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I've read that $V$ and $V^*$ are isomorphic, but not naturally isomorphic (you have to make a choice of basis); $V$ and $V^{**}$ are naturally isomorphic.

I'm only a math freshman, so my knowledge is very limited. I don't understand what is the difference between "naturally isomoprhic" and "not naturally isomorphic". If someone could give an explanation or an example, I'd be grateful (and perhaps a tad wiser).

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    Uhh, didn't you just say it? You have to make a choice of basis – 2017-01-01
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    yes, but I can't think of a concrete example that will illustrate it – 2017-01-01
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    This is not true in general, just FYI. There is a bit of a convoluted post here on natural isomorphisms. – 2017-01-01
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    @CameronWilliams do you care to explain, please? What isn't true in general? – 2017-01-01
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    That $V$ and $V^*$ are isomorphic. – 2017-01-01
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    The precise meaning of natural isomorphism is part of elementary category theory. Try any introduction to the subject. – 2017-01-01
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    I think the really basic idea behind natural and unnatural isomorphisms is whether or not you need to add extra stuff to make it work. That is, a natural isomorphism defines an isomorphism directly from the elements themselves. An unnatural isomorphism relies on passing through something else (like a basis). For instance, you can make the association $\hat{x}(f) = f(x) $ between $x\in V$ and $\hat{x} \in V^{**} $ which doesn't rely on a basis, but to make the association between $V$ and $V^*$ explicit you need a basis. (I'm assuming finite dimensions here.) – 2017-01-01
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    @Giulio your comment above is incorrect and confusing. For finite dimensional spaces $V$ and $V^*$ are isomorphic. That claim has nothing to do with bases. Naturality has something to do with the the morphisms in the category. – 2017-01-01
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    @CameronWilliams your comment above may reflect a heurstic but there is a precise definition of naturality. – 2017-01-01
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    @IttayWeiss yeah that's exactly what I meant it as. Probably should have prefaced with that. My bad. – 2017-01-01
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    Thank you all, but can someone show me in an example why only a proper basis will work with a certain V and V*? I'm having hard time thinking about it – 2017-01-01
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    @blz if you spend a bit of time understanding the proper meaning of naturality, then your confusion will disappear. – 2017-01-01
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    Here's a good example: let $V$ be the set of $n\times m$ matrices with the Euclidean norm. $V^*$ won't really look the same (it is NOT the set of $n\times m$ matrices) so it's hard to see the isomorphism but when you drop down to a basis (they will have the same dimension), it's obvious. However, the relation I gave above between $V$ and $V^{**}$ gives the isomorphism explicitly without referencing a basis. – 2017-01-01
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    I think it should be stressed that $V$ and $V^*$ do not need to be isomorphic if $V$ is infinite dimensional, and that in the infinite dimensional case $V^*$ is the space of continuous linear forms. – 2017-01-01

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I would say canonically isomorphic, not naturally.

This true only for finite dimensional spaces: for any basis $\mathcal B=(e_1,\dots,e_n)$ of the finite dimensional space, you define a dual basis of $V^*$, $\mathcal B^*=(e_1^*,\dots,e_n^*)$ by: $$e_i^*(e_j)=\begin{cases}1&\text{if }\;j= i,\\0&\text{if }\;j\ne i.\end{cases}$$

But the evaluation mapping: \begin{align}V&\longrightarrow V^{**}\\ v&\longmapsto\begin{array}[t]{rl}V&\!\!\!\rightarrow \mathbf K\\\varphi&\!\!\!\mapsto \varphi(v)\end{array}\end{align} is an isomorphism, independent of the choice of any basis.