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From time to time, during my undergraduate lectures on linear algebra appears the following question from the most smart students in the class. I asked to my algebra colleagues but I have not received a satifying answer, so I put it here.

Everybody knows that for a given finite dimensional vector space $E$ the bidual space $E^{\ast\ast}$ is canonically isomorphic to $E$: just exhibit an isomorphism "free of basis choice" to show that.

On the other side, we know that $E$ is not canonically isomorphic to $E^\ast$ (but isomorphic when choosing basis).

My question is the following (maybe here or in a more general fashion): to show that two algebraic objects are canonically isomorphic you just need to exhibit an isomorphism not depending on extra choices, but to show that two algebraic objects are NOT canonically isomorphic I do not know what one needs to do (in particular, I do not know how to prove this fact for $E$ and $E^{\ast}$?

Is there a way to argue on that? And can this be done (for vector spaces) "category-free"? (so this can be explained in an honours linear algebra 1st course)

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    $\begingroup$ If you want a proof that this is impossible, you'd first need a precise statement. Does 'canonical' mean 'natural'? Or does it mean that there is some 'universal formula' that works the same for any finite dimensional vector space without the extra datum of a basis? In the first case, the answer necessarily involves category theory, in the second case logic. In both cases, it would go beyond linear algebra. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 22, 2022 at 21:18
  • $\begingroup$ I mean the second notion, namely the statement (to be proven) should be: it is not possible to define an isomorphism between $E$ and $E^{\ast}$ without the use of a fixed basis in both vector spaces $\endgroup$
    – Johnny Cage
    Commented Apr 23, 2022 at 8:40
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    $\begingroup$ Right. But saying what is not allowed doesn't really work in logic; you have to specify what is allowed. For instance, I don't think you can even define the dual in the first-order language of vector spaces over a fixed field $k$, so you'd need a more elaborate language. I really don't think this is easier than the categorical demand that the isomorphism be $\operatorname{GL}_n$-equivariant. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 23, 2022 at 13:48

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