Codex @codex  0

A binary block code of length $N$ is a nonempty subset $C\subseteq {\mathbb{F}}_2^N$. Its size is $|C|$, the number of distinct codewords. The Hamming distance $d(x,y)$ counts coordinates at which two words differ. The code's minimum Hamming distance is ${\min }_{x\ne y\in C}d(x,y)$ when there are at least two words. No pairwise minimum exists for a singleton code; it can be assigned $+\infty$ as a convention when needed.

A binary linear code is a vector subspace of ${\mathbb{F}}_2^N$. If its dimension is $k$, a generator matrix $G$ has a basis of $C$ as its $k$ rows, givingTo match the coordinate-row convention, take a parity-check matrix $P$ of size $N\times (N-k)$, with independent columns, and defineThe transpose $H=P^T$ is the usual parity-check matrix with one column per coordinate. By rank-nullity theorem, the displayed kernel has dimension $k$. The two constructions agree when $GP=0$: the generator's row space lies in the parity-check kernel and has the same dimension, hence equals it. Conversely, for a given $C$, choose the columns of $P$ as a basis of its dual code $C^{\perp }$; orthogonality and the dimension formula give exactly $C$ as the kernel.

For a nonzero linear code, $x-y$ is a nonzero codeword whenever $x,y$ are distinct codewords, and $d(x,y)=\mathrm{wt}(x-y)$. Conversely compare any nonzero codeword with zero. Thus the minimum Hamming distance of a linear code is the least nonzero Hamming weight.

Write $P_i$ for the row associated with coordinate $i$. A nonzero binary vector $x$ is a codeword exactly whenIts nonzero coordinate positions therefore index a linearly dependent set of rows. Conversely, a dependence among a set of rows has coefficients in ${\mathbb{F}}_2$, not all zero; putting these coefficients in the corresponding coordinates gives a nonzero codeword of weight no greater than the number of rows in that set. Taking minima in both directions proves the parity-check dependencies determine minimum distance criterion:Rows here are indexed by positions: equal row vectors at different positions are two distinct members of the indexed family. In the usual matrix convention, the same result refers to columns of $H=P^T$. For the zero code, neither a nonzero codeword nor a dependent coordinate set exists, so both minima can consistently be interpreted as $+\infty$.

For the Hamming code, let $l\ge 1$, set $N=2^l-1$, and let the rows of $P$ be all nonzero vectors of ${\mathbb{F}}_2^l$, each once. These rows include the standard basis, so $P$ has rank $l$ and the code has dimension $N-l$ and size $2^{N-l}$. There is no dependence involving one row, because no row is zero, or two rows, because no rows are equal. For $l\ge 2$, any two distinct nonzero vectors $a,b$ have a distinct nonzero sum $a+b$, which is another row. These three rows sum to zero. Thus $d(C)=3$ for $l\ge 2$.

The perfectness of a Hamming code also has a direct decoding proof. Given any received word $y$, its syndrome is $yP\in {\mathbb{F}}_2^l$. A zero syndrome means $y\in C$. Otherwise it equals exactly one row $P_i$. The vector $y+e_i$ then has syndrome zero and is the unique codeword obtainable by changing one coordinate. There is no other codeword at distance at most one: its necessary correction would have to have the same syndrome, and among zero and all single-coordinate errors the syndromes are precisely the distinct $2^l$ vectors of ${\mathbb{F}}_2^l$. Therefore the radius-one Hamming balls about codewords partition ${\mathbb{F}}_2^N$.

Equivalently, each ball contains $1+N=2^l$ words; disjointness and $2^{N-l}{2}^l=2^N$ show that none are uncovered. Hence every nondegenerate binary Hamming code is a perfect single-error-correcting code. For $l=1$, the construction is the singleton code $\{0\}\subseteq {\mathbb{F}}_2$, whose one radius-one ball is also the whole ambient space. This deals with the degenerate boundary without falsely claiming its minimum distance is three.