Multiplication of two matrices ? one in GF(2) other in R

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SUMMARY

The discussion focuses on the multiplication of two matrices, H and G, where H contains real numbers and G is defined over the Galois Field GF(2). The user seeks to understand how to compute the product HGm, where Gm is a vector in GF(2) and the multiplication of Gm is performed in GF(2) while H is multiplied using standard real arithmetic. The key insight is that Gm acts as a selection mechanism for the columns of H, resulting in a real-valued output. The distinction between the operations (HG)m and H(Gm) is clarified, emphasizing that they yield different results due to the nature of the operations involved.

PREREQUISITES
  • Understanding of matrix multiplication in real numbers
  • Familiarity with Galois Fields, specifically GF(2)
  • Knowledge of linear algebra concepts, particularly vector spaces
  • Basic understanding of modular arithmetic
NEXT STEPS
  • Study the properties of Galois Fields, particularly GF(2) operations
  • Learn about matrix transformations and their applications in linear algebra
  • Explore the implications of combining different types of matrix multiplications
  • Investigate real-valued vector outputs from binary matrix operations
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Students and researchers in mathematics, particularly those studying linear algebra, matrix theory, and applications of Galois Fields in computational contexts.

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Homework Statement


H is a nxn matrix with elements in {0,1}
G is a nxn matrix with elements in GF(2)
m is a nx1 vector with elements in GF(2).
How can we perceive the output of
HGm where Gm multiplication is in GF(2) and H multiplication is a normal real multiplication.
Actually I want to combine HG transformation into one P transformation. How can I multiply two matrices while elements in one is in GF(2) and other is in R ?
(We can also restrict the entries in H to be one of 0 and 1 but the output can be in R).


Homework Equations





The Attempt at a Solution


 
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the entries in the vector Gm will still be in GF(2) = {0,1}.

so basically the vector Gm acts as a "choice function" picking out which columns of H get summed in the ouput, which will be a real-valued vector.

it makes more sense to do it this way, than to imagine what "HG" means (in this case G acts in a more complicated way, which is then subjected to another selection via m).

in general, (HG)m and H(Gm) won't yield the same results:

[a b]([1 0][1])
[c d]([1 1][1]) =

[a b][1]
[c d][0], which is (a,c)

([a b][1 0])[1]
([c d][1 1])[1] =

[a+b b][1]
[c+d d][1] = (a+2b,c+2d), which is only equal to the first mod 2.
 

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