If (n,k) is a q-ary linear block code of length n and
size q^k, then in order for the code to be self-dual
it is fairly easy to show that n = 2*k. 

Are there any other constraints required of a code
in order for it to be self-dual?

--Randy

Randy Yates wrote:

> If (n,k) is a q-ary linear block code of length n and
> size q^k, then in order for the code to be self-dual
> it is fairly easy to show that n = 2*k. 
> 
> Are there any other constraints required of a code
> in order for it to be self-dual?
> 
> --Randy

Yes there are.  Let G be the generator matrix of a linear
(n, k) code with n = 2k; the dual code is the space of all
vectors orthogonal to every codeword of the code generated
by G.  If H is a parity check matrix for the code generated
by G then H will be a generator matrix for the dual code.
So G generates a self-dual code if and only if G can serve
as its own parity check matrix.  Since G (being a generator
matrix) has full rank, this boils down to the requirement
that GG' = 0.

For the special case of a systematic code G = [I P] this
becomes I + PP' = 0, i.e., P^(-1) = -P'.  Specializing
further to fields of characteristic 2 this becomes P^(-1) = P';
i.e. the code will be self-dual if and only if P is orthogonal.
So picking any square but non-orthogonal P over (say) GF(2)
will yield an (n, k) code with n = 2k that is not self-dual.

Bob Beaudoin