Eigenvector mystery

[JurgenRo]

(10-19-2020 07:19 PM)Thomas Okken Wrote:  

(10-19-2020 07:13 PM)Albert Chan Wrote:  True, but so is M(kv) = w(kv)

For eigenvalue w, all kv are valid eigenvectors for it, as long as k ≠ 0

Also, eigenvalues with multiplicity greater than 1 correspond to subspaces with corresponding dimension, so in those cases the eigenvectors are not uniquely determined even if you normalize their lengths...

Uniqueness is not crucial. It's all about dimension of Eigenspaces (geometrical multiplicity) and to pick a finite set of linear independent vectors to form a base. That there are infinite vectors in a vectorspace is just part of the definition of a vectorspace (scalar multipication) and so it is irrelevant to mention it at all.