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While reading about the Hamming (7,4) code, I saw that it was self dual. After looking up the definition, the dual code has a generator matrix equal to $H^T$, where $H$ is the parity check matrix of the original code.

But these two matrices are of different dimension! How can they be equal to each other?

Also, using the definitions for $G$ and $H$ from the wiki article, they are quite different (after transposing). I'm guessing that there is a large amount of non-uniqueness in these matrices, perhaps even when it comes to their dimensions. What would be the cleanest way to check that it is self-dual?

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    $\begingroup$ Where did you see it was self-dual. It is not. $\endgroup$
    – quid
    Commented Jul 6, 2016 at 20:19
  • $\begingroup$ en.wikipedia.org/wiki/Steane_code - "...using the classical binary self-dual [7,4,3] Hamming code..." $\endgroup$
    – Spine Feast
    Commented Jul 6, 2016 at 20:20
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    $\begingroup$ I think that's just not true then. $\endgroup$
    – quid
    Commented Jul 6, 2016 at 20:23
  • $\begingroup$ How to see that it isn't self dual then? What's the smartest way to approach this sort of problem? I mean other than calculating all the codewords for both codes. $\endgroup$
    – Spine Feast
    Commented Jul 6, 2016 at 20:24
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    $\begingroup$ Is it possible that your source was confusing this with the extended $(8,4,4)$ Hamming code? That code is self-dual. The dual of the $(7,4,3)$ Hamming code is a $(7,3,4)$ code $C$ gotten by shortening the extended Hamming code at some position. The code $C$ is self-orthogonal, in other words it is a subcode of its dual code. As quid pointed out (+1), a self-dual code necessarily has even length. $\endgroup$
    – Jyrki Lahtonen
    Commented Jul 6, 2016 at 20:34

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It is not true that this Hamming code, or any Hamming code for that matter, is self-dual.

A self-dual code must have even length, and its dimension is half its lengths. (This is precisely so that the problem with the dimensions of the matrices you point out does not occur.)

Yet, the length of the code is $7$.

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  • $\begingroup$ So a self dual code would always be of the form $(2n,n)$, is that correct? $\endgroup$
    – Spine Feast
    Commented Jul 6, 2016 at 20:30
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    $\begingroup$ Yes, this is correct. $\endgroup$
    – quid
    Commented Jul 6, 2016 at 20:31

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