Let $\Bbb F_4=\{0,1,w,w^2\}$. I have a code $C \subset \Bbb F_4^6$ where $C=\{(a,b,c,a+b+c,aw^2+bw+c, aw+bw^2+c )|a,b,c \in \Bbb F_4\}$
I found the generator matrix of C: $$\begin{bmatrix}1&0&0&1&w^2&w\\0&1&0&1&w&w^2\\0&0&1&1&1&1 \end{bmatrix}$$
I was able to prove that C is a linear $[6,3]$-code using the generator matrix and a theorem for the properties of a linear code.
I also found that $C$ has minimum distance $4$ by using the fact that it is MDS, since any $k=3$ columns are linearly independent, and thus the minimum distance has to be $n-k+1=6-3+1=4$. Since $d(C)=W(C)$, for a linear code, then the minimum weight of $C$ is also $4$.
Now I have to prove that there are no words of weight $5$ in C. This appears pretty clear to me, since every row in the generator matrix only has weight $4$ and adding rows by multiples still equals a row of weight $4$.
However, is there a way to fully prove this without having to write out the $k^q=3^4=81$ different codewords in $C$? Thank you very much for your help!
I was thinking of using the lemma $w(x-y)=d(x,y)=w(x)+w(y)-2w(x \cap y)$. Since $x-y$ is in $C$ because it is closed under addition, then by the lemma the weight of $x-y$ has to be even, and thus cannot be $5$? But I think this lemma only applied to $\Bbb F_2$.