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Math Help - Linear Algebra in Cryptography

  1. #1
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    Linear Algebra in Cryptography



    So... each letter from A to Z corresponds to a number 0 to 25. Now I want to find the pairs (x,y) such that in \mathbb{Z}_{26},

    \begin{bmatrix} 8 & 3\\ 1 & 7 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} x\\ y \end{bmatrix}

    And we are only told that for (A, A), which corresponds to (0,0), this holds.

    So, how do I find all pairs x, y such that they are unchanged by the multipication? Going through all possible combinations of x and y is a real daunting task. Is there an easy way for finding all the pairs?

    P.S. In \mathbb{Z}_{26},

    K^{-1} = \begin{bmatrix} 7 & 23\\ 25 & 8 \end{bmatrix}.
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  2. #2
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    You need to solve Av=v, that is, (A-I)v = 0 in \mathbb{Z}_{26}. Do you know how to do this?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    You need to solve Av=v, that is, (A-I)v = 0 in \mathbb{Z}_{26}. Do you know how to do this?
    No, I'm not quite sure how to solve this. Could you please explain a bit more? I just got the two equations

    (K-I)v= \left( \begin{bmatrix} 8 & 3\\ 1 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \right) \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 7x+3y\\ x+6y \end{bmatrix} =0

    But don't know how solving the homogeneous system shows the list of elements that are mapped to themselves? How should I continue?
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