In a lot of resources that I have read it is mentioned that the isomorphism between $V$ and $V^*$ is non-canonical, but I was never sure that I properly understood precisely what this means. I haven't studied category theory so I couldn't really understand some other answers on math.stack based on category theory arguments.
I will try to explain how I understand things and maybe someone could comment on that.
Setting
Let $V$ be a finite-dimensional vector space. Its dual $V^*$ is the space of linear maps from $V$ to $\mathbb{F}$: $V^* = L(V,\mathbb{F})$. Now let $v$ be an arbitrary vector from $V$. If I fix a basis $F = (f_1,\ldots,f_n)$ for $V$ then I can write $v = \sum_{i=1}^n [v]_F^i f_i$, for some unique coefficient vector $[v]_F\in\mathbb{F}^n$. Then as far as I understand the "non-canonical isomorphism" that depends on this basis is $v\mapsto\alpha = \sum_{i=1}^n [\alpha]^{F^*}_if^i = \sum_{i=1}^n [v]^i_F f^i$ where $F^* = (f^1,\ldots,f^n)$ is the dual basis w.r.t. $F$ such that $f^i(f_j) = \delta^i_j$. Now let me choose a different basis $G = (g_1,\ldots,g_n)$ for $V$, then the dual basis w.r.t. it is $G^*=(g^1,\ldots,g^n)$ such that $g^i(g_j) = \delta^i_j$. Now I can write the vector $v$ in terms of the basis $G$: $v = \sum_{i=1}^n [v]_G^i g_i$, and I can form a functional $\beta = \sum_{i=1}^n [\beta]^{G^*}_ig^i = \sum_{i=1}^n [v]^i_G g^i$.
For $\alpha$ and $\beta$ to be well-defined (i.e. so that the values $\alpha(u)$ and $\beta(u)$ are fixed for every $u\in V$ irrespective of the coordinate representation), their coordinates must transform in an appropriate manner under a change of basis. Let $P$ be the change of basis matrix such that $g_i = \sum_{j=1}^n f_j P^j_i$ then in order to have $f^i(f_j) =\delta^i_k \implies g^i(g_j) = \delta^i_j$ I need that $g^i = \sum_{j=1}^n (P^{-1})^i_j f^j$ since:
$$g^i(g_j) = \sum_{k=1}^n (P^{-1})^i_k f^k \left(\sum_{l=1}^n P^l_j f_l\right) = \sum_{k=1}^{n}\sum_{l=1}^n(P^{-1})^i_kP^l_jf^k(f_l) = \sum_{k=1}^n(P^{-1})^i_kP^k_j = \delta^i_j.$$
If I make it so that I transform the coordinates as follows: $[\alpha]^{G^*} = [\alpha]^{F^*}P$ and $[v]_G = P^{-1}[v]_F$, then I should get the same result for $\alpha(v)$ regardless of the chosen basis. Of course now $[\alpha]^{G^*}\ne[v]_G^T=[\beta]^{G^*}$ in general showing that $\alpha\ne \beta$.
The Question
Is the "non-canonical isomorphism" referring to the fact that $\alpha\ne \beta$ in the general case? That is, the construction of $\alpha$ was a function of my choice of $f_1,\ldots,f_n$, while the construction of $\beta$ was a function of my choice of $g_1,\ldots,g_n$. I assume that $\alpha$ could miraculously be equal to $\beta$ in some cases (e.g. if $f_1=g_1$ and $[v]^i_F=0=[v]^i_G$ for $i>1$), but in general it is not, so the non-canonicity is referring to the dependence on the choice of basis? If I had a canonical basis, for example $(e_1,\ldots,e_n)$ in $\mathbb{R}^n$, would this then be considered a canonical choice instead?
On the other hand, if I have a non-degenerate bilinear form $B:V\times V\to\mathbb{F}$, I can construct an isomorphism $\gamma_v = B(v,\_)$, that is independent of the basis. So is the word canonic referring to independence of the choice of basis in the initial definition of the functional?
Example
If I take a basis for the space of polynomials of at most degree $1$ then I could choose the monomial basis $(f_1,f_2) = (1,x)$ or the Lagrange basis $(g_1,g_2) = \left(\frac{x-x_0}{x_1-x_0}, \frac{x-x_1}{x_0-x_1}\right)$. The dual bases are given as $(f^1,f^2) = \left(\delta_0, \delta_0 \circ \frac{d}{dx}\right)$ and $(g^1,g^2) = \left(\delta_{x_1},\delta_{x_0}\right)$,
where $\delta_p(f) = f(p)$. Now if I have the vector $v = a\frac{x-x_0}{x_1-x_0} + b\frac{x-x_1}{x_0-x_1}$ then $v = \frac{bx_1-ax_0}{x_1-x_0}+\frac{a-b}{x_1-x_0}x = p\cdot 1 + qx $. This induces the two functionals $\alpha = p\delta_0 + q\delta_0 \circ \frac{d}{dx}$ and $\beta = a\delta_{x_1}+b\delta_{x_0}$. Now $\beta(v) = a^2+b^2$ while $\alpha(v) = p^2+q^2$, and in general those are unequal, thus $\alpha\ne\beta$.
However if I were to take for example the inner product $\langle v, w\rangle = \int_{0}^{1} v(t) w(t)\,dt$ then I could define the "canonical isomorphism" $\alpha_v = \langle v, \_\rangle$. Now a subsequent question is – didn't I just swap the choice of the initial basis for a choice of an inner product? Because I could of course take a number of other inner products that would result in different functionals. In what sense is the choice of inner product more canonical than the choice of basis? Or did I misunderstand this, and a choice of inner product doesn't result in a canonical isomorphism either. But if I go with that and I presuppose that the choice is canonical only when a canonical inner product exists, wouldn't this be the same argument as presupposing that a canonical basis exists? Basically I have trouble making sense of what "canonical" is referring to in the first place.
Canonical Isomorphism to the Double Dual $V^{**}$
The isomorphism $v\mapsto v^{**}(\phi) = \phi(v)$ is considered "canonical", I assume because there is no choice whatsoever in the above definition – I neither choose a basis nor an inner product. So is the above the only thing that "canonical" can refer to? Or is a vector space with a bilinear form isomorphism to $V^*$ also considered "canonical" if I assume that there is some distinguished canonical bilinear form? And similarly a choice $v\mapsto \alpha = \sum_{i=1}[\alpha]^{E^*}_ie^i = \sum_{i=1}^n[v]^i_E e_i$ would be considered "canonical" if there is some distinguished "canonical" basis $E$? Essentially, what do I need for something to be considered canonic?
Edit:
Attempt at a solution of the exercise from coiso
For $(V,\mathbb{F}_2)$ with $\dim = 1$, with $V=\{0,x_1\}$ and $V^*=\{0^*,y^1\}$ I can construct the table:
\begin{equation}
\begin{array}{c|c}
& 0^* & y^1 \\
\hline
0 & 0 & 0 \\
\hline
x_1 & 0 & 1
\end{array}
\end{equation}
If I assign $0\ne x_1\mapsto 0^*$ and $0\mapsto y^1$ then $0^* = (x_1)^* = (x_1+0)^* = 0^* + y^1 = y^1$, and the assignment is not injective hence not an isomorphism. So there is only one valid choice and then $V\cong V^*$.
For $(V,\mathbb{F}_2)$ with $\dim = 2$, with $V=\{0,x_1,x_2,x_{12}=x_1+x_2\}$ and $V^* = \{0^*, y^1, y^2, y^{12}\}$ I can construct the table:
\begin{equation}
\begin{array}{c|c|c|c|c}
& 0^* & y^1 & y^2 & y^{12} \\
\hline
0 & 0 & 0 & 0 & 0 \\
\hline
x_1 & 0 & 1 & 0 & 1 \\
\hline
x_2 & 0 & 0 & 1 & 1 \\
\hline
x_{12} & 0 & 1 & 1 & 0
\end{array}
\end{equation}
I cannot have any other columns in the table as otherwise the functionals won't be linear.
One possible assignment, when taking $x_1,x_2$ as basis vectors for $V$ is $0\mapsto 0^*$ and $x_J\mapsto y^J$. I could of course take $x_1,x_{12}$ instead as a basis, then the dual basis must be $x_1\mapsto y^{12}$ and $x_{12}\mapsto y^2$. Then the only option left is $x_2\mapsto y^1$. I don't get why any of the two assignments should be more canonical, unless I am missing something and the second assignment breaks linearity, which doesn't seem obvious from the table.
Edit 2:
Other Isomorphisms to the Double Dual
Thinking about this I realized that I could define the map $I_{\lambda} : V\to V^{**}$, $v\mapsto I_{\lambda}(V) = \lambda v^{**}$ such that $I_{\lambda}(v)(\phi) = \lambda v^{**}(\phi) = \lambda \phi(v)$ for $\lambda \ne 0$. This map looks like it is bijective, and it is linear since $I_{\lambda}(au + bv)(\phi) = \lambda \phi(au+bv) = a\lambda\phi(u)+b\lambda\phi(v) = (aI(u)+bI(v))(\phi)$, so it must be an isomorphism. Yet I suppose this would not be termed canonical, or will it? To my understanding it is the map $I_{1}(v)(\phi) = \phi(v)$ that is the canonical isomorphism, i.e., where $\lambda=1$. Clearly the above family of isomorphisms $\{I_{\lambda}\,:\,\lambda \ne 0\}$ do not depend on a choice of basis or inner product, and yet I do not think that they would be termed canonical. So I am even more confused now, as clearly "canonical" is not referring to a choice of basis or inner product in this case. Or is it that all $I_{\lambda}$ are considered canonical isomorphisms?
I also found this video, I don't know how relevant it is to the above setting though. From the video it becomes clear that canonical isomorphism is a kind of equality that is true for a certain subset of statements, but not all. So the question is what is canonical referring to in the above cases. Are all $I_{\lambda}$ canonical isomorphisms or is $I_1$ the only canonical isomorphism? Is $v\mapsto B(v, \_)$ canonical if there is only a single non-degenerate bilinear form $B$ given, i.e. $(V,B)\stackrel{canon}{\cong} (V^*, B^*) = (V^*, B(B^{-1}, B^{-1})$ but $V\stackrel{non-canon}{\cong} V^*$, and also $(V,(B_1, B_2)) \stackrel{non-canon}{\cong} (V^*, (B^*_1, B^*_2))$. Is the isomorphism canonical if a single basis is specified but no more, i.e. $(V, E) \stackrel{canon}{\cong} (V^*, E^*)$ but $(V, (E_1, E_2)) \stackrel{non-canon}{\cong} (V, (E_1^*, E_2^*))$? Also in the answers Mikhail Katz mentioned that in differential geometry the map may be considered non-canonical if it is not continuous, but I assume this is slightly different, as I don't need a notion of continuity to talk about the algebraic dual and double dual as far as I know.
Edit 4:
What I Gathered so Far
There are many isomorphisms $V\to V^{**}$ but the canonical one is given by the evaluation functional $\Lambda:V\to V^{**}$:
$$\Lambda(v)(\phi) := \phi(v), \quad \forall \phi\in V^*.$$
According to coiso also the isomorphisms $I_{\lambda}(v)(\phi) = \lambda\phi(v)$ for $\lambda\ne 0$ can be termed canonical, but they are not the canonical isomorphism, which is instead given by the evaluation functional $\Lambda(v)$. As far as I understood the motivation for calling those canonical is because they do not require making a choice of basis, and we care greatly about results being coordinate invariant in linear algebra, differential geometry, physics, etc.
The isomorphism $J : (V,B) \to (V^*,B^*)$ where $J(v) = B(v,\_)$ is also canonical if $B$ is a non-degenerate bilinear form (conjugate-linear in one of the slots in the complex-valued case). If $B$ is degenerate then the latter is not bijective so not an isomorphism, but I guess that restricted to the subspace where it is non-degenerate it will be a bijection and thus a canonical isomorphism on that subspace.
Arturo Magidin made it clear that the isomorphism $J$ is canonical because it is between the vector space along with its inner product/bilinear form $(V,B)$ and its counterpart $(V^*,B^*)$. As far as I understood a map $J_B : V\to V^*$ where $J_B(v) = B(v,\_)$ would not provide a canonical isomorphism between $V$ and $V^*$ by themselves, since one could choose a different bilinear form, which would result in a different identification.
The isomorphism $K_F:V\to V^*$:
$$K_F(v) = \sum_{i=1}^n [v]^i_F f^i \in V^*,$$
where $F = (f_1,\ldots,f_n)$ is a basis for $V$ and $F^*=(f^1,\ldots,f^n)$ is the dual basis for $V^*$ such that $f^i(f_j) = \delta^i_j$, is not a canonical isomorphism between $V$ and $V^*$ in general. This is because it depend on a choice of basis $F$, and as illustrated by my above examples a different basis can result in a different result.
A special case when there is a canonical isomorphism is when $\dim V\leq 2$ and the underlying field is $\mathbb{F}_2$ as given in the solution by coiso. I believe that also a one-dimensional space is canonically isomorphic to its dual (even if it is not over $\mathbb{F}_2$) since the map $K_F$ and $K_G$ actually produce the same functionals irrespective of the choice of $F$ and $G$ (the scalar difference gets canceled out by the change of basis).
Finally, supposedly also $K : (V,F) \to (V^*,F^*)$ where $K(v) = K_F(v)$ should also be a canonical isomorphism since I am now identifying a vector space along with a basis. The latter is more restrictive than providing a non-degenerate bilinear form however.
Best Answer
Yes, something is "canonical" if it does not depend on any choices.
This will depend on what kind of mathematical object you're talking about. For example, for (finite dim) vector spaces $V$ there is no canonical isomorphism $V\cong V^\ast$, but for (finite dim) inner product spaces $V$ there is a canonical isomorphism. An inner product is a positive-definite symmetric bilinear form, but allowing arbitrary nondegenerate bilinear form will do just as well via currying / musical isomorphisms.
I like to think of "canonicity" in game-theoretic terms. Consider a game where you and a teammate are put in separate rooms and given the same mathematical object, and your goal is to both come up with the same construction. For example, you may be both given the same vector space and want to construct an isomorphism $V\cong V^\ast$. You and your teammate are allowed to strategize before going into the rooms, but you don't have any information about the object you will be given other than what kind of mathematical object it will be. Can you and your teammate devise a strategy that guarantees you both come up with the same construction? If your method of construction involves arbitrary choices which the end result depends on, then that will not be such a strategy. If the object you're given is known ahead of time to be an inner product space, then there is such a strategy.
There is no intrinsic sense in which one is more canonical than the other. One could just as well consider "vector space equipped with a basis" as its own kind of mathematical object, and then there is a canonical choice of isomorphism to the dual. However, this is arguably a less "natural" kind of mathematical object. In "nature" (Euclidean space) there is no built-in choice of coordinates but there is a built-in inner product (the dot product which can be defined solely from angles and distances).
Note, by the way, that a basis determines an inner product but not vice-versa. So it is a more "cost-effective" choice to yield a duality (i.e. isomorphism $V\cong V^\ast$). In fact, there is an equivalence between dualities and nondegenerate bilinear forms (via musical isomorphisms).
Responding to edits:
They are all canonical. The fact that $I_1$ is called "the" canonical isomorphism adds into the mix the usage of the word "canonical" for "standard," which is a social thing.
Imagine you and your teammate playing the game are told you will be given a certain kind of mathematical object, and in the strategy session your teammate tells you "hey there's lots of different strategies we have, in fact one strategy for each $\lambda\ne0$, but I've heard when people play this game they tend to use the strategy with $\lambda=1$ so let's do that too."
Remember in this version of the game, you're not given a basis. So I wouldn't call the vectors $x_1,x_2,x_{12}$, I'd call them $x_1,x_2,x_3$, and the dual vectors $x_1^\ast,x_2^\ast,x_3^\ast$.
But notice something. In your labelling, you made $x_1^\ast,x_2^\ast,x_3^\ast$ have respective kernels $\{0,x_2\},\{0,x_1\},\{0,x_3\}$. That means $x_3^\ast$ has kernel associated to $x_3$, but $x_1^\ast,x_2^\ast$ have kernels associated to $x_2,x_1$ respectively. This automatically distinguishes $x_3,x_3^\ast$ as "special."
What's to prevent your teammate from making $x_2,x_2^\ast$ the "special" ones, or $x_1,x_1^\ast$? After all, your labelling of the vectors $x_1,x_2,x_3$ as first, second, third is an arbitrary choice that you made, and there is no guarantee your teammate makes the same choice as you. If they make a different choice of 1/2/3, they get a different isomorphsim.
There are six isomorphism between $V$ and $V^\ast$ (when $V\cong\Bbb F_2^2$), but only one of them is canonical. One way to guarantee your isomorphism is canonical is to describe what $x_i^\ast$ is in a way that doesn't depend on how the other two $x_j,x_k$ elements are labelled...