Let $V= \operatorname{span}\{v_1,v_2,v_3,v_4\}$ be a vector space such that $v_i$ are unit vectors for all $i$ and $v_i.v_j<0$ if $i$ does not equals to $j$.
(i) Show that no two vectors among {$v_1,v_2,v_3,v_4$} are linearly dependent.
(ii) Prove that $\dim V$ is greater than or equals to 3.
My approach: (i) If all the vectors are linearly dependent, this would mean that $v_i=av_j$. However, since |$v_i$| = 1, it does not = a|$v_j$| unless $a = 1$. Thus, it can be proven that no two vectors in the set are linearly dependent.
Please correct me if my approach is incorrect, and may I ask how do I approach the second part of the question?
(i) Assume that exists $i,j\in\{1,2,3,4\}$ such that $v_{i}=av_{j}$. Considering the norm, we have $1=||v_{i}||=||av_{j}||=|a|\,||v_{j}||=|a| \Rightarrow a=-1\,$, since $a=v_{i}\cdot v_{j}<0$. However, we have $v_{i}\cdot v_{k}=a v_{j}\cdot v_{k}=- v_{j}\cdot v_{k} > 0$ for $k\neq i,j$. This is absurd.
(ii) Thanks to (i) $\,\operatorname{dim} V >1$, so it is sufficient to prove that $\operatorname{dim} V \neq 2$.
We assume for absurd that $\operatorname{dim} V = 2$. Let $w= v_{2} - \frac{(v_{2}\cdot v_{1})}{||v_{1}||}\, v_{1} \in V$ such that $w\cdot v_{1}=0$. Then $\operatorname{span}\{v_{1},v_{2},v_{3},v_{4}\}=\operatorname{span}\{v_{1},w\}$, and we can write $$v_{3}=a_{1}v_{1}+a_{2}w\; ,\; v_{4}=b_{1}v_{1}+b_{2}w \;\;\text{with}\; a_{i},b_{i}<0$$ but $0>v_{3}\cdot v_{4}=a_{1}b_{1}+a_{2}b_{2}>0$.
Follow that $\operatorname{dim}V\geq 3$.