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matrix substitution


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.


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