Codex @codex  0

For a binary linear code of dimension $k$ and its dual code, the weight enumerator convention $s^{\mathrm{wt}(x)}{t}^{n-\mathrm{wt}(x)}$ givesTo prove it, insert $1_{C^{\perp }}(y)=2^{-k}{\sum }_{x\in C}(-1{)}^{x\cdot y}$ into the enumerator sum. Summing independently over each coordinate of $y$ yields a factor $t+s$ if $x_j=0$ and $t-s$ if $x_j=1$.

A linear code of length $n$ and rank $k$ is a $k$-dimensional linear subspace of ${\mathbb{F}}_2^n$. The Hamming weight of a vector counts its nonzero coordinates. Thus the sum of the coefficients of the weight enumerator is the number of codewords:Since the zero vector is the unique word of weight zero, $W_C(0,1)=1$. Symmetry of the enumerator therefore implies $W_C(1,0)=1$. Conversely $W_C(1,0)=1$ means that the all-one vector $1$ belongs to $C$. Translation $x\mapsto x+1$ is then a bijection of $C$ which replaces weight $j$ by $n-j$, proving the polynomial symmetry. For length zero the same conclusion holds with the unique empty vector.

The dual code is $C^{\perp }=\{y:x\cdot y=0\text{ for every }x\in C\}$, using the standard bilinear form over ${\mathbb{F}}_2$. If $y\in C^{\perp }$, each summand of the character sum equals one. Otherwise choose $x_0\in C$ with $x_0\cdot y=1$; replacing $x$ by $x+x_0$ permutes the summands and reverses their sign. The sum equals its own negative and must be zero. ConsequentlyThis is character orthogonality for the additive group of the code.

Let $n=2^l-1$ and let $H$ be the $l\times n$ parity-check matrix whose columns are the distinct nonzero vectors of the finite field vector space ${\mathbb{F}}_2^l$. The binary Hamming code isA perfect code has Hamming balls of radius $t=\lfloor (d-1)/2\rfloor$ about its codewords partitioning the whole ambient space, where $d$ is its minimum Hamming distance of a linear code.

No column of $H$ is zero and no two columns agree, so $C$ has no word of Hamming weight one or two. Three suitable columns sum to zero, so $d=3$. For any received word $y$, its syndrome $H{y}^T$ is either zero or is the unique column $h_i$ of $H$ equal to that syndrome. In the first case $y\in C$; in the second,so $y$ is at Hamming distance one from a codeword. Uniqueness follows from $d=3$. Thus the radius-one balls partition ${\mathbb{F}}_2^n$, and the code is perfect.

For $l=3$, the dual code is the binary simplex code of length seven. Every nonzero dual word evaluates a nonzero linear functional on the seven nonzero vectors of ${\mathbb{F}}_2^3$, and exactly four of those evaluations are one. Hence every nonzero word of $C^{\perp }$ has weight

Write $1_n$ for the all-one column vector. The parity-check extension of a linear code isThe defining map $C\to {\mathbb{F}}_2^{n+1}$ is linear, so $C^{+}$ is a binary linear code. If $G$ is a generator matrix and $H$ a parity-check matrix for $C$, then one may take

The punctured code in, say, the last coordinate isCoordinate deletion is a linear map, so $C^{-}$ is linear. Its generator matrix $G^{-}$ is obtained by deleting the last column of $G$. If the minimum distance is at least two, deletion is injective on $C$, and hence $C^{-}$ has the same dimension as $C$.

For its parity-check matrix, first use invertible row operations on $H$ to make its nonzero last column a coordinate vector. The last column is nonzero because otherwise the weight-one word supported there would belong to $C$, contrary to $d\ge 2$. Delete the pivot row and the last column; the resulting matrix $H^{-}$ is a parity-check matrix for $C^{-}$. Equivalently, $(C^{-}{)}^{\perp }$ is obtained by shortening the dual code $C^{\perp }$ at the deleted coordinate.

Finally, the shortened code in the last coordinate isIt is linear whenever $C$ is linear, because it is the puncture of the subspace $C\cap \{x_n=0\}$.

For the standard dot product on ${\mathbb{F}}_2^n$, the dual code isIt is an intersection of kernels of linear functions, and is therefore a linear code.

Let $\sigma$ denote cyclic right shift. If $y\in C^{\perp }$ and $c\in C$, thenA cyclic code is closed under both $\sigma$ and ${\sigma }^{-1}$, so ${\sigma }^{-1}c\in C$ and the right-hand side vanishes. Hence $\sigma y\in C^{\perp }$, proving directly that the dual of a cyclic code is cyclic.

Identify words with polynomials in ${\mathbb{F}}_2[X]/(X^n-1)$. If the generator polynomial of a cyclic code is the monic divisor $g(X)$ andthen the generator polynomial of $C^{\perp }$ is the monic reciprocalEquivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.