MATLAB Solving Linear System with Eigenvalues in Matlab

[member 428835]

Hi PF!

I am looping through a linear system and each time I do I generate a new matrix, call this matrix $A$. When finding the eigenvalues of $A$ in Matlab is use

[a,sigma2M] = eig(A);% a eigenvector and sigma2M matrix of eigenvalues
sigma2(:,ii) = sum(sigma2M);% create matrix with rows of eigenvalues for the iith system

where the ii is the for loop. The plot I get looks bad and you can see how some eigenvalues get mixed up (vertical axis are eigenvalues). Any ideas? I'd really appreciate it!

 

[Orodruin]

Sort the eigenvalues?

 

[member 428835]

Sort the eigenvalues?

Doesn't it automatically do this for me? Is there a way to have this automated?

 

[Orodruin]

Doesn't it automatically do this for me? Is there a way to have this automated?

If it did you would not get the result you are getting. Use the sort function.

I also recommend against using sum to get the eigenvalues out. The matrix contains the eigenvalues on the diagonal so use diag instead. For a square matrix it will give you a vector containing the diagonal elements.

 

[member 428835]

If it did you would not get the result you are getting. Use the sort function.

I also recommend against using sum to get the eigenvalues out. The matrix contains the eigenvalues on the diagonal so use diag instead. For a square matrix it will give you a vector containing the diagonal elements.

Good call on the diag function. I'd like the eigenvalues to remain paired with their corresponding eigenvectors. When I use sort on the eigenvalues, that only change their positions? Do you know of a way to have the eigenvectors align with the eigenvalues?

 

[Orodruin]

If you have the sort function output two return values, i.e.,

[B,I] = sort(A);

then B will contain the sorted list of eigenvalues and I will be a list of indices which is the order of the original indices. You can sort the eigenvectors by using the array I accordingly as an argument of the matrix containing the eigenvectors.

 

[member 428835]

Thanks got it!