Prove or give Counterexample: Is this a Basis.

Question

Prove or give a counterexample: If $v_1, v_2, v_3, v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1,v_2 \in U$ and $v_3, v_4 \notin U$, then $v_1, v_2$ is a basis of $U$.

My guess would be that it's true since we have a list of 4 vectors in $V$ and $U$ is a subspace in $V$ and $v_1, v_2$ should be linearly independent. Not sure how to prove that it is a basis.

Answer 1

Consider $V = \Bbb R^4$ and take the standard basis i.e $v_i=e_i$ for $V$. Let $U$ be the subspace of $V$ with basis $\{(1,0,0,0),(0,1,0,0),(0,0,1,1)\}$. So clearly $v_3, v_4 \notin U$ but $U$ is not 2 dimensional.

Answer 2

$V:=\mathbb{F}^4$.
$U:=\{(x,y,z,0)|x,y,z\in\mathbb{F}\}$.
$U$ is a subspace of $V$.
$v_1:=(1,0,0,0)$.
$v_2:=(0,1,0,0)$.
$v_3:=(0,0,1,1)$.
$v_4:=(0,0,0,1)$.
$v_1,v_2\in U$.
$v_3,v_4\notin U$
$v_1,v_2,v_3,v_4$ is a basis of $V$.