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Creating a matrix that has elements part of a GF

asked 2018-10-26 09:38:11 -0500

Lujmina gravatar image

updated 2018-10-26 10:37:11 -0500

tmonteil gravatar image

I am currently doing some implementation but I have something that I do not seem to find online and bugged me for a few hours:

e = 48;
K = GF(2^e);
KE = GF(2^(e*2));

A = matrix(KE,3,3);
E11 = 24;
E12 = 59;
E21 = 21;
E23 = 28;
E32 = 29;
E33 = 65;
A[0,0] = 2^E11;
A[0,1] = 2^E12;
A[0,2] = 0;
A[1,0] = 2^E21;
A[1,1] = 0;
A[1,2] = 2^E23;
A[2,0] = 0;
A[2,1] = 2^E32;
A[2,2] = 2^E33;

print A[2][1]

When I do this, it print 0, but given that I created it in GF(2^(e*2)), I believe it shouldn't. Because of this, when I try to get the inverse of this matrix, which is invertible, I do not get anything. Please let me know if you have any thoughts.

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answered 2018-10-26 10:39:15 -0500

tmonteil gravatar image

The field KE = GF(2^(e*2)) has characteristic 2, hence every even number (in particular every nontrivial power of 2) will be equal to zero, which is what you got.

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answered 2018-10-26 12:41:11 -0500

rburing gravatar image

updated 2018-10-26 13:04:25 -0500

Did you mean to use the ring Zmod(2^(e*2)) (the integers modulo $2^{2e}$) instead of GF(2^(e*2)) (the finite field of order $2^{2e}$)? They are different things. The matrix is also not invertible over this ring, however.

It can only be invertible in a ring where the determinant over $\mathbb{Z}$,

-1 * 2^81 * 274177 * 67280421310721

is invertible, e.g. modulo primes which do not divide this number.

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Asked: 2018-10-26 09:38:11 -0500

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Last updated: Oct 26 '18