Is the method to prove that a code [n,k,d] punctured at $i$th index gives [n-1, k_i,d_i] code where $k_i\geq k-1$ is correct?

Question

I am self teaching my self error correcting codes from the book Introduction to coding theory by Ron Roth. There is a question that says that suppose a code is given as $[n,k,d]$ where n is the code length, k is the message length and d is the minimum distance of the code. It is over GF(q). now suppose we remove an $i$th index from the codewords and get a new code $C_i$ as $[n-1,k_i,d_i]$. Now the question says that the $k_i\geq k-1$ and to be more explicit there are at least $n-k$ indexes $i$ for which $k_i=k$

My Answer: Suppose that the code word generating matrix for $C_i$ contains some entries with just one $1$ and all other entries zero, now puncturing at such location will lead to all zeros and therefore a $k\times n$ matrix is reduced to $k-1 \times n-1$ matrix, and this will give the $ki=k-1$. Now, as any generator matrix $k \times n$ can have $k \times k$ non singular matrix so total number of such indices removal is maximum $k$ and therefore there are at least $n-k$ indexes $i$ for which $k_i=k$.

As I am still learning coding theory and linear algebra, I want to confirm if my answer is right?

Answer 1

Yes, you have correctly identified the key ingredients in the argument:

• Puncturing at position $i$ is a linear transformation $T_i$ from the code $C$ to the space $GF(q)^{n-1}$. The image $\mathrm{Im}(T_i)=C_i$. By rank-nullity the dimension is $k_i=k-\dim \mathrm{Ker}(T_i)$. As you observed, the kernel is either trivial or $1$-dimensional according to whether there exists words of weight one with the lone non-zero component at position $i$.
• We saw that puncturing at the $i$th position leads to a drop of dimension only when there is a codeword $\vec{x}_i$ with its sole non-zero component at position $i$. Such vectors form a linearly independent set, so there cannot be more than $k$ of them in the code.

Answer 2

This can also be seen by considering its generator matrix in systematic form. It's clear that if we remove one of the columns for the "parity" part ($n-k$ choices), the matrix has still rank $k$.