What is the largest number of mutually obtuse vectors in Rn?

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1. Jan 28, 2016

RandomGuy1

This is my question:

What is the largest m such that there exist v1, ... ,vm ∈ ℝn such that for all i and j, if 1 ≤ i < j ≤ m, then ≤ vivj = 0

I found a couple of solutions online.
http://mathoverflow.net/questions/31436/largest-number-of-vectors-with-pairwise-negative-dot-product
http://math.stanford.edu/~akshay/math113/hw7.pdf (problem 10. But it's basically the same solution as the one in the link above)

I kind of understand the contradiction but I don't get why there won't be a contradiction when you take m = n + 1. This is my first course in linear algebra, so far I've learnt about linear independence, subspaces, orthogonality but I'm not very familiar with things like inner product spaces - only dot products. I need someone to explain this in simpler terms.

2. Jan 28, 2016

Samy_A

I'm referring to the argument in the pdf.

In case b), where M or N is 0, they use the last, until then unused, vector, $v_{n+2}$, to get the contradiction. So we need one vector kept on the side.

Now let's go back to the start of the proof. If m=n+1, and we have to keep one vector on the side, there are only n vectors left to work with. But the argument started with noticing that the n+1 vectors must be linearly dependent. That worked because we had n+1 vectors in an n-dimensional space. With n vectors in an n-dimensional space, you can't be sure that they are linearly dependent. Hence, you don't have the equation marked by (*) in the pdf.

3. Jan 28, 2016

RandomGuy1

I got it. Thank you so much.

4. Jan 28, 2016

Staff: Mentor

It is interesting that the question is quite easy to answer for 90 degrees, but an unsolved problem (Kissing number problem) for 60 degrees.

5. Jan 28, 2016

mathwonk

as often happens your problem is incorrectly stated here. you have to specify non zero vectors to get a contradiction. i.e. a sequence of zero vectors, no matter how long, satisfies the statement you gave above.