Ordering matters less than labeling. And another way to put it is that the affinity or distance matrix A should have the same labels on the rows AND on the columns as were on the rows of the original matrix. Thus, the labels on the rows of Q should be the same as the original labels.
Forming Q' A Q (not QAQ', btw) only gives us the diagonalized form of A which is just the affinity matrix of the eigen-representations. That isn't all that interesting. On Tue, May 24, 2011 at 2:09 PM, Shannon Quinn <squ...@gatech.edu> wrote: > Problem is, I'm not sure if the Lanczos solver or K-Means preserve this > ordering of indices. Does the nth point with label y from the result of > K-means correspond to the nth row of the column matrix of eigenvectors? >
