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Question

While looking over the exercise $3.F-34$ in Linear Algebra Done Right, I encountered the following paragraph

Suppose $V$ is finite dimensional. Then $V$ and $V'$ are isomorphic, but finding an isomorphism from $V$ onto $V'$, generally, requires choosing a basis of $V$. In contrast, the isomorphism from $V$ to $V''$ does not require a choice of basis and thus is considered more natural.

and these questions showed up in my mind:

$1$. Does the word natural just means that we don't need to choose a basis? I have seen the word canonical is used in the similar manner too. Is there a more precise definition for natural or canonical?

$2$. Assuming the answer to question $1$ is Yes, then why there is no natural isomorphism from $V$ onto $V'$?

$3$. I think that there is a relation between the answer to question $2$ and the proof of Riesz representation theorem. So, if we cannot find a natural isomorphism between $V$ and $V'$ then it means that we cannot prove Riesz representation theorem without choosing a basis of $V$. Is this true?


Complementary Informations

The Isomorphism from $V$ onto $V^{''}$.
Suppose $V$ is a finite dimensional vector space. Consider the following map $$ \Lambda(v)(\phi)=\phi(v), \qquad \forall v \in V, \,\, \forall \phi \in V^{'} $$ then $\Lambda$ is an isomorphism from $V$ onto $V''$.

Riesz Representation Theorem.
Suppose $V$ is a finite dimensional linear space equipped with an inner product and $\phi$ is a linear functional on $V$. Then there is a unique vector $v_0 \in V$ such that $$\phi(v) = {\langle v,v_0 \rangle}_{V}, \qquad \forall v \in V$$


Other Related Posts

I found the following posts related to this question on MSE and MO.

Post $1$, Post $2$, Post $3$, Post $4$.

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    $\begingroup$ See here for a related question (and answer) $\endgroup$ – rogerl Aug 22 '16 at 16:26
  • $\begingroup$ The Riesz representation theorem gives a natural isomorphism between a finite-dimensional $\textit{inner product space}$ and its dual. I don't think it's a stretch to say that this is THE distinguishing factor between inner product spaces and arbitrary vector spaces. It is taken advantage of endlessly in, say, Hilbert space theory. $\endgroup$ – Noah Olander Aug 22 '16 at 17:49
  • $\begingroup$ @NoahOlander: So do we have a basis free proof for the Reisz representation theorem I mentioned in the question? If Yes, would you please provide a link or reference? :) $\endgroup$ – H. R. Aug 22 '16 at 17:59
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    $\begingroup$ We don't need a basis to prove the Riesz theorem. Part of the assumptions in the Riesz theorem is the inner product, and thus the spaces it applies to have more structure than just a plain vector space. This additional structure gives you a "natural" (in the sense that one doesn't need a basis) isomorphism (or anti-isomorphism, in case of complex inner product spaces; if the space has a natural conjugation - as e.g. spaces of complex-valued functions often have - then composing this with the Riesz map gives a natural isomorphism for such complex inner product spaces). $\endgroup$ – Daniel Fischer Aug 23 '16 at 13:07
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    $\begingroup$ But if you have only an abstract finite-dimensional vector space over an arbitrary field, you don't have enough structure to define a nontrivial natural map $V \to V'$. $\endgroup$ – Daniel Fischer Aug 23 '16 at 13:08
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There is an interpretation of natural within category theory that allows us to rigorously state that while there is a natural isomorphism from $V$ to $V''$, there is no natural isomorphism between $V$ and $V'$. This interpretation is explained here and here. The second bit is a little too elaborate for me to wrap my head around, but I'll explain what it is about the $V \to V''$ map which is "natural".

Let $\mathcal C$ denote the category whose objects are finite dimensional vector spaces. The morphisms of this category are the linear maps between vector spaces. We define a functor $F:\mathcal C \to \mathcal C$ by $F(V) = V''$ and $F([V \overset{f}{\to}W]) = [V'' \overset{f''}{\to}W'']$. What makes this a functor is that for any $f:V \to W$ and $g:U \to V$, we have $$ F(f \circ g) = F(f) \circ F(g) $$ We define the much simpler identity functor by $$ \DeclareMathOperator{\id}{id} \id(V) = V; \qquad \id([V \overset{f}{\to}W]) = V \overset{f}{\to}W $$ When we say that $V$ is naturally isomorphic to $V''$, we mean that there is a natural isomorphism between the functors $\id$ and $F$. In this case, what this means is that we can assign an isomorphism (invertible morphism) $\eta_V:\id(V) \to F(V)$ to every vector space $V$ in such a way that:

For every $f:V \to W$, we have $\eta_W \circ \id(f) = F(f) \circ \eta_V$

Or, as we can rephrase it in this context (noting $\id$ is just the identity), we need an $\eta_V:V \to V''$ for every $V$ such that

for any $f:V \to W$, $\eta_W \circ f \circ \eta_V^{-1} = f''$

Now, what is this $\eta_X$? Well, it suffices to take $$ \eta_V:V \to V''\\ [\eta(x)](\alpha) = \alpha(x) $$ You know that this map is an isomorphism from the text. Now, we note that for any $\beta \in V''$, there is an $x_\beta$ for which $\alpha(x_{\beta}) = \beta(\alpha)$ for any $\alpha \in V'$, and we have $\eta^{-1}(\beta) = x_{\beta}$. With that in mind, we can see that for any $f:V \to W$ and for any $\beta \in V''$ and $\alpha \in V'$, we have

$$\begin{align} [[\eta_W\circ f \circ \eta_V^{-1}](\beta)](\alpha) &= [[\eta_W\circ f](x_{\beta})](\alpha) \\ &= [\eta_W(f(x_{\beta}))](\alpha) \\ &= \alpha(f(x_{\beta})) \\ &= [\alpha \circ f](x_{\beta}) \\ &= \beta (\alpha \circ f) \\ &= \beta (f'(\alpha)) \\ &= [\beta \circ f'](\alpha) \\ &= [f'' (\beta)](\alpha) \end{align}$$ as required.

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