Showing that a dual basis is a generating set Suppose we have a dual basis $F$* = ($f_1$*,......$f_n$*) of $V$*. and suppose the standard basis $F$ = ($f_1$,......$f_n$) of $V$. I want to show that the $F$* is a basis so i have to show that it is linearly independent and that it is a generating set ( spanning ). 
I know how to show that it is linearly independent but for showing that it is spanning, I thought that i would just write that let a coefficient a in K (where K is the field as it is a $K$ vector space) and then suggest that $a_1$$f_1$* + .....$a_n$$f_n$* = $b_1$$f_1$* + ......$b_n$$f_n$* where $a = b$. In class, they wrote : 
let $g$ be another dual space in $V$*. We let $a_j$ = $g(v_j)$ for $j = 1....n$
then $f$ and $g$ have the same values on the basis $F$ and are therefore equal. 
then $a_{1}$$f_{1}$$(v_j)$ + .....$a_{n}$$f_{n}$ $(v_j)$ = $a_j$ = $g(v_j)$
My question is i dont understand how this shows that it is spanning??
 A: The proof that you wrote above is somewhat confusing, so I'll try to clarify.  Remember that $V^\ast$ is the set of $K$-linear maps from $V$ to $K$.  In other words, $\text{Hom}_K(V,K)$.
Let $F=\{f_1,\cdots,f_n\}$ be a basis of $V$ and $F^\ast=\{f_1^\ast,\cdots,f_n^\ast\}$ be the dual basis.  This means that $f_i^\ast(f_j)=1$ if $i=j$ and $0$ if $i\not=j$.
For $F^\ast$ to be linearly independent, we need that if $a_1f_1^\ast+\cdots+a_nf_n^\ast$ is the zero function, then all the $a_i$'s are zero.  In this case, 
$$
(a_1f_1^\ast+\cdots+a_nf_n^\ast)(f_j)=a_j
$$
which must be zero for this to be the zero function.  Therefore, all $a_j$'s are zero.
To show that $F^\ast$ is spanning, let $g\in V^\ast$.  Then $g$ is a linear function on $V$ and we let $a_j=g(f_j)$.  Our goal is to show that $g=a_1f_1^\ast+\cdots+a_nf_n^\ast$ as functions.  To be equal as functions, they must be the same map (the same inputs give the same outputs).  
Let $b_1f_1+\cdots+b_nf_n$ be an arbitrary element of $V$.  Then
$$
g(b_1f_1+\cdots+b_nf_n)=b_1g(f_1)+\cdots+b_ng(f_n)=b_1a_1+\cdots+b_na_n
$$
by the linearity of $g$.
On the other hand, 
$$
(a_1f_1^\ast+\cdots+a_nf_n^\ast)(b_1f_1+\cdots+b_nf_n)=a_1f_1^\ast(b_1f_1)+\cdots+a_nf_n^\ast(b_nf_n)=a_1b_1+\cdots+a_nb_n.
$$
Therefore, since the functions have the same output for the same input, they are equal.
A: Let $g$ be an arbitrary element in $V^*$, and $v=\alpha_1 f_1+\alpha_2f_2+\cdots+\alpha_n f_n$, $\alpha_k\in K$, $k=\overline{1,n}$,  an arbitrary vector in $V$.
Then due to linearity of $g$ we have:  $g(v)=\sum_{k=1}^n \alpha_kg(f_k)$. We denote $b_k=g(f_k)\in K$, $k=\overline{1,n}$.
On the other hand  $f^*_k(v)=\alpha_k$. Hence we have:
$g(v)=\sum_{k=1}^n b_k f^*_k(v)$,  $\forall\: v\in V$ and thus $g=\sum_{k=1}^n b_k f^*_k$, i.e.
$V^*$ is generated by $f^*_1, f^*_2, \ldots, f^*_n$
