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Recently, I am studying dual space with the book Linear Algebra by Stephen, Arnold, and Lawrence. In the book, it says that $V$ and $V^*$ are isomorphic. Also, it says that $V$ and $V^{**}$ are isomorphic in a "natural" way, which means that there is an isomorphism between $V$ and $V^{**}$ which is independent of the choice of bases. So, my question is

If I denote the relation between $V$ and $V^{**}$ by $V=V^{**}$, can I say $V=V^*$?

Christoph
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Ivy
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    Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *. – Arnaud D. Sep 05 '18 at 08:48
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    Note that we have quite a few questions on the subject already. You might be interested in these two for example : https://math.stackexchange.com/questions/579739/why-it-is-important-for-isomorphism-between-vector-space-and-its-double-dual-spa, https://math.stackexchange.com/questions/1900179/the-isomorphism-of-a-linear-space-with-its-dual-and-double-dual – Arnaud D. Sep 05 '18 at 08:52

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First of all, the isomorphisms $V\cong V^*$ and $V\cong V^{**}$ only hold (in general) for finite dimensional $K$-vector spaces. If $V$ does not admit a finite basis, the isomorphisms may fail.

For an isomorphism $V\to V^*$ you have to specify how a vector $v\in V$ should act as a linear form $V\to K$. It turns out that this amounts to specifying a non-degenerate bilinear form $\langle-,-\rangle\colon V\times V\to K$ and send $v\in V$ to $\langle v,-\rangle\in V^*$, i.e., the linear form sending $w$ to $\langle v,w\rangle$. Another way to achieve the same thing is by picking a basis $e_1,\dots,e_n$ and declaring it to be orthonormal, hence defining a bilinear form with $\langle e_i, e_j\rangle = \delta_{ij}$. In this case you get a dual basis $e_i^* = \langle e_i, -\rangle$.

To summarize, in order to construct an isomorphism $V\to V^*$, you have to either choose a basis of $V$ or a non-degenerate bilinear form.

Now for an isomorphism $V\to V^{**}$, you have to specify how a vector $v\in V$ acts as a linear form $V^*\to K$. So given a linear form $\omega\in V^*$, how does $v\in V$ act on $\omega$ to produce an element of $K$? In this situation it is just natural to just apply the linear form $\omega$ to the vector $v$. Hence, the isomorphism $V\to V^{**}$ sends $v\in V$ to the linear form $V^*\to K$ with $\omega\mapsto \omega(v)$.

We see that we did not need to make any choices in writing down this isomorphism $V\to V^{**}$.

To answer your question, if you use "$V=V^{**}$" to denote the identification of $V$ and $V^{**}$ under the above natural isomorphism, you can not write "$V=V^*$" in the same spirit, since any isomorphism $V\to V^*$ requires you to make some (arbitrary) choices.

Christoph
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  • So, there are different kinds of isomorphism. Is that right? – Ivy Sep 05 '18 at 10:28
  • I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them. – Christoph Sep 05 '18 at 10:36
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    The difference is that every isomorphism between $V$ and $V^$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^{}$ that maps $v \in V$ to $v'' \in V^{}$ where $v''(\omega) = \omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^{*}$ - for example, we could define $v''$ by $v''(\omega) = 2\omega(v)$, but the factor of 2 is then an arbitrary choice. – gandalf61 Sep 05 '18 at 11:27