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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 $v_i^* = f_i$?

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  • $\begingroup$ Yes, $\{ v_i \}$ will be the dual basis to $\{ f_i \}$. $\endgroup$ Dec 4, 2020 at 3:03
  • $\begingroup$ Sorry, can you elaborate? I am only given $f_1,...,f_n$ basis of $V^*$, how should I construct the $v_i$? $\endgroup$ Dec 4, 2020 at 3:06
  • $\begingroup$ Do you know what is the canonical isomorphism $V \to V^{**}$? $\endgroup$
    – azif00
    Dec 4, 2020 at 3:06
  • $\begingroup$ Yes, taking $v$ to $E_v$ $\endgroup$ Dec 4, 2020 at 3:07

2 Answers 2

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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 $\{v_1^*,\dots, v_n^*\}=\{f_1,\dots, f_n\}$.

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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 $V$.

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