Let $V$ be a finite dimensional vector space, then given a basis for $V$ constructing an isomorphism $V \rightarrow V^*$ is easy, but how about the reverse direction? Given an explicit isomorphism $\phi : V \rightarrow V^*$ can we fully recover a basis for $V$?

Denoting $v \in V$ denote $\phi(v) = \phi_v$. Take the elements $e_i \in V$ satisfying $\phi_{e_i}(e_j) = \delta_{i,j}$. It's easy to check that these elements are linearly independent but do they form a basis?

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    $\begingroup$ If $V$ has dimension $n$ then any $n$ independent vectors form a basis. This is the finite-dimensional-vector-spaces-are-magic theorem... $\endgroup$ – David C. Ullrich Aug 28 '15 at 14:23
  • $\begingroup$ Yup, so the question becomes, for any isomorphism are there $n$ such vectors $e_i$ satisfying this condition. $\endgroup$ – Nostromo Aug 28 '15 at 14:31
  • $\begingroup$ $\phi^{-1}$ is an isomorphism $V^* \to V$. Also $\phi^*$ is an isomorphism $V^* \to V^{**}$. So this really isn't even a different problem! $\endgroup$ – Hurkyl Aug 28 '15 at 14:34
  • $\begingroup$ My last comment (deleted) didn't make sense. I don't see how this is the same problem in disguise, can you elaborate? $\endgroup$ – Nostromo Aug 28 '15 at 15:14
  • $\begingroup$ The answer to my original question is no. The non-degenerate bilinear form corresponding to the isomorphism $\phi$ won't satisfy the frobenius axioms in general. It's known that bases of vector spaces correspond with special commutative frobenius algebras. $\endgroup$ – Nostromo Aug 28 '15 at 15:26

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