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 <ilhanpo...@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.peti...@csi.cuny.edu<mailto:louis.peti...@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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