Prove that sum of matrices equals zero 
The matrix $A$ has size $3 \times 3$ and we know that for any column
  vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
  $A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.

So if 
$$A = \begin{pmatrix}
    a_{11}  & a_{12} & a_{13}\\
    a_{21}  & a_{22} & a_{23}\\
    a_{31}  & a_{32} & a_{33}
\end{pmatrix}$$
and
$$v = \begin{pmatrix}
    x\\
    y\\
    z  
\end{pmatrix}$$
orthogonality of $Av$ and $v$ brought me to the equation
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$
and the matrix $A^{T} + A$ equals
$$A^{T} + A = \begin{pmatrix}
    2a_{11}  & a_{12} + a_{21} & a_{13} + a_{31}\\
    a_{12} + a_{21}  & 2a_{22} & a_{23} + a_{32}\\
    a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33}
\end{pmatrix}$$
However I don't see how then prove that $A^{T} + A = 0$.

Since all vectors $v$ are orthogonal to $Av$ then $v^TAv=0$ and likewise $(Av)^Tv=v^TA^Tv = 0$ and so adding them together we get $v^TAv + v^TA^Tv = v^T(A+A^T)v=0$ and finally since $v$ was arbitrary we must have $A+A^T=0$.
