[seqfan] Re: A generalization of the matrix permanent: question
Creighton Kenneth Dement
creighton.k.dement at mail.uni-oldenburg.de
Wed Apr 8 17:04:38 CEST 2009
> Well, if you sum over the alternating group, you get the average (mean)
> of the permanent and the determinant.
>
> Summing over the cyclic group just includes each term in the matrix in
> a single product, one product for each modular diagonal; I doubt that
> there's much interest there. There may be some value in summing over
> the dihedral group.
>
> (As to your question, I have no idea if there's any literature on this.)
>
> Franklin T. Adams-Watters
I find this answer extremely interesting. I'm only a math enthusiast,
however, I'll try to answer this question to the best of my abilities.
In a way, just about anything you do with floretions has to do with
"summing up over generalized determinants". Interested readers can have a
look, below. (If there are parts which are still unclear, my Floretion
2009 draft paper can be made available on individual request).
********Definition******
For X = A'i + B'j + C'k + Di' + Ej' + Fk' + G'ii' + H'jj' + I'kk' + J'ij'
+ K'ik' + L'ji' + M'jk' + N'ki' + O'kj' + P'ee', define:
ves(X) = A+B+C+D+E+F+G+H+I+J+K+L+M+N+O+P
VES(X) = X
jesleft(X) = A+B+C
JESLEFT(X) = A'i + B'j + C'k
jesright(X) = D+E+F
JESRIGHT(X) = Di' + Ej' + Fk'
jes(X) = A+B+C+D+E+F
JES(X) = A'i + B'j + C'k + Di' + Ej' + Fk'
les(X) = G+H+I+J+K+L+M+N+O
LES(X) = G'ii' + H'jj' + I'kk' + J'ij' + K'ik' + L'ji' + M'jk' + N'ki' +
O'kj'
tes(X) = P
TES(X) = P'ee"
i-base(X) = A, j-base(X) = B, k-base(X) = C, base-i = D ...
********End Definition******
********** Theorem I - "Wave Equation"
For any X, Y and Z the following holds:
If X = JES(X), then X^2 = LES(X^2) + TES(X^2)
If X = JES(X), then X^3 = JES(X^3)
***********
Now, consider the real vectors X = (A,B,C), Y = (D,E,F), Z = (G,H,I) in
the space R^3.
We have the operations:
-add
-perform scalar multiplication
-calculate the scalar (dot) product
-vector product
-3X3 determinant
The same results are achieved with floretions. The next theorem re-writes
X, Y, and Z above as "pure" quaternions; abusing notation by using the
same symbols for convenience.
********** Theorem II
If X = JESLEFT(X), Y = JESLEFT(Y), Z = JESLEFT(Z) (traditionally, these
would be called "pure" quaterions) then
X scalar-product Y = -tes(X*Y)
X vector-product Y = JESLEFT(X*Y) ( = JES(X*Y) )
det(X,Y,Z) = -tes(X*Y*Z)
**********End Theorem
In a nutshell, the last property means that a "Cramer's Rule, Quaternion
Style" exists. In particular, assuming we know that (a(0), a(1), ...) is a
3rd order linear recurrence of the form a(n) = q*a(n-1) + r*a(n-2) +
s*a(n-3), we can determine the coefficients q, r, and s using
quaternions alone. There is also a "Cramer's Rule, Floretion Style", but
that would take us too far adrift.
As mentioned, just about anything you do can be interpreted as "summing up
over generalized determinants". For ex.,
for X = (A,B,C), Y = (D,E,F), Z = (G,H,I) as above, we have
tes(X*Y*Z) = +1.0AFH +1.0BDI +1.0CEG -1.0BFG -1.0AEI -1.0CDH
i-base(X*Y*Z) = +1.0BDH -1.0AEH +1.0CDI -1.0ADG -1.0BEG -1.0AFI -1.0CFG
j-base(X*Y*Z) = +1.0CEI -1.0BEH -1.0CFH -1.0BDG -1.0ADH +1.0AEG -1.0BFI
k-base(X*Y*Z) = -1.0CEH -1.0CDG -1.0CFI +1.0AFG -1.0ADI -1.0BEI +1.0BFH
Thanks for reading- I will stop here for fear of writing too much.
Sincerely,
Creighton
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