I'm thinking of in $\Bbb R^n$, at most how many vectors can we have any two of them's dot product are less than 0
In $\Bbb R^3$, I guess that there are at most $4$ vector but I can't prove it.(at least algebrically)
What approach can you suggest to deal with this?
Remark: In a previous edition of this question I was asking about vectors with dot product less than $1$ instead of $0$. From the answers below, there would be infinitely many of such vectors.
On
There are infinitely many vectors in $\mathbb R^n$ such that any pairwise product will be less than one.
Take, for example, the set $$S=\{v\in\mathbb R^n| v\cdot v = 1\}$$
So each vector in $S$ has a norm less than $1$.
Then, any $v,w\in S$ satisfy the equation $$v\cdot w \leq |v||w| = 1$$
For example, here are, for $n\geq 2$, sixvectors that belong to the set $S$:
$$v_1 = \left(1,0,0,\dots,0\right)\\ v_2 = \left(\frac{2}{\sqrt{5}},\frac{1}{\sqrt5},0,\dots,0\right)\\ v_3 = \left(\frac{\sqrt3}{\sqrt{5}},\frac{\sqrt2}{\sqrt5},0,\dots,0\right)\\ v_4 = \left(\frac{\sqrt2}{\sqrt{5}},\frac{\sqrt3}{\sqrt5},0,\dots,0\right)\\ v_5 = \left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt5},0,\dots,0\right)\\ v_6 = \left(0,1,0,\dots,0\right)$$
On
Consider $a = (1,0,…,0), b= (-1,0,…,0)$. This gives a dot product of $-1<0$, and then we have an uncountably infinite number of such pairs with the same dot product, one for each $\lambda∈ℝ\setminus \{0\}$: $$(\lambda,0,…,0)\text{ and }\left(\frac{-1}{\lambda},0,…,0\right)$$
One can deduce there are many others, from the equality $a· b = |a| |b| \cos \angle ab$ (just choose an obtuse angle)
We can find $n+1$ such vectors on $\mathbb R^n$, for example by taking the vertices of an origin-centric simplex (triangle in case $n=2$, tetrahedron in case $n=3$, etc.).
We cannot find $n+2$ such vectors. Assume otherwise, i.e., for some $n\ge 0$ we have $v_1,\ldots, v_{n+2}\in\mathbb R^n$ with $\langle v_i,v_j\rangle <0$ for $i\ne j$. Note that $v_{n+2}\ne0$ because $\langle v_{n+2},v_1\rangle <0$. Consider the hyperplane $V$ orthogonal to $v_{n+2}$ and project all other vectors into this hyperplane. The projected vectors are of the form $u_i=v_i+c_iv_{n+2}$ where $0=\langle u_i,v_{n+2}\rangle=\langle v_i,v_{n+2}\rangle+c_i\left|v_{n+2}\right|^2 $ implies $c_i=-\frac{\langle v_i,v_{n+2}\rangle}{\left|v_{n+2}\right|^2}>0$. Then $$\langle u_i,u_j\rangle =\langle u_i,v_j\rangle +c_j\langle u_i,v_{n+2}\rangle =\langle u_i,v_j\rangle=\langle v_i,v_j\rangle+c_i\langle v_{n+2},v_j\rangle<0$$ so that by identifying the hyperplane $V$ with $\mathbb R^{n-1}$ we found $n+1$ vectors in $\mathbb R^{n-1}$ with pairwise negative dot product. Together with the observation that we cannot find two vectors with negative dot product in $\mathbb R^0$, this makes a proof by induction that $n+2$ such vectors in $\mathbb R^n$ are not possible for any $n$ at all.