Does every complex $n$ by $n$ matrix have at least one eigenvector? Like is it possible for a complex square matrix to have zero eigenvectors?

Question

I know that the question has been answered regarding whether every $n$ by $n$ matrix has $n$ number of eigenvectors but what about at least one?

Answer 1

Yes. Eigenvalues are roots of the characteristic polynomial, so by the fundamental theorem of algegbra, there exist $n$ [possibly complex] eigenvalues, which must correspond with some eigenvectors, so there does exist at least one eigenvector.

Answer 2

Since $\Bbb C$ is algebraically closed, the characteristic polynomial, $\operatorname {det}(A-x I)$, has a root $\lambda $. Then $A-\lambda I$ is singular, so there's an eigenvector.