Prove that if $A^{2}=O,$ then 0 is the only eigenvalue of $A$ Getting Started: You need to show that if there exists a nonzero vector $\mathbf{x}$ and a real number $\lambda$ such that $A \mathbf{x}=\lambda \mathbf{x},$ then if $A^{2}=O, \lambda$ must be zero.
(i) $A^{2}=A \cdot A,$ so you can write $A^{2} \mathbf{x}$ as $A(A \mathbf{x})$
(ii) Use the fact that $A \mathbf{x}=\lambda \mathbf{x}$ and the properties of matrix multiplication to show that $A^{2} \mathbf{x}=\lambda^{2} \mathbf{x}$
(iii) $A^{2}$ is a zero matrix, so you can conclude that $\lambda$ must be zero.