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I know that this is an isomorphism. I can see why the function is injective, but not why it is surjective.

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Note that $\dim(V^{**}) = \dim(V)$. By the rank-nullity theorem, any injective map between vector spaces of the same dimension is necessarily surjective.

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  • $\begingroup$ For a given $\psi \in V^{**}$, how can one determine a vector $v \in V$ with $\Phi(v)=\psi$? $\endgroup$ Mar 7, 2017 at 2:42
  • $\begingroup$ Here's one approach: let $e_1,\dots,e_n$ denote a basis of $V$. Let $\alpha_1,\dots, \alpha_n$ denote the associated dual basis of $V^*$. Then $$ v = \sum_{i=1}^n \psi(\alpha_i)e_i $$ Defining an explicit inverse will ultimately depend (in some way) on the fact that we're considering a finite dimensional space. $\endgroup$ Mar 7, 2017 at 2:45
  • $\begingroup$ Thanks! For beginners like me, here is how to derive this (could you please point out if the following has errors?): We're looking for $v \in V$ with $\psi(f)=f(v)$ for every $f \in V^{*}$. For $f=\sum_{i=1}^{n}\lambda_i.\alpha_i$ and $v=\sum_{j=1}^{n}\mu_j.e_j$ we have: $(\sum_{i=1}^{n}\lambda_i.\alpha_i)(\sum_{j=1}^{n}\mu_j.e_j)=\sum_{i=1}^{n}\lambda_i.\psi(\alpha_i)$, therefore $\sum_{i=1}^{n}\lambda_i.\mu_i = \sum_{i=1}^{n}\lambda_i.\psi_(\alpha_i)$ . Since this is satisfied for $\mu_i = \psi(\alpha_i)$, the unique $v$ we are looking for is $v=\sum_{i=1}^{n}\psi(\alpha_i).e_i$. $\endgroup$ Mar 7, 2017 at 16:55

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