Ternary self orthogonal codes have weight divisible by $3$

Question

I'm not too sure where to start with this.

I've tried to take a codevector $\underline{c} \in C$ and look at $\underline{c} \cdot \underline{c}$, which I know must be zero modulo $3$ as the code is self dual. But I can't get anything fruitful from this.

I've looked at the binary case(showing codevector hasve even weight) and this works out as $\underline{c} \cdot \underline{c}$ is the weight of the codevector, but in the ternary case this isn't true!

Answer 1

Let $\underline{c}=(a_0,...,a_n)$ and $a_{i_1},...,a_{i_k}$ be the non-zero positions (So that $w(\underline{c})=k$). Therefore $a_{i_l}^2 \equiv 1 \ \mbox{mod}(3)$. Then $0 \equiv \underline{c} \cdot \underline{c}\equiv a_{i_1}^2+...+a_{i_k}^2 \equiv k \cdot 1 \ \mbox{mod}(3) $.

We then conclude that $k \equiv 0 \ \mbox{mod}(3)$.