Why does "polyeig" returns eigenvectors that have smaller size than number of eigenvalues?
5 次查看(过去 30 天)
显示 更早的评论
I tried to solve an eigenvalue problem with " polyeig ". It correctly evaluated eigenvalues, however eigenvectors are not same as correct answer, which are obtained using " eig ". Do you know why " polyeig " returns smaller size eigenvectors?
This matter is clearly noted in the explanation of " polyeig ": [X,e] = polyeig(A0,A1,...,Ap) also returns matrix X, of size n-by-n*p, whose columns are the eigenvectors.
2 个评论
采纳的回答
David Goodmanson
2018-8-2
编辑:David Goodmanson
2018-8-2
Hi Masoud,
Your matrix A is 402x402. Denote Gamma by G and Omega by W, both 201x201. Let each eigenvector of A be the concatenation [u; v] where u and v are each 201x1. The system of equations is
Wu + Gv = bu (1)
-G'u - Wv = bv (2)
with eigenvalue b.
Working out polyeig for u gives the formulas you got. Both polyeig(A0,A21,A2) and eig(A) have 402 eigenvalues, and each polyeig eigenvector u is 201x1 as you noted. To obtain the second half of the eigenvector of A, you can invert (1) and use
v = G\(bu-Wu)
3 个评论
David Goodmanson
2018-8-2
编辑:David Goodmanson
2018-8-2
Hi Masoud,
I'm taking the point of view that it doesn't really matter what Gamma and Omega are as long as they are complex square matrices of the same size. The code below is basically a copy of yours once the matrices have been created. All the eigenvalues come out the same between the two methods. So it appears to work. But your case appears to come out differently ?!
If in your case Gamma has a large condition number, there could of course be some problems.
n = 201;
Gamma = 2*rand(n,n)-1 +i*(2*rand(n,n)-1);
Omega = 2*rand(n,n)-1 +i*(2*rand(n,n)-1);
A=[Omega Gamma;-Gamma' -Omega];
A0=Gamma'-((Omega/Gamma)*Omega);
A1=(Omega/Gamma)-(Gamma\Omega);
A2=inv(Gamma);
[X1 e1] = polyeig(A0,A1,A2); % e1 is a vector
[X2 e2] = eig(A); % e2 is a diagonal matrix
e2vec = diag(e2);
figure(1)
plot(e1)
hold on
plot(e2vec,'o')
hold off
更多回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!