# Why my symbolic result of eigenvalue doesn't match my numeric result?

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JingChong Ning on 31 Jan 2023
I have a set of three points, . As well as a matrix
[2, 0, 0]
[0, 2, 4*z2]
[0, 4*z2, 4*y2 + 2]
I need to find the eigenvalue of this matrix both symbolically and numerically after we substitude the y and z value in the matrix with the values in the provided points. However, if I plug in the numeric y and z value into the symbolic result I got from using
eigen = eig(eqn2ma)
It is different from first substituting the y and z into the matrix, then taking its eigenvalue.
This is the code
syms x y x1 y1 a x2 y2 z2 lamb
%{
%Q2
t=@(x,y) power(atan(x.*y),3)+power(sin(x),2);
star = 2*pi;
x = 0:0.01:star; % define range and mesh of x and y which will be shown in figure
y = -1:0.01:1;
[X, Y] = meshgrid(x, y);
figure
surf(X, Y, t(X,Y))
f = power(atan(x1*y1),3)+power(sin(x1),2);
f2 = diff(f,x1) == 0;
f3 = diff(f,y1) == 0;
[sx1,sy1] = vpasolve(f2,f3);%find the values of x and y of the minimum
extreme_values = subs(f, {x1,y1}, {sx1,sy1})%solve for the minimum
%}
%
%Q3
eqn = (x2^2)+2*x2+(y2^2)+2*y2*(z2^2)+(z2^2);
eqn1 = diff(eqn,x2)==0
eqn2 = diff(eqn,y2)==0
eqn3 = diff(eqn,z2)==0
[sx,sy,sz] = vpasolve(eqn1,eqn2,eqn3) %find the inflection points
vari = {x2, y2, z2};
eqn1ma = [diff(eqn,x2) diff(eqn,y2) diff(eqn,z2)];
eqn2ma = [a a a; a a a; a a a];
eqn2sol = zeros(3,3);
for i = 1:3
for j=1:3
eqn2ma(i,j)=diff(eqn1ma(1,j),vari(i));%create Hessian
end
end
eqn2ma
lambdia = [lamb 0 0; 0 lamb 0; 0 0 lamb];
eqn2malamb = eqn2ma-lambdia;
lamb0 = det(eqn2malamb);
S = solve(lamb0, lamb)
eigen = eig(eqn2ma)
for i = 1:3
for j = 1:3
for k = 1:3
eqn2sol(j,k)=subs(eqn2ma(j,k), {x2,y2,z2}, {sx(i,1),sy(i,1),sz(i,1)});
end
end
eqn2sol %ouput Hessian
eigenvalules = eig(eqn2sol)%find eigenvalue
end
Could you tell me why?
##### 2 CommentsShowHide 1 older comment
Torsten on 31 Jan 2023
If you subs the {sx,sy,sz} in the symbolic expression "eigen", you will get the same numerical eigenvalues as you get from the line "eigenvalules = eig(eqn2sol)".
So I don't understand your point.

Sulaymon Eshkabilov on 31 Jan 2023
The reason for having three different eigen values can be explained with the followings:
Eigenvalues1 = eig([2 0 0; 0 2 0; 0 0 2]) % @ (-1 0 0)
Eigenvalues1 = 3×1
2 2 2
Eigenvalues2 = eig([2 0 0; 0 2 4*-sqrt(.5); 0 4*-sqrt(.5) 4*-.5+2]) % @ (-1 -.5 -sqrt(.5))
Eigenvalues2 = 3×1
-2.0000 2.0000 4.0000
Eigenvalues3 = eig([2 0 0; 0 2 4*sqrt(.5); 0 4*sqrt(.5) 4*-.5+2]) % @ (-1 -.5 sqrt(.5))
Eigenvalues3 = 3×1
-2.0000 2.0000 4.0000

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