matrix substitution

*To*: mathgroup at smc.vnet.net*Subject*: [mg67021] matrix substitution*From*: Roger Bagula <rlbagulatftn at yahoo.com>*Date*: Wed, 7 Jun 2006 05:09:35 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

In certain complex 2by2 groups there are equivalent 4by4 representations whicg you can get by the substitutiuon: 1->{{1,0},{0,1}} I->{{0,1},{-1,0}} An example; I used the code : a = {{1, 0}, {0, 1}} b = {{0, 1}, {-1, 0}} c = {{0, 0}, {0, 0}} i = MatrixForm[{{c, a}, {-a, c}}] j = MatrixForm[{{c, b}, {b, c}}] k = MatrixForm[{{b, c}, {c, -b}}] e = MatrixForm[{{a, c}, {c, a}}] q[t_, x_, y_, z_] = ExpandAll[e*t + x*i + j*y + k*z] to get: i = {{0, 0, 1, 0}, {0, 0, 0, 1}, {-1, 0, 0, 0}, {0, -1, 0, 0}} j = {{0, 0, 0, 1}, {0, 0, -1, 0}, {0, 1, 0, 0}, {-1, 0, 0, 0}} k = {{0, 1, 0, 0}, {-1, 0, 0, 0}, {0, 0, 0, -1}, {0, 0, 1, 0}} e = {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} ( Try e^2-i^2-j^2-k^2) But I got the result by hand after the fact. There should be a way to do the derivation directly, but my skills aren't up to it, yet. My current problem is quotient group made from the golden mean characteristic: a=Table[x^n,{n,0,1}] b=Table[PolynomialMod[a[[n]]*a[[m]],x^2-x-1],{n,1,2},{m,1,2}] MatrixForm[b] The result is a quotient group for the golden men. Which gives a multipilcation table like:1->e, x->I {{e,I},{I,e+I}} which I can do the same sort of substitution on, but I'd like to have a more effective method! The benefit of the 4b4 matrices is that the result is in real numbers without and complex numbers.