Suppose $v_3\in V$ but $v_3 \notin W$. Prove or disprove: $\{v_1,v_2,v_3\}$ is a basis for $V$.

Question

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 $v_1,v_2,v_3$?

Answer 1

Your justification is right, but the calculation is wrong. $v_1$ and $v_2$ must have four components, so you need to express $x_4$ in terms of $x_1$ and $x_3$. Otherwise things will not work out. If you just plug any value for $x_4$, (say 0) in the equations, you are likely that a combinations of your $v_1$ and $v_2$ do not make both equations equal to 0. You should then get $v_1=(1,1,0,-3)$ and $v_2=(0,2,1,-3)$. You should be able to verify that $av_1+bv_2$ verifies both equations.

Now to prove your problem, you might just need to show that $v_3$ is linearly independent of $v_1, v_2$.