Assembling the eigensolutions of a system of differential equations into a matrix results in what some have coined as the "fundamantal matrix". If the determinant of this matrix is NONZERO then all of the solutions are considered linearly independent.
I have obtained both the eigenvalues & eigenvectors from a state-space matrix representing a system of differential eqs. I suspect that the eigenvectors are linearly independent, but when I take the determinant of the assembled fundamental matrix I get a value that is NONZERO, but extremely small. This can be seen by running the code below after placing the contents of matlab.mat into your WORKSPACE.
TF=zpk(SIMULINKbode.values)
TF2=tf(SIMULINKbode.values);
[V,D] = eig(SIMULINKbode.values(2,1).a);
det(V)
The 2 norm of V = 1. Hence, the magnitude of the individual elements of V are all less than unity. So taking the determinant means multiplying a bunch of elements with magnitudes less than UNITY which inherently yields a very small number. So that makes it difficult to ascertain whether determinant is NONZERO and small due to that fact, or if the determinant is actually = 0 but numerical error prevents the determinant being IDENTICALLY = 0.
Does anyone have any suggestions how to address this? Is there some threshold of significant figures that I can say that the determinant is either ZERO or NONZERO?
It should be noted there is some level of error in V due to error in assessing the eigenvalues of SIMULINKbode.values(2,1).a. The code below demonstrates that the error in the eigenvalues is on the order of 10^-11. So that error will propagate through to the eigenvectors. What level of significant figures for that error I do not know, nevertheless, it exists. So this makes it confusing when interpreting the result of the determinant.
[row,col,v] = find(D);
v(:,2)=cell2mat(TF(2,1).p);
vdiff=[v(:,1)-v(:,2)];
