Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: EIG function and unexpected complex modes Date: Wed, 15 Jun 2011 02:19:14 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 13 Message-ID: <it94r2$t1o$1@newscl01ah.mathworks.com> References: <it85cc$1to$1@newscl01ah.mathworks.com> <it87h9$9k8$1@newscl01ah.mathworks.com> <it8sml$a6j$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-05-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1308104354 29752 172.30.248.37 (15 Jun 2011 02:19:14 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Wed, 15 Jun 2011 02:19:14 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:731976 "Dario " <dario.donatiello@gmail.com> wrote in message <it8sml$a6j$1@newscl01ah.mathworks.com>... > I'm Dario and I'm working at the same project of Alessandro. As expected, sometimes we have two eigenvectors with the the same eigenvalues (for example two symmetric modes). > > For example: > considering the problem [V D] = eig(M\K); > at some point in the diagonal of the D matrix we encounter two eigenvalues with the same real part of the form: > l1 = a + 0.0000000i > l2 = a - 0.0000000i > The corrisponding two eigenvectors are complex conjugated. It doesn't seem that a scalar value could make the imaginary part negligible. - - - - - - - - - - - If you have two eigenvalues that are (nearly) equal, then any linear combination of their two eigenvectors will also be (nearly) an eigenvector. If they are (nearly) complex conjugates of one another, then the common real part is one such linear combination and the difference of their imaginary parts is another. Each of these linear combinations contains only real components and constitutes a valid eigenvector (which you can normalize if desired.) Roger Stafford