# 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.