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Regarding Generator Matrices of MDS codes?

Started by Unknown July 19, 2006
Hi There,

I am a bit confused.  If I recall correctly, I have read/learned that
any submatrix of a generator matrix is invertible.  If one considers
the non-systematic Reed-Solomon code generated by a Vandermonde matrix,
this is true.

However, if one considers the "Reed-Solomon" code generated by [I | C],
where I is a k X k identity matrix, and C is a (k X n-k) Cauchy matrix,
not every submatrix is invertible.  However, every k X k submatrix will
be invertible.

Could someone clarify?

Also, the second code that I discussed, is that a Reed-Solomon code?
In literature I see that some people regard it as a Reed-Solomon code.
The two definitions that I know of for Reed-Solomon codes are either
the Vandermonde generator matrix (Reed and Solomon's original work) or
the n-k consecutive roots generator polynomial?

Your time and effort will greatly be appreciated
Jaco

<jaco.versfeld@gmail.com> wrote in message 
news:1153314649.845377.152090@m79g2000cwm.googlegroups.com...
> If I recall correctly, I have read/learned that > any submatrix of a generator matrix is invertible.
It is not true in general that any submatrix of a generator matrix is invertible. The statement is not true for Reed-Solomon (more generally, MDS) codes unless the assertion includes the further qualification that the submatrix is of size k x k. So, either your recollection is faulty or you read/learned something that is incorrect.
> However, if one considers the "Reed-Solomon" code generated by [I | C], > where I is a k X k identity matrix, and C is a (k X n-k) Cauchy matrix, > not every submatrix is invertible. However, every k X k submatrix will > be invertible.
>Could someone clarify?
It is an axiom of the (much maligned in the U.S.) scientific method that a theory that does not fit the facts must be discarded or modified. You have a fact here that is counter to your theory that every submatrix of a generator matrix is invertible. What do you want to do with your theory?
Thank you very much.

Kind Regards,
Jaco