Find $A$ when $A^2$ is a $2\times 2$ zero-matrix and $A$ is symmetric. 
$A^2 = 
       \begin{pmatrix}
        0 & 0 \\
        0 & 0 
        \end{pmatrix}$ and $A$ is symmetric. Find the matrix $A$.

Well, I know that $A = A^T$ and $A^2 = 
       \begin{pmatrix}
        0 & 0 \\
        0 & 0 
        \end{pmatrix}
$. I'm not sure how to continue from here. Any suggestions?
Btw, I thought of something: $A^2 = AA = A A^T = I$ but it is wrong.

$\left(\matrix{a & b\\ b & d}\right)^2=\left(\matrix{a^2+b^2 &  b(a+d)\\b(a+d) & b^2+d^2}\right)=\left(\matrix{0 & 0\\ 0 & 0}\right)\to b=0$ and $a^2=d^2=0$ or $a=-d$ and $b^2=-a^2$, ie, A is zero
