I'm given the following information:
Consider the two equations $$x_1+2x_2-x_3+x_4=0 (*),$$ $$2x_1+x_2+x_3+x_4=0 (**).$$ Let $V$ be the set of all solutions of $(*)$ and $W$ be the set of all solutions of both $(*)$ $\textit{and}$ $(**)$. Thus $W$ is a 2-dimensional subspace of the 3-dimensional subspace $V$ of $\mathbb{R}^4$.
I'm asked to first find a basis $\{v_1,v_2\}$ for $W$ and then to solve the problem in the title. Here's what I've done for the basis:
We must first find the set of all solutions to these two equations. Setting them equal and reducing, we get $-x_1+x_2-2x_3=0$. Thus, $x_2=x_1+2x_3$. Then, $(x_1,x_2,x_3)=(x_1,x_1+2x_3,x_3)=x_1(1,1,0)+x_3(0,2,1)$. Therefore, $v_1=(1,1,0), v_2=(0,2,1)$ is our basis.
I'd like to verify that this is correct, and I'm looking for advice on the proof of the title problem. My initial thought was that it is false, because $v_3$ could be a linear combination of $v_1$ and $v_2$. But I see that is impossible because then it would be an element of $W$. So I'm inclined to think that it's true, as any element of $V$ that is not in $W$ is certainly linearly independent with the basis for $W$. Does this then imply that $V$ must be spanned by $v_1,v_2,v_3$?