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(All vector spaces mentioned here are finite.)

Suppose we're given the isomorphism $\Phi:V\to V^{**}: v \mapsto (\Phi_v:V^*\to \mathbb{K}:\phi \mapsto \phi(v))$. Consider a basis $(\varepsilon_i)_{i\in I}$ of $V^*$. Claim that there exists a $(e_i)_{i\in I}$ of $V$ such that $(\varepsilon_i)_{i\in I}$ is its dual basis.

My book says that this claim can be shown by considering the isomorphism $\Phi$ and by looking at the dual basis of $(\varepsilon_i)_{i\in I}$ in $V^{**}$. I'm not really sure what this last bit means. How do I find this dual basis in $V^{**}$. How can I use it to show the claim?

Thanks!

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    $\begingroup$ What do you understand by a dual basis for a given basis? $\endgroup$ – k.stm May 29 '19 at 14:19
  • $\begingroup$ Given $\beta=\{v_1,\dots,v_n\}$, its dual is $\beta^*=\{f_1,\dots,f_n\}$ such that $f_i(v_j)=\delta_{ij}$. $\endgroup$ – Zachary May 29 '19 at 14:23
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Given a basis $w_1,\ldots, w_n$ of (finite-dimensional) $W$, we can find a basis $\phi_1,\ldots,\phi_n$ of $W^*$ such that $\phi_i(w_j)=\delta_{ij}$ (the dual basis). Now let $W=V^*$ and take the dual basis $\phi_1,\ldots,\phi_n$ of $\epsilon_1,\ldots,\epsilon_n$ in $W^*=V^{**}$, transport it to $V$ via the canonical isomorphism (producing $e_1,\ldots,e_n$) and verify that $$\epsilon_i(e_j)=\psi_j(\epsilon_i)=\delta_{ji}=\delta_{ij}. $$

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