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I am trying to do the following problem for practice:

Let $Q = \{ \textbf{u}_1,\textbf{u}_2,\textbf{u}_3,\textbf{u}_4\}$ is a basis for a vector space $V$. Show that the vectors $\textbf{v}_1$,$\textbf{v}_2$,$\textbf{v}_3$ are linearly independent in $V$ is and only if their coordinate vectors $(\textbf{v}_1)_S$,$(\textbf{v}_2)_S$,$(\textbf{v}_3)_S$ are linearly independent in $R^4$

What I was thinking about doing is talking about the transformation matrix $P_{S\rightarrow R^4}$ and similarly the transformation matrix $P_{R^4\rightarrow S}$ and come up with an argument that since these matrices are inverses of one another then they must be independent within their own space, and and then say any vector multiplied by either matrix must stay linearly independent if they are in their original space.... i.e if vectors in $V$ are linearly in dependent in $V$ then when multiplied by $P_{S\rightarrow R^4}$ they must still stay linearly independent. I feel like that is the argument that would be needed to prove both sides of this statement, but I am unsure if
1) it's the correct argument
2) if it is correct how would I show it more extensively.
Any help would be greatly appreciated. Thank you.

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  • $\begingroup$ I'm not sure if there is a convienient way to talk about matrices in some abstract vector space that is not $\mathbb{R}^4$, so i'd rather view your $P_{S \rightarrow \mathbb{R}^4}$ as a linear transformation instead of a matrix. I'm not sure what you mean by "matrices independent in their own space". What you write next is however indeed correct: when you apply an isomorphism to a linearly independant family, it will stay linearly independant. I will leave in the answears a way I would write the proof down. Hopefully it will be to your liking. $\endgroup$ – Keen Sep 29 '17 at 18:24
  • $\begingroup$ @Keen What I meant by that comment was that their determinant would be zero, since they have to be inverses of one another... Sorry about the confusion, should have worded it a little bit better $\endgroup$ – Robert Sep 29 '17 at 18:28
  • $\begingroup$ I'm assuming $V$ is a vector space over $\mathbb{R}$ and not some other field? Then you have an isomorphism $V \to \mathbb{R}^{4}$ given by assigning coordinates. Should be straightforward from there. $\endgroup$ – Morgan Rodgers Sep 29 '17 at 18:38
  • $\begingroup$ couldn't 3 vectors lie in a 3d subspace of $R^4$ with common 4th component? $\endgroup$ – f5r5e5d Sep 29 '17 at 18:57
  • $\begingroup$ @f5r5e5d I don't understand your comment. But yes, 3 linearly independent vectors will necessarily span a 3-dimensional subspace of $\mathbb{R}^{4}$. $\endgroup$ – Morgan Rodgers Oct 2 '17 at 5:04
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Let's call $s$ the isomorphism that sends $u_1, \cdots , u_4$ onto the standard basis in $\mathbb{R}^4$. Let us assume that $v_1, v_2,v_3$ is linearly independant. Then if $\lambda_1 s(v_1)+\lambda_2 s(v_2)+ \lambda_3 s(v_3)=0$, then $\lambda_1 v_1 +\lambda_2 v_2 +\lambda_3 v_3 =0$ by aplying $s^{-1}$.Therefore $\lambda_1 =\lambda_2= \lambda_3=0$. Hence $s(v_1),s(v_2),s(v_3)$ are linearly independent. Same kind of proof goes for the other implication, we just apply s instead of $s^{-1}$.

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