Eigenvector calculation; to get dominant eigen vector

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Yunus Emre Bülbül
Yunus Emre Bülbül 2017 年 10 月 27 日
編集済み: Yunus Emre Bülbül 2017 年 10 月 27 日
I have a question regarding to eigenvector. The below simple script results in wrong eigenvectors.
mscalar=100; %kips/g
%sqrt(k/m) is calculated as 55.187
%kscalar
kscalar=mscalar*55.187^2;
k=kscalar*[2 -1 0 0 0;-1 2 -1 0 0;0 -1 2 -1 0;0 0 -1 2 -1;0 0 0 -1 1];
m=mscalar*[1 0 0 0 0;0 1 0 0 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1];
[V,D]=eig(k,m);
V
sqrt(D)
  2 件のコメント
Birdman
Birdman 2017 年 10 月 27 日
Are you trying to verify
k*V=m*V*D
Yunus Emre Bülbül
Yunus Emre Bülbül 2017 年 10 月 27 日
編集済み: Yunus Emre Bülbül 2017 年 10 月 27 日
No actually. I need eigenvectors also to be calculated from the eigenvalues obtained from the analysis which are correct. But at the end k*V=m*V*D will be verified.

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回答 (1 件)

John D'Errico
John D'Errico 2017 年 10 月 27 日
編集済み: John D'Errico 2017 年 10 月 27 日
The immediate answer is that eigenvectors are defined only to within a constant multiplier. Multiply an eigenvector by ANY constant, and it still satisfies the classic relationship
A*v = lambda*v
eig returns eigenvectors normalized to have unit norm, which is pretty standard. But in fact, there can always be an arbitrary factor of -1 in there. Just flip the signs on some of your eigenvectors, and nothing changes. The result is still completely valid. So sometimes you just get the wrong sign. People are always confused by that.
Note: If some of the eigenvalues had multiplicity greater than 1, there are other reasons why you can get differing eigenvectors. But that does not happen here.

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