If $M$ is a monomial matrix with entries only from ${0,-1,1}$, then $C$ is self dual if and only if $CM$ is self dual.

Question

I am interested to prove the theorem here but I am stuck. Can you please help me? Here it is

Let $\mathcal{C}$ be a code over $\mathbb{F}_q$. Then

1. If $M$ is a monomial matrix with entries only from $\{0,-1,1\}$, then $\mathcal{C}$ is self-dual if and only if $\mathcal{C}M$ is self-dual.
2. If $q=3$ and $\mathcal{C}$ is equivalent to $\mathcal{C}_1$, then $\mathcal{C}$ is self-dual if and only if $\mathcal{C}_1$ is self-dual.
3. If $q=4$ and $\mathcal{C}$ is equivalent to $\mathcal{C}_1$, then $\mathcal{C}$ is Hermitian self-dual if and only if $\mathcal{C}_1$ is Hermitian self-dual.

Now, when a code $\mathcal{C}$ is self-dual if and only if $\mathcal{C}$ is equal to its dual $\mathcal{C}^\perp$. A monomial matrix is a square matrix with exactly one nonzero entry each row and column.

So for number 1, what I did is by assuming the $\mathcal{C}$ is self-dual, then I would $\mathcal{C} = \mathcal{C}^\perp$. Then $\mathcal{C}M = \mathcal{C}^\perp M$. How can I be able to attain the equality that $\mathcal{C}M = (\mathcal{C}M)^{\perp}$ so that I can say that $\mathcal{C}M$ is self-dual?

For number 2, I am also confused that what is it something to do with the dimension of the field $\mathbb{F}$? If $\mathcal{C}$ is equivalent to $\mathcal{C}_1$, then there exists a monomial matrix $M$ and an automorphism $\gamma$ of the field such that $\displaystyle{\mathcal{C}_1 = CM\gamma}$. Then if $\mathcal{C}$ is self-dual, then I would $\mathcal{C} = \mathcal{C}^\perp$, but what it is something to do with arriving $\mathcal{C}_1 = \mathcal{C}_1^\perp$. What is in the dimension of the field?

For number 3, same case and confusion of number 2.

Can you please help me with that? I would surely appreciate the help.

Answer 1

Hints:

1. The condition implies that $MM^T=I_n$. If $\mathcal{C}$ is self-dual then its generator matrix $G$ satisfies $GG^T=0$. What can you say about $GM$?
2. $\mathcal{C}$ and $\mathcal{C}'$ are equivalent if and only if there exists a monomial matrix $M$ such that $\mathcal{C}'=\mathcal{C} M$. When $q=3$, $\Bbb{F}_q=\{0,1,-1\}$. Therefore the claim follows from $1$.
3. With $q=4$ we have $\Bbb{F}_4=\{0,1,\omega,\overline{\omega}\}$ with $\omega^2=\overline{\omega}=\omega+1=\omega^{-1}$. This time you need to show that $M\overline{M}^T=I_n$.