K-dimensional subspaces of GF(p)^{2k} with zero intersection

[jdm]

I am trying to construct a set of k-dimensional subspaces of
GF(p)^{2k} such that all of the subspaces are disjoint except for the
zero vector. I know that such a set can always be constructed with
three subspaces contained in it, and have established an upper bound
of p^{k} + 1 subspaces.

I currently believe that this upper bound can always be achived for
any prime p and natural number k, but am having difficulty proving it
and would be interested to know if anyone could help me. This would
enable me (assuming the result is correct) to significantly strengthen
a result in a paper I am writing on nonlinear secret sharing.

Thanks,

James McLaughlin.

[Ilya Zakharevich]

[A complimentary Cc of this posting was sent to
jdm
<james.d.m...@googlemail.com>], who wrote in article <gbpej9$ph4$1...@news.acm.uiuc.edu>:

> I am trying to construct a set of k-dimensional subspaces of
> GF(p)^{2k} such that all of the subspaces are disjoint except for the
> zero vector.

The case k=2 is trivial. The case k=2 implies the general case via
field extension: consider GF(k^p)^2.

Hope this helps,
Ilya

[victor_me...@yahoo.co.uk]

On 29 Sep, 03:30, jdm <james.d.mclaugh...@googlemail.com> wrote:
> I am trying to construct a set of k-dimensional subspaces of
> GF(p)^{2k} such that all of the subspaces are disjoint except for the
> zero vector. I know that such a set can always be constructed with
> three subspaces contained in it, and have established an upper bound
> of p^{k} + 1 subspaces.

I already answered this on sci.math. Consider
V = GF(p)^{2k} as a 2-dimensional vector space over GF(p^k).
The 1-dimensional GF(p^k)-subspaces of V suffice.

Victor Meldrew
"I don't believe it!"