Eigenvector calculation; to get dominant eigen vector

6 次查看（过去 30 天）
Yunus Emre Bülbül
Yunus Emre Bülbül 2017-10-27
编辑： Yunus Emre Bülbül 2017-10-27
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 个评论
Yunus Emre Bülbül
Yunus Emre Bülbül 2017-10-27
编辑：Yunus Emre Bülbül 2017-10-27
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.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

John D'Errico
John D'Errico 2017-10-27
编辑：John D'Errico 2017-10-27
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.

类别

在 Help Center 和 File Exchange 中查找有关 Linear Algebra 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by