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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 $𝑣_1,𝑣_2,𝑣_3,𝑣_4$ is a basis of 𝑉 and π‘ˆ is a subspace of 𝑉 such that $𝑣_1,𝑣_2βˆˆπ‘ˆ$ and $𝑣_3,𝑣_4βˆ‰π‘ˆ$, then $𝑣_1,𝑣_2$ is a basis of π‘ˆ.

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 $v_nβˆ‰span(v_1,v_2)$ 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.

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  • $\begingroup$ Note: $(1,1)$ and $(1,-1)$ are not in the $1$-dimensional space spanned by $(1,0)$, but their sum is $\endgroup$ – J. W. Tanner Jul 25 '19 at 16:21
  • $\begingroup$ can you already think of a counter example in say, $\Bbb{R}^2$? also could you be more specific about what you're looking for, because I'm not sure what you mean by "I want to more clearly understand... " Because to me a counterexample already shows you that this is a misconception which you should just erase from memory $\endgroup$ – peek-a-boo Jul 25 '19 at 16:23
  • $\begingroup$ Got it, I think I just got hung up on the misconception that $v_3, v_4$ not existing in $U$ did not imply that all their linear combinations also did not exist in $U$ $\endgroup$ – Richard K Yu Jul 25 '19 at 16:25
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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 $U$.

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  • $\begingroup$ I think this helps clear it up. The idea that I started to see with the counterexamples is that the dimension of individual vectors is not necessarily the same as the dimension of their sum. $\endgroup$ – Richard K Yu Jul 25 '19 at 16:32

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