TITLE: Prove that if $v_1,v_2,...,v_r$ form a linearly independent set of vectors in $V$...
QUESTION [2 upvotes]: Let $S$ be a basis for an n-dimensional vector space $V$. Prove that if $v_1,v_2,...,v_r$ form a linearly independent set of vectors in $V$, then the coordinate vectors $(v_1)_S,(v_2)_S,...,(v_r)_S $form a linearly independent set in $R^n$, and conversely.

REPLY [6 votes]: We can easily prove that the map
$$\Phi:V\rightarrow \Bbb R^n,\quad x\mapsto(x_1,\ldots,x_n)$$
where $(x_1,\ldots,x_n)$ are the component of $x$ in the basis $S$, is an isomorphism of vector spaces, hence the desired result follows immediately.
Added To prove that $\Phi$ is an isomorphism of vector spaces you should show:

$\forall x,y$ with component $(x_1,\ldots,x_n)$ and $(y_1,\ldots,y_n)$ respectively in $S$ and $\forall\lambda\in \Bbb R$ we have
$$\Phi(\lambda x+y)=\lambda\Phi(x)+\Phi(y)$$
$\Phi$ is a bijective map.

and to prove that the given vectors are linearly independent notice:
$$\lambda_1v_1+\cdots+\lambda_r v_r=0\overset{\hspace{3mm}\Phi ,\Phi ^{-1}}{\Longleftrightarrow } \lambda_1(v_1)_S+\cdots+\lambda_r (v_r)_S=0$$
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