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Say we have a vector space $V$, and $V^{*}$ is the dual space for $V$, and $V^{**}$ is the dual space for $V^{*}$.

There is a theorem saying:

$\omega:V \to V^{**}, a \mapsto (\phi \mapsto \phi(a))$ is an isomorphism and $\omega=f:V \to V^{**}:a_i \mapsto \psi_i$, where $A:=(a_1,\dots,a_n)$ is a basis of $V$, $\Phi:=(\phi_1,\dots,\phi_n)$ is the dual basis for $A$ and $(\psi_1,\dots,\psi_n)$ is the dual basis for $\Phi$.

Consequently, if I have another basis of $V$, say $(\widetilde a_1,\dots,\widetilde a_n)$, and the dual one to it $(\widetilde \phi_1,\dots, \widetilde\phi_n)$ and the dual one to the dual one $(\widetilde \psi_1,\dots, \widetilde\psi_n)$, then $f=\widetilde f:\widetilde a_i \mapsto \widetilde \psi_i$.

My lecture notes treat this result ($f=\widetilde f$) as something not trivial and important. I am not sure why.

After some thinking my guess is that the non-triviality of the result is based on the fact that in general $g:a_i \mapsto \phi_i \ne \widetilde g:\widetilde a_i \mapsto \widetilde \phi_i$.

In words, if I create an isomorphism between $V$ and $V^{*}$ by mapping a basis to its dual one, then what I get depends on the basis I have used (the same goes for $V^{*}$ and $V^{**}$). But if I create an isomorphism between $V$ and $V^{**}$ by mapping a basis to the dual basis of its dual basis, then the original basis suddenly does not matter any longer. And that indeed sounds non-trivial to me.

Am I getting this right?

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    $\begingroup$ In general, $V$ is not isomorphic to $(V^*)^*$, see here. $\endgroup$ – Dietrich Burde Aug 20 '17 at 16:59
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    $\begingroup$ Ok, ... I guess I've forgotten to mention that $\dim(V) < \infty$. $\endgroup$ – Sergey Zykov Aug 20 '17 at 17:04
  • $\begingroup$ No, no, "I am not sure why" applies to the finite dimensional case. $\endgroup$ – Sergey Zykov Aug 20 '17 at 19:20
  • $\begingroup$ Let's work under the assumption that the space is finitely dimensional, then the theorem holds. Given that, do you find my understanding of the theorem's importance from above correct? $\endgroup$ – Sergey Zykov Aug 20 '17 at 19:28
  • $\begingroup$ It is one aspect of many, I think. The theorem is important for many other reasons, too. $\endgroup$ – Dietrich Burde Aug 20 '17 at 19:30

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