Answer
See below
Work Step by Step
Given $\{v_1, v_2\}$ be a linearly independent set in a vector space $V$
$v,w$ are linearly independent if $av+bw=0 \rightarrow \alpha=\beta=0 \forall a,b \in R$
then $av+bw\\
=a(\alpha_1+v_2)+b(v_1+\alpha v_2)\\
=v_1(a\alpha +b)+v_2(a+\alpha b)\\
=0$
Since $v_1,v_2$ are linearly independent, then we have
$a \alpha +b=a+\alpha b=0\\
\rightarrow b=-\alpha a$
Substituting: $a+\alpha (-\alpha a)=0\\
\rightarrow a(-\alpha^2+1)=0\\
\rightarrow -a(\alpha +1)(\alpha -1)=0\\
\rightarrow a=0\\
b=0\\
\alpha=\pm 1$
Hence, the vectors $v$ and $w$ are linearly independents when $\alpha \ne \pm 1$
Substituting: $\alpha =1\\
\rightarrow av+bw\\
=(a+b)v_1+(a+b)v_2\\
=(a+b)(v_1+v_2)=0$
