversely, given a basis say $(f_1,f_2,...,f_n)$ of $V^*$, I expect to prove that there exists a basis of $V$ say $(y_1, y_2,...,y_n)$ such that $f_i = y^*_i$ for all $i$.

Please help me. 

This is an application of the fact that the map $\omega_V\colon V\to V^{**}$ defined by
$$
\omega_V(x)=\hat{x},
\qquad
\hat{x}(f)=f(x),\text{ for }f\in V^*
$$
is an isomorphism when $V$ is finite dimensional. (Prove $\omega_V$ is injective.)
Thus you can consider the dual basis of $\{f_1,\dots,f_n\}$ in $V^{**}$ and take the elements in $V$ that map to it.
