A linear code $C$ is self-orthogonal if it is contained in its dual code, that is $C\subseteq C^{\perp}$
I want to proof that every codeword of $C$ has even weight
What I got by far:
Supposing that $C$ is a binary linear code, I can consider $x\in C$, so $xx\equiv 0 \pmod{2}$
...
How can I follow the proof?
You have a word $w=(a_1,a_2,\ldots,a_n)$ where $a_i\in\{0,1\}$. If it's in a self-orthogonal code, then $w\cdot w=0$ over the field of two elements. But $w\cdot w=a_1^2+a_2^2+\cdots+a_n^2$ which equals the number of $1$s in $w$, that is the weight of $w$. So $w\cdot w=0$ in $\Bbb F_2$ iff $w$ has even weight.
