@jeffreyNgodfrey Since the d+1 vectors are strictly negatively correlated with the (d+2)-th vector, it seems like, writing down the formula for dot product taking the (d+2)-th vector as one of the basis vector, that after projection the vectors are even more negatively correlated, no?
@jeffreyNgodfrey actually, now I wonder if this apparently too simple solution works: suppose the d+2 vectors exist. Project the d+1 first vectors to the space orthogonal to the (d+2)-th vector. The d+1 vectors remain strictly neg. correlated, but on a space of dim d+1. Conclude by induction??
@jeffreyNgodfrey Actually, thinking more about it, one can do without the randomness. Prove that there is no (d+2) vectors in R^d whose pairwise dot product is strictly negative.
PSA from your neighborhood @NeurIPSConf SAC:
Reviewers: We write papers to advance knowledge. The purpose of experiments is to empirically evaluate hypotheses. If you ask for an experiment on a particular benchmark, please be clear about what hypothesis you want investigated.
@uPicchini "imagine a world where [...] you say, ‘our university is thinking of promoting x from this position to this position on the basis of his or her research activity’ and then everybody in the world is invited to comment." - - > probably much better than the current system 😅