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Considered the following inner products:

$(1)$ $\langle x,y \rangle = \sum_{t=1}^{n}x_{t}y_{t}$

$(2)$ $\langle x,y \rangle_{c} = \sum_{t=1}^{n}x_{t}\bar{y}_{t}$

consider the following surfaces:

$\underline{Surface (a)}$: $\langle x, x \rangle = 1$

$\underline{Surface (b)}$: $\langle x, x \rangle = \mathbb{i} = \sqrt{-1}$

$\underline{Surface (c)}$: $\langle x, x \rangle_{c} = 1$

In each of the above surfaces, how many points can one place so that the inner product (defined in both $(1)$ and $(2)$) between any pair of the points is purely imaginary of form $0 + \mathbb{i}r$ where $\mathbb{i}=\sqrt{-1}$ and $r \in \mathbb{R}$ and how many points are there so that the pairwise product is purely real of form $r \in \mathbb{R}$?

The case when we seek the pairwise inner product(both $(1)$ and $(2)$ to be purely real is infinite for surfaces $(a)$ and $(c)$ (Just restrict your sphere to have purely real coordinates and search among those points).

Likewise, the case when we seek the pairwise inner product(both $(1)$ and $(2)$ to be purely imaginary is infinite for surface $(b)$ (Just restrict your sphere to have purely imaginary coordinates and search among those points).

What happens in the following combinations?

$\underline{A}$:$(b)$ when we seek pure imaginary inner products (both $(1)$ and $(2)$).

$\underline{B}$:$(a)$ and $(c)$ when we seek pure real inner products (both $(1)$ and $(2)$).

$\underline{A}$ has been shown to have finitely many points ($O(n)$ atmost) by unknown(google) below.

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1 Answer 1

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It's easy to see you can't have infinitely many points: there would be two that are within $\epsilon>0$ of each other, and thus would have inner product very close to $\langle z,z\rangle=1$ (which would then not be purely imaginary).

Since $\Re\langle u,v\rangle$ forms a genuine inner product on $\mathbb C^n$, two vectors whose inner product is purely imaginary would be orthogonal. Thus you can have at most $2n$ such vectors. In the other direction, it's easy to see that the following is a collection of $2n$ vectors where every pairwise inner product is purely imaginary: $$ (1,0,\ldots,0),(i,0,\ldots,0),(0,1,0,\ldots,0),(0,i,0,\ldots,0),\ldots $$

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I looked at example $c)$ here on page $1$. It gives example of $x=1$ and $y=i$ and provides inner product which is purely imaginary and states that $x$ and $y$ are not orthogonal. faculty.iu-bremen.de/stoll/teaching/LinearAlgebra2-2007-Spring/… –  Turbo Oct 10 '11 at 3:43
    
Your example isn't quite right; note that the problem wants a set of vectors such that the pairwise inner product is strictly non-zero. –  ARupinski Oct 10 '11 at 3:44
    
It seems just a bit odd that we have infinite points with pairwise real inner products and only finitely many imaginary inner products. –  Turbo Oct 10 '11 at 3:56
    
I think you are correct. –  Turbo Oct 10 '11 at 3:57
    
I am including a variant of the inner product on which as well I am interested. –  Turbo Oct 10 '11 at 4:02

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