Restriction of dual basis to subspaces. Let V be a finite dimensional vector space and f_1,..,f_m be a linearly indepedent in V*. Prove that the following two statements are equivalent.
(i)f_1,..,f_m is a basis of V*.
(ii)for every subspace W of V, the restrictions of f_1,..,f_m to W spans W*
Here is what I have so far.
(i)=>(ii). We want to show that for a $T\in W^*$, $T=c_1*f_1+...+c_m*f_m$. Let $w\in W\subseteq V$. Let $\{v_1,..,v_m\}$ be a basis of V. Since w is also in V, we can write $w=a_1*v_1+...+a_m*v_m$ for some scalars $a_i$. 
LHS:$T(w)=a_1*T(v_1)+...+a_m*T(v_m)$
RHS: $(c_1*f_1+...+c_m*f_m)(w)=c_1*f_1(w)+...+c_m*f_m(w)= c_1*a_1+...+c_m*a_m.$
Therefore if we let $T(v_i)=c_i$, then LHS=RHS which implies f_1,..,f_m spans W*.
I don't know how to do (ii)=>(i)
Thanks
 A: Part 2.  To show: If V is a vector space of dimension m and  {$f_1, f_2, ... f_m$} = S  span $W^*$ for every subspace W of V, then S is a basis of $V^*$.
If dim(V) = m then dim($V^*$) = m also.
Let $W_1$ be a subspace of V with dimension m - 1, By hypothesis S spans $W_1^*$  So there must be m-1 independent elements S, call them $S_1$ = {$f_1, f_2, ...  f_{m-1}$} that form a basis for $W_1^*$.
There are m distinct subspaces $W_i$ of V with dimension m - 1, and for each of these S spans $W_i^*$.  However each $W_i$ is different and therefore each $W_i^*$ is different -- that is, no two of them can have the same basis. (This is because $W_i^{**} = W_i$.) So in each case we must pick a different set of m-1 elements of S to form a basis for $W_i^*$.
Let U be a subspace of dimension 1 such that $W \bigcap U$ = 0.  Then some element g of S must be the basis for $U^*$.  There must be at least one of the  $W_i^*$, call it W whose basis does not include g. We have W $\bigcup$ U = V so that $V^*$ =$(W \bigcup U)^*$ = $W^* \bigcup U^*$.  So the basis for $V^*$ is the union of the basis for W and g; and that is S.
