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The double dual space is a linear map $V^{\ast\ast}:V^\ast\to F$, where $F$ is the underlying field of the vector space. If so, why are able to substitute $V^{\ast\ast}$ with $V$ here. Is it because they're isomorphic?

Tell me if anything else is wrong since I'm just learning this.

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  • $\begingroup$ only true for finite dimensional vector spaces. $\endgroup$ – H. Gutsche Jun 27 '18 at 2:25
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It's not just because they're isomorphic. For a finite-dimensional space, $V$ and the single-dual $V^\ast$ are isomorphic as well, but we don't usually identify them together. The difference is that the isomorphism $V\to V^{\ast\ast}$ is a "natural" isomorphism. What this means exactly is a bit technical, but essentially it boils down to the fact that each choice of a basis gives you a different isomorphism $V\to V^\ast$, but the isomorphism $V\to V^{\ast\ast}$ is the same map regardless of what basis you choose.

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Yes, if the dimension is finite.

We have the natural map $\Lambda: V \to V^{\ast \ast}$ given by $x\mapsto \Lambda_x$, where $\Lambda_xf := f(x)$ for all $f\in V^\ast$.

This map is linear and injective. If you suppose $\dim V < \infty $ then $\dim V= \dim V^\ast = \dim V^{\ast \ast}$ and therefore $\lambda$ is a isomorphism.

In other words, if the dimension of $V$ is finite then we can naturally identify $V$ and $V^{\ast \ast}$.

Obs.: If the dimension is infinity then $\lambda$ is not isomorphism.

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