Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the usual theory of codes, a code $C$ of length $n$ of dimension $d$ over a finite field $F$ is a linear subspace $C$ of $\mathbb{F}^n$ of dimension $d$ normed by the Hamming metric. In this sense, we can view codes as metric linear spaces.

In the usual sense two codes $C_0$ and $C_1$ of the same length over the same field $F$ are said to be equivalent if there is a non-singular diagonal matrix $D$ and permutation matrix $P$ such that $$DP\cdot C_0=C_1$$ One can check that usual properties of linear codes are preserved in this way.

The problem I found is the following, instead of taking the previous approach, I considered a conceptual approach that given two codes $C_0$ and $C_1$ of the same length over the same field $F$ I consider an equivalence of codes the following: a map $$f:C_0\rightarrow C_1$$ such that $f$ is an isometry with respect to the Hamming distance induced in $C_0$ and $C_1$ and $f$ being an $F$-linear isomorphism.

I am aware that these two approaches may not be equivalent since in the second one we are not considering the ambient space. So in some sense a positive answer to the equivalence of both concepts mean that properties of codes are intrinsic not depending on the ambient space.

So my question is if the two previous definition of equivalence are equivalent. Or rephrasing it: Given a Hamming distance preserving $F$ linear mapping

$$f:C\rightarrow F^n$$

of a code $C$ of length $n$ into its ambient space, can be always be extended to a Hamming distance preserving $F$-linear isomorphism $\hat{f}:F^n\rightarrow F^n$?

share|cite|improve this question
up vote 1 down vote accepted

Yes. This is exactly the MacWilliams equivalence theorem (sometimes also called extension theorem).

share|cite|improve this answer
Thank you very much. I have found a proof in – Josué Tonelli-Cueto Jun 26 '13 at 1:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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