$G_{24}$ is a self-dual [24, 12, 8] binary code. The weights of all vectors in $G_{24}$ are multiples of 4.

Proof. The rows in $G$ have length 24. Since the $12\times 12$ identity matrix is contained in $G$, the 12 rows of $G$ are linearly independent. Therefore, $G_{24}$ has dimension 12, so it is a $[24\text{,}12\text{,}d]$ code for some $d$. The main work will be to show that $d=8$. Along the way, we’ll show that $G_{24}$ is self-dual and that the weights of its codewords are $0(\mathrm{m}\mathrm{o}\mathrm{d}4)$.

Of course, it would be possible to have a computer list all $2^{12}=4096$ elements of $G_{24}$ and their weights. We would then verify the claims of the theorem. However, we prefer to give a more theoretical proof.

Let $r_1$ be the first row of $G$ and let $r\ne r_1$ be any of the other first 11 rows. An easy check shows that $r_1$ and $r$ have exactly four 1s in common, and each has four 1s that are matched with 0s in the other vector. In the sum $r_1+r$, the four common 1s cancel mod 2, and the remaining four 1s from each row give a total of eight 1s in the sum. Therefore, $r_1+r$ has weight 8. Also, the dot product $r_1\cdot r$ receives contributions only from the common 1s, so $r_1\cdot r=1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1=4\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$.

Now let $u$ and $v$ be any two distinct rows of $G$, other than the last row. The first 12 entries and the last 11 entries of $v$ are cyclic permutations of the corresponding parts of $u$ and also of the corresponding parts of the first row. Since a permutation of the entries does not change the weights of vectors or the value of dot products, the preceding calculation of $r_1+r$ and $r_1\cdot r$ applies to $u$ and $v$. Therefore,

1. $wt(u+v)=8$

2. $u\cdot v\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$.

Any easy check shows that (1) and (2) also hold if $u$ or $v$ is the last row of $G$, so we see that (1) and (2) hold for any two distinct rows $u\text{,}v$ of $G$. Also, each row of $G$ has an even number of 1s, so (2) holds even when $u=v$.

Now let $c_1$ and $c_2$ be arbitrary elements of $G_{24}$. Then $c_1$ and $c_2$ are linear combinations of rows of $G$, so $c_1\cdot c_2$ is a linear combination of numbers of the form $u\cdot v$ for various rows $u$ and $v$ of $G$. Each of these dot products is 0 mod 2, so $r_1\cdot r_2\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$. This implies that $C\subseteq C^{\perp }$. Since $C$ is a 12-dimensional subspace of 24-dimensional space, $C^{\perp }$ has dimension $24-12=12$. Therefore, $C$ and $C^{\perp }$ have the same dimension, and one is contained in the other. Therefore, $C=C^{\perp }$, which says that $C$ is self-dual.

Observe that the weight of each row of $G$ is a multiple of 4. The following lemma will be used to show that every element of $G_{24}$ has a weight that is a multiple of 4.

The rows of the $12\times 12$ matrix $B$ formed from the last 12 columns of $G$ are linearly independent mod 2. The rows of the $11\times 11$ matrix $A$ formed from the last 11 elements of the first 11 rows of $G$ are linearly dependent mod 2. The only linear dependence relation is that the sum of all 11 rows of $A$ is 0 mod 2.

Proof. Since $G_{24}$ is self-dual, the dot product of any two rows of $G$ is 0. This means that the matrix product $G{G}^T=0$. Since $G=[I|B]$ (that is, $I$ followed by the matrix $B$), this may be rewritten as

which implies that $B^{-1}=B^T$ (we’re working mod 2, so the minus signs disappear). This means that $B$ is invertible, so the rows are linearly independent.

The sum of the rows of $A$ is 0 mod 2, so this is a dependence relation. Let $v_1=(1\text{,}\dots \text{,}1{)}^T$ be an 11-dimensional column vector. Then $A{v}_1=0$, which is just another way of saying that the sum of the rows is 0. Suppose $v_2$ is a nonzero 11-dimensional column vector such that $A{v}_2=0$. Extend $v_1$ and $v_2$ to 12-dimensional vectors $v_1^{{}^{\prime }}\text{,}{v}_2^{{}^{\prime }}$ by adjoining a 0 at the top of each column vector. Let $r_{12}$ be the bottom row of $B$. Then

This equation follows from the fact that $A{v}_i=0$. Note that multiplying a matrix times a vector consists of taking the dot products of the rows of the matrix with the vector.

Since $B$ is invertible and $v_{\mathrm{i}}^{{}^{\prime }}\ne 0$, we have $B{v}_{\mathrm{i}}^{{}^{\prime }}\ne 0$, so $r_1\cdot v_{\mathrm{i}}^{{}^{\prime }}\ne 0$ Since we are working mod 2, the dot product must equal 1. Therefore,

Since $B$ is invertible, we must have $v_1^{{}^{\prime }}+v_2^{{}^{\prime }}=0$, so $v_1^{{}^{\prime }}=v_2^{{}^{\prime }}$ (we are working mod 2). Ignoring the top entries in $v_1^{{}^{\prime }}$ and $v_2^{{}^{\prime }}$, we obtain $v_2=(1\text{,}\dots \text{,}1)$. Therefore, the only nonzero vector in the null space of $A$ is $v_1$. Since the vectors in the null space of a matrix give the linear dependencies among the rows of the matrix, we conclude that the only dependency among the rows of $A$ is that the sum of the rows is 0. This proves the lemma.

$G_{23}$ is a linear [23,12,7] code.

Proof. Clearly each codeword has length 23. Also, the set of vectors in $G_{23}$ is easily seen to be closed under addition (if $v_1\text{,}{v}_2$ are vectors of length 24, then the first 23 entries of $v_1+v_2$ are computed from the first 23 entries of $v_1$ and $v_2$) and $G_{23}$ forms a binary vector space. The generating matrix $G^{\prime }$ for $G_{23}$ is obtained by removing the last column of the matrix $G$ for $G_{24}$. Since $G^{\prime }$ contains the $12\times 12$ identity matrix, the rows of $G^{\prime }$ are linearly independent, and hence span a 12-dimensional vector space. If $g^{\prime }$ is a codeword in $G_{23}$, then $g^{\prime }$ can be obtained by removing the last entry of some element $g$ of $G_{24}$. If $g^{\prime }\ne 0$, then $g\ne 0$, so $wt(g)\ge 8$. Since $g^{\prime }$ has one entry fewer than $g$, we have $wt(g^{\prime })\ge 7$. This completes the proof.