How to calculate an exact solution: non-linear equation with multiple variables

In order to diagonalization of 3x3 matrix, I'm using 'fsolve' command. (there are 3 solutions for each matrix element of diagoanl)

However, it shows the deviation between the calculation solution and the real solution. Even it is reduced by change the initial value.

It means fsolver could show the exact value, I guess...

(I tried setting the option such as 'FunctionTolerance', 'StepTolerance', there was no effect.)

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Coef=readmatrix('coef.txt');

v1=0;

T_eff=Coef(v1,1); A_eff=Coef(v1,2); eps_eff=Coef(v1,3); alp_eff=Coef(v1,4); B_eff=Coef(v1,5); D_eff=Coef(v1,6); p_eff=Coef(v1,7); q_eff=Coef(v1,8); A_D_eff=Coef(v1,9); gamma_eff=Coef(v1,10); H_eff=Coef(v1,11);

Eg=zeros(100,3);

for J=0:1:50;

x=N*(N+1);

W11=T_eff-A_eff+eps_eff-alp_eff+B_eff*(x+1)+0.25+q_eff*x+0.5*(1-1)*(q_eff+p_eff)-A_D_eff*(x+1)-D_eff*(x^2+4*x+1)-H_eff*(x^3+9*x^2+9*x+1);

W12=-(B_eff+(0.25-0.5)*q_eff+(0.125-1/4)*p_eff-0.5*A_D_eff-2*D_eff*(x+1)-0.5*gamma_eff+H_eff*(3*x^2+10*x-1))*sqrt(2*x);

W13=-(2*D_eff+0.25*q_eff-2*H_eff*(3*x-1))*sqrt(x*(x-2));

W21=W12;

W22=T_eff-2*eps_eff+B_eff*(x+1)+0.25*q_eff*(1-1)*x+0.5*(q_eff+p_eff)-D_eff*(x^2+6*x-3)+H_eff*(x^3+15*x^2-5*x+5);

W23=-(B_eff+0.25*q_eff+0.125*p_eff+0.5*A_eff-2*D_eff*(x-1)-0.5*gamma_eff+H_eff*(3*x^2-2*x+3))*sqrt(2*(x-2));

W31=W13;

W32=W23;

W33=T_eff+A_eff+eps_eff+B_eff*(x-3)+0.25*q_eff*(x-2)+A_D_eff*(x-3)-D_eff*(x^2-4*x+5)+H_eff*(x^3-3*x^2+5*x-7);

E_mat=@(E)det([W11-E(1) W12 W13; W21 W22-E(2) W23; W31 W32 W33-E(3)]);

E0=[W11,W22,W33];

options=optimoptions('fsolve','Display','iter','FunctionTolerance',0,'StepTolerance',0);

[E,fval]=fsolve(E_mat,E0,options);

E_ans(J+1,:)=det([W11-E_B W12 W13; W21 W22-E_B W23; W31 W32 W33-E_B]); %test line for checking the deviation

end

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No Solution Found

fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the default value of the function tolerance.

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determinant error is below just 1e-5, however, I want to exact value.

Answer 1

[Torsten]

Use "eig", not "fsolve".

[JongGu Han]

Thank you for answer. I have one more question.

I get eigenvalues using eig function, and then substitute that value in 3by3 matrix mentioned above. This matrix performed determinante and show some strange results, huge error such as 1e4..1e5..1e6.. not zero.

code exmaple:

Matrix=[W11 W12 W13; W21 W22 W23; W31 W32 W33];

[eig_vec,eig_val]=eig(Matrix);

Det_err=det([W11-eig_val(1,1) W12 W13; W21 W22-eig_val(1,2) W23; W31 W32 W33-eig_val(1,3)];

Is there some problem?

[Torsten]

You have one equation

det([W11-E(1) W12 W13; W21 W22-E(2) W23; W31 W32 W33-E(3)]) = 0

for three unknowns E(1), E(2) and E(3).

I doubt this is what you want.

The three eigenvalues solve

Det_err1 = det([W11-eig_val(1,1) W12 W13; W21 W22-eig_val(1,1) W23; W31 W32 W33-eig_val(1,1)] = 0
Det_err2 = det([W11-eig_val(1,2) W12 W13; W21 W22-eig_val(1,2) W23; W31 W32 W33-eig_val(1,2)] = 0
Det_err3 = det([W11-eig_val(1,3) W12 W13; W21 W22-eig_val(1,3) W23; W31 W32 W33-eig_val(1,3)] = 0

[JongGu Han]

oh, it's poor mistake.

Thank you.