linear independency in equation of linear span

Question

we got the following vectors: $$v_1, v_2, w_1, w_3 \in V$$ $V$ is a vector space so that $\DeclareMathOperator{Sp}{Sp}\Sp\{v_1,v_2\} = \Sp\{w_1,w_2\}$

it's also defined that $\{v_1,w_2\}$ is linear independent.

1. Prove that the group $\{v_1,v_2\}$ is also a linear independent.

I thought about proving it by contradiction:

if $\{v_1,v_2\}$ is linear dependent, than we can get rid of one of the vectors.

for example $v_2$ so that $\{v_1\}$ will be linear independent. but it is a contradiction to the given definition which says that $\{v_1,w_2\}$ is linear independent.

that's how I thought about how to solve it but i'm not exactly sure how to prove it in my answer.

Answer 1

Here my argumentation: Let $W:=Sp\{v_1,v_2\} = Sp\{w_1,w_2\}$. Since $v_1,w_2\in W$ and $\{v_1,w_2\}$ is linearly independent, $W$ has dimension 2. Thus $v_1$ and $v_2$ must be linearly independent because they span $W$.

Your argumentation is right. Note, that if $\{v_1,v_2\}$ linearly dependent, then $W$ has dimension 1. Thus there cannot be two linearly independent vectors in $W$...

Answer 2

Ok, let $W := Sp(\{v_1, v_2\}) = Sp(\{ w_1, w_2\})$. Assume that $\{ v_1, v_2\}$ is linearly dependant. Then we can write $v_2 = \delta v_1$, so $W = Sp(\{v_1\})$. We consider the equation $$\alpha v_1 + \beta w_2 = 0 \; .$$ Since $w_2 \in Sp(\{w_1, w_2\}) = Sp(\{v_1\})$, we can write $w_2 = \gamma v_1$ for some $\gamma \in K$. This gives us $$ 0 = \alpha v_1 + \beta w_2 = (\alpha + \beta \gamma) v_1 \; ,$$ so this equation is fullfilled for $\alpha = 1$ and $\beta = -\gamma^{-1}$, which contradicts to the fact, that $\{ v_1, w_2\}$ is linearly independant.