On Sat, Feb 25, 2023 at 4:11 PM Louis Petingi <louis.peti...@csi.cuny.edu> wrote:
> 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. > There are multiple rules that could get this result. You could multiply all of the eigenvectors by `sqrt(n)`. Alternately, you could multiply each eigenvector `v` by `1/v[np.nonzero(v)[0][0]]` (i.e. the first nonzero value in the eigenvector). Both would give you the desired result for the 0-eigenvalue eigenvector, but different results for every other eigenvector. Both are simple and coherent rules, but quite likely neither are what you are looking for. Remember, each eigenvector can be scaled independently of any other, so establishing the desired result for one eigenvector does nothing to constrain any other. If you want help finding the rule that will help you in your research, you'll have to let us know what you are going to use the magnitudes _for_. The information you've given so far (where the eigenvectors come _from_) doesn't actually help narrow anything down. The only application of eigenvector magnitudes of graph Laplacians that I am aware of is the Fiedler vector, and that actually requires unit eigenvectors. -- Robert Kern
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