[Numpy-discussion] non normalised eigenvectors
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 ___ NumPy-Discussion mailing list -- numpy-discussion@python.org To unsubscribe send an email to numpy-discussion-le...@python.org https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: arch...@mail-archive.com
[Numpy-discussion] Re: non normalised eigenvectors
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 Sent: Saturday, February 25, 2023 11:46 AM To: Discussion of Numerical Python 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 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 ___ NumPy-Discussion mailing list -- numpy-discussion@python.org<mailto:numpy-discussion@python.org> To unsubscribe send an email to numpy-discussion-le...@python.org<mailto:numpy-discussion-le...@python.org> https://mail.python.org/mailman3/lists/numpy-discussion.python.org/<https://nam02.safelinks.protection.outlook.com/?url=https%3A%2F%2Fmail.python.org%2Fmailman3%2Flists%2Fnumpy-discussion.python.org%2F&data=05%7C01%7CLOUIS.PETINGI06%40CUNY907.mail.onmicrosoft.com%7C836b31ea969149d5531908db17507377%7C6f60f0b35f064e099715989dba8cc7d8%7C0%7C0%7C638129406524877235%7CUnknown%7CTWFpbGZsb3d8eyJWIjoiMC4wLjAwMDAiLCJQIjoiV2luMzIiLCJBTiI6Ik1haWwiLCJXVCI6Mn0%3D%7C3000%7C%7C%7C&sdata=5zj2Nd3JCW8wWgBmvZ%2BekSP0sd31RlneOTZGIiPbR%2Fg%3D&reserved=0> Member address: ilhanpo...@gmail.com<mailto:ilhanpo...@gmail.com> ___ NumPy-Discussion mailing list -- numpy-discussion@python.org To unsubscribe send an email to numpy-discussion-le...@python.org https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: arch...@mail-archive.com
[Numpy-discussion] Re: non normalised eigenvectors
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 Sent: Saturday, February 25, 2023 2:28:45 PM To: Discussion of Numerical Python Subject: [Numpy-discussion] Re: non normalised eigenvectors On Sat, Feb 25, 2023 at 2:11 PM Louis Petingi 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 ___ NumPy-Discussion mailing list -- numpy-discussion@python.org To unsubscribe send an email to numpy-discussion-le...@python.org https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: arch...@mail-archive.com
[Numpy-discussion] Dear Robert
As you mentioned this is a generalization of the Fiedler eigenvector. When applying spectral clustering, and you want to find the two clusters then the Fiedler eigenvector tells you how to partition the vertices (bipartition) so the normalized cut is minimized. The concept can be generalized to k clusters by applying K-mean to the first k eigenvectors. The two partitions can be determined by grouping the vertices of the corresponding negative values of the eigenvector for one cluster and the other cluster are vertices corresponding to the non-negative values. I can use perhaps the normalized eigenvectors, but I am not sure if this is correct thus, I prefer using the magnitudes. In theory it may work as for example the entries of the Fielder eigenvector are divided by the norm. The eigenvalues and eigenvector calculators (in internet) find the magnitude of the vector. The project is aimed to show that spectral clustering works for images that are graphs where the pizels are the vertices. We are having some issues with K-mean but if we get the Fieldler eigenvector we can test how K-mean works for two partitions. Best Louis From: Robert Kern Sent: Saturday, February 25, 2023 4:22 PM To: Discussion of Numerical Python Subject: [Numpy-discussion] Re: non normalised eigenvectors On Sat, Feb 25, 2023 at 4:11 PM Louis Petingi mailto: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 ___ NumPy-Discussion mailing list -- numpy-discussion@python.org To unsubscribe send an email to numpy-discussion-le...@python.org https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: arch...@mail-archive.com
[Numpy-discussion] Re: Dear Robert
Hi Robert I read somewhere that we can use the unit vector times a scalar for the Friedler eigenvector. Thus, the question is that for the first k-eigenvectors do we multiply the corresponding unit vectors them by the same scalar? That said my feeling is that when applying k-mean on the first k eigenvectors will give you different clusters as when you multiply the unit eigenvectors by different scalars the k-dimensional coordinates will be different and the distances between the points will also change. We could try starting with the unit vectors and multiply by different scalars and see what we get. As you mentioned k-mean is a heuristic and perhaps there is a graph-theoretical approach to find the k clusters by generalizing Fiedler result. Thanks, and I'll let you know Louis From: Louis Petingi Sent: Saturday, February 25, 2023 5:25 PM To: Discussion of Numerical Python Subject: [Numpy-discussion] Dear Robert As you mentioned this is a generalization of the Fiedler eigenvector. When applying spectral clustering, and you want to find the two clusters then the Fiedler eigenvector tells you how to partition the vertices (bipartition) so the normalized cut is minimized. The concept can be generalized to k clusters by applying K-mean to the first k eigenvectors. The two partitions can be determined by grouping the vertices of the corresponding negative values of the eigenvector for one cluster and the other cluster are vertices corresponding to the non-negative values. I can use perhaps the normalized eigenvectors, but I am not sure if this is correct thus, I prefer using the magnitudes. In theory it may work as for example the entries of the Fielder eigenvector are divided by the norm. The eigenvalues and eigenvector calculators (in internet) find the magnitude of the vector. The project is aimed to show that spectral clustering works for images that are graphs where the pizels are the vertices. We are having some issues with K-mean but if we get the Fieldler eigenvector we can test how K-mean works for two partitions. Best Louis From: Robert Kern Sent: Saturday, February 25, 2023 4:22 PM To: Discussion of Numerical Python Subject: [Numpy-discussion] Re: non normalised eigenvectors On Sat, Feb 25, 2023 at 4:11 PM Louis Petingi mailto: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 ___ NumPy-Discussion mailing list -- numpy-discussion@python.org To unsubscribe send an email to numpy-discussion-le...@python.org https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: arch...@mail-archive.com
