Proving that there exists a basis for a given dual basis

Question

I need some guidance with the following proof:

Let V be a finite dimensional vector space, and V* its dual.
Let C = $(f1, ... , fn)\subset{V*}$ be a basis for V*.
Let $w\in{V*}$.
Prove that there exists $B\subset{V}$ such that C is dual for B.

Here's what I have so far:
Since C is a basis, then its elements are linearly independent and are not zeroes. But I can't figure out the use of "w", as well as how to continue proving this.

Thanks in advance for your help!

Answer 1

The dual basis $f_1^*, \dotsc, f_n^*$ in $V^{**}$ of $C=\{f_1, \dotsc, f_n \}$ gives rise to a basis $B$ of $V$ via the canonical isomorphism $j:V \to V^{**}$. The dual of $B$ is $C$.

Proof: We compute:

$$\delta_{ab}=f_a^*(f_b)=j(j^{-1}(f_a^*))(f_b)=f_b(j^{-1}(f_a^*)),$$

hence the elements $f_1, \dotsc, f_n$ form a dual basis of the basis $j^{-1}(f_1^*), \dotsc, j^{-1}(f_n^*)$.

Answer 2

Define $i_V \colon V \to V^{**}$ as follows: For $w \in V^*$, $v \in V$ let $$ i_V(v)(w) = w(v) $$ that is $i_V(v)$ is "evaluation at $v$". Then $i_V$ is linear and one-to-one: If $v \ne 0$, extend $v$ to a basis $B$ of $V$ and define $w \in V^*$ by $w(v) = 1$, $w(b) = 0$, $b \in B \setminus \{v\}$. Then $$ i_V(v)(w) = w(v) = 1 $$ hence $i_V(v) \ne 0$. So $i_V$ is one-to-one.

As $V$ is finite-dimensional, $\dim V^{**} = \dim V$, so $i_V$ is an isomorphism.

Now let $C^*$ be the dual basis of $C$ in $V^{**}$, that is with $C^* =(f_1^*, \ldots, f_n^*)$ we have $$ f_i^*(f_j) = \delta_{ij}, \qquad 1 \le i,j \le n $$ Define $B := i_V^{-1}[C]$. Then, as $i_V$ is an isomorphism, $B$ is a basis of $V$ and with $B = (v_1, \ldots, v_n)$ we have $$ f_j(v_i) = i_V(v_i)(f_j) = f_i^*(f_j) = \delta_{ij}$$