Hi Thanks Very simply one of the solutions for the zero eigenvalue is the 1 eigenvector. If I get back this 1 vector, for the 0 eigenvalue then the other eigenvectors will be in the right format I am looking for. Once again, the 1 vector is the normalized eigenvector * norm.
Best Louis Petingi Get Outlook for iOS<https://aka.ms/o0ukef> ________________________________ From: Robert Kern <robert.k...@gmail.com> Sent: Saturday, February 25, 2023 2:28:45 PM To: Discussion of Numerical Python <numpy-discussion@python.org> Subject: [Numpy-discussion] Re: non normalised eigenvectors On Sat, Feb 25, 2023 at 2:11 PM Louis Petingi <louis.peti...@csi.cuny.edu<mailto:louis.peti...@csi.cuny.edu>> wrote: Thank you for the reply. I am working with the Laplacian matrix of a graph which is the Degree matrix minus the adjacency matrix. The Laplacian is a symmetric matrix and the smallest eigenvalue is zero. As the rows add it to 0, Lx=0x, and 1 is the resulting vector. The normalized eigenvector is the 1 vector divided by the norm. So if a have 10 vertices of the graph the normalized eigenvector is 1/sqrt(10). I do understand that a scale * normalized eigenvector is also a solution but for the purpose of my research I need the normalized eigenvector * norm. I apologize, but I'm going to harp on this: you are using the word "the" as if there is one unique magnitude that we could report. There is none. If you have a use for a specific convention for the magnitudes, you'll have to point us to the literature that talks about it, and we might be able to give you pointers as to how to get the magnitudes that your research requires. For the 0 eigenvalue the norm of the eigenvector is easy to figure out but not for the other eigenvalues. That is what I meant by the original eigenvector and sorry for the confusion the confusion. Most eigenvalues/eigenvalues calculators will give you 1 for first eigenvector I'm afraid that doesn't really narrow anything down for us. -- Robert Kern
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