# Prove or give Counterexample: Is this a Basis (from a more abstract perspective)

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 𝑈 is a subspace of 𝑉 such that $$𝑣_1,𝑣_2∈𝑈$$ and $$𝑣_3,𝑣_4∉𝑈$$, then $$𝑣_1,𝑣_2$$ is a basis of 𝑈.

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 $$v_n∉span(v_1,v_2)$$ 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.

• Note: $(1,1)$ and $(1,-1)$ are not in the $1$-dimensional space spanned by $(1,0)$, but their sum is – J. W. Tanner Jul 25 at 16:21
• 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 – peek-a-boo Jul 25 at 16:23
• 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 $U$ – Richard K Yu Jul 25 at 16:25

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 $$U$$.