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.

  • $\begingroup$ Suppose one has a dependence in the coordinate vectors. What does this imply? $\endgroup$ – BlueBuck Apr 23 '14 at 21:37
  • $\begingroup$ 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 $\endgroup$ – 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: $$\lambda_1v_1+\cdots+\lambda_r v_r=0\overset{\hspace{3mm}\Phi ,\Phi ^{-1}}{\Longleftrightarrow } \lambda_1(v_1)_S+\cdots+\lambda_r (v_r)_S=0$$

  • $\begingroup$ Im sure this is a valid proof, however, I do not have experience with maps and isomorphisms. Can this be phrased in a more basic way. $\endgroup$ – Al Jebr Apr 24 '14 at 3:45
  • $\begingroup$ I am an undergraduate and L.A. and multivariable calc are the highest math courses I have taken. $\endgroup$ – Al Jebr Apr 24 '14 at 3:45
  • $\begingroup$ I edited my answer. $\endgroup$ – user63181 Apr 24 '14 at 6:48

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