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.
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 $s^{-1}$.