I was working on a solution to this question (linked below), but wanted to understand it from a more abstract perspective (not considering an explicit counterexample) because I got it wrong initially.
Prove or give Counterexample: Is this a Basis.
Problem Statement
Prove or give a counterexample: If $π£_1,π£_2,π£_3,π£_4$ is a basis of π and π is a subspace of π such that $π£_1,π£_2βπ$ and $π£_3,π£_4βπ$, then $π£_1,π£_2$ is a basis of π.
Proof.
Suppose that $v_1,v_2$ is a basis of $U$. Then $span(v_1,v_2) = U$ and $v_1,v_2$ are linearly independent, by the definition of basis.
We know $v_1,v_2$ is linearly independent because removing vectors from a linearly independent list results in a shorter linearly independent list of vectors. In this case, start with $v_1,v_2,v_3,v_4$ and remove vectors to get that $v_1,v_2$ is linearly independent.
So, we just have to check if $v_1,v_2$ spans $U$.
If $v_1,v_2$ spans $U$, then no vector $v_nβspan(v_1,v_2)$ exist such that $v_n \in U$.
I reasoned here that the only vectors in the larger space $V$ that were not in $U$ were $v_3$ and $v_4$, meaning that $v_n \in span(v_3,v_4)$.
Here is my mistake: I reasoned that because $v_3 \notin U$ and $v_4 \notin U$, then $v_n \notin U$.
From this, I incorrectly concluded $span(v_1,v_2) = U$ and that $v_1,v_2$ was a basis of $U$
From the counterexample, I realized it was possible for $v_3+v_4
\in U$ despite $v_3 \notin U$ and $v_4 \notin U$.
I want to more clearly understand how two linearly independent vectors ($v_3,v_4$ in this example) can not be in some subspace, while their sum can exist in that subspace.
Thanks.
Best Answer
Is a concrete example enough? Consider $V=\mathbb{R}^2$ with the usual vector space structure. The vectors $(0,1)$ and $(1,0)$ are linearly independent. Neither belongs to the subspace $U = \{(x,x) \mid x \in \mathbb{R}\}$ determined by the line $y=x$, but the sum $(0,1)+(1,0)=(1,1)$ certainly belongs to $U$.