@acer 

i updated my previous reply

and

a := MatrixMatrixMultiply(eigenvector1 ,  eigenvector1);

b := MatrixMatrixMultiply(eigenvector2 ,  eigenvector2);

a and b are different if eigenvector1 and 2 have different sign

quantum event is eigenvector

because projection of eigenvector is equal to itself

when deal with quantum logic with some fuzzy operation, it will have different result if sign are different

assume all calculation done in the same software, can i assume same choice of solution from a set of different sign solution for all different input eliminate the effect of sign different?

in software A , if a1 'value is -1 , a2 's value is 1, -1*1 = -1

in software B, if a1's value is 1 , a2 's value is -1 , -1*1 = -1

though A and B software use different set of solution, the result of matrix operation can be the same ?

if software A one of eignenvector using the choice of itself and another eigenvector using the choice of software B's choice,

it will have problem in result not the same in software A and software B

i think out a problem caused by sign different must result in different result

this is that it can not do power of matrix such as A^2, which is A*A

if comare A*A in different software A and B , for example , extreme library and maple, they must result in different result

i put a variable in matrix and apply your method, can not further calculate at eigenvectorpre

InputMatrix3 := Matrix([[x,30.8,30.5],[30.8,30.5,0],[30.5,0,0]]);
NewInput3 := MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3);
G := Matrix(NewInput3^%T . NewInput3, shape=symmetric):
evals:=Eigenvalues(G);
eigenvectorpre := seq( NullSpace(CharacteristicMatrix(G,evals[i])), i=1..3 );
eigenvector := Matrix([[eigenvectorpre[3][1][1], eigenvectorpre[2][1][1], eigenvectorpre[1][1][1]],[eigenvectorpre[3][1][2], eigenvectorpre[2][1][2], eigenvectorpre[1][1][2]],[eigenvectorpre[3][1][3], eigenvectorpre[2][1][3], eigenvectorpre[1][1][3]]]);

@acer 

because directsearch got error value then i debug to see

{seq(seq((NewMatrix3 . NewInput3(1..-1,i))[j]=(v[i]* NewInput3(1..-1,i))[j], j=1..3), i=1..3)}

whether correct or not.

then it is wrong,

all because matrix has one value is a variable , not a decimal number

and eigenvector can not calculate variable matrix

another reason is accuracy problem, there are two kinds of solution which different in sign

but sign different could result in different result after doing multiply operation between eigenvectors

quantum event will have two more different solutions

then i think should we absolute values of the eigenvectors in order to operate quantum event?

then i try your method above

https://drive.google.com/file/d/0B2D69u2pweEvODRCVGZ6RkZnUjA/edit?usp=sharing

it has sign different , and i do not know which column is the first, second or third,

they are different from eigenvector function,

how to do it the same as maple?

then i validate your method , it is correct,

it is the third kind of solutions i have ever seen.

@Mac Dude 

https://drive.google.com/file/d/0B2D69u2pweEvakM1LUQ2R0dqZEU/edit?usp=sharing

i validate extreme library solution and maple 15 solution,

they are all correct though not exactly zero

but i got another question

if there exist more than one set of solution, how do i know choose which set is correct?

because if using eigenvector multiply operation on another eigenvector, there will be two kinds of solutions

https://drive.google.com/file/d/0B2D69u2pweEvN09kYTk4cHdmZ28/edit?usp=sharing

there will be two kinds of solution if doing quantum event operation

a := MatrixMatrixMultiply(eigenvector1 ,  eigenvector1);

b := MatrixMatrixMultiply(eigenvector2 ,  eigenvector2);

a and b are different if eigenvector1 and 2 have different sign

quantum event is eigenvector

because projection of eigenvector is equal to itself

when deal with quantum logic with some fuzzy operation, it will have different result if sign are different

assume all calculation done in the same software, can i assume same choice of solution from a set of different sign solution for all different input eliminate the effect of sign different?

in software A , if a1 'value is -1 , a2 's value is 1, -1*1 = -1

in software B, if a1's value is 1 , a2 's value is -1 , -1*1 = -1

though A and B software use different set of solution, the result of matrix operation can be the same ?

if software A one of eignenvector using the choice of itself and another eigenvector using the choice of software B's choice,

it will have problem in result not the same in software A and software B

@Mac Dude 

https://drive.google.com/file/d/0B2D69u2pweEvd054Ti1VdTE1NlE/edit?usp=sharing

after solve, all results are zero [0,0,0]

is solving the system (A - lambda)*x = 0 the correct way to find eigenvector?

i search in books, they are all write in this.

but when i solve this, all are zeros.

if not this way, what is the correct way to find eigenvectors?

what algorithm do maple use to find eigenvector?

@mehdi jafari

can this be like taylor expression
just want to get coefficient of a polynomial^n

observe
taylor((a*y^2+b*y+c)^x, x = 0, 10);
taylor((a*y^2+b*y+c)^x, x = 0, 20);

i see it use ln(a*y^2+b*y+c)

is it really simple for n?

@Carl Love 

which math tools or theory can help to guess the structure?

@Carl Love 

if given general form a^i + b^j, how to do?

@Markiyan Hirnyk 

Hi ,

i upload again, and tested in maple and sure that it can run.

can you take a look?

https://drive.google.com/file/d/0B8F2D27rfQWgVXE1alN0V3JWU1U/edit?usp=sharing

@Markiyan Hirnyk 

after further more fit to describe the relationship, it still can not get the correct answer

f1 := ExpectCommutateLeft[1,1] - RealCommutateRight[1,1];
f2 := ExpectCommutateLeft[2,2] - RealCommutateRight[2,2];
f3 := ExpectCommutateLeft[3,3] - RealCommutateRight[3,3];
f4 := ExpectCommutateLeft[1,2] - RealCommutateRight[1,2];
f5 := ExpectCommutateLeft[2,1] - RealCommutateRight[2,1];
f6 := ExpectCommutateLeft[1,2] + ExpectCommutateLeft[2,1];
f7 := ExpectCommutateLeft[1,3] + ExpectCommutateLeft[3,1];
f8 := ExpectCommutateLeft[2,3] + ExpectCommutateLeft[3,2];
f9 := ExpectCommutateLeft[3,1]*2 + ExpectCommutateLeft[3,2];
f1 := Re(f1);
f2 := Re(f2);
f3 := Re(f3);
f4 := Re(f4);
f5 := Re(f5);
f6 := Re(f6);
f7 := Re(f7);
f8 := Re(f8);
f9 := Re(f9);
g1 := 25.3+3-x;
g2 := x-35.3+3;
with(DirectSearch):
startrange := 25.3+3;
endrange := 35.3-3;
constr := [x = startrange .. endrange];
CompromiseProgramming([f1, f2, f3, f4, f5, f6, f7, f8, f9], {g1 <= 0, g2 <= 0}, strategy = globalsearch);

@Markiyan Hirnyk 

https://drive.google.com/file/d/0B8F2D27rfQWgVXE1alN0V3JWU1U/edit?usp=sharing

use f1 := Re(f1), can solve this, however the answer is not the expected one

i know the answer of the problem, then i pretend unknown it and try to find this unknown

however a wrong answer, the reason i use this is commutative property

with(LinearAlgebra): InputMatrix3 := Matrix(3, 3, {(1, 1) = 30.15, (1, 2) = 30.3, (1, 3) = 29.95, (2, 1) = 30.3, (2, 2) = 29.95, (2, 3) = 0, (3, 1) = 29.95, (3, 2) = 0, (3, 3) = 0}); InputMatrix3b := Matrix(3, 3, {(1, 1) = 30.3, (1, 2) = 29.95, (1, 3) = 29.95, (2, 1) = 29.95, (2, 2) = 29.95, (2, 3) = 0, (3, 1) = 29.95, (3, 2) = 0, (3, 3) = 0}); InputMatrix3c := Matrix(3, 3, {(1, 1) = 29.95, (1, 2) = 29.95, (1, 3) = 29.6, (2, 1) = 29.95, (2, 2) = 29.6, (2, 3) = 0, (3, 1) = 29.6, (3, 2) = 0, (3, 3) = 0}); Old_Asso_eigenvector := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3)): Old_Asso_eigenvector2 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3b), InputMatrix3b)): Old_Asso_eigenvector3 := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3c), InputMatrix3c)): CommutateLeft:= MatrixMatrixMultiply(Matrix(Old_Asso_eigenvector[2], shape = hermitian), Matrix(Old_Asso_eigenvector2[2], shape = hermitian)); CommutateRight := MatrixMatrixMultiply(Matrix(Old_Asso_eigenvector2[2], shape = hermitian), Matrix(Old_Asso_eigenvector[2], shape = hermitian)); Reason := CommutateLeft - CommutateRight; ExpectedInputMatrix3 := Matrix(3, 3, {(1, 1) = x, (1, 2) = 30.3, (1, 3) = 29.95, (2, 1) = 30.3, (2, 2) = 29.95, (2, 3) = 0, (3, 1) = 29.95, (3, 2) = 0, (3, 3) = 0}); FirstEigenValue := solve(Determinant(ExpectedInputMatrix3-Matrix([[lambda1, 0, 0], [0, lambda1, 0], [0, 0, lambda1]])), lambda1)[1]; # find back eigenvalue from eigenvector SecondEigenValue := solve(Determinant(ExpectedInputMatrix3-Matrix([[lambda1, 0, 0], [0, lambda1, 0], [0, 0, lambda1]])), lambda1)[2]; # find back eigenvalue from eigenvector ThirdEigenValue := solve(Determinant(ExpectedInputMatrix3-Matrix([[lambda1, 0, 0], [0, lambda1, 0], [0, 0, lambda1]])), lambda1)[3]; v:=[ FirstEigenValue, SecondEigenValue, ThirdEigenValue]; NewMatrix3 := Matrix([[x1,x2,x3], [x4,x5,x6], [x7,x8,x9]]); simplify(solve({seq(seq((NewMatrix3 . ExpectedInputMatrix3(1..-1,i))[j]=(v[i]*ExpectedInputMatrix3(1..-1,i))[j], j=1..3), i=1..3)}, {seq(x||i, i=1..9)})); assign(%); ExpectCommutateLeft:= MatrixMatrixMultiply(Matrix(NewMatrix3, shape = hermitian), Matrix(Old_Asso_eigenvector2[2], shape = hermitian)); RealCommutateRight := MatrixMatrixMultiply(Matrix(Old_Asso_eigenvector2[2], shape = hermitian), Matrix(NewMatrix3, shape = hermitian)); f1 := ExpectCommutateLeft[1,1] - RealCommutateRight[1,1]; f2 := ExpectCommutateLeft[2,2] - RealCommutateRight[2,2]; f3 := ExpectCommutateLeft[3,3] - RealCommutateRight[3,3]; g1 := 25.3-x; g2 := x-35.3; with(DirectSearch): startrange := 25.3; endrange := 35.3; constr := [x = startrange .. endrange]; CompromiseProgramming([f1, f2, f3], {g1 <= 0, g2 <= 0}, strategy = globalsearch); CompromiseProgramming([f1, f2, f3], {g1 <= 0, g2 <= 0}, optimizer = globaloptima); CompromiseProgramming([f1, f2, f3], constr, pointrange = constr, optimizer = globalsearch); CompromiseProgramming([f1, f2, f3], {g1 <= 0, g2 <= 0});

Correct answer is 30.15

@Markiyan Hirnyk 

https://drive.google.com/file/d/0B8F2D27rfQWgVXE1alN0V3JWU1U/edit?usp=sharing

f1 := (-6.56889552306723*10^(-16)-0.*I)*(-101.*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+(3.138380888*10^18*I)*x+(2.619683546*10^12*I)*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+1.010000*10^6*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)+40400.*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)-6.645906620*10^20*I-(1.047873418*10^17*I)*x^2)/(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)+(6.09968908424774*10^(-19)-0.*I)*(-5.373044975*10^19*x^2+1.137808203*10^22*x+1.794005000*10^18*x^3+4.485012500*10^13*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+2.00665*10^5*x^2*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)-1.0495750*10^7*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)-(1.035769269*10^18*I)*x^3-(8.470997541*10^21*I)*x-(1.587528229*10^15*I)*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+(2.589423174*10^13*I)*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+61206.*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+(9.452241876*10^19*I)*x^2-6.12060000*10^8*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)+4.027419413*10^23*I)/(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)+(6.56889552306723*10^(-16)-0.*I)*conjugate((-101.*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+(3.138380888*10^18*I)*x+(2.619683546*10^12*I)*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+1.010000*10^6*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)+40400.*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)-6.645906620*10^20*I-(1.047873418*10^17*I)*x^2)/(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3))+(-6.09968908424774*10^(-19)-0.*I)*conjugate((-5.373044975*10^19*x^2+1.137808203*10^22*x+1.794005000*10^18*x^3+4.485012500*10^13*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+2.00665*10^5*x^2*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)-1.0495750*10^7*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)-(1.035769269*10^18*I)*x^3-(8.470997541*10^21*I)*x-(1.587528229*10^15*I)*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+(2.589423174*10^13*I)*x*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+61206.*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(2/3)+(9.452241876*10^19*I)*x^2-6.12060000*10^8*(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3)+4.027419413*10^23*I)/(5.457930000*10^10*x-3.59400000*10^8*x^2-7.294897545*10^11+8.000000*10^6*x^3+60000.*sqrt(6.599230918*10^9*x^3-1.475596013*10^12*x^2-1.07640300*10^8*x^4+4.213249283*10^13*x-4.387559363*10^15))^(1/3));

@Markiyan Hirnyk 

thanks, directsearch is great.

if want to understand the algorithm or write into other language programming

how to minimize this function?

can steepest descent algorithm replace this minimize the penalty which has max?

@mehdi jafari 

how to get the Drift of this process if it is process?

Drift(PDF(X,t));

@mehdi jafari 

just practice an example 3.1 after google this

http://www.scirp.org/journal/PaperInformation.aspx?paperID=8918

@Markiyan Hirnyk 

one have large values, another have small values

after plot them, can not see the shape of line for compare

how plot two lines for compare?

that means can it have two different y axis, 

or

two diagram overlapped together to display shape of two lines

failed after try

plotcompare(sol[1](x), sol[2](x), x=0..25);

sol[1] and sol[2] are vector