TITLE: Geometric multiplicity equal to 0
QUESTION [5 upvotes]: Is it possible for an eigenvalue of endomorphism to have geometric multiplicity equal to 0? 
I would be grateful if anyone who has an answer to this question would care to explain.

REPLY [9 votes]: No -- if $\lambda$ is an algebraic eigenvalue at all, then by definition $A-\lambda I$ has determinant zero, which means that the equation $(A-\lambda I)x=0$ has at least one nontrivial solution for $x$. This solution is an eigenvector, so the eigenspace must have dimension at least $1$.