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The answer to this question of mine provided me with the fact that every isomorphism $$ \phi: K^{n} \rightarrow V, $$ where $V$ is an arbitrary vector space of finite dimension $n$ over the field $K$, has the form

$$\begin{eqnarray*} & \phi\left(x_{1},\ldots,x_{n}\right)= x_{1}\vec{v}_{1}+\ldots+x_{n}\vec{v}_{n}, \end{eqnarray*}$$ where $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$ is some basis of $V$.

Now I was wondering, is it possible to exhibit an isomorphism from which one can't immediatly guess the basis $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$ in which its written in ?

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Consider this analogy: Every linear mapping $f$ from, say $\mathbb{R}^2$ to $\mathbb{R}^2$ , has the form $$ f(x_1,x_2)=(t_1x_1+t_2x_2)\vec{e}_{1} +(t_3x_1+t_4x_2)\vec{e}_{2},$$

where $t_1,t_2,t_3,t_4$ are some real numbers.

A rotation of $45^{\circ}$ counter-clockwise direction is a linear mapping, but it is not obvious , what the $t_1,t_2,t_3,t_4$ should be. (Only after some tinkering one gets that they are $\cos \frac{\pi}{2},-\sin \frac{\pi}{2},\sin \frac{\pi}{2},\cos \frac{\pi}{2}$)

Nonetheless for the typical linear mappings like
$$ f(x_1,x_2)= (x_1,x_2)$$ we won't have such a "form-less" interpretation like saying "it's a rotation", since here it is obvious, that $t_1=1,t_2=0,t_3=0,t_4=1$.

(A different example of a linear mapping for which the $t_1,t_2,t_3,t_4$ aren't obvious to guess is the differentiation operator, that maps every polynomial of degree at most $1$ to its derivative)

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No guess is required, you can compute the basis from $v_i = \phi(e_i)$. For example, if $f : \mathbb{R}^2 \to \mathbb{R}^2$ is the rotation you mentioned, we have (by elementary trigonometry) $f(e_1) = (\sqrt{2}/2,\sqrt{2}/2)$ and $f(e_2) = (-\sqrt{2}/2,\sqrt{2}/2)$. Your second example $f=\mathrm{id}$ is the rotation by 0°.

EDIT: I think now I've understood your question. Well, there are tons of examples and occur everywhere in algebraic-tasty branches of mathematics. I won't attempt to start a list here, since this won't lead anywhere. Just an example:

Consider the map $S^1 \to S^1$, $z \mapsto z^p$ which winds the circle $p$ times around itsself. On homology $H_1(-,K)$, this induces a $K$-linear map map $K \to K$ (after having fixed the canonical generators of $H_1(S^1,K)$). But which one? It turns out that this is $x \mapsto px$. Actually this is an easy example, since one can verify this guess immediately from the definitions. But for more complicated spaces and maps, this kind of writing down a linear map on homology, cohomology or homotopy groups becomes terribly difficult. Often it isn't enough to track down the definitions, but rather one has to use sophisticated tricks. And this becomes even more advanced when all these vector spaces are pieces of a spectral sequence, whose differentials you would like to understand.

Another, more elementary example: It is known that $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$. Now consider the isomorphism $\mathbb{C} \times \mathbb{C} \to \mathbb{C} \times \mathbb{C}$ which swaps the entries, $(a,b) \mapsto (b,a)$. What do you think this map corresponds to on the tensor product $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$? Is it really the one which swaps the tensor factors? It turns out that it corresponds to $\mathrm{id} \otimes c$ (or $c \otimes \mathrm{id}$, depending on choices), where $c$ denotes the complex conjugation!

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    I think you have misunderstood me: I don't want do compute the basis (that is already explained in the answer I linked). I want to know if there are isomorphisms that can be **presented** in such a way that the don't tanke the form $$\begin{eqnarray*} & \phi\left(x_{1},\ldots,x_{n}\right)= x_{1}\vec{v}_{1}+\ldots+x_{n}\vec{v}_{n}, \end{eqnarray*}$$ where $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$ is some basis2012-05-21
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    Of course there are such isomorphisms. You have mentioned them. But on the other hand, you can immediately express them via the basis. So what is the question?2012-05-21
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    Like I tried to explain in my analogy section, I'm looking for isomorphisms, that have a different **representation**, aside from being representable in the form where the basis is written out. Like in the linear-mapping-analogy: **although** I **could** write a rotation out, by exhibiting $t_1,\ldots,t_4$, it **also** has a different way to be represented (namely by saying "it's a rotation).2012-05-21
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    (As I further went on in my analogy, this is not trivial, since most linear mapping don't have a "second meaning" like that, so the representation via the $t_1,\ldots,t_4$ is **all I have**. I'm looking for isomorphisms, where writing them out in the basis is **not** all I have - hence the title "names of isomorphism". And since it seemed to me, that, at least in the case of linear mappings, for those linear mappings that have a "second meaning", one has to work a little bit, to find out the $t_1,\ldots,t_4$, I added, if the linear mapping is given by it's "second meaning", I also [...]2012-05-21
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    added to the title "base-less isomorphism", meaning isomorphisms for which one can't (without a little bit of work) immediatly guess their basis, since they are given via their "second meaning", so one would have to compute their basis (in the way you indicated in your answer). I hope this wasn't to much information..if it's confungs, just stick to the "I'm looking for isomorphisms, where writing them out in the basis is **not** all I have"-part)2012-05-21
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    Ah, I see. Yes, that is exactly what I meant! Would you maybe also have a slightly easier example ? I must admit, I don't know what "homology" is...2012-05-21
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A finite-dimensional vector space $V$ is isomorphic to its dual $V^*$, but the isomorphism depends on the choice of a basis in $V$.

However, there is a canonical isomorphism between $V$ and its double dual $V^{**}=(V^*)^*$ given by $v \mapsto (f \mapsto f(v))$, which does not depend on a basis.