Show that the matrix is a generator matrix of a MDS codeLet a 1 ,...

Cory Patrick

Cory Patrick

Answered

2022-06-24

Show that the matrix is a generator matrix of a MDS code
Let a 1 , , a n be pairwise distinct elements of F q and k n. I have to show that
1. The matrix [ 1 1 0 a 1 a n 0 a 1 2 a n 2 0 0 a 1 k 1 a n k 1 1 ] is a generator matrix of an [ n + 1 , k , n k + 2 ] MDS code.
2. if q is a power of 2, then the matrix [ 1 1 0 0 a 1 a q 1 0 a 1 2 a q 2 0 1 ] is a generator matrix of an [ q + 2 , 3 , q ] MDS code.
I don't even know where to start, the things we covered in the lecture are not so many, but what I thought could be useful, but don't know how to apply are:
Theorem: Let C be an [n,k] code with generator matrix G and parity check matrix H. The following are equivalent:
1. C is MDS
2. All ( n k ) × ( n k ) full size minors of H are invertible
3. All k × k full size minors og G are invertibleMaybe I could use this theorem, part (3), but I'm not sure how to show that all the minors og G are invertible.
Should I maybe, from definition of MDS, show that the distance is indeed d ( C ) = n k + 2, and then conclude that the code is MDS? But aren't the elements arbitrary?
Should I maybe look for a parity check matrix H and do something?

Answer & Explanation

Jovan Wong

Jovan Wong

Expert

2022-06-25Added 23 answers

Step 1
Use characterization (3) of your theorem and remember the formula for the determinant of a Vandermonde matrix!
Step 2
Come up with different cases for the k × k minor, depending on the number of the involved "special" columns (the 1 or 2 rightmost columns of the generator matrix). Then, you can either directly apply Vandermonde or you have to do some transformation first.
For the second case of the [ q + 2 , 3 ] MDS code, there are cases without Vandermonde. But here, the minors are of fixed size 3 × 3, so just approach those cases by a direct computation of the determinant.

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