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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?

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  • $\begingroup$ Wouldn’t the characteristic polynomial have degree 5? $\endgroup$ – Randall Mar 14 '18 at 3:36
  • $\begingroup$ Yeah, you are right! The maximum degree must be 5. $\endgroup$ – Shane Dizzy Sukardy Mar 14 '18 at 3:38
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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.

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  • $\begingroup$ Oh... So, do you hand the solution, Mr Brian? $\endgroup$ – Shane Dizzy Sukardy Mar 14 '18 at 3:50

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