Prove there doesn't exist [10,5,4] linear binary and self-dual code. I have a problem with the following question:
Prove there doesn't exist [10,5,4] linear binary and self-dual code.
I understood that the problem is that the minimal distance is too high for such code, but I couldn't find an easy way to prove why. I tried using the fact that the generator matrix is the same as the parity check matrix and use them to use the minimal distance, but no way was succesful.
I also tried to get G to its standart form [I|A] and deduce about the weights of the rows of A and all the possibilites, but it didn't work also.
Thanks for the help.
 A: Assume such a code $C$ exists. It has a generator matrix $G=(I|A)$, with a fixed 5x5 block $A$. By general results $C$ then has a parity check matrix $H=(A^T|I)$. Because $C$ is self-dual, $H$ also has to be generator matrix. Therefore $H$ and $G$
must be row-equivalent. Therefore we must have $H=A^TG$, so $A$ has to be invertible
and its transpose must be equal to its inverse. 
This means that the rows of $A$ are orthogonal to each other. But all the rows of $A$ must have an odd weight $\ge3$. If one of the rows of $A$ has weight five, then this is an immediate contradiction, because that row will not be orthogonal to the
other rows. After all they then share an odd number of common 1s. So all the rows of $A$ must have weight $3$ exactly. Here there are at least two ways to proceed. 
I) An easy induction then proves that the weights of all words of $C$ must be divisible by four. Thus $C$ is doubly even, and its length has to be a multiple of eight by the result cited by azimut contradicting the assumption.
II) We can also observe that any pair of rows of $A$ must have exactly two ones in common. There are undoubtedly many ways of seeing that there does not exist a set of five orthogonal vectors of length five and weight 3. For example, if some subset of three such vectors shares the same doublet of 1s, then there is no room for a fourth vector (easy to see). If no such subset exists, then we can find four such vectors, but there combined support has size four only, and it is impossible to find a fifth.
As pointed out in the comments, Gleason's theorem also forces the existence of a word of weight six leading to the contradiction described in the second paragraph. The point of including approach II was to avoid using Gleason's theory altogether.
