The binary [7,4,3] Hamming code and its single-error syndromes

The last four rows of $G$ form $I_4$, so its columns are independent and [L1] gives $\dim C=4$. The first three columns of $H$ form $I_3$, so for every $y\in F_2^3$ the input $(y_0,y_1,y_2,0,0,0,0)$ maps to $y$. Also, a linear combination of the rows of $H$ has its first three entries equal to its three coefficients, so the rows are independent and [F5] gives rank $3$. For $x=(a,b,c,d)$, $Gx=(a+b+d,a+c+d,b+c+d,a,b,c,d)$; applying the three rows of $H$ gives $(0,0,0)$. Thus $HG=0$ and $C\subseteq \ker H$.

The map $H:F_2^7\to F_2^3$ is onto by its first three columns, so [F3] gives $\dim \ker H=7-3=4$. The subspace $C$ also has dimension $4$ by step 1.1; [F4] now gives $C=\ker H$.

The columns of $H$ are $100,010,001,110,101,011,111$. They are all nonzero and pairwise distinct. A weight-one vector has syndrome one of these columns; a weight-two vector has syndrome the sum of two distinct columns, which is nonzero in characteristic two. Thus $C$ has no nonzero word of weight below $3$. The word $e_0+e_1+e_3$ has syndrome $100+010+110=000$ and weight $3$, so it lies in $C$ and $d(C)=3$.

Multiplying $H$ by the zero word and each unit vector gives $s(0000000)=000$, $s(1000000)=100$, $s(0100000)=010$, $s(0010000)=001$, $s(0001000)=110$, $s(0000100)=101$, $s(0000010)=011$, and $s(0000001)=111$; each nonzero value is the corresponding column of $H$. These eight syndromes are distinct, so [L3] guarantees unique correction of any one bit flip.