In a finite dimensional vector space, if $0$ is an eigenvalue and the only eigenvalue of a linear operator, is that operator nilpotent?
There is this post which shows the other direction.
Prove that the only eigenvalue of a nilpotent operator is 0?
I would think the question would be posed as "iff" to the extent the answer to my question is affirmative.
To the extent that is not the case, I would please appreciate an example to that effect.
Thanks
Best Answer
If your field is algebraically closed (i.e. if we're including complex eigenvalues/eigenvectors), then the answer is yes.
If $0$ is the only eigenvector of the operator $A$, then $A$ has characteristic polynomial $p(x) = x^n$. By the Cayley-Hamilton theorem, $A^n = 0$.
On the other hand: if we're only including real eigenvalues, then we can say that the operator $$ \pmatrix{0&-1&0\\1&0&0\\0&0&0} $$ has zero as its only eigenvalue but fails to be nilpotent.