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Let $V$ be a vector space of dimension $n$. I know that if $\{x_1,...,x_n\}$ is a basis of $V$ then we can give easily a basis $\{f_1,...,f_n\}$ of the dual space $V^*$ by setting $f_i(x_j)=\delta_{ij}$.

My question now is:

If we have a basis $\{f_1,...,f_n\}$ of $V^*$, is it possible to give a basis $\{x_1,...,x_n\}$ of $V$ such that $f_i(x_j)=\delta_{ij}$?

What do you think?

Thank you.

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  • $\begingroup$ Assuming that $n < \infty$ then yes. In fact, there is a canonical isomorphism $V^{**} := (V^*)^* \cong V$. $\endgroup$ – Travis Willse Apr 9 '16 at 18:43
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You can always find such $x_j$'s. (not only when the dimension of $V$ is finite.) Since $f_1, \dots, f_n$ are linearly independent in $V^\ast$, we have that for all $i \in \{1, \dots, n \}$, $$ \left(\bigcap_{j \neq i} {\ker{f_j}} \right) \setminus \ker{f_i}\, \neq \emptyset $$ In this post you can find a proof why this is true. So you can choose vectors in these intersections and then you scale them.

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Yes, in finite dimension, $V$ is $(V^*)^*$ that this the bidual of $V$ is $V$.

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