Question

If C is a linear code, prove that either every codeword has even Hamming weight or exactly half of the codewords have even Hamming weight.

Short answer

In a binary linear code, either all code words or exactly half of the code words have even Hamming weight.

Step-by-step solution

Conceptual Introduction

Coding theory is the study of a code's characteristics and how well-suited it is to various applications. Data compression, cryptography, error detection, and correction, as well as data transport and storage, all depend on codes.

Observation

In order to check the validity of the statement, first, consider the following observation:

Observation 1:

Assume x and y be two vectors in $F_{\mathrm{n}}^2$. The Hamming weight of $\mathrm{x}+\mathrm{y}$ is even if and only if the Hamming weight of both vectors is odd.

Assume that C is a binary linear code and ${\mathrm{v}}_1,{\mathrm{v}}_2,....,{\mathrm{v}}_{\mathrm{k}}$is the basis for C (to put it another way, every codeword in C is a linear combination of ${\mathrm{v}}_1,{\mathrm{v}}_2,....,{\mathrm{v}}_{\mathrm{k}}$).

Explanation

If the Hamming weight of all${\mathrm{v}}_1,{\mathrm{v}}_2,....,{\mathrm{v}}_{\mathrm{k}}$ is even, then by observing, all codeword in C have even Hamming weight. Assume, there exist $t\ge 1$vectors in the basis with odd Hamming weight. Every codeword in C is a linear combination of ${\mathrm{v}}_1,{\mathrm{v}}_2,....,{\mathrm{v}}_{\mathrm{k}}$and may thus be taken as a binary string of length k (with 1 in the $\mathrm{i}-\mathrm{th}$coordinate).

Now each codeword has Even Hamming weight if and only if the number of the vectors from ${\mathrm{v}}_1,{\mathrm{v}}_2,....{\mathrm{v}}_{\mathrm{i}}$in the combination is even.

The proving of statement

That is, if and only if the number of 1's in coordinates is even, a codeword c has Even Hamming weight.

This equals the number of binary strings of length t with even number of 1's number of binary strings of length $\mathrm{k}-\mathrm{t}$. This equals

$\begin{array}{rcl}\frac{2^{\mathrm{t}}}{2}\times 2^{\mathrm{k}-\mathrm{t}}& =& 2^{\mathrm{k}-\mathrm{t}+\mathrm{t}-1}\\ & =& \frac{2^{\mathrm{k}}}{2}\end{array}$

As a result, it may be stated that in a binary linear code, either all codewords or exactly half of the codewords have even Hamming weight.