Errors in using null command due to truncation error
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Hi,
I'm trying to find eigenvectors of a 9-by-9 square matrix, corresponding to its eigenvalues. The matrix consists of components with complex numbers and one symbolic 'a', so I found nine eigenvalues ('a'), via solving the determinant of the matrix. For some eigenvalues, I used the 'vpa' command since, without 'vpa', they are obtained as a form of 'root(eqn, z, integer)'. Here the issue seems to arise. Due to the truncation error, the 'null' shows empty eigenvectors corresponding to the eigenvalues. FYI, I don't know how to assign a specific variable in 'eig' and it takes forever to run. Is there a breakthrough other than Gauss elimination method with suppressing close-to-zero values?
clc;
close all;
clear;
syms a
R = 0.5234;
r = 0.0054;
s = 0.0084;
for p = 1:3
M = [a*R*r 1i*r 0 0 0 0 a*R*8.78+1i*(8.78-p*R) 0 0;
0 0 a*R*s 1i*s 0 0 0 a*R*79.88+1i*(79.88-p*R) 0;
0 0 0 0 a*R*0.4542 1i*0.4542 -(a*R*291.06+1i*(291.06+p*R)) -(a*R*291.06+1i*(291.06+p*R)) 0;
a*R*8.78+1i*(8.78-p*R) 0 0 0 0 0 0 0 a*R;
0 a*R*8.78+1i*(8.78-p*R) 0 0 0 0 0 0 1i;
0 0 a*R*79.88+1i*(79.88-p*R) 0 0 0 0 0 a*R;
0 0 0 a*R*79.88+1i*(79.88-p*R) 0 0 0 0 1i;
0 0 0 0 a*R*291.06+1i*(291.06+p*R) 0 0 0 a*R;
0 0 0 0 0 a*R*291.06+1i*(291.06+p*R) 0 0 1i];
Det = det(M);
DetEqn = Det == 0;
EigenVal1 = solve(DetEqn,a);
EigVal = vpa(EigenVal1);
for j=1:rank(M)
M_temp = subs(M,a,EigVal(j));
EigVec(:,j) = null(M_temp)
end
end
Answers (1)
Walter Roberson
on 17 Aug 2023
2 votes
Your code assumes that the null space is the same size each time, but most of the time the null space is empty. You cannot store an empty vector into a definite vector location.
You need to decide what you want to do when the null space is empty.
6 Comments
syms a
R = 0.5234;
r = 0.0054;
s = 0.0084;
for p = 1:3
M = [a*R*r 1i*r 0 0 0 0 a*R*8.78+1i*(8.78-p*R) 0 0;
0 0 a*R*s 1i*s 0 0 0 a*R*79.88+1i*(79.88-p*R) 0;
0 0 0 0 a*R*0.4542 1i*0.4542 -(a*R*291.06+1i*(291.06+p*R)) -(a*R*291.06+1i*(291.06+p*R)) 0;
a*R*8.78+1i*(8.78-p*R) 0 0 0 0 0 0 0 a*R;
0 a*R*8.78+1i*(8.78-p*R) 0 0 0 0 0 0 1i;
0 0 a*R*79.88+1i*(79.88-p*R) 0 0 0 0 0 a*R;
0 0 0 a*R*79.88+1i*(79.88-p*R) 0 0 0 0 1i;
0 0 0 0 a*R*291.06+1i*(291.06+p*R) 0 0 0 a*R;
0 0 0 0 0 a*R*291.06+1i*(291.06+p*R) 0 0 1i];
Det = det(M);
DetEqn = Det == 0;
EigenVal1 = solve(DetEqn,a);
EigVal = (EigenVal1);
for j=1:rank(M)
M_temp = subs(M,a,EigVal(j));
EV = null(M_temp);
if isempty(EV)
EigVec(:,j,p) = sym(NaN(size(EV,1),1));
else
EigVec(:,j,p) = EV;
end
end
end
format long g
EigVec = double(EigVec)
Walter Roberson
on 17 Aug 2023
Note that in this above code, the null() calculation is working on the symbolic solutions, so there is no truncation error going on.
Seung Hyeop Hyun
on 18 Aug 2023
Torsten
on 18 Aug 2023
I don't know why you talk about "eigenvalues", but I agree that if "a" gives det(M(a)) = 0, null(M(a)) should be at least 1-dimensional and not empty.
You did not take into account that you use floating point constants and that some of the calculations take place in floating point instead of as symbolic numbers.
When you use symbolic numbers consistently then the problem does not show up.
Q = @(v) sym(v);
syms a
R = Q(5234)/Q(10)^4;
r = Q(54)/Q(10)^4;
s = Q(84)/Q(10)^4;
n8_78 = Q(878)/Q(10)^2;
n79_88 = Q(7988)/Q(10)^2;
n_4542 = Q(4542)/Q(10)^4;
n291_06 = Q(29106)/Q(10)^2;
for p = 1:3
M = [a*R*r 1i*r 0 0 0 0 a*R*n8_78+1i*(n8_78-p*R) 0 0;
0 0 a*R*s 1i*s 0 0 0 a*R*n79_88+1i*(n79_88-p*R) 0;
0 0 0 0 a*R*n_4542 1i*n_4542 -(a*R*n291_06+1i*(n291_06+p*R)) -(a*R*n291_06+1i*(n291_06+p*R)) 0;
a*R*n8_78+1i*(n8_78-p*R) 0 0 0 0 0 0 0 a*R;
0 a*R*n8_78+1i*(n8_78-p*R) 0 0 0 0 0 0 1i;
0 0 a*R*n79_88+1i*(n79_88-p*R) 0 0 0 0 0 a*R;
0 0 0 a*R*n79_88+1i*(n79_88-p*R) 0 0 0 0 1i;
0 0 0 0 a*R*n291_06+1i*(n291_06+p*R) 0 0 0 a*R;
0 0 0 0 0 a*R*n291_06+1i*(n291_06+p*R) 0 0 1i];
Det = det(M);
DetEqn = Det == 0;
EigenVal1 = solve(DetEqn,a);
EigVal = EigenVal1;
for j=1:rank(M)
M_temp = subs(M,a,EigVal(j));
EV = null(M_temp);
if isempty(EV)
EigVec(:,j,p) = sym(NaN(size(EV,1),1));
else
EigVec(:,j,p) = EV;
end
end
end
EigVec
format long g
EigVec = double(EigVec)
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