An informal proof that $K^n =(K^n)^{**}$ or transposing two times does nothing I was lately confronted with the following "proof" that $K^n$ is isomorphic to $(K^n)^{**}$: $K^n$ contains column vectors, transposing them we get an element in $(K^n)^{*}$ and transposing again we get an element in $(K^n)^{**}$; but the second transposition is actually our original element, so $K^n =(K^n)^{**}$.
How can this proof be made rigorous ? What I'm unsure of, is whether the second transposition indeed directly gives "our original element", because the elements of  $(K^n)^{**}$ actually are maps that take functionals as argument and return scalars, but, as this previous question of mine shows, the second transposition actually gives our initial vector again, not a map (taking functionals and returning scalars), so identifying a vector with such a map isn't trivial and needs a proof, I think.
Therefore I suspect that the above argument isn't really a proof; more something like a mnemonic for remembering that $K^n =(K^n)^{**}$, because it seems to me that only after one has formally proved that these spaces are biejctive (by using a bijection between $K^n$ and $(K^n)^{*}$ and then an analoguous one between $(K^n)^{*}$ and $(K^n)^{**}$ -- or even shorter the natural bijection) one is entitled to say that the second transposition gives our original element - not the other way around!
 A: If $V$ is a vector space over the field $F$ then its double dual $V^{**}$ is certainly different from $V$ as a set. However, when $V$ is finite dimensional the map $V\to V^{**}$ associating to each $v\in V$ the linear functional $\tilde{v}\colon V^*\to F$ defined by $\tilde{v}(f)=f(v)$ is a natural isomorphism of vector spaces. Here natural means that the map $V\to V^{**}$ does not depend on a choice of basis. To further appreciate the concept of naturality we need some category theory (it is said that category theory was invented, at least in part, to make the notion of naturality rigorous). Notice that $V$ and $V^*$ are also isomorphic, but to specify an isomorphism you need to choose a basis of $V$.
A: It's not exactly correct.  First of all, what exactly are you trying to prove?  Just that $K^n$ and $(K^n)^{\ast\ast}$ are isomorphic?  If you have a rigorous proof that elements of $(K^n)^\ast$ are given by row matrices so that the transpose gives an isomorphism $K^n \to (K^n)^\ast$, then you already have that because you've proved that $V \simeq V^\ast$ for any finite dimensional vector space $V$ so $V \simeq V^\ast \simeq V^{\ast\ast}$.
Second, you are right that it doesn't give the details about how various objects are identified so on that grounds you can call it not rigorous.  It's not too hard to make it rigorous but you'll see that in doing so you actually use the thing to be proved:
Observe what's actually happening when you say that the transpose gives the dual.  Indeed $(K^n)^\ast$ is linear maps $K^n \to K$.  While they can be represented as matrices, they are not equal to matrices as objects.  Let $C = K^n$ be the set of $n \times 1$ matrices with entries in $K$ (column matrices) and let $R$ be the set of $1 \times n$ matrices with entries in $K$ (row matrices).  Now, what you are really doing when you say that transpose gives a map $C \to C^\ast$ is


*

*Giving an isomorphism $C^\ast \simeq R$ (by identifying a map with the matrix that represents that map).

*Giving a map $t\colon C \to R$ (the transpose)

*Declaring the map $C \to C^\ast$ to be defined as the composition $C \overset{t}{\to} R \simeq C^\ast$ of the two above.


Now for the second transpose we


*

*Give an isomorphism $C^{\ast\ast} \simeq C$ by saying that $x \in \mathbb C$ represents the map $C^\ast \to K$ which sends $f\colon C \to K$ to $f(x)$.

*Give a map $t'\colon R \to \mathbb C$ (the transpose)

*Declare the map $C^\ast \to C^{\ast\ast}$ to be the composition $C^\ast \simeq R \overset{t'}{\to} C \simeq \mathbb C^{\ast\ast}$.


If you do this and then you compute the composition $C \to C^\ast \to C^{\ast\ast} \simeq C$ of the two maps from above you'll find it's the identity and all you've done is take the transpose twice.  And now you can see why this transpose argument is weird, because you're using $C^{\ast\ast} \simeq C$ in the argument.
