Let $V$ be an arbitrary vector space of finite dimension $n$ over the field $K$. It is known that in that case $ V\simeq K^{n}. $
The canonical isomorphism which achieves this is \begin{eqnarray*} & \phi:K^{n}\rightarrow V\\ & \left(x_{1},\ldots,x_{n}\right)\mapsto x_{1}\vec{v}_{1}+\ldots+x_{n}\vec{v}_{n}, \end{eqnarray*} where $\left(\vec{v}_{1},\ldots,\vec{v}_{n}\right)$ is a fixed basis of $V$.
Of course this isn't the only isomorphism - different basis give different isomorphisms. But may there also be others ? In other words: Given an arbitrary isomorphism $\varphi:K^{n}\rightarrow V$ is there always a basis $\left(\vec{u}_{1},\ldots,\vec{u}_{n}\right)$, such that in this basis $\varphi$ looks like above, i.e. $ \varphi\left(x_{1},\ldots,x_{n}\right)=x_{1}\vec{u}_{1}+\ldots+x_{n}\vec{u}_{n}\ ? $
If not, could you give an example ?