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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 ?

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It is wrong to call this isomorphism "canonical", because it depends on the choice of basis for $V$. Change the basis, the isomorphism changes. It might be best to remark that this is the "standard way of proving they are isomorphic", but "canonical" in this context has a very specific meaning which is not met in this case. (Compare the fact that every finite dimensional vector space is canonically isomorphic to its double dual, but only non-canonically isomorphic to its dual.) –  Arturo Magidin May 18 '12 at 2:38
    
@ArturoMagidin Do you mean with <"canonical" in this context has a very specific meaning> that the word "canonical" generally means "base-less" in the context of linear algebra, but in other areas of mathematics means something different ? –  temo May 21 '12 at 8:49
    
"base-less" is a good approximation of the meaning in linear algebra; and in other areas it does mean something different. E.g., in the context of category theory, morphisms induced by universal properties are often called "canonical". E.g., the "canonical morphism into the quotient" in group theory. The isomorphisms between arbitrary $n$-dimensional $K$-vector spaces and $K^n$ are not canonical in either meaning. –  Arturo Magidin May 21 '12 at 14:39

2 Answers 2

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Yes, there is always such a basis - specifically, given an isomorphism $\phi:K^n\to V$, then letting $$\vec{u}_i=\phi(0,\ldots,\underbrace{1}_{i\text{th place}},\ldots,0)$$ we have that $$\phi(x_1,\ldots,x_n)=x_1\phi(1,0,\ldots,0)+\cdots+x_n\phi(0,\ldots,0,1)=x_1\vec{u}_1+\cdots+x_n\vec{u}_n.$$ The set $\{\vec{u}_1,\ldots,\vec{u}_n\}$ is a basis because $$x_1\vec{u}_1+\cdots+x_n\vec{u}_n=\phi(x_1,\ldots,x_n)=0\in V$$ if and only if $(x_1,\ldots,x_n)=(0,\ldots,0)\in K^n$, because $\phi$ is an isomorphism.

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Yes. Let $e_1, \dots, e_n$ denote the standard basis on $K^n$, i.e. $$e_1 = (1,0,0,\dots,0), e_2 = (0,1,0,\dots,0), \dots, e_n = (0,0,0,\dots,1).$$ Given an isomorphism $\phi : K^n \rightarrow V$, let $v_i = \phi(e_i)$. Then $v_1, \dots, v_n$ is a basis on $V$ (why?) and $$\phi(x_1,\dots,x_n) = x_1 \phi(e_1) + \cdots + x_n \phi(e_n) = x_1 v_1 + \cdots + x_n v_n.$$

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