Find eigenvalue of $A$, where $A^2+2A=O$ Find eigenvalue of $A_{3\times3}$, where $A^2+2A=O$ , $r(A)=2$ and $A^{T}=A$.
$A\xi=\lambda\xi$, so I got ${\lambda}^2+2\lambda=0$, which means $\lambda_1=0$ and $\lambda_2=-2$ are the eigenvalues of $A$ and $r(A)=2$ means that $\lambda_3\ne 0$. 
But the answer indicates that $\lambda_3=-2=\lambda_2$ without reasoning. Any hints or solutions would be helpful.

An eigen value $\lambda$ satisfies $Ax = \lambda x$, where $x$ is a non-zero vector. Hence, if $x$ is an eigenvector, we have
$$A^2x + 2Ax = \lambda^2x + 2\lambda x = \lambda(\lambda+2)x = 0$$
which gives us $\lambda = 0,-2$.
