Re: matrix substitution
- To: mathgroup at smc.vnet.net
- Subject: [mg67476] Re: matrix substitution
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Tue, 27 Jun 2006 03:14:54 -0400 (EDT)
- References: <e665nv$n43$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
My original substitution used the Russian:
"1"->{{1,0},{0,1}}
"I"-> {{0,1},{-1,0}}
for 1 and I to get a 4by4 quaternion.
It occurred to me that using
"1"->{{I,0},{0,-I}
"I"- >{{0,I},{I,0}}
also gave determinant one type results: everywhere perpendicular to the
other 4by4 quaternions.
I put the two substitutions together to get an eight element 4by4 group
that appears to
be octonion like:
a = {{I, 0}, {0, -I}}
b = {{0, I}, {I, 0}}
c = {{0, 0}, {0, 0}}
s[1] = {{c, a}, {-a, c}}
s[2] = {{c, b}, {b, c}}
s[3] = {{b, c}, {c, -b}}
s[0] = {{a, c}, {c, a}}
q[t_, x_, y_, z_] = ExpandAll[s[0]*t + x*s[1] + s[2]*y + s[3]*z]
Dimensions[s[0]]
w = Table[Flatten[Table[{Flatten[Table[s[k][[n, m]][[1, i]], {n,
1, 2}, {i, 1, 2}]], Flatten[Table[s[k][[n,
m]][[2, i]], {n, 1, 2}, {i, 1, 2}]]}, {m, 1, 2}], 1], {k, 0, 3}]
MatrixForm[w]
w[[1]]
MatrixForm[Table[w[[n]].w[[m]], {n, 1, 4}, {m, 1, 4}]]
Clear[a, b, c, s]
a = {{1, 0}, {0, 1}}
b = {{0, -1}, {1, 0}}
c = {{0, 0}, {0, 0}}
s[1] = {{c, a}, {-a, c}}
s[2] = {{c, b}, {b, c}}
s[3] = {{b, c}, {c, -b}}
s[0] = {{a, c}, {c, a}}
Dimensions[s[0]]
v = Table[Flatten[Table[{Flatten[Table[s[k][[n, m]][[1, i]], {
n, 1, 2}, {i, 1, 2}]], Flatten[Table[s[k][[n, m]][[2, i]], {
n, 1, 2}, {i, 1, 2}]]}, {m, 1, 2}], 1], {k, 0, 3}]
MatrixForm[v]
v[[1]]
MatrixForm[Table[v[[n]].v[[m]], {n, 1, 4}, {m, 1, 4}]]
ss[n_] := If[n â?¤ 4, v[[n]], w[[n - 4]]]
o = Table[ss[n].ss[m], {n, 1, 8}, {m, 1, 8}]
MatrixForm[o]