IMC 2021 · Day 2 · P8

Hint: Play with the Gram matrix of these vectors.

$2n$ is the maximal number.

An example of $2n$ 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 $2n + 1$ vectors in the set.

Consider the Gram matrix $A$ with entries $a_{ij} = e_i \cdot e_j$. Its rank is at most $n$, its eigenvalues are real and non-negative. Put $B = A - I_{2n+1}$, 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 $$c_{ii} = \sum_{i \ne j \ne k \ne i} a_{ij} a_{jk} a_{ki}.$$ The problem statement implies that in every summand of this expression at least one factor is zero. Hence $\operatorname{tr} C = 0$. Let $x_1, \dots, x_m$ be the positive eigenvalues of $B$, their number is $m \le n$ as noted above. From $\operatorname{tr} B = \operatorname{tr} C$ we deduce (taking into account that the eigenvalues between $-1$ and 0 satisfy $\lambda^3 \ge \lambda$): $$x_1 + \dots + x_m \ge x_1^3 + \dots + x_m^3$$ Applying $\operatorname{tr} C = 0$ once again and noting that $C$ has eigenvalue $-1$ of multiplicity at least $n + 1$, we obtain $$x_1^3 + \dots + x_m^3 \ge n + 1.$$ It also follows that $$(x_1 + \dots + x_m)^3 \ge (x_1^3 + \dots + x_m^3)(n+1)^2.$$ By Hölder's inequality, we obtain $$(x_1^3 + \dots + x_m^3) m^2 \ge (x_1 + \dots + x_m)^3,$$ which is a contradiction with $m \le n$.

Let $P_i$ denote the projection onto $i$-th vector, $i = 1, \dots, N$. Then our relation reads as $\operatorname{tr}(P_i P_j P_k) = 0$ for distinct $i, j, k$. Consider the operator $Q = \sum_{i=1}^{N} P_i$, it is non-negative definite, let $t_1, \dots, t_n$ be its eigenvalues, $\sum t_i = \operatorname{tr} Q = N$. We get $$\sum t_i^3 = \operatorname{tr} Q^3 = N + 6 \sum_{i<j} \operatorname{tr} P_i P_j = N + 3 (\operatorname{tr} Q^2 - N) = 3 \sum t_i^2 - 2N$$ (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 $(t_i - 2)^2 (t_i + 1) = t_i^3 - 3 t_i^2 + 4 \geqslant 0$, thus $-2N = \sum t_i^3 - 3 t_i^2 \geqslant -4n$ and $N \leqslant 2n$.