Find a set of vectors with special property Given vectors $\alpha_1, \dots , \alpha_{m}$ in an $n$-dimensional Euclidean space, such
that $(\alpha_i, \alpha_j) \leq 0$ for $i\neq j$ and $(a_i,a_i)\neq 0$. Find the maximum value of $m$.

The answer is $2n$.
It's easy to know $m\geq 2n$. Let $e_1,\dots,e_n$ be the orthonormal basis of $V$. And take $e_{n+i}=-e_i.$ Then $(e_i,e_j)\leq 0,i\neq j.$
But I don't know how to show $m\leq 2n.$
 A: Let us prove by induction that there are at most $2n$ such vectors. First for a vector $v$, we define the vector $v'$ as the vector without the first coordinate. For example, $v=(1,2,3)$, then $v'=(2,3)$.
Now suppose we have $k$ nonzero vectors $v_{1},\cdots,v_{k}$ in $\Bbb{R}^n$ with pairwise non-positive inner product. We can assume $v_{1}=(1,0,\cdots,0)$. Then $v_{2},\cdots,v_{k}$ have non-positive first coordinate, so $v_{2}',\cdots,v_{k}'\in \Bbb{R}^{n-1}$ also have pairwise nonpositive inner product (otherwise if $\langle v_{i}',v_{j}'\rangle> 0$, then $\langle v_{i},v_{j}\rangle=$ (product of the first coordinates of $v_{i}$ and $v_{j}$)+$\langle v_{i}',v_{j}'\rangle>0$). However, some $v_{i}'$ here can be zero, so we assume $A\subseteq \{v_{2}',\cdots,v_{k}'\}$ is the set of all nonzero vectors.
Thus, we can use the induction hypothesis, we know $A$ has at most $2n-2$ cardinality. Since there is no $v_{i}, v_{j}$ such that $v_{i}'=v_{j}'$ (because $\langle v_{i}',v_{j}'\rangle\leq 0$ as we showed), there are at most $2n-2$ vectors in $\Bbb{R}^n$ which are nonzero after deleting the first coordinate.
We also have at most 2 vectors in $\Bbb{R}^n$ which are zero after deleting the first coordinate. The first one is $v_{1}$, the second one is $v_{0}=(a,0,\cdots, 0)$ with $a<0$. If we have the third one $t=(b,0,\cdots, 0)$,  then one of $\langle t,v_{1}\rangle$, $\langle t,v_{0}\rangle$ is positive.
Finally in total there are at most $2n-2+2=2n$ vectors.
A: I seem to be unable to undelete my previous answer.
It is straightforward to see that for $n=1$ there at at most $2n$ vectors satisfying the conditions.
Suppose it is true for $1,..,n$.
Further, suppose we have $a_1,...,a_{2(n+1)+1}$ points in $\mathbb{R}^{n+1}$ that satisfy the conditions. This will lead to a contradiction.
For $j \ge 2$ we can write $a_j = \lambda_j a_1 + b_j$, where $b_j \bot a_1$. Since $\langle a_j, a_1 \rangle \le 0$, we see that $\lambda_j \le 0$.
Also, $\langle a_i, a_j \rangle = \lambda_i \lambda_j \|a_1\|^2+\langle b_i, b_j \rangle$, and so $\langle b_i, b_j \rangle \le 0$ for $i \neq j$.
Note that the $b_j $ all lie in the $n-1$ dimensional subspace $a_1^\bot$.
The inductive hypothesis shows that $b_2,...,b_{2(n+1)+1}$ (there are $2n+2$ elements) must have at least
two zero elements, without loss of generality suppose $b_2,b_3$ are zero. Then
$a_2 = \mu_2 a_1, a_3 = \mu_3 a_1$ for some non zero $\mu_2,\mu_3$ and a pair of the points $a_1,a_2,a_3$ must violate the non positive inner product condition.
