a wrong description in KPCA
line 239~240
# if u is an eigenvector of Phi(X)Phi(X)'
# then Phi(X)'u is an eigenvector of Phi(X)Phi(X)'
I now prove it simply.
Phi(X)Phi(X)'u = \lambda u
Phi(X)'Phi(X)Phi(X)'u = \lambda Phi(X)' u
Phi(X)'Phi(X) (Phi(X)'u) = \lambda (Phi(X)' u)
suggest as follows
# if u is an eigenvector of Phi(X)Phi(X)'
# then Phi(X)'u is an eigenvector of Phi(X)Phi(X)'
--> should be
# if u is an eigenvector of Phi(X)Phi(X)'
# then Phi(X)'u is an eigenvector of Phi(X)'Phi(X)
a wrong description in KPCA
line 239~240
I now prove it simply.
suggest as follows
--> should be