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. 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


Best

Louis Petingi
Professor of Computer Science
College of Staten Island
City University of NY
From: Ilhan Polat <ilhanpolat@gmail.com>
Sent: Saturday, February 25, 2023 11:46 AM
To: Discussion of Numerical Python <numpy-discussion@python.org>
Subject: [Numpy-discussion] Re: non normalised eigenvectors
 
Could you elaborate a bit more about what you mean with original eigenvectors? They denote the direction hence you can scale them to any size anyways.

On Sat, Feb 25, 2023 at 5:38 PM <louis.petingi@csi.cuny.edu> wrote:
Dear all,

I am not an expert in NumPy but my undergraduate student is having some issues with the way Numpy returns the normalized eigenvectors corresponding to the eigenvalues. We do understand that an eigenvector is divided by the norm to get the unit eigenvectors, however we do need the original vectors for the purpose of my research. This has been a really frustrated experience as NumPy returns the normalized vectors as a default. I appreciate any suggestions of how to go about this issue. This seems to be a outstanding issue from people using Numpy.

Thanks

LP
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