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.

Since you are given a polynomial $P(A)$ such that $P(A)=0$, you know that the only possible eigenvalues of $A$ are roots of $P$.  Since your matrix is symmetric, it is diagonalizable, so this is an easy way to prove this to yourself.  Think about the case where $A$ is rank 2 and diagonal (diagonalization reduces it to this case) - your matrix looks like 
$$
A = \left(\begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\0 & 0 & 0\end{array}\right)
$$
If you compute $A^2+A$, you'll see that both $\lambda_1$ and $\lambda_2$ must satisfy $\lambda^2+\lambda = 0$, i.e. $\lambda = 0$ and $\lambda = -2$ are the only options.
The deeper theory here is related to the minimal polynomial.
