Suppose $A$ is a $5 \times 5$ matrix satisfying $A^{2018} = 0$. How many eigenvalues of $A$?
I think this is one of the problem about Cayley-Hamilton Theorem in eigenvalues. If $A^{2018} = 0$, then the characteristic polynomial must be in form of $\lambda^{2018} = 0$, thus there is only one eigenvalues exist. Is this argument true? Could you convince me to solve the problem?
Let $\lambda$ be an eigenvalue of $A$. Then $A\vec{v}=\lambda\cdot\vec{v}$ for some $\vec{v}\neq\vec{0}$. It follows that $\vec{0}=A^{2018}\vec{v}=\lambda^{2018}\vec{v}$ so that $\lambda^{2018}=0$.
Matrices $N$ satisfying $N^k=0$ for some $k$ are called nilpotent matrices and the argument above shows that the only eigenvalue of a nilpotent matrix is zero.