eig return complex values

Hello,

I'm trying to find the eigenvalues and eigenvectors of an invertible matrix. The eig function returns me complex values.

But the matrix is invertible: I invert it on Pascal.

How to explain and especially how to solve this problem please?

The matrix I am trying to invert is the inv(C)*A matrix, from the attached files.

Thanks,

Michael

[Matt J]

Invertible matrices can have complex eigenvalues. That in itself is not a sign of a problem.

[Michael cohen]

Correct, I meant that it is a diagonalizable matrix !

[Matt J]

A diagonalizable matrix can also be invertible with complex eigenvalues. A simple example is C=1i*eye(N).

In any case, I do not see the complex eigenvalues for the matrix you've provided.

load(websave('t','https://www.mathworks.com/matlabcentral/answers/uploaded_files/869735/matrix_C.mat'))

isreal(eig(C))

ans = logical
   1

[Michael cohen]

Thank you, but in fact it is the matrix_invC.A.mat that I try to diagonalize :)

[Matt J]

That matrix is not symmetric, so there is no reason to think it will have real eigenvalues.

Answer 1

[Torsten]

Use

E = eig(A,C)

instead of

E = eig(inv(C)*A)

or

E = eig(C\A)

[Michael cohen]

Thank you very much,

What is the difference between these two functions? Matlab documentation speaks of "generalized eigenvalues" for eig(A,C), what does that mean?

Moreover, I don't have quite the same results when comparing the real part of the first method eig(C\A) and the one you propose, is it normal ?

Thanks in advance

[Matt J]

Moreover, I don't have quite the same results when comparing the real part of the first method eig(C\A) and the one you propose,

There is essentially no difference in the results of the two methods:

load matrices;
E0=sort(  real(eig(C\A)) );
E=sort( eig(A,C) );

I=1:248;
plot(I,E0,'x', I,E,'--'); legend('eig(C\A)','eig(A,C)','interpreter','none')

[Torsten]

@Matt J

Although negligible, eig(A,C) produces no imaginary parts.

@Michael cohen

E = eig(A,C) solves for the lambda-values that satisfy

A*x = lambda*C*x (*)

for a vector x~=0.

If C is invertible, these are the eigenvalues of inv(C)*A (as you can see by multiplying (*) with

inv(C) ).

[Michael cohen]

Wouah, thank you very much. It’s very clear and allow us to solve our problem 🙏

Answer 2

[Matt J]

It turns out that B=C\A does have real eigenvalues in this particular case, but floating point errors approximations produce a small imaginary part that can be ignored.

load matrices

E=eig(C\A);

I=norm(imag(E))/norm(real(E))

I = 3.3264e-18

So just discard the imaginary values,

E=real(E);

[Michael cohen]

Thank you very much @Matt J for all those explanations !

[Matt J]

You 're welcome but please Accept-click one of the answers.