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)