On Sat, Feb 25, 2023 at 8:17 PM Louis Petingi
wrote:
> Hi Robert
>
> Just a follow up. I was able (my student) to get the 1 vector from the 0
> eigenvector. Even though the normalized or this set of eigenvectors will
> work we could try the two sets. Not sure if multiplying the unit vectors by
>
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
On Sat, Feb 25, 2023 at 5:33 PM Louis Petingi
wrote:
> 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
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
On Sat, Feb 25, 2023 at 4:11 PM Louis Petingi
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
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
On Sat, 25 Feb 2023 at 20:09, Louis Petingi
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,
On Sat, Feb 25, 2023 at 2:11 PM Louis Petingi
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,
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
On Sat, Feb 25, 2023 at 11:39 AM 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
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 wrote:
> Dear all,
>
> I am not an expert in NumPy but my undergraduate student is having some
> issues with the
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
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