The Exercise

A classical linear code is often handed over as a generator matrix $G$, an $n \times k$ matrix whose $k$ columns span the code inside the space of words of length $n$. The exercise asks to show that the code is weakly self-dual, contained in its own dual, if and only if $G^{T} G = 0$.^✦

Codes and Their Duals

Fix a field $K$. A linear code $C$ of length $n$ over $K$ is a subspace of $K^n$, and its elements are its codewords. The code $C$ generated by an $n \times k$ matrix $G$ over $K$ is the linear code of length $n$ given by the column space of $G$, the set of codewords $Gx$ obtained by encoding each message $x \in K^k$ of $k$ symbols: $$C = \{\, Gx : x \in K^k \,\}.$$(1)

Two words $x, y \in K^n$ are orthogonal when their dot product vanishes, $$x \cdot y = \sum_{i=1}^{n} x_i y_i = 0 .$$(2) The dual code $C^{\perp}$ of a linear code $C$ is the linear code of all words orthogonal to every codeword of $C$, $$C^{\perp} = \{\, y \in K^n : x \cdot y = 0 \text{ for all } x \in C \,\}.$$(3)^✦ A linear code $C$ is weakly self-dual if it is contained in its dual, $$C \subseteq C^{\perp} .$$(4)^✦

The Claim

What Is Checked

The claim is stated over the code generated by an arbitrary $n \times k$ matrix $G$ over $K$ and over weak self-duality (4), with the dual (3) taken under the dot product of (2).