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# To get dominant eigen vector

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Commented: Savas Erdim on 6 Apr 2021 at 1:11
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?
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But Chen's answer seems to be faster as directly I'd get the eigen value and the corresponding vector too.

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### Accepted Answer

Chaowei Chen on 28 Aug 2011
[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.
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Illya on 14 Jul 2014
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'.

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### More Answers (2)

Riley Manfull on 9 Feb 2018
[A,B]=eigs(C,1)
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Savas Erdim on 6 Apr 2021 at 1:11
That worked very well. Thank you Riley!

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Christine Tobler on 12 May 2017
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.
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