0
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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: $$ (1,0,\ldots,0),(i,0,\ldots,0),(0,1,0,\ldots,0),(0,i,0,\ldots,0),\ldots $$

$\endgroup$
6
  • $\begingroup$ 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/… $\endgroup$
    – Turbo
    Oct 10, 2011 at 3:43
  • $\begingroup$ 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. $\endgroup$
    – ARupinski
    Oct 10, 2011 at 3:44
  • $\begingroup$ It seems just a bit odd that we have infinite points with pairwise real inner products and only finitely many imaginary inner products. $\endgroup$
    – Turbo
    Oct 10, 2011 at 3:56
  • $\begingroup$ I think you are correct. $\endgroup$
    – Turbo
    Oct 10, 2011 at 3:57
  • $\begingroup$ I am including a variant of the inner product on which as well I am interested. $\endgroup$
    – Turbo
    Oct 10, 2011 at 4:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.