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    Orthogonality of a Generator Matrix of a Binary Block Code

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    4. Let be the generator matrix of a binary block code .

    a) Show that each row of G is orthogonal to itself and to each of the other rows of G.
    b) Show that each codeword in is orthogonal to itself and to every other codeword in .
    Note: Use part a) and the fact that each codeword is a linear combination of the rows of G.
    c) What is the dimension of ? Explain
    d) What is the dimension of ? Explain
    e) Show that is a self-dual code, that is show that = .
    Can you explain what does orthogonally means and self-dual code?

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    https://brainmass.com/math/matrices/orthogonality-generator-matrix-binary-block-code-98055

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    Orthogonality
    -------------
    Originally, this concept means at right angles
    or 90° or perpendicular. Then using vectors
    we find that they are orthogonal if their
    dot product is zero. For example,
    vector a = [ 1 0 3 ]
    vector b = [-3 7 1 ]
    take the dot product,
    a · b = 1×-3 + 0×7 + 3×1 = -3 + 0 + 3 = 0

    Now we extend the concept to binary codes. But
    you must remember the laws for binary codes are
    slightly different.
    0 + 0 = 0 0 × 0 = 0
    0 + 1 = 1 0 × 1 = 0
    1 + 0 = 1 1 × 0 = 0
    1 + 1 = 0 1 × 1 = 1
    Notice that 1+1=0
    + can be interpreted as + modulo 2: whenever
    you have a 2, you always reset to 0
    if you know electronics or computer science,
    you can interpret this + as XOR (exclusive or)

    × is similar to our ordinary ×
    it can be interpreted in electronics or computer
    science as AND.

    Once you understand this, you can do part a).
    [ 1 0 0 0 0 1 1 1 ]
    G = [ 0 1 0 0 1 0 1 1 ]
    [ 0 0 1 0 1 1 0 1 ]
    [ 0 0 0 1 1 1 1 0 ]
    Denote the rows by r1, r2, r3 and r4. We can
    calculate all the dot products, for example

    r1·r1
    = 1×1 + 0×0 + 0×0 + 0×0 + 0×0 + 1×1 + 1×1 + 1×1
    = 1 + 0 + 0 + 0 + 0 + 1 + 1 + 1
    = 0
    (remember, every time you get 1+1 it's reset to
    zero. if you get 4 ones, an even number, the
    final result is zero.)

    r1·r2
    = 1×0 + 0×1 + 0×0 + 0×0 + 0×1 + 1×0 + 1×1 + 1×1
    = 0 + 0 + 0 + 0 + 0 + 0 + 1 + 1
    = 0 (two ones, even number, so total is 0)

    By the same type of calculation, you can check
    that r1·r3=0, r1·r4=0, and also
    r2·r1=0, r2·r2=0, r2·r3=0, r2·r4=0,
    r3·r1=0, r3·r2=0, r3·r3=0, r3·r4=0,
    r4·r1=0, r4·r2=0, r4·r3=0, r4·r4=0.

    Generator and Linear Combination
    --------------------------------
    For part b, you need to understand the concept
    of Generator and Linear ...

    Solution Summary

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