Let $u_1,...,u_k$ vectors in $\mathbb R ^ n$ such that for every i,j we have $u_i \cdot u_j < 0$, and I want to show that $k\leq n+1$.
I tried somehow use the fact that each vector seperates the whole plane into to parts that only one contain those which inner product is negative, and show that the intersection is 0 for every $k>n+1$ but with no success.
Can anyone give me a hint?
Let us go for an induction.
For $n=1$ this is true.
If $\forall u_1\dots u_k \in \Bbb R^n\ \ [\forall i\neq j \ \ u_i\cdot u_j < 0] \implies k\le n+1$: consider $\forall u_1\dots u_k \in \Bbb R^{n+1}$. Assume $k\ge 2$.
Let $F = u_k^\perp$ (note that $u_k\neq 0$). Let $P$ the orthogonal projection on $F$.
Let $v_i = P_F u_i = u_i - \frac{u_i\cdot u_k}{|u_k|^2} u_k $. Then \begin{align} v_i\cdot v_j &= u_i\cdot u_j - \frac{u_i\cdot u_k}{|u_k|^2} u_k \cdot u_j - \frac{u_i\cdot u_k}{|u_k|^2} u_k \cdot u_i + \frac{u_i\cdot u_k}{|u_k|^2} \frac{u_j\cdot u_k}{|u_k|^2} |u_k|^2 \\&= u_i\cdot u_j - \frac{(u_i\cdot u_k)(u_k \cdot u_j )}{|u_k|^2} <0 \end{align}because $u_{i}\cdot u_k < 0, u_{j}\cdot u_k < 0$. $$ k - 1 \le n + 1 $$