Geometric Multiplicity and eigenvalue

Question

Can the geometric multiplicity of an eigenvalue $\lambda$, (dimension of eigenspace of $\lambda$) ever be $0$? I know if that was the case then the eigenspace would consist of the zero vector. So I would be correct to say that the geometric multiplicity could be $0$? I was told that it could never be zero. May someone elaborate?

Answer 1

If you had to say something like that, then that would have to be about a value $\lambda$ which is not a root of the characteristic polynomial (and thus not an eigenvalue). Those non-eigenvalues can with reasonably good conscience be said to have geometric (and algebraic) multiplicity $0$. I have never actually heard anyone say something like that, though.

As to actual eigenvalues, no, they can never have geometric multiplicity $0$. That's simply a consequence of the fact that for a linear map $A$, if $\lambda$ is a number such that $A-\lambda I$ has determinant $0$, then $A-\lambda I$ must have a non-zero kernel, and thus the eigenvalue $\lambda$ of $A$ has a strictly positive geometric multiplicity.

Answer 2

By the usual definition, $\lambda$ is not an eigenvalue unless there exists a corresponding eigenvector.

On the other hand, in infinite-dimensional Banach spaces the spectrum of a linear operator can contain things other than eigenvalues. I think that's the closest you can come to what you're asking about.