International Mathematics Competition

Solution 1. $2 n$ is the maximal number. An example of $2 n$ vectors in the set is given by a basis and its opposite vectors. In the rest of the text we prove that it is impossible to have $2 n+1$ vectors in the set. Consider the Gram matrix $A$ with entries $a_{i j}=e_{i} \cdot e_{j}$. Its rank is at most $n$, its eigenvalues are real and non-negative. Put $B=A-I_{2 n+1}$, this is the same matrix, but with zeros on the diagonal. The eigenvalues of $B$ are real, greater or equal to -1, and the multiplicity of -1 is at least $n+1$. The matrix $C=B^{3}$ has the following diagonal entries $$c_{i i}=\sum_{i \neq j \neq k \neq i} a_{i j} a_{j k} a_{k i}$$ The problem statement implies that in every summand of this expression at least one factor is zero. Hence $\operatorname{tr} C=0$. Let $x_{1}, \ldots, x_{m}$ be the positive eigenvalues of $B$, their number is $m \leq n$ as noted above. From $\operatorname{tr} B=\operatorname{tr} C$ we deduce (taking into account that the eigenvalues between -1 and 0 satisfy $\lambda^{3} \geq \lambda$ ): $$x_{1}+\cdots+x_{m} \geq x_{1}^{3}+\cdots+x_{m}^{3}$$ Applying $\operatorname{tr} C=0$ once again and noting that $C$ has eigenvalue -1 of multiplicity at least $n+1$, we obtain $$x_{1}^{3}+\cdots+x_{m}^{3} \geq n+1$$ It also follows that $$\left(x_{1}+\cdots+x_{m}\right)^{3} \geq\left(x_{1}^{3}+\cdots+x_{m}^{3}\right)(n+1)^{2}$$ By Hölder's inequality, we obtain $$\left(x_{1}^{3}+\cdots+x_{m}^{3}\right) m^{2} \geq\left(x_{1}+\cdots+x_{m}\right)^{3}$$ which is a contradiction with $m \leq n$. Solution 2. Let $P_{i}$ denote the projection onto $i$-th vector, $i=1, \ldots, N$. Then our relation reads as $\operatorname{tr}\left(P_{i} P_{j} P_{k}\right)=0$ for distinct $i, j, k$. Consider the operator $Q=\sum_{i=1}^{N} P_{i}$, it is non-negative definite, let $t_{1}, \ldots, t_{n}$ be its eigenvalues, $\sum t_{i}=\operatorname{tr} Q=N$. We get $$\sum t_{i}^{3}=\operatorname{tr} Q^{3}=N+6 \sum_{i<j} \operatorname{tr} P_{i} P_{j}=N+3\left(\operatorname{tr} Q^{2}-N\right)=3 \sum t_{i}^{2}-2 N$$ (we used the obvious identities like $\operatorname{tr} P_{i} P_{j} P_{i}=\operatorname{tr} P_{i}^{2} P_{j}=\operatorname{tr} P_{i} P_{j}$ ). But $\left(t_{i}-2\right)^{2}\left(t_{i}+1\right)=t_{i}^{3}-3 t_{i}^{2}+4 \geqslant 0$, thus $-2 N=\sum t_{i}^{3}-3 t_{i}^{2} \geqslant-4 n$ and $N \leqslant 2 n$.