To get dominant eigen vector

In Matlab/Octave, [A B] = eig(C) returns a matrix of eigen vectors and a diagonal matrix of eigen values of C. Even though the values may be theoretically real, these are given to be complex with very low imaginary values. Due to this, the eigen values are not put in a decreasing order. Hence to find the eigen vectors corresponding to dominant eigen values, some calculations are required, which take up processing time in a big loop. Is there a remedy to this to find dominant eigen vectors?

[Bjorn Gustavsson]

I can't for the life of me believe that finding the largest few real eigenvalues (or the eigenvalues with the larges magnitude or whatever sorting you might need to do on the eigenvalue matrix) might be all that computational intensive compared to the calculations of the eigen-decomposition. That surely has to be a very minor time wasted on doing something like:

[EigvSorted,idx4ES] = sort(diag(real(EigV)));

[Pannir Selvam Elamvazhuthi]

I agree Bjorn. Thanks.

[Pannir Selvam Elamvazhuthi]

But Chen's answer seems to be faster as directly I'd get the eigen value and the corresponding vector too.

Answer 1

[Chaowei Chen]

[U,S,V]=svd(C) gives you the singular value decomposition of C. i.e., C=U*S*V'

where the singular values S are in decreasing order. Therefore, the most dominant eigenvector is U(:,1), for example.

[Pannir Selvam Elamvazhuthi]

Thanks Chen.

[Illya]

Almost correct: SVD is works fine only for PSD case.

If C is not PSD, there is no way to use SVD to get eigenvectors. Even using svd(C*C) ou svd(C*C') will not produce correct eigenvectors. Try, for example,

C=[0 1 1;0 1 1;1 0 0];

first with eigs()

 [V,D]=eigs(C); 
 lambda=D(1,1), v=V(:,1),
 C*v-lambda*v

 % lambda =
 % 
 %     1.6180
 % 
 % 
 % v =
 % 
 %    -0.6479
 %    -0.6479
 %    -0.4004
 % 
 % 
 % ans =
 % 
 %   1.0e-015 *
 % 
 %     0.2220
 %          0
 %     0.4441

And then with svd():

 [U,S,V]=svd(C); v=U(:,1)
 C*v-lambda*v

 %  v =
 % 
 %    -0.7071
 %    -0.7071
 %          0

 %ans =
 %
 %    0.4370
 %    0.4370
 %   -0.7071

or, even using squared a argument:

 [U,S,V]=svd(C*C); v=U(:,1)
 lambda=sqrt(S(1,1))
 C*v-lambda*v

 % v =
 % 
 %    -0.6280
 %    -0.6280
 %    -0.4597
 % 
 % 
 % lambda =
 % 
 %     1.6529
 % 
 % 
 % ans =
 % 
 %    -0.0497
 %    -0.0497
 %     0.1319

 The squared version is much closer, but still far away from the right   answer. The 'pseudo PSD' version (svd(C*C')) would produce the same U vectors as svd(C), which *is the eigenvector of C*C' but not an igenvector of C*.

However, all the above code would produce correct results for some PSD matrix, e.g. C1=C*C'.

Answer 2

[Riley Manfull]

[A,B]=eigs(C,1)

[Savas Erdim]

That worked very well. Thank you Riley!

Answer 3

[Christine Tobler]

I realize this is an old post, but this might be helpful to others:

One reason for EIG to return complex values with very small imaginary part, could be that A is very close to, but not exactly, symmetric. In this case,

 [U, D] = eig( ( A + A')/2 );

will make EIG treat the input as symmetric, and always return real eigenvalues.