On Sat, Feb 25, 2023 at 8:17 PM Louis Petingi <[email protected]> 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 > a scalar is the correct approach. > Yeah, that essentially amounts to multiplying all of the eigenvectors by `np.sqrt(n)`. I don't imagine it will do much to change the results of K-means or other clustering algorithm unless if it has absolute distance thresholds embedded in it. But it's absolutely a free choice for you to precondition things for your clustering algorithm. -- Robert Kern
_______________________________________________ NumPy-Discussion mailing list -- [email protected] To unsubscribe send an email to [email protected] https://mail.python.org/mailman3/lists/numpy-discussion.python.org/ Member address: [email protected]
