Let $V$ be a vector space over the field $K$ and $V^*=\mathcal{L}(V,K)$ its dual space. We can prove that $V$ is naturally isomorphic to its double dual $V^{**}$, but why does every isomorphism between $V$ and its dual $V^*$ depend on the choice of basis? We certainly use dual basis, but the number of element of a basis i.e. the dimension is not basis-dependent. For example, in Linear Algebra by Serge Lang, I have found this:

Let $V$ be a vector space over $K$ with a non-degenerate scalar product, $\langle\cdot,\cdot\rangle:V\times V\rightarrow K$. Let $v\in V$, the map $L_v$ such that \begin{equation} V\ni u\overset{L_v}{\longrightarrow}\langle u,v\rangle \end{equation} is a linear functional, thus an element of $V^*$.

The map such that \begin{equation} V\ni v\rightarrow L_v \end{equation} is an isomorphism (between $V$ and its dual). This is proved by showing that this map is linear, injective (because of non-degeneracy) and surjective ($dimV=dimV^*$). So, how does this depend on the choice of basis? It is true that we used the dual basis at the beginning, but as I said above, every basis would give us the same answer as for the dimension of the space, that is what the author used in the last proof.

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    $\begingroup$ Yes, if you fix an inner product you get an isomorphism between the space and its dual. However, there is no natural choice of inner product on a vector space. $\endgroup$
    – lulu
    Aug 1, 2020 at 18:29
  • $\begingroup$ So what you're saying is that the fact this isomorphism is not natural is not because of the dual basis but because of the scalar product? $\endgroup$
    – Feynman_00
    Aug 1, 2020 at 18:38
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    $\begingroup$ I didn't mention a basis. I'm just saying that you got your isomorphism by adding structure, so there was nothing natural about it. $\endgroup$
    – lulu
    Aug 1, 2020 at 18:40
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    $\begingroup$ May be it's time to define what is natural... $\endgroup$ Aug 1, 2020 at 18:45
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    $\begingroup$ Given any basis $\beta = \{ v_1, \cdots, v_n\}$ there is an unique inner product $\langle \cdot, \cdot\rangle_\beta$ which makes that basis orthonormal. The set of inner product is in one-to-one correspondence with the set of basis. $\endgroup$ Aug 1, 2020 at 19:35

1 Answer 1


It's not natural because the identification between elements of the vector space and its dual depend upon the basis and, importantly, the correspondence is not maintained upon a change of basis. The latter is easiest to see by thinking about how the components of vectors and convectors change in opposite senses under a change of basis. In your case, note that your map defines dual vectors with respect to the basis dual to that of the original space which is a basis dependent identification.

This is not the case for the correspondence between the dual of the dual and the original space, where a basis-independent identification can be made. In this way, although (for finite dim vector spaces) any vector space of the same dimension is equivalent to the original space, the identification $(V^*)^* \cong V$ is natural because everyone will agree on the correspondence between vectors regardless of the basis they're using.


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