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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.

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    $\begingroup$ Your guess is wrong: $U$ could be three dimensional. $\endgroup$ – uniquesolution Sep 18 '15 at 8:56
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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.

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