Given $\{v_1,v_2\}$ linearly independent $v_1,v_2\in\mathbb{R}^n$ , $\exists u\neq0\in\mathbb{R}^n$ such that
$u\perp \{v_1-v_2,v_1+v_2\}$, prove $n\geq3$
My friend's attempt (because mine was with flaws)
We will tag - $A=Sp\{v_1,v_2\}$
from the given: $u\cdot (v_1-v_2)=0$
$u\cdot (v_1+v_2)=0$
$\implies$ $u\cdot (v_1-v_2)=u\cdot (v_1+v_2)$ $\iff$ $u\cdot v_1- u \cdot v_2= u \cdot v_1+ u \cdot v_2$
$2u\cdot v_2=0$ $\implies$ $u \cdot v_2 = 0$ $\implies$ $u \cdot v_1 = 0$\ $\implies$ $u \perp$ ,
{$v_1$,$v_2$} $\implies$ $u \perp A$ , $u \in A^\perp$, $u\neq0 \implies$
$\dim (A^\perp)\geq 1$
it's known that $\mathbb{R}^n=A\bigoplus A^\perp\implies 3\leq\dim(A) + \dim (A^\perp)=\dim (\mathbb{R}^n)\implies n \geq3 $.