Firstly, let me remind you the definition of two equivalent linear codes. Let $\alpha \in \Bbb F_q^*$. We define the following maps:
For any permutation $\sigma \in S_n$, we define the map \begin{align*} f_\sigma: \Bbb F_q^n \longrightarrow \Bbb F_q^n, \ (x_1,\dots,x_n) \longmapsto (x_{\sigma(1)},\dots,x_{\sigma(n)}) \end{align*} and \begin{align*} f_i^{\alpha}: \Bbb F_q^n \longrightarrow \Bbb F_q^n, \ (x_1,\dots,x_{i-1},x_i,x_{i+1},\dots,x_n) \longmapsto (x_1,\dots,x_{i-1},\alpha x_i,x_{i+1},\dots,x_n) \end{align*} where $i\in \{1,\dots,n\}$.
Definition. Let $C_1,C_2\leq \Bbb F_q^n$ be two $q$-ary linear codes of length $n$. We say that $C_1$ is equivalent to $C_2$ and we write $C_1\sim C_2$, if there is a map $h:\Bbb F_q^n \longrightarrow \Bbb F_q^n$, which is a composition of maps of the form $f_\sigma$ and $f_i^\alpha$, such that $h(C_1)=C_2$.
Question. Let $C_1\sim C_2$ be two equivalent codes. By definition both these codes must be of length $n$ and of same size $M$. But what about their dimension? Should it be tha same? And if not, could you give a counter example?
Thank you.
Let $C_1 \sim C_2$. Then by definition there is a map $h: \mathbb{F}_q^n \to \mathbb{F}_q^n$ such that $h(C_1) = C_2$. Note that this map $h$ is in fact the composition of maps that are all linear isomorphisms. Thus $h$ itself is a linear isomorphism. We thus get the even stronger statement that $C_1 \cong C_2$ are isomorphic as $\mathbb{F}_q$-vector spaces. In particular, their dimensions must be equal.