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I have been studying about these topics. Noether's First isomorphism theorem states that if $f:V\to W$ is linear, then there exists an isomorphism between $V/\text{Ker}(f)$ and the subspace $\text{Im}(f)$. I have seen in some books how they show that $V$ and $V^*$ are isomorphic in finite dimension using the bases and the fact that they have the same dimension. But is there an alternative way to show the isomorphism of $V$ and $V^*$ using Noether's Theorem? Both results make me believe that they have some relation, but I am not sure, and I can't find any information about it either.

JadMON2k1
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No, there's no way to show $V \cong V^*$ using the first isomorphism theorem -- at least, not without involving dimension somehow. The reason for this is quite simple: the first isomorphism theorem holds for any and all vector spaces, whether finite-dimensional or infinite-dimensional (in fact, it holds much more widely: there is a version of this theorem for groups, monoids, rings, ... - it can be shown that it holds in all equational theories).

But $V \cong V^*$ only ever holds if $V$ is finite-dimensional! See https://en.wikipedia.org/wiki/Dual_space#Infinite-dimensional_case for examples, and Why are vector spaces not isomorphic to their duals? for further discussion.

Johannes Kloos
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  • I am working in the finite dimensional case, so somehow I will have to use this fact. My question is whether I can adapt the Isomorphism Theorem to this case to prove (in an alternative way to the usual books way, with bases and all that) the isomorphism between a vector space and its dual. – JadMON2k1 Apr 20 '23 at 06:30
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    Well, the first isomorphism theorem really says "if you have a map $\varphi: M \to N$, you can turn it into an isomorphism $M/\ker \varphi \cong \im \varphi$". You'd have to find that map $\varphi$ somehow, and since $V \cong V^$ crucially relies on the existence of bases, you'd be hard-pressed to find such a $\varphi$ without dealing with bases. Another fact that works against you here is that the isomorphism $V \cong V^$ is not canonical - there are many "equally good" isomorphisms, depending on choice of basis. Contrast to $V \cong V^{**}$, where there is an obvious construction. – Johannes Kloos Apr 20 '23 at 07:30