# Linear Algebra in Cryptography

• March 22nd 2011, 04:26 PM
demode
Linear Algebra in Cryptography
http://img138.imageshack.us/img138/4062/13177071.png

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 $\mathbb{Z}_{26}$,

$K^{-1} = \begin{bmatrix} 7 & 23\\ 25 & 8 \end{bmatrix}$.
• March 22nd 2011, 04:42 PM
Defunkt
You need to solve $Av=v$, that is, $(A-I)v = 0$ in $\mathbb{Z}_{26}$. Do you know how to do this?
• March 22nd 2011, 11:04 PM
demode
Quote:

Originally Posted by Defunkt
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?