fixing dual basis to find basis of vector space

Question

Given a finite-dimensional vector space $V$ with basis $\{v_1,...,v_n\} \subset V$, we also fix a basis for dual space $V^*$ by $\{v_1^*,...,v_n^*\} \subset V^*$ where: $v_i^*(v_j) = 1$ if $j = i$ and $0$ otherwise. If on the other hand, we are given the vector space $V$ and dual space $V^*$ and a basis $\{f_1,...,f_n\}$ of $V^*$, is it possible to find $v_1,...,v_n \in V$ such that $v_i^* = f_i$?

Answer 1

Yes, it is. The key point is that for finite-dimensional spaces, there is a natural isomorphism $\theta:V\to V^{**}$, given by $\theta(v)=\text{evaluation at $v$} = \left(f\mapsto f(v)\right)$. If you want, I can elaborate more on this, but this is a common isomorphism, so you should be able to read up more on this easily.

Now, taking this for granted, consider the dual basis $\{f_1^*,\dots, f_n^*\}$ of $V^{**}$. Then, because $\theta:V\to V^{**}$ is an isomorphism, we can consider the vectors $v_i:= \theta^{-1}(f_i^*)$. Now, it's a matter of unwinding definitions to see that $\{v_1,\dots, v_n\}$ is a basis for $V$, whose dual is $\{v_1^*,\dots, v_n^*\}=\{f_1,\dots, f_n\}$.

Answer 2

Yes. Consider the dual basis $\phi_1,\dots,\phi_n$ of $f_1,\dots,f_n$. Now, since $v \mapsto E_v$ is surjective, we have that $\phi_i = E_{v_i}$ for some $v_i \in V$ (for $i=1,\dots,n$). Finally, observe that $$\forall i, j \in \{1,\dots,n\} : \quad f_i(v_j) = E_{v_j}(f_i) = \phi_j(f_i) = \delta_{ij},$$ which means that $v_1,\dots,v_n$ is the desired basis for $V$.