If a vector space $V$ is spanned by $\{v_1,v_2\}$ then any three vectors in $V$ is linearly dependent

Question

Let $V$ be a vector space spanned by $\{v_1,v_2\}$.

Claim: Any set consisting of three vectors in $V$ is linearly dependent.

Proof: Let $w_1,w_2,w_3 \in V$.

Call the set consisting $\{w_1,w_2,w_3\}$ be $A$.

So, we're trying to show that $A$ is linearly dependent.

By definition of linear dependence, we want to show that there exists $c_i \neq 0$ such that $c_1w_1+c_2w_2+c_3w_3=0$, where $c_i \in \mathbb{R}$.

We can equivalently write the equation above as follows since $\{v_1,v_2\}$ spans $V$.

$c_1(a_{11}v_1+a_{12}v_2)+c_2(a_{21}v_1+a_{22}v_2)+c_3(a_{31}v_1+a_{32}v_2)=0$ where $a_{ij} \in \mathbb{R}$.

$\iff v_1(c_1a_{11}+c_2a_{21}+c_3a_{31})+v_2(c_1a_{12}+c_2a_{22}+c_3a_{32})=0$

Stuck here

If I knew that $\{v_1,v_2\}$ is linearly independent then I can proceed the proof perfectly fine and it will all work out. However, I am not given this information thus, I am stuck as to how I can proceed from what I have.

Any suggestion/tip would be appreciated! Thank you in advance.

Edit #1: It seemed unclear from the answers that I'm getting in what I'm trying to ask. Sorry about that. My question would be do I need the information that $\{v_1,v_2\}$ is linearly independent? None of the comments/answer seems to help me guide in what my question is...

Answer 1

Let consider the two cases

• $v_1$ and $v_2$ not linearly independent
• $v_1$ and $v_2$ linearly independent

For $v_1$ and $v_2$ not linearly independent we have

• $v_2=av_1$

and thus

• $w_1=bv_1$
• $w_2=cv_1$
• $w_3=dv_1$

For $v_1$ and $v_2$ linearly independent, suppose wlog that $w_1$ and $w_2$ are linearly independent then from that

• $w_1=a_1v_1+a_2v_2$

• $w_2=b_1v_1+b_2v_2$

we obtain for $\begin{bmatrix}c_1&d_1\\c_2&d_2\end{bmatrix}=\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}^{-1}$

• $v_1=c_1w_1+c_2w_2$

• $v_2=d_1w_1+d_2w_2$

and thus

• $w_3=e_1v_1+e_2v_2=f_1w_1+f_2w_2$

Answer 2

You don't need $v_1,\ v_2$ to be linearly independent. By the linear dependence lemma, if you have a linearly dependent list, then there exists a vector in that list which can be removed without changing the span of the list. Repeat this procedure until you have a linearly independent list; this list has the same span as the original list.

Thus, we can break the problem down into cases. Either:

1. $v_1,\ v_2$ is linearly independent;
2. $v_1$ is linearly independent (or $v_2$; without loss of generality, pick the first); or
3. we end up with an empty list.

In case 2, each $w_i$ can be written as a multiple of $v_1$; the proof in this case is then simpler than in case 1. Case 3 is a bit pathological; it means both $v_1$ and $v_2$ were zero vectors (otherwise the procedure would have stopped at case 2: a nonzero vector is always linearly independent by itself). This forces $V$ to be the trivial vector space, i.e., $V = \{0\}$. Then $w_1 = w_2 = w_3 = 0$, which leads to $A$ being linearly dependent.