# Prove that if $v_1,v_2,…,v_r$ form a linearly independent set of vectors in $V$…

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.

• Suppose one has a dependence in the coordinate vectors. What does this imply? – BlueBuck Apr 23 '14 at 21:37
• This would imply that the vectors $v_1,v_2,...,v_r$ can be written as a linear combination of some vector in that set – Al Jebr Apr 23 '14 at 21:41

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: $$\lambda_1v_1+\cdots+\lambda_r v_r=0\overset{\hspace{3mm}\Phi ,\Phi ^{-1}}{\Longleftrightarrow } \lambda_1(v_1)_S+\cdots+\lambda_r (v_r)_S=0$$