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Repost from SCI.p.Research−Matrix Determinant as Path Integral? Repost from SCI.p.Research−Matrix Determinant as Path Integral? Source: http://sci.tech−archive.net/Archive/sci.physics/2007−08/msg00265.html • From: Douglas Eagleson <eaglesondouglas@xxxxxxxxx> • Date: Fri, 03 Aug 2007 05:29:37 −0700 I will top or front comment. The original post follows. It is difficult to convince people as it is a modern technique of matrice. ALL matrix as a third abstracted relation of set element to location is termed matrice. So inclusive of all graph theory was the matrice. A study of the entire works of Hamilton is a good introduction to this topic. Hamilton literaly used quaterions to talk about his quest. A matrix has a number locations associated with the set element. That is the property unique to a matrix. It make it different from a number without location relationship. I can likely answer the question of correct usage of matrice, but open the floor to debate. A method was suggested in the original post. And the rules for speculation prevented the moderators of ..research from obtaining an answer. Here are the basic benefits of the method. 1. Infinite renormalized. A single benefit appears. A means with renormal as operated element appears. A selection of element size as opposed to sovled equation, causes the solution. This is a major, major benefit. So I will speculate about this topic to begin with. An integral as opposed to the path integral will be used now. All integral have solved elements called dx. This is simply a usage of the basic formula. Normally dx is caused an infinite limit and the function solved. In matrice all dx are given a constant, is it the average dx for the integral top and bottom limits of integration, or a true differential constant of third relative change over the span? Sorry it is the second usage. A perfect integral is obtainable in matrice. Repost from SCI.p.Research−Matrix Determinant as Path Integral? 1 Repost from SCI.p.Research−Matrix Determinant as Path Integral? This is the goal, I claim it obtained. y=f(x) For any function define the matrice as integral. answer= determinante(cross product)matrix matrix= |dx1 area| |dx2 area| |dx3 area| |dx4 area| This is Hamilton's matrix usage. Hamilton in my thinking discovered it in western lands. A cross product of the matrix determinante with the original matrix causes!!!!!!!!!!!!!!!!!!!!!!! It defines his matrice. Without discussing this great discovery here is the means of integral answering. Matrix eigen as answer. AN Eigen is the term for the left hand side. It is not an eigenvalue, but is derived from one. SELECT Eigen. And with Hamilton's matrice the selection ALWAYS appears a sufficincey of dx limits! For y=f(x) place the eigenvalues equal to the functions y at the dx1,2,3,4 integral limits!. A selected eigenvalue causes element area to equate. A literal search will obtain a correct determinate, and thereffor integral. Sufficiency is not the size of the Eigen, it is the limit selected. So automatic limit number appearance is built in. That is the usage. Beleive it or not. The original poster wonders of determinate theory, and this answer distills the wonder. Lets get this exactly clear: And eigenvalue set will cause four function values, each an x value of Repost from SCI.p.Research−Matrix Determinant as Path Integral? 2 Repost from SCI.p.Research−Matrix Determinant as Path Integral? the four dx elements. When an independent four set matchs, a visual match, happens the Eigen is the integral. Independent means to have each x be the x−axis value for the given function eqi−division of limits of integration. Eigen as a term is a new term dervied from eigenvalue and is technically a third order of logic use of matrix location to solve. *******************Original post******* This post is a little dreamy and may not make sense, so please bear with it. On the other hand, there might be something to it. The other day I noticed that the determinant of a matrix looks like a sum−over−histories. You can think of it as the sum of all the paths from the top to the bottom, and if you use complex numbers then you even have interference, but with the problem that you have the alternating {+1 and −1} in the products. There is also a thing called the Permanent [see wikipedia] that doesn't have alternating signs, it just uses +1 the entire time. You can also define all sorts of things with other numbers on the unit circle like {1,i,−1,−i}, etc. At first I thought it was interesting but not−so−great, until i remembered that the determinant is a group homomorphism. That means the kernel is mapped to the identity element of the image group. Now I often call the identity element the "do nothing" element, because it does just that: nothing. For instance, 2*1 = 2, or 2+0=2, etc. Now the shortest "distance" between 2 points is a geodesic, and if its light, then its a "null geodesic": it does nothing, and everything can be defined in reference to it. So fermats principle of least time, the concept of the geodesic, and of the fundamental frequency, and of quickest path to a win in a stochastic game, and least action all seem to be describable by the identity Repost from SCI.p.Research−Matrix Determinant as Path Integral? 3 Repost from SCI.p.Research−Matrix Determinant as Path Integral? element. For instance, you can think of the fourier transform of the number 1: You have all negative and positive frequencies and they cancel out to a dirac delta function. Its the identity element of an additive group. You wouldnt get much energy coming over the radio at that frequency. But the path integral cancels out to those same things. Perhaps the determinant is some kind of path integral that is related to fermions, the permanent is one for bosons, and the other similar structures are for anyons. I found some references on the internet, but they seemed sort of lofty and mysterious. Here are 3 questions that are possible a little more practical: if you have a system with a continuous spectrum, then 1. The matrix has a dimension that is continuous, so can the determinant be defined? If it converged, it would be an integral of infinite products. 2. If it could be defined, would the resulting continuous−dimensional structure still be a group? [set of matrices with certain types of determinant = group.] 3. Do you need lebesgue theory or just the more familiar riemann integrals? If you are an expert on this stuff, please explain. JHS . Repost from SCI.p.Research−Matrix Determinant as Path Integral? 4